THE ELEMENTS OF PHYSICAL CHEMISTRY BY J. LIVINGSTON R. MORGAN, PH.D., M Professor of Physical Chemistrv, Columbia University. THIRD EDITION, REVISED AND ENLARGED. FIRST THOUSAND. UN1VERG OF S NEW YORK: JOHN WILEY & SONS. LONDON : CHAPMAN & HALL, LIMITED. 1905. Copyright, 1899, 1902, 1905, BY J. L. R. MORGAN. ROBERT DRUMMOND, PRINTER, NEW YORK. PREFACE TO THE FIRST EDITION. THE object of this book is to present the elements of the entire subject of Physical Chemistry in one volume, together with the important and but little known appli- cations of it to the other branches of chemistry. Many persons have found it difficult to obtain a comprehensive outline of the subject, owing to the length of time which it has been necessary to spend upon the separate volumes devoted to special portions of it. To all such this volume may be of value. It is especially intended as a text-book for either class- work or self-instruction, and although the calculus is used in the derivation of some of the laws, still much can be done without any training in the higher mathe- matics. In general, references are given, so that any one wishing to make an extended study of any special portion may do so with little difficulty. The amount of the subject included, however, embraces that which is likely to be useful to all chemists, and is that which vi PREFACE TO THE SECOND EDITION. to those studying alone, as well as to others who have already studied the subject but not yet attempted to apply it. J. L. R. M. HAVEMEYER LABORATORIES, December, 1901. PREFACE TO THE THIRD EDITION. As a basis for the revision of this work I have been so fortunate as to have a copy of the former edition contain- ing suggestions and criticisms from the hand of Prof. Ostwald. For his great kindness in having volunteered to take this trouble, and for many other favors, both during my student days and since, I am indeed very grateful, and I cannot allow this opportunity to pass without expressing to him, as far as words may suffice, my sincerest thanks. Further, I wish to acknowledge the influence of his " Vorlesungen uber Naturphilosophie," some points of which I have attempted to apply in this edition. The ideas I have had in mind in making this revision may be summarized as follows: To bring the subject- matter up to date; to distinguish sharply between hypothesis and fact, avoiding the former as far as is possible; and to accentuate the physical meaning of the results of mathematical reasoning, i.e., to employ vii Vlii , PREFACE TO THE THIRD EDITION. mathematics simply as a means to an end, and to make the matter as intelligible as possible, even to the non- mathematical reader. In few words the motive, if I may so express myself, of this edition is to treat the subject as something which can be applied to other branches and to illustrate the application and methods of application by the use of numerous problems. In this way it is hoped that the subject, instead of being merely a more or less interesting theoretical study, as is too often the case, may appeal to the student as a tool by the aid of which actual results may be obtained. This plan is the expression of the need of my own students, and I assume consequently that it will be welcome to others who not only wish to know the subject, but also to use it. J. L. R. M. COLUMBIA UNIVERSITY, May, 1905. CONTENTS. CHAPTER I. PACE INTRODUCTORY REMARKS i i. Physical chemistry. 2. Energy. 3. The factors of energy. 4. Atomic and molecular weights. CHAPTER II. THE GASEOUS STATE 9 5. Definition of a gas. 6. The gas laws. 7. The specific gravity of gases. 8. Methods of determining the specific grav- ity. 9. Abnormal vapor-densities. Dissociation. 10. Vol- ume, pressure, and concentration, n. Variation from the gas laws. The equation of Van der Waals. 12. Specific heat. The first principle of thermodynamics. 13. Determination of the specific heat of gases. 14. The second principle of thermo- dynamics. 15. The cycle. Entropy. 1 6. "The factors of heat energy. CHAPTER III. THE LIQUID STATE 61 17. Distinction between liquids and gases. 18. Connection between the gaseous and liquid states. 19. Vapor-pressure and boiling-point. 20. The heat of evaporation. 21. The relation between vapor-pressure, heat of evaporation, and temperature. 22. Refraction of light. 23. Surface-tension. 24. Molecular vreight in the liquid state. Critical temperature. CONTENTS. CHAPTER IV. PAGE THE SOLID STATE 91 25. Remarks. 26. Atomic heat. Law of Dulong and Petit. 27. Changes in the state of aggregation. CHAPTER V. THE PHASE RULE 104 28. Object of the phase rule. 29. The phase rule. 30. Deri- vation of the phase rule. 31. The equilibrium of water in its phases. 32. The phase rule and the identification of basic salts. CHAPTER VI. SOLUTIONS 116 33. Definition of a solution. 34. Gases in liquids. 35. Liq- uids in liquids. 36. Solids in liquids. 37. Osmotic pressure. 38. Electrolytic dissociation or ionization. 39. Solution-pres- sure. 40. Vapor-pressures of solutions. 41. The relation be- tween osmotic pressure and the depression of the vapor-pressure. 42. Increase of the boiling-point. 43. Depression of the freez- ing-point. 44. Division of a substance between two non-misci- ble solvents. Depressed solubility. 45. Solid solutions. 46. Col- loidal solutions. 47. The molecular weight in solution. CHAPTER VII. THERMOCHEMISTRY 200 48. Definition. 49. Application of the principle of the con- servation of energy. 50. The heat of formation. 51. Chemical changes at a constant volume. 52. Chemical changes at a con- stant pressure. 53. Relation between results for constant volume and constant pressure. 54. The effect of temperature. 55. The thermal reactions of electrolytes. CONTENTS. xi CHAPTER VIII. PAGE CHEMICAL CHANGE. A. Equilibrium 222 56. Reversible reactions. 57. The law of mass action. 58- Equilibrium in homogenous gaseous systems. 59. Equi- librium in non-homogenous systems. 60. Dissociation of a solid into more than one gas. 61. Equilibrium in liquid systems. 62. The effect of temperature upon an equilibrium. The varia- tion of the constant of equilibrium with the temperature. B. Chemical Kinetics 261 63. Application of the law of mass action. 64. Reactions of the first order. 65. Catalytic action of hydrogen ions. Catal- ysis. 66. Reactions of the second order. 67. Reactions of the third order. 68. Incomplete reactions. 69 Reactions be- tween solids and liquids. 70. Speed of reaction and temperature. C. Application of the Law of Mass Action to Electrolytes Ionic Equilibria 281 71. Organic acids and bases. The Ostwald dilution law. 72. Acids, bases, and salts which are ionized to a considerable extent Empirical dilution laws. 73. Heat of ionization. ^jta. 74. Solubility or ionic product. 75. Hydrolytic dissociation. Hydrolysis. 76. Determination of the ionization constant from observations of increased solubility. 77. Ionic equilibria. 78. The color of solutions. 79. The action of indicators. 80. General analytical reactions. CHAPTER IX. ELECTROCHEMISTRY. A. The Migration of the Ions 364 81. Electrical units. 82. Faraday's law. 83. The migra- tion of the ions. 84. Determination of the relative velocity of migration. B. The Conductivity of Electrolytes 374 85. The specific conductivity. 86. Molecular and equivalent conductivity. 87. Determination of electrical conductivity. 88. Ionic conductivities. General rules. 89. The conductivity of organic acids. 90. The absolute mobility of the ions. xii CONTENTS. PAGE 91. The basicity of an acid. 92. The conductivity of neutral salts. 93. The ionization of water. 94. The temperature coefficient of conductivity. 95. Conductivity of difficultly solu- ble salts. 96. The dielectric constant and dissociating power. Other solvents than water. 97. The conductivity of mixtures of substances having an ion in common. C. Electromotive Force 399 98. Determination of the electromotive force. 99. Types of cells. 100. Chemical or thermodynamical theory of the cell. 101. The osmotic theory of the cell. 102. Mathematical ex- pression of the osmotic theory of the cell. 103. The measure- ment of the potential difference between a metal and a solution. 104. The heat of ionization. 105. Concentration cells. 106. Dissociation by aid of the electromotive force. 107. Electrolytic solution pressures. 108. Cells with inert electrodes. 109. Processes taking place in the cells in common use. D. Electrolysis and Polarization . 440 no. Decomposition values, in. -Theory of polarization. 112. Primary decomposition of water. 113. Electrolytic sepa- ration of metals by graded electromotive forces. CHAPTER X. PROBLEMS . 453 TABLES o 485 APPENDIX 497 INDEX . 503 ELEMENTS OF PHYSICAL: ^ CHEMISTRY. CHAPTER I. INTRODUCTORY REMARKS. i. Physical Chemistry is that branch of the science oj chemistry which has for its object the study of the laws governing chemical phenomena. Other titles for this subject are also in use (General Chemistry, Theoretical Chemistry), but, since the sub- jects treated lie in that border-land between Physics and Chemistry, and since many purely physical methods are used, the term Physical Chemistry is the most de- scriptive one, and is to be preferred. The subject of Physical Chemistry is usually divided into two parts: Stoichiometry and Chemical Energy, the latter term in- cluding the laws of affinity and all other allied subjects. As it is difficult, however, to make a sharp distinction, we shall consider both together, dividing the subject only into minor portions in the form of chapters. 2 ELEMENTS OF PHYSICAL CHEMISTRY. 2. Energy. Since much of our work has to do with the different forms of energy, it will be well first to recall some points which later we shall use constantly. Force is whatever changes, or tends to change, the motion of a body by altering its direction or its magni- tude* : The - unit . of force is the dyne. A dyne acting upon a gram, for one second will give it the velocity of i centimeter per second, and n dynes, n centimeters per second. At Washington a body falling freely for one second acquires the velocity of 980.10 centimeters, therefore the force of gravitation of i gram at Wash- ington is equal to 980.1 dynes. Work. The unit of work is that work which is done when unit force is overcome through unit distance. This is called the erg. We have then dynes X centimeters = ergs. The maximum work to be obtained from any process is that amount which can be produced under ideal con- ditions. Energy is work or anything which can be transformed into work, or produced from work. Mechanical energy is of two kinds, kinetic and distance the former being due to motion, the latter to position. As all forms of energy can be transformed the one into the other, the general unit of energy is the erg. Since it is impossible to re- move all the energy from a body, we have no way of INTRODUCTORY REMARKS. 3 determining how much is contained in it. We can only measure the excess of energy which a body contains in a given state over that which it contains in a certain standard state. This is determined by allowing a body to go from the given state to the standard state in such a way that the difference in energy all appears in a measurable form, for example, as heat. If we measure this heat, we have the energy of the body in the given state, as compared to that in the normal state. 3. The factors of energy. Experience has shown that the possibility of a transfer of energy does not depend upon the amount of the energy in question, but rather upon the intensity of the energy. For this reason the amount of energy is usually expressed as a product of two factors. The one factor determines the amount of energy which can and must be absorbed by a body in going from one state into another. This is called the capacity jactor of the energy. Upon the other factor depends the possibility of the transfer of the energy from one body to another in such a way that if the value of this factor for both the bodies is the same no transfer takes place. This is called the intensity j actor of the energy. Bodies witK the same' value for the intensity factor are in equilibriufe, i.e., no finite transfer of energy- takes place between them. If the intensity factor has a different value in two states of a substance, an exchange of energy will take place between them until the intensity 4 ELEMENTS OF PHYSICAL CHEMISTRY. factor has become of the same value for both and equi- librium is established. In general the capacity factor can be distinguished from the intensity factor by the fact that in all changes the sum of the capacity factors is a constant, which is not true for the intensity factor. Examples. For heat energy the capacity factor is called entropy, while the temperature is the intensity factor. If two bodies at the same temperature are brought in contact, there is no change in them. Should the temperatures be different, however, there is an exchange of heat energy of such a nature that the temperatures become the same. Bodies containing different amounts of heat, when brought together, remain unchanged pro- vided their temperatures are the same. The old unit of heat energy, the calorie (18), is equal to 41,830,000 ergs. Volume energy is equal to the work required to cause a decrease of volume, or that which is done by an increase of volume. The intensity factor is the pressure, while the volume is the capacity factor. Imagine a gas enclosed in a cylinder which is provided with a movable partition. If, on each side of this partition, we have the gas under different pressures, then the partition will move to the side with the smaller pressure, until the two become equalized. As long as the pressure is the same the volumes may have any relation without causing the parti- tion to move. Pressure is expressed in dynes per square centimeter, and its unit is that pressure under which an INTRODUCTORY REMARKS. 5 increase of volume equal to i cubic centimeter will do the work equal to i erg. The pressure of the atmos- phere, then, is equal to 76X980.1 Xi3. 5953= 1,012,681, i.e., a. little over a million of these units. For kinetic energy, equal to i/2Mv 2 , the capacity factor is the mass, and the intensity factor is the square of the velocity. i/2Mv 2 y it is true, may also be divided into the factors Mv(= momentum) and v( = velocity). In this case the momentum is the capacity factor, the velocity the intensity factor. For electrical energy the amount of electricity is the capacity factor, the electromotive force being the intensity factor. In the same way all other energies may be divided into factors, and it has been found that the study of the different energies is much simplified by the process. In general, then, for all energies we have (i) E=ci, where c is the capacity factor and i the intensity factor, E representing the energy itself. If the energy is increased by an infinitesimal amount, then dE = d(ci) = cdi + idc; and from this we may find the equations for c, i, dc y and di. They are: 6 ELEMENTS OF PHYSICAL CHEMISTRY. 7 t 1 (2) c = -j-r(dc=o, c = const.); (3) di = (dc = o, c = const.) ; (4) i=-r-(di = o, i = const.); dE (5) dc=*r-(dio, i = const.). If in a system there are two kinds of energy which are so related that a change in the one causes a corresponding change in the other, i.e., one is a .function of the other, then, if the two kinds of energy are E\ and 2, dE 1 =dE 2 or d(ciii)=d(c 2 i^\ i.e., if c= const. (6) Cidii = c 2 di 2 . This method of deriving an equation giving the relation between two kinds of energy in one system is very useful, and we shall have occasion to use it later. 4. Atomic and molecular weights. Since we are to make very extensive use of these two factors in our later work, it is quite necessary here to make clear the exact INTRODUCTORY REMARKS. 7 meanings of the words, as we shall employ them. A glance at the practical application of the words shows that there is nothing hypothetical about them, other than their derivation. It is unfortunately true, however, that from this one early acquires the idea that the actual results of an analysis or chemical reaction are dependent not only upon the practical work, but also upon the molecu- lar or atomic hypothesis. This idea is so strong, indeed, that it is not uncommon to find people assuming that any direct proof contrary to the hypothesis would actually affect the practical results themselves. It is not possible to discuss this question in detail here, so the reader must be referred elsewhere. The definition of the two terms, however, will serve to show that they are simply expres- sions of experimentally determined facts, and in such a sense they are used throughout this book. The combining weight of any element is the weight which combines with 16 units of weight of oxygen, or some multiple or submultiple of that; or combines with the combining weight of some other element, the value of which is expressed hi terms of oxygen. The combining weight is identical with the atomic weight, when there is but one combining ratio; and is usually the least common multiple of the combining weight in case there is more than one combining ratio. The molecular or formula weight in the gaseous state (that in other states will be defined later) expresses the 8 ELEMENTS OF PHYSICAL CHEMISTRY. number of grams of gas which at o C. and i atmosphere pressure occupy the same volume as two combining weights of oxygen (16X2 = 32 gr.), i.e., approximately 22.4 liters (see p. 12), or a corresponding volume at any other temperature or pressure. Since the factor for the conversion of grams into ounces (av.) (0.0353) * s the same as that for the conversion of liters into cubic ieet, this relation also holds for ounces (av.) and cubic feet. In other words, the molecular weight of any gas, expressed in ounces (av.), occupies the volume of 22.4 cubic feet. CHAPTER II. THE GASEOUS STATE. 5. Definition of a gas. A gas is distinguished by its property of indefinite expansion. In other words, a gas is limited in volume only by the walls of the vessel which contains it. 6. The gas laws. These are the result of experience as to the behavior of gases in general under varying conditions. Boyle's (Mariotte's) law: At constant temperature the volume oj any gas is inversely proportional to the pres- sure under which it exists. If v represents the volume of any weight of a gas at the pressure p, and v\ the volume of the same amount of the gas at the pressure pi, then (7) pv = piVi= constant for constant temperature. If the temperature is different for the two cases the product pv is no longer equal to the product p\v-\_. The effect of a change in temperature upon the volume of a gas is given by the law oj Charles (Dalton, Gay-Lussac); The volume of any gas increases by the 9 io ELEMENTS OF PHYSICAL CHEMISTRY. 1/273 P art l M S volume at o C. jor every increase o] Us temperature equal to i C. If the temperature were decreased continuously from o, we would finally reach a point at which the volume is equal to zero. Provided that the law of Charles holds for such a low temperature, this point will be 273 centi- grade degrees below o C. This point is called the absolute zero, and the temperature measured from it in centigrade degrees is the absolute temperature. This is designated by the letter T, the temperature according to the centigrade scale being distinguished by the letter t. Between these two terms, then, we have the relation r=/ + 27 3 , or *=r-273. The volume of a gas is thus proportional to its abso- lute temperature, provided its pressure remains constant; and, by Boyles' law, its pressure is proportional to the absolute temperature, when its volume remains con- stant. From this we can now find the value of the product pv for any temperature. We have vocT(p = const.); -(r = const.); hence T v f THE GASEOUS STATE. 1 1 and jT v = k P or (8) pv = kT. But at o C. (80) pov = k2 73, and combining (8) and (So), with elimination of the con- stant k, we find 273 This term is a constant for any one gas, how- ever, for />o = iO33 grams per square centimeter (i.e., 76X13.6) and ^o is the volume occupied by a definite weight of the gas under this pressure at the tempera- ture of o C. If v Q is the volume of i gram of gas at o, 76 cms., we have where v is the volume of i gram of gas at the tempera- 12 ELEMENTS OF PHYSICAL CHEMISTRY. ture T and the pressure p, 'and r is the specific gas con- i slant oj the gas. Or, comparing such quantities of different gases as will give the volume of 22.4 liters at o C. and i atmos- phere pressure (i.e., comparing the molecular weights), we find (9) pV=RT(\.z., ^ for i mole* = 12 J, where R is the molecular gas constant, which is the same for all gases, r multiplied by the molecular weight of the gas, then, must be equal to R. Equation (9) is the equation oj state for gases. In it p, the pressure, is to be expressed as a weight in grams upon the square centimeter, V is the volume of one mole in cubic centimeters, and T is the absolute tem- perature in centigrade degrees, i.e,, T = 1 + 273. The molecular gas constant, R, may be calculated as follows : In tf-, Fo = 22400 cc ., hence ^2*400X1033 273 273 = 84800, where the weight of i gram per square centi- meter is taken as the unit. To transform this into abso- lute units it is then only necessary to multiply it by 980.1 (page 2). * From here on We shall call the molecular Weight in grams the mole, as has been proposed by Prof. Qstwald. A mole of any gas at o and 76 cms. pressure occupies 22.4 liters, which is the average result of many determinations. THE GASEOUS STATE. 13 Or, if po is given in atmospheres and FO in liters, we have R=-7p= - =0.0821 liter-atmospheres, and R= ^ =8.3iXio 7 ergs, when p is given in dynes and V in cc. Dalton's law refers to the pressure exerted by the single gases in a mixture of gases. The pressure exerted upon the walls of a vessel con- taining a mixture oj gases is equal to the sum of the pres- sures which the single gases would exert were they alone in the vessel. The exact meaning of this will be seen from the following example: When into a closed ex- hausted vessel we introduce i gram of gas it will exert a certain pressure upon each square centimeter of the surface. If now we introduce a second gram of the same or a different gas, it will exert exactly the same pressure upon the vessel that it would have exerted had the first gram not been there. Upon the walls, however, with the same gas, the pressure will be doubled. One law still remains to be considered which has had and still has great value in chemistry; it is Avogadro's law or hypothesis: All gases under the same conditions oj pressure and temperature contain in unit volume the same number oj gram molecules. In its original form this relation referred to the hypothetical molecules them- 14 ELEMENTS OF PHYSICAL CHEMISTRY. selves (not to gram molecules), and hence was hypothet- ical. In the form given- above, however, it is simply expressive of the definition of molecular weight already given (pp. 7, 8). As a matter of fact the original form has always been used of late as though it did refer to gram molecules, so that the hypothetical character has been lost, and the relation may well be designated as Avogadro's law. The advantage of the statement in this form is that it accentuates the difference between hypothesis and fact, and shows plainly that the relation itself is free from all hypothesis. By determining the density of any gas referred to oxygen it is very easy, then, to determine its molecular or formula weight. Since equal volumes under like conditions contain the same number of moles, the weights of these volumes will be related as the molecular weights, the basis of which is that of oxygen, i.e., 32. The term formula weight, which we shall use as identical with molecular weight, is to be preferred to the latter, for the formula shows distinctly the relation between atomic weight (i.Q., the combining ratio) and molecular weight (i>e., the number of combining or atomic weight which occupy the volume of 22.4 liters at o C. and 760 mm. Hg pressure). Thus H represents the atomic weight of hydrogen (i.e., 1.008), while H2 is the molecular or formula weight (i.e., 2X1.008 gr.); and for compounds this is much more convenient, especially since the mole- THE GASEOUS STATE. 15 cular or formula weight is not a fixed quantity for all conditions, but varies with these. - Thus we have FeCls and Fe2Cl6, according to the conditions, and a glance at the formula shows at once the molecular weight. 7. The specific gravity of gases. The specific gravity of a gas is the weight of the unit of volume, but as this is always very small it is customary to find in these terms the value of a certain gas taken as a standard, and then to express the density of other gases in terms of this gas, taken as unity. As the composition of air varies, oxygen has been chosen as the chemical standard. Since this is very easily obtained in the pure state and its combining weight with other elements can be accurately determined, its advantages as a standard are apparent. One liter of O weighs 1.4290 grams at o and 76 centimeters pressure. Its specific volume, i.e., the volume of i gram at any temperature and pressure, is _ 0.0014290 Its density, i.e., the weight of i cc. at any temperature and pressure, is the reciprocal of the specific volume, i.e., d = - =0.0014290 -7 grams. v - 76(1+1/275) G i6 ELEMENTS OF PHYSICAL CHEMISTRY. The molecular volume, since 02 = 32, is If w is the weight of a gas, and g*is the weight of an equal volume, under like conditions, of the standard gas, i.e., O, the specific gravity of the gas is w The relation in weight between hydrogen and oxygen is as 1:15.88; between hydrogen and air is as 1:14.48. In this way knowing the density referred to one, that referring to either of the others can be determined. Referred to air the density can be employed to find the molecular weight by aid of the formula M = 0.001 293X2 2400^ = 28.96^, where 0.001293 is the weight of i liter of air at o and 76 cm. Hg pressure, and A is the density of the gas based upon air as a standard. 8. Methods of determining the specific gravity. Here we shall consider the subject both for gases and for the vapors generated from substances by heat, since the determination of the latter is of importance for the ascertaining of the molecular weight. For convenience we shall divide the subject into three groups of methods, THE GASEOUS STATE. 17 but shall consider only the theory of the methods, assuming that the student will look up the practical details in one of the many laboratory manuals. A. The weight oj a certain 'volume of the gas. This may be used both for gases and for the vapors given off by substances under high temperatures. a. For gases. The weight of a certain volume is found by weighing first a balloon filled with the gas, and then exhausting this and weighing it alone. In this way we can find the weight of the volume of gas, and by comparing this with the weight of the same volume of oxygen ascertain the density. P. For gases generated from substances by heat (Dumas) . Here the process is slightly different. The substance is placed in a weighed flask with a slender neck and the whole heated at a constant temperature until the sub- stance is all transformed into gas, when the neck of the flask is fused together. After cooling, the flask is weighed and the neck cut off so that the volume of the flask can be found from the weight of water it will contain. We have then the weight of a certain number of cubic centi- meters of the gas, which, when compared with the weight of an equal vohime of oxygen at that temperature, will give the density referred to that gas, and then by calcu- lation that based upon hydrogen or air. This method of Dumas has been further modified, so that it is possible to work under diminished pressure *8 ELEMENTS OF PHYSICAL CHEMISTRY. and consequently at lower temperatures, which is of great importance in case the substance decomposes at a higher temperature. The flask for this purpose is the same as that used before except that the neck is con- nected with an air-pump and manometer. When the pressure has reached the desired point the temperature is increased until the substance goes into the gaseous form, when, as above, the neck is sealed. The calcula- tion is the same as before except that the pressure under which the gas exists is equal to that of the barometer minus that of the manometer. B. The volume occupied by a certain weight. By these methods we avoid the error which is always present when the weight of a volume of gas is determined. The substance, in solid or liquid form, is weighed and then transformed into the gaseous state, the volume which it then occupies being measured. The method of Gay- Lussac, as improved by Hofmann, is intended only for the determination of the specific gravity of the gases formed from solid or liquid substances. A weighed amount of substance is brought into the Torricelli vacuum of a barometer-tube, which is surrounded by a tempera- ture bath, and the volume of the gas measured. The gas is under a pressure equal to that of the atmosphere minus that of the column of mercury in the tube. This volume of gas, reduced to o C. and 76 cms. of Hg, weighs just what the original substance did, and this weight THE CASEOUS STATE. 19 compared with that of an equal volume of the standard gas will give the density. The method of Victor Meyer differs from that of Hofmann in that the gas is formed in a flask and displaces the air, the volume of this at a lower temperature being measured in a burette connected with the flask. This volume is of course the same as that the substance itself would assume at that temperature, provided that no con- densation takes place, for the coefficient of expansion of all gases is the same. The method under this group which may be used for gases in general is to pass a determined volume of the gas over or through a weighed substance which will absorb it all. In this way the difference in weight of the substance before and after the passage of gas gives the weight of the determined volume of gas. Thus oxygen may be absorbed by glowing metallic copper, the copper oxide formed giving the weight of copper plus oxygen. C. The value of this method (Bunsen's), which is adapted only to gases, is that but very small amounts of substance are needed for the experiment. It depends upon the relation of the velocity of outflow of a gas through a small aperture to the specific gravity of the gas. The formula is derived as follows: Let p be the differ- ence between the pressure under which a gas exists and that of the atmosphere, and v be the volume of the gas which flows out in the unit of time. The total work done 20 ELEMENTS OF PHYSICAL CHEMISTRY. by the change in volume is then pv. Since this work is used to force the gas out, it is equal to the increase oi kinetic energy of the gas, i.e., where c is the velocity of outflow of the gas, and m is its mass. If equal volumes of different gases are consid- ered, flowing through the same-sized openings under the same pressure, then or or, since the masses of equal volumes are proportioned to their densities The apparatus used by Bunsen is very simple. A glass tube having a very fine opening at one end, below which is a stop-cock, is clamped into a vessel of mercury. In this tube is a piece of glass rod to act as a float, while on the outside there are two marks close together. The gas is now passed into the tube at a certain pressure, and the tube lowered into the mercury until the float is just at the lower mark. The cock is then opened and the time necessary for the passage of the float from the lower THE GASEOUS STATE. 21 to the upper mark observed. Since the time is inversely proportioned to the velocity, we have By carrying out the experiment for one gas and then for oxygen we can find the density of the gas in terms of oxygen. The results by this method, and this is true in a lesser degree for all methods for molecular weight, are not exact, but are simply intended to indicate what number of combining weights of an element, or what weight of a compound, represents the formula weight. For this reason very great accuracy is not necessary, for it is only a question of finding whether the combining weight or the simplest ratio, as found by analysis, is to be multiplied by the factor i, 2, 3, etc. 9. Abnormal vapor densities. Dissociation. The specific gravity or density of the gases generated by heat from substances is found in many cases to be too small, i.e., the molecular weights thus determined are smaller than the theoretical values. And, further, the molecular weight is found to vary with the temperature, pressure and some other factors. These results, naturally, must be obtained with greater accuracy than those mentioned above if the process is to be followed at all closely. Since, according to our definition of molecular weight, i mole occupies 22.4 liters at o and 760 mm. Hg pressure, a smaller density would represent a decom- 22 ELEMENTS OF PHYSICAL CHEMISTRY. position of the substance employed. Thus if a substance decomposes in such a way that from each molecule of the gas there are two others formed, then the vapor density must be one half what it should be. For each volume of the undecomposed gas we shall have two volumes of the gases formed from it at the same tem- perature, since by Avogadro's law we have the same number of molecules in equal volumes. These two volumes, however, will weigh the same as the original one volume, so that the weight of one volume of the mixed gases will be one half that of one volume of the undecomposed gas. St, Clair Deville called this de- composition dissociation. Thus NKUCl dissociates ac- cording to the scheme where the sign <= means that the reaction may go in either direction, according to the conditions. That these two gases, NHs and HC1, are actually present in the vapor of NH 4 C1 can be shown in the following way (Pebal and Than) : A lump of solid NH 4 C1 is placed in a tube upon an asbestos plug, and the temperature of the tube raised. The NH 4 C1 now volatilizes and dis- sociates in NHa and HC1. Since NHa is the lighter gas, it diffuses more rapidly through the asbestos plug than the HC1, consequently on one side "of the plug we shall have an excess of HC1 and on the other an excess THE GASEOUS STATE. 23 of NH 3 . The presence of these two products may be shown by passing a current of dry air through the parts of the tube on each side of the plug and then over moistened litmus paper, which will show the nature of the gases present. Since the density of the substance, in case of a dis- sociation, decreases to an extent dependent upon the temperature it is an important thing to be able to find the degree of the dissociation at any one temperature, i.e., to determine the extent of the reaction (p. 22) form- ing the products on the right. This can be found from the relation of the vapor density to the dissociation, i.e., the dependence of both upon the number of moles. If, for example, a is the percentage of the gas dissociated, i.e., the degree of disso- ciation, and we start with i mole of the gas, then i a is the undissociated portion. If there are n moles of the products formed from i mole of the gas, the total number of moles present at any time is (i - a) + na = i + (n i)a. The ratio, then, of i to i + (n i)a will be the same as that of the vapor density as it is, to the vapor density as it should be, i.e., 24 ELEMENTS OF PHYSICAL CHEMISTRY. Where d u is the vapor density as it should be (i.e., as it is without dissociation), and d d is what it actually is. The degree of dissociation, then, is d u dd (n-i)dd If a substance dissociates completely into two products, its vapor density is 1/2 what it should be; if into three, 1/3, etc. Examples. NH 4 C1 v.d= nearly 1/2 what it should be; hence NH 4 C1 = NH 2 CO 2 NH 4 *;.4 by heat becomes brownish, red, due to the for- mation of 2NG>2. At 500 C. this color disappears en- tirely, for then 2NC>2 breaks down into 2NO and O2, both of which are colorless. When a gas composed of a heavy metallic and a light gaseous element is dissociated by heat in an open tube, and cooled suddenly, the heavy element will crystallize out. This is due to the fact that the two gases go through the tube with a different velpcity and the sud- den cooling forms a compound of what is left, and the heavy element, being in excess, remains to a certain extent in the free state. An example of this is given by Marsh's test. The specific heat of a dissociating gas is very large owing to the fact that the heat supplied increases the dissociation as well as the temperature. When the 26 ELEMENTS OF PHYSICAL CHEMISTRY. dissociation is complete, however, the specific heat be- comes constant. Thus we have for acetic acid gas: / 129 160 200 240 280 Sp. h't 1.50 1.27 0.95 0.64 0.480 The specific heat of a dissociating gas also varies greatly with the pressure, as does the dissociation. The conduction of heat by a dissociating gas is very much larger than by a gas not capable of dissociating. The heat at the one side causes dissociation, and the products of dissociation go to the colder portion, where they unite, giving up heat and causing other molecules to dissociate, which in their turn diffuse and unite, etc. This process continues until the gas is all dissociated, when its conduction of heat becomes normal. Experiment shows that a dissociation takes place to a smaller extent when one of the products of the dissocia- tion is already present. Thus PCls is almost entirely undissociated in presence of an excess of chlorine or PCls, i.e., its vapor density under such conditions is found to be 104.5 m place of the calculated value 104. Later, under Chemical Change, we shall find a formula giving the relations between the original and final sub- stances in a chemical reaction for a constant temperature, and another showing the effect of a change in temperature, but here it is simply necessary to note that dissociation depends upon temperature, pressure and presence of THE GASEOUS STATE. 27 the products of the dissociation. The following tables will serve to show how great the effect of temperature and pressure is upon the dissociation of a gaseous substance: DISSOCIATION OF NITROGEN TETROXIDE, N 2 O 4 . (Density of N 2 O 4 = 3.18; of NO 2 +NO 2 =i.59; air =i.) Temp. Sp. Gr. of Gas. Percentage Dis 26. 7 2.65 19.96 35- 4 2.53 2 5- 6 5 39. 8 2.46 29.23 49. 6 2.27 40.04 60. 2 2.08 52.84 70. o .92 65-57 80. 6 .80 76.61 90. o .72 84-83 100. I .68 89-23 in . 3 65 92.67 121. 5 .62 96.23 135- .60 98.69 154. 58 IOO.OO DISSOCIATION OF PC1 5 . (Density PC1 6 = 7.2; PC1 3 +C1 2 =3.6; air-i.) Temp. 182 190 200 230 250 274 288 300 Density. Percentage Dissociation. 5.08 41.7 4-99 44-3 4-85 48.5 4.30 67.4 4.00 8o.O 3-84 87.5 3.67 96.2 3-65 97-3 28 ELEMENTS OF PHYSICAL CHEMISTRY. DISSOCIATION OF N 2 O 4 . (Equal Temperatures, Varying Pressures.) Temp. Pressure. Density (air = i). i8.o 279 mm. 2.71 17.3 i8.s 136 " 2.45 29.8 20. O 301 " 2.70 17.8 20. 8 153.5 " 2.46 29.3 The molecular weights (see definition, p. 7) of some of the elements in the gaseous state are given below and will serve to show the importance of the process of dis- sociation, and how dependent the molecular weight is upon the temperature. ARSENIC. tC. M. 644 309 As 4 = 3oo 670 308 As 2 =-isc 1715 X 57 1736 160 PHOSPHORUS. 313 128 500 126.1 P 4 =I2 4 1484 105 P 2 = 62 1678 93-3 1708 9i IODINE. 253 285 1330 162 We see from these that the higher the temperature the lower the molecular weight. Iodine at high tempera- tures is monatomic, as are mercury between 446-! 731, cadmium at 1040, tin at 1400, and sodium and potassium. THE GASEOUS STATE. 29 And sulphur in the gaseous state, according to the temperature, may be S$, S*i or $2, the latter being the condition at 800 and above. Mixtures of these three forms can also exist, at the lower temperatures, the molecular weight of these lying between the two extremes, Ss and 2; at higher temperatures, 1900-2000 C., the $2 is observed to break down further into the form of 25 (Nernst), although Biltz and Meyer found the formula weight to be 63.5 at 1719 C. Among compounds we find the chlorides of iron and aluminium, Fe2Cl6 and A^Cle only to exist at com- paratively low temperatures, being present in the forms FeCla and Aids at higher ones. 10. Volume, pressure and concentration. Thus far we have only used density to show the number of gram molecules present at any time. It is also possible, how- ever, to express this in other ways. The V in equation (9), it will be remembered, was used to designate the number of liters of space in which i mole exists, and consequently the volume of a gas or of a mixture of gaseous substances will give the number of moles present, provided the temperature and pressure be known. Since volume and pressure for any one temperature are inversely proportional, the number of moles can also be determined by the pressure at a known tempera- ture and volume. At times density, volume' and pressure can be employed with equal facility; at others, however, 30 ELEMENTS OF PHYSICAL CHEMISTRY. one of these is often found to be simpler in application than the others. This term concentration is used to express the number oj moles oj substance per liter, and is designated throughout this work by the letter c. ^ The relations of these quantities to one another is not difficult to express. Since the concentration of i mole per liter at o gives the pressure 22.4 atmospheres, and a pressure at other temperatures which is proportional to the absolute temperature, concentrations (pressures) can be transformed readily in pressures (concentrations). And the same is true of volumes under constant pressure. As it is quite necessary to become familiar with the use of these terms (and especially in their application to dissociation phenomena) in our later work, the following examples are given: At 190 C. the reaction PC1 5 PC1 3 + C1 2 goes to the right to such an extent that 44.3% of the original quantity of PICs is decomposed, i.e., a, the degree of dissociation, is 0.443. Starting with i mole of PC1 5 at 190 and atmospheric pressure, and assuming that no dissociation takes place, the volume occupied by the gas would be 22.4 --- liters. Since we lose 0.443 f tne penta- chloride and gain a like fraction of a mole of the tri- chloride and of chlorine, the final number of moles will be equal to (1-0.443)4-2X0.443, or 1.443. As the volume (at constant temperature and pressure) 5 is pro- THE GASEOUS STATE. 31 portional to the number of moles present, and this has gone from i to 1.443, the final volume will be 1.443 times the original volume, i.e., 1.443X22.4 liters. In case the volume remains constant, i.e., the internal pressure increases (assuming that the increase of pressure has no effect upon the dissociation) the pressure-change will be equal to the volume-change when the pressure remains constant. Since the original (i.e., undissociated) volume of the mole of PC1 5 at 190 273 + 100 , is 22.4 - - liters at atmospheric pressure, the final pressure in the volume of i liter will be I.443X 22.4 - atmospheres. Of this value, 1.443, the / O fraction 10.443 * s due to tne Pdo> -443 to PCla and 0.443 to tne chlorine. The partial pressures at equili- 270 -\- IQO brium, then, are 0.557X22.4 --- atmospheres of PCls, and 0.443X22.4 -- each for the trichloride -73 and the chlorine (Dalton's law). To express the quantity of these three substances in terms of concentration (i.e., in moles per liter) it is only necessary to divide the final number of moles of each substance by the total volume of the system. Thus of the mole of PCls with which we started we have but 10.443 mole at equilibrium, and gain 0.443 m ole of 32 ELEMENTS OF PHYSICAL CHEMISTRY. each of the two constituents. Since the final volume is 2*7 3 + IOO 1.443X22.4 - liters, the final concentrations are 273 1-0.443 273 + I 9 1.443X22.4-^^ Q-443 273 + 190 1.443X22.4-^- :C PC1 3 = ^chlorine. Here, naturally, the volume changes. In case the volume remains constant it is to be treated in the same way, for it is still the final volume. In the above 273+190 case it would be 22.4 . Example. 10 gr. of PC^ at atmospheric pressure (in a cylinder, for instance, which is provided with a movable piston) are heated to 250 C., at which temperature a is equal to 0.8. Find the partial pressure and concen- tration of each of the three gases present at equilibrium. If the PC\5 were undissociated at this temperature 10 273 + 2^0 , the 10 gr. would occupy -~ 22.4 liters, V l *o / o v 280.3 being the molecular, molar, or formula weight of the PCls. Since a = 0.8, at equilibrium we must have 1.8 -^ moles of the mixture of gases (i.e., ((i a) + 2a)n, where n is the number of moles present), in the final THE GASEOUS STATE. 33 volume 1.8 ( 22.4 -- ) liters, the total pressure \28o.3 273 / remaining constant. The partial pressures, then, are 1-0.8 0.8 0.8 ppch = I g i PPCI, =g and /> ch iorine = :j-g atmospheres. As there are i -0.8 moles of PC1 5 , 0.8 moles of PC1 3 , and 0.8 moles of chlorine the final concentrations (moles per liter) are - for PC1 5 , and . 28o. 3 273 0.8 *(; 10 273 + 250^ 22.4 each for PCla and chlorine. ^280.3 273 / Since in this case 2 moles are formed from each original mole decomposed, the volume of the substance on the right is greater than the volume of that on the left, and the reaction must go backward (i.e., toward the left), when the external pressure is increased. This law is general, and is expressed by Le Chatelier as fol- lows: "Any change in the factors of equilibrium from the exterior is followed by a reverse change within the system." In other words, pressure in the above case favors the formation of the substance with the smaller volume. (That a dissociating system is an equilibrium is shown by the fact that the direction of the reaction is dependent upon the direction of the temperature-change). Such a reaction as 2HI = H 2 +l2, on the other hand, '*' U-IMT 34 ELEMENTS OF PHYSICAL CHEMISTRY. where the same number of moles exist on either side (i.e., a reaction unaccompanied by any volume-change) will be uninfluenced by a change in pressure, as indeed is observed by experiment. ii. Variation from the gas laws. The equation of Van der Waals. The gas laws are simply limiting laws, and while they hold in general for pressures up to 2 atmospheres, above this value differences are observed. In the case of hydrogen the product pv is always higher than it should be. This was discovered by Natterer and ascribed by Budde to the fact that the molecules them- selves occupy some of the volume. If this correction for the volume is b y then, instead of (9), we have p(V-b)-RT, where b is a constant for each gas. From the formula RT b = -- + V P Budde calculated the value of b and found it to be equal to 0.00082, at which point it remained constant, for pressures varying from 1000 to 2800 meters of mercury. In the case of hydrogen this volume-correction is satis- factory; for other gases, however, the variation is some- what different. In general for these the compressibility at low pressure is much greater than is accounted for THE GASEOUS STATE. 35 by the Boyle-Mariotte law, reaches a minimum and then increases so that the product pV passes through the value given by the law. This shows that there is some factor in the behavior of these gases which is absent or negligibly small in the case of hydrogen. Van der Waals ascribed this to the mutual attraction of the molecules, which acts in the same direction as the pres- sure, making that larger than it seems to be.- If p is the pressure and a the specific attraction, then (10) which is the equation of Van der Waals. Here a is the molecular attraction when i mole occupies the volume of i cc., and b is the volume occupied by the molecules themselves, p and a are expressed most conveniently in terms based on po = i, where this is the original pressure in atmospheres when the gas is confined in the volume VQ. V and b are then expressed in terms of VQ. This law, although deduced by aid of hypothetical assumptions, will hold even in the face of evidence showing these assumptions to be false. For the terms b and a are determined by experiment (b and T^ expressing the discrepancy between the observed facts and the conclusions from the simple gas laws) ; and the worst that can befall them is a loss of name. 36 ELEMENTS OF PHYSICAL CHEMISTRY. By this law we can understand just how gases vary from Boyle's law. For small pressure and large volume the term ^ and b disappear in contrast to p and F, and the gas follows the simple law. The effect of the value of a and b upon the product pV is shown by (10) in the form or, for constant temperature, a ab When V is large and p is small ^ and bp disappear as compared with -~; when = (~^ + bp\ we have the minimum of the variation of pV and Boyle's law holds. These constants have been determined for the dif- ferent gases from the curves for pV } and a few of the values found are given below: 002=0.00874 802 = 0.039 Air =0.03 7 H = 0.0000 THE GASEOUS STATE. 37 Air = 0.0026 002=0.0023 ["=0.0067 The extent of the variation in the value of pV at con- stant temperature is shown in Figure i. If pV is con- A0 60 80 W 120 W 160 180 SOO 220 3M f60 W 300 FIG. i. NITROGEN. stant, the curve would be a straight line parallel to the />-axis (pV being the axis of ordinates). The higher the temperature, the more the gas behaves like hydrogen, The effect of the term 7^ is shown by the difference between the curve for N and for H. This difference can perhaps be shown in a more striking 3& ELEMENTS OF PHYSICAL CHEMISTRY. way by a comparison of the 'numerical values^ of pV. The following table is for N at 22, the values having been found by Amagat. p is given in atmospheres. p pv p PV I. I .0000 109.17 0.9940 27.29 0.9894 126.90 1.0015 62.03 0.9858 251-13 1.0815 80.58 0.9875 430.77 I . 2966 We see by this that up to about 12 atmospheres pV decreases in value, then follows Boyle's law, and then varies just like hydrogen. When a and b have been once determined for any pressure they may be used for all pressures. For ethylene at 20 by Van der Waals' equation, we have Obs. Calc. 31-58 914 %5 84.16 399 392 110.47 454 45 6 282.21 941 940 398-71 1248 "54 where p is in atmospheres, pV is multiplied by 1000, a = 0.00786, and = 0.0024. 12. Specific heat. The first principle of thermo- dynamics. When heat energy is applied to a body the temperature rises. The ratio of the amount of heat supplied to the consequent rise in temperature is called the capacity of the body for heat, i.e., . The value of THE GASEOUS STATE. 39 this term depends naturally upon the original temperature of the body, its pressure, etc. The s-pecific heat of any substance is the capacity jor heat of the unit of mass, i.e., _ m dt' It has been observed that the specific heat of a gas depends upon the conditions under which it is determined. If the gas is allowed to expand under its previous pressure the specific heat, C PI is different from that obtained when the pressure . varies and the volume remains constant, c v . Before considering the reasons for this, and finding the relation between the two values, it will be necessary for us to inquire into the nature of heat energy and its possible transformations. Mayer in 1841 was the first to develop the subject of the theory of heat and its transformations, or Thermo- dynamics, .as we know it to-day. Before that date heat was considered as an actual substance, which could be made to enter or leave a body. Mayer first recognized it as an energy, which could be obtained from any other energy or turned into the latter; his principal work was to determine the equivalent of heat energy, i.e., a factor by which heat energy can be given in units of mechanical energy. He was led to this conclusion by the point already mentioned, that the specific heat of gas for con- 40 ELEMENTS OF PHYSICAL CHEMISTRY. stant pressure is always larger than the specific heat of the same for constant volume. His reasoning was as follows: &ince the specific heat at constant pressure is always greater than that for constant volume, there must be some condition in the former state which absorbs this extra heat. In the case of constant pressure the volume increases, and work must be done to overcome the atmospheric pressure, which would keep the volume constant. It is a logical consequence, then, to consider that this extra heat is simply the amount necessary to do the mechanical work of expansion. In other words, a certain number of calories are found to be equal to a definite number of mechanical units; this value for one calorie is the mechanical equivalent oj heat. This term may be calculated as follows: The difference between the heats necessary to raise the temperature of i gram of air i C. under the two conditions is by experiment c p c v = o.o6g2 cal. This 0.0692 cal. is the heat which is equivalent to the work necessary to expand i gram of air 1/273 f its volume at o. Imagine i gram of air at o enclosed in a tube with a cross-section of i square centimeter. It will occupy the space of 773.3 cms., if the pressure is 76 cms. oijpflg, for i gram of air occupies under these conditions 773-3\cc., which is the specific volume of air. An increase of temperature of i C. will expand THE GASEOUS STATE. \ 41 this volume 1/273, *- e -> 2 -^3 cms - The weight of the atmosphere, 1033 grams, will then be raised >through this distance. The work necessary to do this i. 1033 X 2.83 = 2923.4 gr.-cms., which is equivalent to 0.0692 calorie. For one calorie, then, we have 2923.4 = 42245 gr.-cms., which is the mechanical equivalent oj heat. Joule transformed a known amount of mechanical work into heat by friction, and found from the heat developed that i cal. =42355 gr.-cms.* The great consequence . of Mayer's work is what is known as the first principle of thermodynamics. Accord- ing to this the energy of any isolated system is constant, i.e., when energy seems to disappear it is simply trans- formed into another form. If, for example, a gaseous body is heated, its internal energy is increased; if the * A liter-atmosphere is the work necessary to force the weight of the atmosphere through a liter of space. We have then loX 100 i L. A. = ' 2 ^ =24.2? cals., 42600 42600 being the mechanical equivalent now used, the average of a number of late determinations. R, the molecular gas constant, which is equal to 84800 gr.-cms., or 0.0821 L. A., is equal to 2 cals. 4 2 ELEMENTS OF PHYSICAL CHEMISTRY. volume increases, however, a certain amount of this heat is used to overcome the atmospheric pressure, and the increase of the internal energy is less than would other- wise be the case. If U is the internal energy and the amount of heat dQ is supplied to the body, then dQ where dW is the amount of work done by the body by virtue of the heat absorbed, and dU is the corresponding increase in the internal energy. If we consider the work as overcoming a resistance it is more readily handled. Suppose the gas to increase its volume, the pressure remaining constant; then for i mole dW=pdV, where p is the intensity factgr and V the capacity factor of volume energy. Substituting this value of dW in the former equation, we have (u) dQ where dU, dQ, and pdV are expressed in absolute units. If dU and pdV are given in mechanical units, then (n) becomes dU+pdV where A is the mechanical equivalent of heat, i.e., the value of i calorie. We shall use the form (n), however, THE GASEOUS STATE. 43 always remembering that dQ, dU, and pdV are expressed in the same units. The internal energy U of a gas may be considered as a function of pressure and volume, of pressure and temperature, or of volume and temperature. Of these we shall only consider the case where temperature and volume are variable. We have then dU The term ~T<7~dV, however, is equal to zero, for the internal energy of a gas, according to Gay-Lussac's experiment, remains the same after a change in volume provided no external work is done. By this experiment it was shown that the expansion of a gas into a vacuum does not change the temperature of the system ; although there is a slight increase in the temperature of the exhausted vessel and a slight decrease in the other, they compensate one another. Equation (u) becomes then If V is constant, i.e., dV = o, then 44 ELEMENTS OF PHYSICAL CHEMISTRY. This term, -T~T, the increase in the internal energy caused by an increase of temperature, is, however, the molecular specific heat for constant volume; hence (for constant volume). If the pressure remains constant and the volume varies, i.e., if dV does not equal zero, then pdV does not disap- pear, and we have (12) dQ or dT~ v dT' This term, -7=,, under these conditions is the specific heat at constant pressure, i.e., By differentiating (9), p remaining constant, we have pdV=RdT, or dV THE GASEOUS STATE. 45 Substituting this in (13), we find (14) C P = C V +R, when C P and C v refer to the specific heats for i mole of gas, and R is the molecular gas constant. If in the system no heat is absorbed or given up by conduction or radiation, .i.e., dQ=o, we have an adia- batic process. From (9), by complete differentiation, we have pdV+Vdp = RdT. If in (12) we eliminate dT, by this equation, remembering that we obtain (15) dQ= Since in an adiabatic process dQ=o, = j And, representing the ratio of the specific heats, T=T-, by k (for air = 1.41), we have Vdp+kpdV-o, 46 ELEMENTS OF PHYSICAL CHEMISTRY. or And by integrating between the limits p, V, and p\ 9 FI, we have log, p + k log, V = log, p 1 + k log, Fi, i.e., = ^_ Fl -' o or, in another form, From (17) we see that for an adiabatic process the volume changes less for a change in pressure, or the pressure changes more for a change in volume, than is the case when the temperature remains constant (i.e., an isothermal process). For rapid compressions or expansions, then, i.e., where no equalization of tempera- ture is possible, #!:#:: 7*: Fi*. If the process is very slow and the temperature remains constant, then we have (Boyle's law) UNIVERSITY THE GASEOUS + Boyle's law holds also of course for an adiabatic change after the temperature produced by the change of volume has been reduced to the original one. In equation (17) we know that and piV hence pV T_ piV^Ti But, by (17) V Whence, by substitution, V p By substituting the value of 77- instead of that of , we obtain The temperatures to be obtained by the expansion of a gas at the temperature T can be found from(i8). Examples of this are given in Table I. 48 ELEMENTS OF PHYSICAL CHEMISTRY. TABLE I. TEMPERATURES CAUSED BY ADIABATIC EXPANSION OF AIR. =1.4. p (atmospheres). t (initial from T). I' (final from TI). ioo 71. 5 -201. 5 200 58. 5 -213. 5 300 52 . O 221. O 400 47. 9 -225. o 500 44. 8 -228. 2 The temperatures are those of the air as it expands. Owing to the specific heat of the vessels used, it is im- possible to reach these temperatures in anything in contact with the gas, although very low temperatures may be thus obtained. Equation (14), i.e., C P -C,=R, shows that for all gases the difference between the molecu- lar specific heats is R = 2 cals. This equation can also be derived in the following simple way. The difference between the molecular heat at constant pressure and that at constant volume is obviously equal to the heat equivalent of the work of expansion. Since the volume- increase resulting from a temperature-increase of i C. is 1/273 times the volume at o C., i.e., , according / \j to the notation used above, and the amount of the in- crease is dependent only upon the temperature interval and not upon the actual temperature, the work consumed THE GASEOUS STATE. 49 in the expansion produced by an increase of i is . But ~ -=R (page 12), hence Mc P -Mc v = C p -C =R, where R is expressed in calories. Naturally the terms C p and C v are to be determined at the same temperature. That this relation holds is shown by the following list of experimentally determined values (Clausius-Ostwald). MOLECULAR SPECIFIC HEATS. Name. Const. Pres. Const. Vol. Ratio (fc). Oxygen 6.96 Nitrogen 6 . 93 Hydrogen 6 . 82 Chlorine 8 . 58 Bromine 8 . 88 Nitric oxide (NO) 6.95 Carbon monoxide , 6.86 Hydrochloric acid 6. 68 Carbon dioxide 9.55 Nitrous oxide (NzO) 9-94 Water 8.65 Sulphur dioxide 9. 88 4.96 .40 4.83 41 4.82 .41 6.58 30 6.88 .29 4-95 .40 4.86 .41 4.68 43 7-55 .26 7-95 25 6.65 .28 7.88 .25 The ratio k ( = J is used to find the number of combining weights contained in a formula weight of a gas. This is purely an empirical relation, it having been observed that known monotomic gases, mercury for example, give = 1.667, diatomic ones about 1.4 etc.; the 50 ELEMENTS OF PHYSICAL CHEMISTRY. greater the number of atomic weights to the formula weight, the smaller the value of the relation . An example of this application is given by Ramsay in the work on argon, where the combining weight, owing to the non-existence of any compounds with this, is un- known. 13. Determination of the specific heat of gases. The specific heat for constant volume c v cannot be de- termined directly with accuracy, since the vessel con- taining the gas absorbs so much more heat than the gas itself. Even under the most favorable conditions the ratio of the absorption of heat by the gas to that by the vessel is 1 155. From the specific heat for constant pressure and the ratio k y c v can, however, be found indirectly. C v The specific heat at constant pressure is determined by passin'g a certain volume of the gas, heated under constant pressure, to a certain temperature, through the worm of a calorimeter and observing the consequent increase in the temperature of the water. We krfow then the number of calories which causes a certain tern- < perature to exist in the known volume of the gas; and from this data it is easy to calculate the specific heat, c p . The value of the molecular specific heat for constant pressure for all gases seems to converge toward the value THE GASEOUS ST^TE. St 6.5 at the absolute zero, We have in general the relation where a is a constant for each gas and is the larger the more complex the formula. For a we find the following Values: for H2, N2, O 2 , and CO, a =-.001, NH$= 010071, CO 2 ^= 0.0084, C 6 H 6 = o.o5io, ether = 0.0738, H 2 O = 0.0089, C 2 H 4 =0.0137, CHCla = 0.0305, C 2 H 5 Br = 0.0324, CH 3 COOC 2 H 5 = 0.0674, C 3 H 6 O =0.0403, N 2 O =0.008^ This formula may be used in the following way to find the value c p . For CeHe, a =0.0510; hence at 50, since CM = C we have 6.5+0.0510X323 -- =0.295, while Wiedemann found 0.299 experimentally. The ratio = can be found easily by the method c v of Clement and Desormes. A glass balloon, holding about 20 liters, is provided with a brass stop-cock and a manometer. From this vessel the air is partially rarefied and the pressure ob- served by aid of the manometer. Let this initial pres- sure be pQ and the atmospheric pressure be P. If the cock is now opened for half a second air will rush in until the external and internal pressures are the same. As the air goes in, however, heat will be developed, 52 ELEMENTS OF PHYSICAL CHEMISTRY. which, as it is not removed, will increase the tempera- ture of the gas according to the law of adiabatic com- pression (p. 46). By the process we have increased the pressure from p Q to P. If the initial specific volume is v and the final one is v, then k can be determined from the equation The final specific volume is not known yet, however. To find this we wait until the flask and air have been reduced to the temperature of the surrounding air and again measure the pressure. If this pressure is p at the temperature /, then or po \po/ and log P-log log ^-log^o' In one experiment with air P = 1.0036, #o = o-9953> and p =1.0088 atmospheres; hence = 1.3524. 14. The second principle of thermodynamics. By the first principle we have found that a certain amount THE GASEOUS STATE. 53 of heat is equivalent to a certain amount of work. If we consider, however, the possible means of transform- ing heat into work we find that a certain condition must be fulfilled, i.e., the heat must go from a warmer to a colder body. In other words, heat can only produce work by going from a higher to a lower temperature. This condition has been expressed in other words. For example, heat of itself can never go from a colder to a warmer body without work being used upon it. A per- petual motion oj the second kind is impossible. A device to produce a perpetual motion of the second kind would be a machine which would run, for example, from the heat of the sea. It is related to the second just as an ordinary perpetual motion is related to the first principle. Since heat can only be transformed into work by a transference of heat from a higher to a lower tempera- ture, it is important to know the relation which exists between the amount of heat transferred as heat and that transformed into work. For this investigation it is necessary that all the work done be external work, which can be readily observed, and then compared with the known amount of heat. We employ for this purpose a process which was origi- nated in 1824 by Sadi Carnot and which is known as the Carnot cycle. The arrangement is such that the final state is identical with the initial one, i.e., before 54 ELEMENTS OF PHYSICAL CHEMISTRY. and after the operation the body contains exactly the same amount of energy. Then the relation between the heat transferred and the work done is very readily ob- tained. Naturally no heat must be lost by radiation or conduction, or the final result will be incorrect; for this reason the process is an ideal one and cannot be realized practically. Since all transference of heat and work is assumed to take place without differences in temperature and pressure the process may go in either direction, i.e., it is reversible. This condition of rever- sibility is never obtained in practice, so that the relation which we find is the limit under the most favorable conditions. 15. The cycle. Entropy. We assume the process to take place in four steps: i. Assume an ideal gas enclosed in a cylinder with a movable piston at a certain temperature and pressure. The cylinder is now placed in a heating-bath at the temperature 7\, and the volume is allowed to increase under a constant pressure which is just greater than that of the atmosphere. By this expansion the gas will cool. Here, however, we .assume heat to be absorbed by it to such an extent that the temperature remains constant. If the heat absorbed by the gas is Qi t its initial volume Vi, and its final one v 2 , and the constant temperature TI, then the work done by the gas will be THE GASEOUS STATE. 55 equal to jf 2 pdv. Since the temperature remains con- stant, equation (12) becomes rT or, since p=~ , or (20) Gi-rJ, lo&S 2. The gas is next allowed to expand adiabatically until the temperature falls to T 2 . For this, the new volume being v%, we have the relation 3. Next the pressure is increased until the volume decreases to v 4 , heat being removed to the amount Q 2y so that the temperature remains constant at T 2 . The work done here by the gas is J *pdv, we have then (22) 4. Finally, the gas is compressed adiabatically until the original volume v\ and the original temperature T\ 56 ELEMENTS OF PHYSICAL CHEMISTRY. are reached. This can be arranged by choosing v of the proper size. We have for this relation I*J V -1\ TI W We have thus carried the gas through a series of changes, and have finally the same state as that from which we started. The amount of heat Qi is absorbed at the higher temperature TI, and a smaller amount Q% is given up at the lower temperature T^ and a certain amount has been transformed into work. The amount of heat which is equivalent to the work done is equal to Q = Qi -Q 2 . The heat Q 2 has simply been transferred from the temperature TI to T 2 . The relation between Qi and Q 2 is given by equations (20) and (22) By (21) and (23), however, . -=-, Le, log,-=log,-; hence THE GASEOUS STATE. 57 i.e., the amounts of heat absorbed and liberated are pro- portional to the absolute temperatures. The amount of heat transformed into work is Q = (?i~ (?2- We have then Qi-Q2T l -T 2 and The heat transformed into work by any reversible process is to that transferred from the higher to the lower tem- perature as the difference in temperature is to the lower absolute temperature. Or from the other standpoint, The work in calories necessary to transfer a certain amount of heat from one temperature to a higher one by a reversible process is to the amount of heat as the temperature interval is to the final high absolute temperature. For T 2 = o, Q 2 will equal o and Q\ Q 2 = Qi, i.e., at the absolute zero all heat is transformed into work. For Ti T 2 . all higher temperatures ^ is smaller than i. 1 2 If the difference in temperature T\ T 2 is very small we can substitute for it dT, and as the difference between 5 ELEMENTS OF PHYSICAL CHEMISTRY. the amounts of heat will then also be small, we may substitute dQ for Qi Qz' we find then dQ_dT Q~ T' If now we consider the heat liberated as negative and that absorbed as positive, i.e., Q2 negative and Q\ positive, then 1,62 or or where there are as many terms as there are tempera- tures. Q is the amount of heat absorbed or emitted. Changing the sign of summation to that of integration, and we have /dO T =o. This is the analytical expression of the second princi- ple for reversible processes. In words ft means that for any REVERSIBLE process the transformation oj heat into work takes place in such a way that the sum of the amounts THE GASEOUS STATE. 59 of heat absorbed and liberated, each divided by its corre- sponding temperature, is equal to zero. If T remains constant dQ=o, i.e., just as much heat is liberated as is absorbed, and no heat is transformed into work. /j/~\ -~, found as above, was called by Clausius, the entropy. If this term is represented by s, then dQ When dQ = o, ds = o, i.e., when a substance by a re- versible change neither gives up nor absorbs heat its entropy remains constant. The two terms vary to- gether and have the same signs. For this reason adiabatic changes are also called isentropic. T always decreases in any change, for heat can only go from a dQ warmer to a colder body. ~r =s therefore increases during any change, i.e., the entropy tends toward a maximum. 16. The factors of heat energy. We have already found that the temperature is the intensity factor of heat energy. The capacity factor, entropy, is the ratio of the heat absorbed or liberated at constant temperature 60 ELEMENTS OF PHYSICAL CHEMISTRY. to the absolute temperature of the process. Therefore entropy times temperature is equal to heat energy We have in general, then, for all chemical processes ?-*;. dQ-Tds. CHAPTER in. THE LIQUID STATE. 17. Distinction between liquids and gases. Liquids are distinguished from gases by the fact that they pos- sess a volume of their own, which, although dependent upon pressure and temperature, cannot be changed to any extent by them. From the standpoint of experiment all we can say of a liquid, as compared to a gas, is that it contains less energy than the former, for energy is always absorbed by a liquid when it is being transformed into a gas. On hypothety ical grounds it is said that^ the attraction betw^fti tne molecules of a liquid is greater than between those of a gas. - ififtcfrth ty&tQ Tpf &&Wi Since the volume in the gaseous state is always greater than that in the liquid state, a greater temperature is necessary to cause a liquid to boil when the pressure over it is increased. And, conversely, a liquid boils more readily (i.e., at a lower temperature), when the pressure over it is decreased. (See Le Chatelier's theorem, p. 33). 61 62 ELEMENTS OF PHYSICAL CHEMISTRY. 18. Connection between the gaseous and liquid states. If at constant temperature a gas is subjected to a con- stantly increasing pressure, its state may change in one of two ways according to the conditions : a. This case has already been considered under gases. The volume at first changes more rapidly than the pres- sure, next in the same ratio, and finally more slowly. When the pressure becomes very high a further increase has but a slight effect upon the volume. b. Here the relation between pressure and volume is quite different, although the first step is the same, i.e., the volume changes more rapidly than the pressure. The ratio here, however, in contrast to a, increases con- tinually. When a certain pressure is reached a new phenomenon is observed: the gas is no longer homo- geneous; one part has separated which behaves differ- ently from the rest, i.e., the gas is partly liquefied. For a constant temperature this liquefying pressure remains constant, while the volume decreases steadily, i.e., for one pressure we have a whole series of volumes. An in- crease in the external pressure has no lasting effect upon the internal pressure (which still remains as it was), but it causes the volume to decrease more rapidly. Only after the whole gas has been liquefied can the pressure be increased. Then, however, we shall compress only the liquid, just as in the last stage of a we compressed thq gas. THE LIQUID STATE. 63 The condition which causes a gas under compression to follow a or b is the temperature. If this is above a certain point, which depends upon the nature of the gas, process a will be followed ; if below this point, process b. This was the first recognized by Andrews in 1871. If, for example, we compress CC>2 gas, keeping the tem- perature at o, the volume changes more rapidly than the pressure, and at a pressure of 35.4 atmospheres the gas condenses to a liquid. The higher the temperature, under 3o.92, the higher the pressure must be to cause condensation; but above 3o .Q2 no condensation is possible. Thus: t p 31, impossible to liquefy it. 30 .92 = 73.6 atmospheres. 30= 73-o 130.1 = 48.9 t p 2i=2i.5 atmospheres. -4o=n.o -59 -4 = 4-6 - 7 o.6 = 2.3 -78= 1.2 Andrews called the temperature 30. 9 2 the critical temperature. Correspondingly we call the pressure which is necessary to liquefy the gas, at the critical temperature, the critical pressure (73.6 atmos. at 3O.9), and the volume which the gas or liquid occupies (d t = d g ), under these two conditions, the critical volume. The best method of showing the behavior of a gas under compression is by plotting a curve in a system of coordinates where the pressures are laid out upon the ELEMENTS OF PHYSICAL CHEMISTRY. axis of ordinates, and the volumes upon the axis of abscissae. In these two curves the horizontal parts, which show constant pressure for varying volume, in- 4.05 iOO 95 90 45 SO 15 -10 65 -60 -55 -50 FIG. 2. dicate that the gas liquefies. The vertical parts refer to the liquid state, while all others refer to the gaseous state. Fig. 2 shows the behavior of CO 2 and air, the ordi- nates being pressures in atmospheres. At 31.! CO 2 there is no horizontal part, i.e., no varying volume for THE LIQUID STATE. 65 constant pressure; the gas does not liquefy, since it is above its critical temperature. For all temperatures below 31. i the horizontal part is present, i.e., the gas liquefies* under a sufficient pressure. Under all these pressures air remains a gas and behaves as CC>2 does when above 3iC. When a solution is heated to its critical temperature it is observed that there is no separation of solvent and solute, and that apparently the gaseous solvent has the same power to dissolve the substance that the liquid one had. 19. Vapor-pressure and boiling-point. The vapor- pressure of a liquid is the pressure to which, at any fixed temperature, it will evaporate in an exhausted space. If the vapor thus formed is removed, naturally more vapor will form, and by successive removals of portions in this way the entire amount of liquid may be evaporated. To cause a liquid to evaporate contin- uously, then, it is necessary to remove the pressure of the vapor above it. This may be done as above, or the same condition may be attained by making the exhausted space large enough to contain the vapor of all the liquid and yet not reach a high enough pressure to prevent further evaporation. This latter condition is indeed obtained in the open air when the vapor pressure of the liquid is increased to such an extent by heat that it exceeds the counter-pressilre of the atmosphere. The temperature 66 ELEMENTS OF PHYSICAL CHEMISTRY. necessary for this is called the boiling-point of the liquid, the boiling-point designating that temperature at which liquid and gas can exist together in all proportions. Every external pressure corresponds to a certain temperature, and these two magnitudes increase and decrease together. At the critical temperature the significance of vapor- pressure is naturally lost. For every pressure under the critical one, however, there is a certain temperature at which gas and liquid may exist together in all propor- tions: this temperature is the boiling-point at that pres- sure. Correspondingly, for eveVy temperature under the critical one, there is a certain pressure at which gas and liquid may exist in equilibrium in all proportions: this is the vapor- pressure at that temperature. Vapor-pressure of a liquid may be determined in one of several ways. One method of theoretical interest which gives good results is by J. Walker. A current of dry air is passed through the liquid which is held at constant temperature and the loss in weight of the liquid observed. The air in passing through the liquid will absorb an amount of vapor proportional to the vapor- pressure of the liquid. If V is the volume of air which in passing through the liquid absorbs the weight of I mole, then in that volume we have the relation pV = RT '(where p is the vapor-pressure). If v is the volume of air which absorbs x grams of the THE LIQUID STATE. 67 liquid with the molecular weight M in the gaseous state, then we have or The term v here represents the total volume, i.e., that of vapor and air, but that of the latter may be taken as a rule as the volume of the other is small in comparison. If we compare two liquids at the same temperature, passing equal volumes of air through them, L-?L Ml. p~M X x" or x M' where M and M' are the molecular weights of the sub- stances in gaseous form. This equation may also be used for one liquid alone, that is in the form Here naturally we must assume that the saturated vapor behaves as a more permanent gas would, i.e., obeys the 68 ELEMENTS OP PHYSICAL CHEMISTRY. gas laws. The only difficulty in the use of these formulas is the necessity o] having the same dimensions on either side. If p be expressed in atmospheres, V in liters and R in liter-atmospheres little difficulty will be experienced. >v oc and M are then found in grams, the ratio -TT being equal to the number of moles going info the gaseous state. Although this method has been used by several investigators, apparently with success, it has lately been the subject of a polemic by Carveth-Fowler and Perman (J. Phys. Chem., 8 and 9, 'o4~'o5). 20. The heat of evaporation. For the transformation of a liquid into its gaseous form a considerable amount of heat is necessary. There are two causes for this absorption of heat. The volume must be increased against atmospheric pressure, and the liquid must be transformed into the gaseous state. The former is of the least value. Its amount may be calculated from pV = RT = 2T cals.;* i.e., for every mole of water- vapor formed from liquid water 2 X (2 73 + 100) =746 cals. are used for the expansion. Since the heat of evapo- ration of i mole (18 grams) of H^O to steam at 100 is 9650 cals., and the amount necessary for the expansion * Where V represents the volume of the gas, which is so large that that of the liquid may be neglected. pV then represents the actual work of expansion against the constant atmospheric pressure, p, i.e., is equiva- lent to p(V g -Vi). THE LIQUID STATE. 69 is but 746 cals., it will be observed that the difference in energy -content of the two states is very considerable. The ratio of the molecular heat of evaporation (= mo- lecular weight multiplied by the heat of evaporation of i gram) to the absolute boiling-point has been found experimentally by Trouton to be a constant for a very large number of substances, and equal to approximately 21, i.e., Mw By aid of this equation unknown heats of evaporation may be calculated (with a possible error of 5%) from the molecular weight and the boiling-point. By this formula it is also possible to find the molecular weight of a liquid, provided we know the latent heat of evapo- ration. Longuinine found in this way that liquid acetic acid molecules are twice the size of the gaseous ones, i.e., instead of finding 21, using the molecular weight as found from the gaseous state, he obtained about 13. If the molecular weight used is doubled, 26 instead of 21 is found, showing that undoubtedly an association does take place. Van Laar explains the maximum density of water at 4 as dependent upon association which increases with decreasing temperature. The liquid itself without association would increase in density; the association 7o ELEMENTS OF PHYSICAL CHEMISTRY. causes the density to decrease. At 4 these two influence just compensate each other, and the liquid has its max: mum density.* Below this point the influence of th association is the stronger, and consequently the volurn increases. Young and Thomas have also found an empirics relation from which molecular weights of liquids ma be determined. It is MP k D k T k ~ where M is the molecular weight of the liquid, P k th critical pressure, D k the critical density, and T k th critical temperature (absolute). We have for CCl 22XD k XT k 22X0.556X556 M = - n - = - = I c ?Ii Pk 44-9 Theoretical = 154, and for acetic acid , 93> Theoretical=6o . We see then from this that acetic acid is associated eve at the critical state. * 1 8 gr. H 2 O going into the state (H 2 O) 2 would increase in volume b 8.44 cc., 46 gr. C 2 H 6 O in (C 2 H 8 O) 2 by 20 c., etc. Van Laar also explair v the increase in volume when alcohol and water are mixed to a mutu< breaking down of the associated forms (H 2 O) 2 and THE LIQUID STATE. 71 21. The relation between vapor-pressure, heat of evaporation, and temperature. An equation showing the relation between these three quantities can be found as follows: Equation (6) gives the relation between any two kinds of energy when in equilibrium. We have SdT=cdi=vdp, or, since S can only be determined as a difference, (Si-S2)dT=(vi-v 2 )dp, where Si is larger than S 2 and refers to the state which absorbs heat when formed (gaseous). Since S varies with the amount of heat, we may write (01 C?2\ W } we may substitute -~, where w is the latent heat of vaporization for i gram of liquid. We have then w dp T(vi-v 2 ) dT T(v 1 -v 2 ) = df w = ~dp' Since the volume of gas is large compared to that of the liquid, we may neglect the latter, and, provided the gas obeys Boyle's law, find the former from pV-RT, F= . 7 2 ELEMENTS OF PHYSICAL CHEMISTRY. We have then, changing w to /, i.e., the molecular latent heat, and vi to F, which is the volume of i mole of the gas, JL_dp_ ,1 2T 2 dT ^f~ 2== df ~pT = Jp' It is necessary to remember in using these equations that 2 (=R) and / are in calories and p in mechanical dp units, -j~ gives us the change in vapor-pressure for i dT of temperature, and -j gives the change in boiling- point for unit of external pressure. To get a correct re- sult, then, care must be taken with the dimensions (see pp. 68 and 43). Example. The boiling-point of ethyl iodid at 73.3 cms. is 72.2 C., / = 7597; find change in vapor-pressure for i C. We have 2X(345-4) 2 Another relation involving the heat of evaporation has been derived by Crompton (Proc. Chem. Soc., 17, 61, 1901) and gives very good results. Assume that in a saturated vapor the equation pV=RT holds, and that it is possible at constant temperature and by compression alone to reduce its volume V g to that occupied by the liquid F/, without any change in state occurring and THE LIQUID STATE. 73 the gas continuing to obey the law throughout the com- pression. The work done in causing this change of volume will be equal, then, to /v g RT fW- and as no change in temperature takes place heat equiva- lent to this will be given out. The gas will now occupy the volume which would be occupied by the liquid, but of itself it will not retain its volume. In order to obtain it in such a state that it will do this it is necessary to remove its ability to expand. By removing an amount of energy equal to that expended in compressing the y material, viz., RT log/rr, we shall, however, also re- move this power of expansion. The total heat given out, then, by the production of the liquid from the vapor y is 2R T log/TT, and this is equal to the latent heat of evaporation of the molecular amount, i.e., y or, since the ratio of the molecular volumes -*?- under normal conditions is the same as that of the specific 74 ELEMENTS OF PHYSICAL CHEMISTRY. volumes, or inversely as the specific densities,, and using Brigg's logarithms, we have . -4343 Thus, since for CC>2 at 7^ = 248, ^=1.10, d g = 0.044, M = 44, we find 7^ = 71.91 calories, where the experi- mental value is 72.23. Any association in the liquid state naturally fails to be accounted for in this equation. For liquids which are shown in other ways to be associated (i.e., apparently larger in molecular weight than in the gaseous state) the calculated value is found to be too low, the molecu- lar aggregations on liquefaction apparently involving the evolution of heat. This formula, according to Mills, gives very accurate values at temperatures at which the vapor-pressure is high, but values which are too large for those temperatures giving low vapor-pressures; the divergence being greater than one calorie only for a few cases. (J. Phys. Chem., 8, p. 593, 1904.) The term =- of Trouton's law (p. 69) is equal according to this equation to 2R(\og e V g log e Fj), which for many liquids gives values lying between 22.8 and 23.9. The heat of evaporation, however, also varies with the temperature. The expression for this variation is THE LIQUID STATE. 75 derived as follows : (i) Allow i mole of liquid to vaporize at the boiling temperature T, and then heat this volume, holding it constant, to the temperature T + t. The gas in expanding has done the work pV=RT and has ab- sorbed the heat C v t in going to the higher temperature, while the latent heat of evaporation is IT- (2) We reach the same final state by first heating the liquid to T+t, by which the heat Me is absorbed, and then allow the gas to form at T+t. The work of expansion here is pV=R(T + f), and the latent heat l T+t . By these two processes we have reached the same final state from the same initial one, so by the principle of the conservation of energy the loss in energy is the same in both cases. We have then -(l T + Cj)=RT+Rt-(l T + t +Mct), or and, by (14), dl dw c ~ c = i.e., the latent heat of evaporation increases for each degree by an amount equal to the specific heat of the gaseous state at constant pressure, minus the specific heat of the liquid. Since the specific heat of the gas- is 76 ELEMENTS OF PHYSICAL CHEMISTRY. always less than that of the liquid, the increase is negative, i.e., the heat of evaporation always decreases with in- creasing temperature, c for benzene at 50 is 0.45, c p for 50 (see p. 51) =0.295, hence ^ = 0.295-0.45= -0.155, for which 0.158 has. been found by experiment, the difference being within the experimental error. The substance must have the same molecular weight in both the gaseous and liquid state if correct results are to be obtained by this formula. If associated in liquid form, the result here will be too small, owing to the fact that Mc = C will be smaller than the proper C. It has been found experimentally and can be proved theoretically that the specific heat in the liquid state and that in the gaseous state bear a constant ratio to each other, so that unless the molecular weight is different in dw the two states, -j~ is the same for all temperatures. For ethyl alcohol there seems to be an association at 10; we have / = 220.9 at > 2 1 1. 2 at 10, and 220.6 at 20. 22. Refraction of light. La Place, in determining the velocity of light in different media, found the relation 7^2 j -j = constant for all temperatures when n is the refraction of light and d is the density of the medium. For many liquids this holds true, approximately, THE LIQUID STATE. v 77 but Gladstone and Dale found empirically a relation which holds more exactly and for many more liquids than the former. This is n-i -j- = constant for any one liquid. Further, in the case of a mixture, the constant is found to be equal to the sum of those for the ingredients. We have then H I HI I W/2 I where p is equal to 100, and pi and p 2 represent the percentages by weight of the two ingredients. By aid of this it is possible to make an optical analysis. Example. An unknown mixture of amyl and ethyl alcohols gives n = 1.3822, ^=0.8065. What is the per- centage composition of each in the mixture? For amyl alcohol n\ = 1.4057, di =0.8135; f alcohol #2 = i -3606, or^=( =-) , where y is the factor of association. Later work by Ramsay and Aston has resulted in substituting for the above relation for y another according to which y = i ' (i + fit) ) , when /* is a constant depend- ing upon the nature of the liquid. // is not as yet settled which of these relations is correct, if either, so that appar- ently it is not yet possible to accurately determine y from the variation from the normal value 0} k. An example of the use of these two equations will per- haps make them clearer. The radius of a capillary tube is 0.01843, at 4 6 > ^ ( tne height of CC1 4 in the tube) = 11.643 cm -j 5 ( tne density) = 1.5420, hence x = \ hrs Xp8o. = 22.9 dynes per centimeter. Since the critical tempera- ture =283 and k = 2.i2, x(Mv)$ = 4g2.6 and M = i^. 82 ELEMENTS OF PHYSICAL CHEMISTRY. 2.12 Forwaterat 10 K = o.Sj y hence ^ = 3.81 (from -^r- = y$) , 0.07 consequently the association is such that the average molec- ular weight of liquid water is 3.81X18. This could be caused by a mixture of associated and simple molecules. The results given below will show the molecular weights of some of the more common non-associated substances, and the factors of association of some of those which are associated, as determined by both formulas. (Ramsay and Shields, Zeit, f. phys. Chem., 12, 431-755, 1893); Ram- say and Aston, 'Proc. Roy. Soc., 56, 162, 1894.) NON-ASSOCIATED LIQUIDS. CSg. Carbon disulphide. t #(dynes per sq. cm.) x(Mvft. yXM i9-4 33-58 515.4 46. i 29.41 461.4 1.07X76 N2O 4 . Nitrogen peroxide. i.6 2 9-5 2 461.9 19. 8 26.56 423.5 1.01X92 SiCl 4 . Silicon tetrachloride. i8.9 16.31 383.9 45- 5 13-66 329.8 i. 06X169. < PC1 3 . Phosphorus trichloride. i6.4 28.71 562.3 46. 2 24.91 499.8 1.02X137.2 CC1 4 . Carbon tetrachloride i .01 X 153 . ( C 6 H . Benzene i .01 X 78 (C 2 H 5 ) 2 O. Ethyl ether 0.99X74 C,H 6 C1. Chlorbenzene 1.03X112^ C 6 H 5 NH 2 . Aniline i .05X93 C 5 H 5 N. Pyridin 0.93X79 Hg.* Mercury (liquid) 171 * Kistjakowski, J. russ. phys. chem. Ges., 34, pp. 70-90. This value ii found from surface-tension assuming the critical temperature to be about 700 C THE. LIQUID STATE. ASSOCIATED LIQUIDS BY FORMULA y = \~jr-) ALCOHOLS. Name. 16-46 yXM 46-78 yXM 78-132 yXM Methvl 3.43X 32 3.24X 32 2.8oX 32 Ethvf 2.74X46 2.43X46 I .07X46 Propvl 2 25X60 2 31X60 Isopropvl 2 86X60 2 72X60 Butyl '. . . I 04X74 I 72X74 I 76X74 Isobutyl Arnvl . . 1.95X74 1.97X88 1.86X74 1.69X88 1.64X74 I.S7X88 Allvl I 88X58 I 86X58 Glvcol 2 92X62 2 48X62 > 12X6-* Formic ACIDS. 3 61X46 3 13X46 Acetic 3 . 62 X 60 3 t2X6o 2 77X6O Propionic .77X74 .78X74 88X74 Butyric 58X88 73X88 69X88 Isobutyric . . . 4.^X88 82X88 77X88 Valerianic Capronic .36X102 -46X 116 .37X102 47X 116 .70X102 49X116 WATER. t * *U/tO* yXM 10 20 30 40 50 60 70 80 90 73-21 71.94 70.60 69. 10 67.50 65.98 64.27 62.55 60.84 58.92 502 . 9 494-2 485-3 476.0 466.0 456.7 446.4 436.0 425-9 414.3 3.81X18 3-68X18 3-44X18 3.18X18 3-13X18 3.00X18 2.96X18 2.83X18 2.79X18 2 66X18 100 57-15 403-7 2.61X18 110 120 130 I40 55-25 53-30 51-44 49.42 392-3 380.4 369.2 356-8 2.47X18 2.47X18 2.32X18 8 4 ELEMENTS OF PHYSICAL CHEMISTRY. FACTOR OF ASSOCIATION BY FORMULA >'= - (i+ /*/) . Methyl Alcohol. Ethyl Alcohol. Water. / y t y t y -89.8 2.65 -89.8 2.03 o .707 + 20 2.32 + 20 1.65 20 .644 70 2.17 40 59 40 .582 90 2 . II 60 5 2 60 5 2 3 110 2.06 80 .46 80 463 I 3 .00 IOO 39 IOO 405 *5 94 120 33 120 346 170 .89 140 .27 140 .289 1 80 .86 1 60 .21 190 83 1 80 15 200 .81 200 .09 2IO .78 210 .06 220 75 22O 1.03 230 I .00 Measurements have also been made by Bottomley in this way of the factors of association of fused salts, using , 2.12 if the equation y = ] r- His results are as follows; POTASSIUM NITRATE, KNO 3 . t X (Mi,)* K yXM 338 406 109.8 105.8 i57i 1539 0.471 9.55X101 349 414 106.4 loo. 7 1521 1485 0.584 7.49X101 341 407 106.4 104.6 J 55i I 5 I 9 0.485 9.145X101 THE LIQUID STATE. SODIUM NITRATE, NaNO 3 . f X (AfiO* K yXM 339 106.4 405 101.8 1374 1308 0.500 8.73X85 3 2 9 405 no. 8 106.5 1398 1369 0.381 13-13X85 344 III .0 1403 -453 10.17X85 For liquefied gases Baly and Donnan found slightly lower values for k than 2.12. Their values are: Substance. 15 J: Oxygen Y J-9 1 ? Nitrogen 2 . 002 Argon 2 . 020 Carbon monoxide i . 996 Oxygen and nitrogen apparently cut the temperature axis at points which are practically coincident with their critical temperatures, while argon and carbon monoxide cut the axis at temperatures lower than their critical values (9 for argon and 5 for carbon monoxide.) For liquefied carbon dioxide Verschaffelt has found the value = 2.222, and for nitrous oxide, = 2.198, in both cases the temperature axis being cut 6 below the criti- cal value. These results, together with the one for liquid mercury given above, show that this method in its approximate form (i.e., for molecular weight determinations) is appli- cable to all liquids. It has been shown by Ramsay to hold also for liquid mixtures. 86 ELEMENTS OF PHYSICAL CHEMISTRY. 24. Molecular weight in the liquid state. Critical temperature. It will be seen, thus, that in reality we have but two methods for determining the molecular weight in the liquid state, viz., the relations of Young and Thomas, and that of Eotvos and Ramsay and Shields. And as the former is only applicable to the critical state, the number is reduced to one when any range of tempera- ture is to be employed. It is true that we can tell from div the ratio - for different temperatures, whether the ratio of the molecular weights in the liquid and gaseous states remain constant, but further than that the relation is valueless in this respect. In fact even the ratio between the two molecular weights is not always easy of interpre- tation, for both may vary, as will be seen from the follow- ing results for acetic acid. FACTORS OF ASSOCIATION. t Liquid State. Gaseous State. 20 2.13 .98 40 2.06 87 60 99 79 80 .92 73 100 .86 .70 120 79 .68 140 .72 .67 The relation of Trouton (p. 69) also indicates whether the substance is associated or non-associated, but its results are not very accurate, so that the method by surface- THE LIQUID STATE. 87 tension is actually the only method which is accurate and applicable to any very great extent. We can now define molecular weight in the liquid state ; in other words, we can obtain a relation for liquids which may be used in an analogous way to the one com- monly employed for gases. It is at once apparent that, according to our present use of the term, the molecular weight of a liquid at any temperature is that weight in grams which at that temperature will occupy such a volume V (in cc.) in the formula xV$ = k(r d) that k is equal approximately to 2.12 ergs. For this purpose it is necessary to know the surface-tension at that temperature and the term T, i.e., the difference between the critical temperature of the liquid and the temperature of the experiment. As it is not an easy matter to determine the critical temperature experimentally this definition is not so convenient for use as the one given below, although in essence both express the same facts. By the determination of the surface-tension and specific volume of a liquid at two different temperatures, how- ever, the difficulty of the above definition is avoided, and, at the same time, it is possible to determine the critical temperature of the liquid with considerable accuracy. Applying the formula already used for the same liquid at two temperatures, which do not lie too far apart, we obtain ELEMENTS OF PHYSICAL CHEMISTRY. and which when combined give where A(xV*) denotes a finite change in the term corresponding to a finite change of temperature. In this form it is unnecessary to know the critical -temperature- of the liquid. By a slight rearrangement we obtain --2.12, i.e., the temperature coefficient oj the molecular surface- tension for all liquids is 2.12 ergs. According to this, then, the molecular weight in the liquid state is that weight in grams which at that temperature gives such a volume that a change in temperature of i involves an amount of surface-work equal to 2.12 ergs. This is similar to a definition which might be used for the gaseous state, i.e., the one expressive of the amount of external work in- volved by the change in temperature of i mole of gas i. This, according to p. 49, is equal to R, and we may say the molecular weight in the gaseous state is that weight in grams occupying such a volume that a tempera- ture-change of i will involve the external work R, i.e., 2 calorics, 0.0821 liter atmospheres, or 8.3iXio 7 ergs. THE LIQUID STATE. 89 As already mentioned this J-formula may be used to determine the critical temperature. To do this we substitute for V its value, assuming the gaseous molecular weight, and thus find the specific value of k for this liquid (i.e., the value actually found in place of 2.12). Dividing the value of x(Mv)* by this new value for k and adding 6 as well as the temperature of observation to this quotient we obtain the critical temperature. In other words, we find k by the J-formula and substitute it in the original form x(Mv)* = k(T d). An example will make this quite clear. At i9-4, for CS 2 , # = 33.58, at 46.!, # = 29.41 dynes. We have then for x(Mv)* at the two temperatures and \* X 29.41, 461.4. / 76 \* I Vl.22' 223 Where 1.264 an d 1.223 are the densities at the two tempera- d(x(Mv)*) . 515.4461.4 54.0 tures. v ,' is then equal to J7 46.1 19.4 26.7 2.022, i.e., the value for k in this specific case. The critical temperature of CS2 is then equal to ^ - +6 + 19.4 = 28o.3 C. There is a slight uncertainty here on account of the value 6 for d, for this is not the same for all liquids; the possible error is small, however. po ELEMENTS OF PHYSICAL CHEMISTRY. For T less that 35 the formula does not seem to hold. And this is also true for associated liquids, only non- associated liquids allowing this determination of critical temperature to be made. Some of the critical tempera- tures determined in this way by Ramsay and Shields (I.e.) are given below and will serve to show the value of this method, which has completely replaced the older and less accurate relation of Schiff, based upon the empirical formulae x t = XQfit and at 2 =a^ 2 ^ f t, where x is the surface-tension, a is height of liquid standing in a capil- lary tube of J mm. radius, and /? and /?' are the respec- tive temperature coefficients. CRITICAL TEMPERATURES. Substance. Calculated. Observed. Carbon disulphide, CS 2 ............ 280 .3 275 273 277.7 271.8 279.6 Nitrogen peroxide, N 2 O 4 .......... * 226.4 171.2* Silicon tetrachloride, SiCl 4 ........ 213-8 230 221 Phosphorus trichloride, PC1 3 ....... 290 . 5 285 . 5 Benzene, C 6 H 6 ................... 288 288. 5 Chlorbenzene, C 6 H 6 C1 ............. 359-7 360 Ethylacetic acid, CH 3 COOC 2 H 6 . . . . 250.5 251 Ethyl ether, (C2H 5 ) 2 O ............ 191.8 194 . 5 Carbon tetrachloride, CC1 4 ........ 282. i 283.2 Paraldehyde, C 6 H 12 O 3 ............. 310.5 * The calculated value here does not account for the dissociation into The value 171.2 is the critical temperature of a mixture of NjC^ and NOj. CHAPTER IV. THE SOLID STATE. 25. Remarks. A solid is a substance which pos- sesses a form of its own, i.e., its shape is not dependent upon that of the vessel in which it is. Just as we find that a substance contains less energy in the liquid state than it does as a gas, so we find that a solid contains less energy than a liquid, i.e., in going into the liquid state, it absorbs heat-energy. 26. Atomic heat. Law of Dulong and Petit. The atomic weight of all elements have approximately the same capacity for heat. A few have been found to give different values, but in general for all elements (except boron, silicon, and carbon) we find that atomic heat = atomic weight X specific heat = 6.34. This is the law of Dulong and Petit. By the aid of it it is possible to determine the atomic weight from 91 92 ELEMENTS OF PHYSICAL CHEMISTRY. the experimentally observed specific heat, i.e., for i gram, we have 6.34 atomic weight = r^ r 7- 7 \ specific heat (i.e., tor i gram; According to the law of Joule (1844) the heat capacity of a compound is equal to the sum of the heat capacities of its constituents. The specific heat of a compound, then, will show the approximate number of atomic weights of which it is composed; and this represents the only method we at present possess for determining the molecular weight in the solid state. In other words, the molecular weight of a compound in the solid state may be defined as the weight which has a heat capacity ap- proximately equal to the sum of the atomic heats, as determined from the ratios found by analysis. This definition is not as convenient in form as it might be, but then its importance is not so great as those for the gaseous and liquid states. Whether the relations as expressed above are correct or not is impossible to say; at any rate this method is the basis of the formula weights in the solid state as we use them. An example will serve to make the relation clearer. Analysis shows that lead chloride contains 103.5 P ar ts of lead to 35.45 parts of chlorine. The question then arises as to the value of the atomic weight of lead, whether 103.5, 20 7> or 3 IO -55 m other words, is the formula weight of lead chloride PbCl, PbCl 2 , or PbCl 3 ? THE SOLID STATE. 93 Since the specific heat of lead chloride is 0.0664, accord- ing to the above relations, we have: Molecular Molecular Heat Sum of Weight. =M X o.o 664. Atomic Heats. PbCl i3-5 + 35-45 9-3 2+6.4=12.8 PbCl 2 207.0+70.9 18.5 3X6.4=19.2 PbCl 3 310.5+106.3 27.7 4X6.4 = 25.6 Since the formula PbC^ gives a molecular heat ap- proximately equal to three times the atomic heat (6.4), this is assumed to be the correct formula, i.e., the molec- ular weight is 207. + 70.9 = 277.9. 27. Changes in the state of aggregation. The most important process concerning solids, for our purpose, is their formation from the liquid state, or the formation of the liquid state from them. If a liquid is cooled very carefully it is possible to reduce its temperature below the solidifying-point and yet have no solid formed. If the crystal of the sub- stance is thrown in, crystallization takes place imme- diately, and the temperature rises rapidly to the true solidifying-point. Besides this change in temperature we have also one in volume, since the specific volume in the one state differs from that in the other. When a gas condenses and goes into the liquid state heat is given off. In the same way heat is also given off when a solid is formed from a liquid, and this is the cause of the rise in the temperature spoken of above. The heat which is liberated by a liquid solidifying (or 94 ELEMENTS OF PHYSICAL CHEMISTRY. absorbed by a solid liquefying) is called, for i gram of substance, the latent heat of solidification (or fusion). This heat, referring to i mole of substance (heat for the molecular weight), is called the molecular heat of solidification (fusion). The effect of pressure upon the solidifying-point may be derived by the consideration of equation (6), just as we did for liquids, p. 71, or in the following way: Starting with i gram of a liquid at the temperature T, occupying the volume v } and under its own vapor-pres- sure p, cool 4T, i.e., to T 4T and assume it to be completely transformed into the solid state. During this process an amount of heat equal to w calories at the temperature T AT will be evolved, the volume will change from v\ to v 2 , the vapor-pressure from p to p dp, and an amount of work equal to (v\v^{pAp) will be involved. Now heat the solid thus obtained to T and allow it to melt at the pressure p to form the original volume, v\. This change in volume will be the same as the previous one, except that it has the opposite sign. The heat absorbed at T is so nearly equal to w that the difference may be neglected when AT is small, and the work in- volved is p(viv 2 ). The total work of the whole pro- cess, i.e., after the original state is once more attained, is Jp(vi-v 2 ), i.e., p(vi-v 2 )-(p-Ap}(vi-v 2 ) J and by the second law (page 58). THE SOLID STATE. 95 (24) or Jp _ w AT T(vi-v 2 ) = J = w Vi being the specific volume of the liquid, 1/2 that of the solid substance, and w the latent heat of fusion for i gram. This formula is not only of use in case of a change in the state of aggregation, but also for any reversible change, for example, that taking place between allotropic forms of the same substance. Thus at 95.6 C. the rhombic form of sulphur is transformed into the mono- clinic form. The reaction is perfectly reversible, an in- crease of heat increasing the amount of the monoclinic form, a decrease increasing that of the rhombic. The change in volume being 0.014 cm - an d the heat of trans- formation per gram 2.52 cals. we have, AT TJv (273 + 95.6)0.000014 4p~ w ' 2.52X0.041 ~' 5 ' where the term w is transformed into liter-atmospheres, and the result given in degrees per atmosphere, i.e., an increase of pressure of i atmosphere increases the transformation temperature by o.o5, a result which agrees well with the experimental value. 96 ELEMENTS OF PHYSICAL CHEMISTRY. If the specific volume of the solid is greater than that of the liquid, i.e., if the solid floats upon the liquid (ice in water), the freezing-point will be depressed by pressure. If, on the other hand, the specific volume of the solid is less than that of the liquid, i.e., if the solid sinks, the solidifying-point will be raised by increased pressure. In the former case the pressure will tend to keep the sub- stance in the liquid form, since that has the smaller volume, and a lower temperature will be necessary to produce the solid. In the latter case the formation of the solid will be aided by the pressure, and consequently the temperature need not be lowered to such an extent. This is simply another application of the principle of Le Chatelier (page 33). The actual effect of pressure upon the solidifying-point, however, is not large. Thus an increase of pressure up to 8.1 atmospheres depresses the freezing-point of water but 0.059 c - Equations (24) and (25) give us also the heat of subli- mation, i.e., of the direct transition from the solid to the gaseous state, if the volumes represent those of solid and gas; or the volume of gas alone may be used by substituting for v the value found from pV = RT, changing w to /, i.e., the gram heat to the molecular heat. The sublimation-pressure of a solid is to be considered just as the vapor-pressure (i.e., evaporation-pressure) of a liquid. It is noticed that some solids sublime while THE SOLID STATE. 97 others melt and evaporate. The reason for this difference is the fact that when the sublimation-pressure exceeds the pressure of the atmosphere sublimation takes place and thus keeps the solid from melting. When heated in closed vessels, however, the pressure becomes great enough to prevent continuous sublimation, so that after a small portion sublimes and raises the pressure, the remainder will melt. This can be shown experimentally by heating iodine, which usually sublimes and does not melt, under sulphuric acid, the pressure of which prevents sublimation and allows fusion to take place. Since at the solidifying- or freezing-point the solid and liquid exist in contact in all proportions, the vapor- pressure of the solid must be equal to that of the liquid, both at the same temperature. That this must be so follows from the following process : Imagine in the annular- shaped vessel, Fig. 4, water at 6, ice at a, and the vapor FIG. 4. of the two at c : all being at the temperature of o C. If the vapor-pressure over the water at b is greater than 9^ ELEMENTS OF PHYSICAL CHEMISTRY. that over the ice at a, a distillation from b to a must take place. By the evaporation at b heat will be absorbed, and by the condensation of this vapor to ice at a, under the diminished pressure, heat will be liberated. A layer of ice will thus be formed at b, while the heat liberated will melt an amount of ice at a. We would have, then, an exchange of heat at the same temperature. If this were possible, perpetual motion would be possible; since, however, the latter is impossible, the vapor-pressure of the solid must be the same as that of the liquid, both having the same temperature. The latent heat of fusion of a solid varies with the temperature just as does that of evaporation, and the expression for it may be derived in the same manner. We assume here that all change takes place at constant volume, i.e., that no external work is done, (i) One gram of solid is allowed to melt at its fusion temperature T by which the heat absorbed is w^, and the resulting liquid heated to T + t, which absorbs the heat at. (2) The gram of solid is first heated to T + t, which absorbs the heat c s t, and then allowed to melt, absorbing the heat wr+f Since the loss in energy by the two processes must be the same, we have and THE SOLID STATE. 99 By this we sec that if the specific heat of the solid is larger than that of the liquid state, the heat of fusion decreases with increasing temperature. In the case of H^O r/ = i, ^=0.5; hence jy, = o.5, i.e., the heat of evaporation changes by 0.5 cals. for a change in temperature of i. In this case there is a decrease in heat with a decrease of temperature, i.e., temperature and heat increase together. If the sign of Aw 77^ were minus this relation would be just the contrary, and the heat would decrease with an increase in tempera- ture. The latent heat of fusion for water, then, is Thus when freezing at 4, caused by pressure of 530 atmospheres, w w = f jS. From the relation JT(cic 8 )=w we can find the change in temperature necessary to reduce the latent heat of fusion to zero, i.e., the temperature at which it passes through the value zero and changes its sign. The variation in the latent heat of sublimation with the temperature can be calculated by the formula used for the variation of the heat of vaporization, the specific heat of the solid being substituted for that of the liquid (76.) Although in the derivation on page 94 we assumed that loo ELEMENTS OF PHYSICAL CHEMISTRY. all the liquid solidifies at the temperature TAT, this could not happen instantaneously. There is, in fact, a definite relationship between the number of degrees of overcooling and the fraction of liquid which as a result separates at once to form solid. That this must be sc is obvious after a moment's thought. Since the freezing- point of a liquid is that point at which solid and liquid can exist together in all proportions, it must be the final point attained after any overcooling which leaves a portion of the liquid. According to this, then, only as much solid can separate after a definite overcooling a< will just bring the system from its overcooled point tc its freezing-point. Since the latent heat represents the heat which is evolved when i gram of solid is formed, anc the specific heat of the system will regulate the rise ir temperature resulting, it is obvious that the fraction o solid which can be formed by an overcooling of i mus be equal to the relation of specific to latent heat. Fo: a greater overcooling, then, we have where / is the fraction of liquid separating as solid. Thi term c here, although referring in reality to the averag< specific heat of the system liquid-solid between the temperatures / and At may be replaced by the specific heai of the liquid between these limits ; and iv, which really THE SOLID STATE. 101 refers to the latent heat at the temperature / At may be replaced by the latent heat at /. This possibility is due to the fact that the change in the specific heat of the system, as well as the variation in the latent heat per degree, depends upon the specific heats ?D., the, two states Jj " ' ' * ' so that the ratio remains constant iw matter whyat ,th amount of liquid transformed into the solid state. If, for example, the fraction / of liquid solidifies, owing to the overcooling J/, we have as specific heat of the system (if)ci+}c t , and the heat necessary to raise it J/ is A[(I/)CI +/ C J- The latent heat at / J/, however, according to p. 98, is j[w+M(ci c s )] where / is the fraction of the amount of liquid becoming solid and thus giving out / times the latent heat. The decrease in the amount of heat necessary to heat the system is the decrease in the amount of heat delivered by the solid- ification is and inspection shows tkat the two are equal, hence the specific heat in the liquid state and the latent heat at the freezing-point may be used throughout to give the fraction of liquid separated. A specific example may perhaps make this clearer. For water we have ci = i, 102 ELEMENTS OF PHYSICAL CHEMISTRY. c s =o.5, while w at o is 80 cals. When overcooled to 10 C. how much of the liquid is separated as ice? i. e., 125 grams of liquid from each liter will solidify. To a die.Qk this.va^lue we shall calculate the specific heat of the system and see that the amount of heat given out at 10 is sufficient to raise the temperature to o. The specific heat of the system is (i J) i + JX .5, i.e., *g*. The latent heat at -10 is (80-10X0.5) =75 cals. The ratio, specific heat to latent heat, then, as before, is 75 8o' The heat of sublimation (pp. 96 and 99) at the me' ting- point must naturally be equal to the sum of the heat of fusion and the heat of evaporation, i.e., where p is the vapor-pressure of the solid. But where P is the vapor-pressure of the liquid. THE SOLID STATE. 103 By combination, then, we obtain, by aid of (9), T AT $ \AT AT' where $ is the vapor-pressure for both liquid and solid at the melting-point. As an example of the use of this formula we shall calculate the difference in the tem- perature coefficients of the vapor-pressure in the solid and liquid states. For benzene at the fusion temperature 5.6 the heat of fusion is 30.18 cals. and p is equal to 35.5 mm., and we have RT*/A AP Ap AP 30.18X35-5X78 ~Af~AT = 2(273 + 5.6)2 = while by experiment it is found to be Ap AP ~ = 2 ' 428 ~ I ' 9 5 = ' 523 CHAPTER V. THE PHASE RULE.* 28. Object of the phase rule. Thus far we have con- sidered the three possible states of aggregation, gases, liquids, and solids, as systems existing alone, and at most have studied the quantitative relations of these as such, and their mutual transformations the one into the other. Although all the quantitative relations possible are thus expressed, there are certain further qualitative relations, regulating the equilibrium existing between the states, which have not been tauched upon. These relations are summed up in the phase rule, as derived by Willard Gibbs, which is the subject of this chapter. In order to understand quite clearly the object of this rule, the following systems may be considered: When water is placed in the Torricellian vacuum of a barometer the mercury is depressed and the depression increases with the temperature of the water. At any one tempera- ture, however, the pressure is independent of the amount of water present. Water and its vapor, then, can exist * For further details on this subject see Findlay's "The Phase Rule and its Applications," of which I have made free use in this Chapter. 104 THE PHASE RULE. 105 permanently side by side, i.e., in equilibrium when, at a given temperature, the pressure has a definite value. If a salt solution is substituted for the water, however, the relation is different. It is true that an increase of temperature still increases the pressure; but the pressure is no longer independent of the volume, for the greater the volume (i.e., the greater the amount of water forming water- vapor), the smaller is the pressure. The presence of solid salt, however, in contact with the solution, causes the pressure to remain constant, irrespective of the actual volume. Further, by cooling water ice is formed, and so long as this is present with the water the application (or removal) of heat varies neither the temperature nor the vapor- pressure of the system. Thus it is only at one definite pressure and one definite temperature that ice, water and water- vapor can exist together in equilibrium. In the case of a salt solution, on the other hand, we may have ice in contact with the solution at different temperatures and pressures. The phase rule enables us to decide the necessary con- ditions for equilibrium in cases of this sort, as well as in all others where an analogous equilibrium is possible. 29. The phase rule. Before formulating the law itself it will be necessary to define the terms used by it. A phase is a physically distinct, homogeneous and mechanic- ally separable portion of the system; it need not, however, io6 ELEMENTS OF PHYSICAL CHEMISTRY. be chemically simple. A gaseous mixture or a solution may form a phase; but a heterogeneous mixture of solids represents as many phases as there are substances present. The dissociation of calcium carbonate thus gives two solid phases, calcium carbonate and calcium oxide, and one gaseous phase, carbon dioxide. It is to be remembered that but one gaseous phase may be present, for all gases are miscible in all proportions. The components of a system are those constituents which in concentration can undergo independent variation in the different phases. Thus, in the above example, water, not hydrogen and oxygen, is the component ; and although calcium carbonate, oxide and carbon dioxide are constituents of the equilib- rium, only two need be considered as components, for the amount of the other is not independent of the amounts of these, but dependent according to the chemical reaction Thus taking CaO and CO 2 as the components, the compo- sition of each phase can be easily expressed. In general, the components are to be chosen from the constituents at equilibrium, and are chosen as the smallest number of such constituents necessary to express the composition of each phase participating in the equi- librium, zero and negative quantities of the components being permissible. The degree of freedom is the number of variable factors, temperature, pressure and concentration of the com- THE PHASE RULE. 107 ponents which must be arbitrarily fixed in order that the condition of the system may be perfectly denned. Thus a gas has two degrees of freedom, for it is necessary to fix two of the conditions, temperature, pressure or volume, to define it; a liquid and its vapor has but one, for equi- librium at a certain temperature exists only at a certain * pressure; while a system of a liquid, its solid and gas, has no degree of freedom, for equilibrium can only exist at a certain temperature and pressure, for otherwise one of the phases will disappear. Gibbs' phase rule defines the condition of equilibrium by the relation between the number of coexisting phases and components. According to it a system made up oj n components in n + 2 phases can only exist when pressure temperature and composition have definite fixed values- a system oj n components in n + i phases can exist so long as only one o) the factors varies; and a system of n com- ponents in n phases can exist while two oj the factors vary. In other words, the degree oj freedom is expressed by the equation P + F = C + 2, or F=C+2-P where P designates the number of phases, C the number of components, and F the degree of freedom. Since in the equilibrium of CaCOs we have two com- ponents (CaO and CO 2 ) and three phases "(two solid, CaO and CaCO 3 , and i gaseous, CO 2 ) the degree of freedom is 2^23 = 1. This is a univariant system, 108 ELEMENTS OF PHYSICAL CHEMISTRY. then, just as water and water- vapor (i.e., F= i + 2 2 = 1) and so in each we have a definite pressure for each tem- perature within certain limits. This formula for the degree of freedom is used as a system of classification for equilibria, and accordingly we have invariant, uni- variant, bivariant, multivariant systems, the members of each group behaving similarly. 30. Derivation of the phase rule. Although this deri- vation is not the original one of Gibbs, it may serve to show how such a relation can be deduced without hy- pothesis and solely by mathematical reasoning. Imagine an invariant equilibrium consisting of y phases of n components. Consider one phase alone. This phase will contain, then, a certain amount of all the n components. It may be a gas or a liquid, for all components go into the gaseous form and into solution, at least to a slight extent. In this phase which we are considering let the concen- trations of the n components be c\c% . . . c n . Since the equilibrium is invariant, the composition of this phase will be altered by a change of concentration, temperature, or pressure of the system. A change in one of these will cause a corresponding change in the other two. This we may express by the equation . . . c n ,p, T=o, where F is any function of the variables. THE PHASE RULE. 109 The composition of one phase, however, determines that of all others wm'ch are in equilibrium with it; for \ all phases which are in equilibrium with one must be in equilibrium with each other, and this is only possible for a certain ratio of concentration between the con- stituents. Thus for a certain liquid phase we have a certain gaseous one in equilibrium with it and perhaps a solid. It follows, then, that the composition of all the phases is a certain function of the same variables. For each phase, then, we have an equation of the form F(ClC 2 C n ,p, T)=0. Since there are y phases, we must have y equations of this form. There are, however, n + 2 variables in each equation, so that if y = n + 2, i.e., if we have two more phases than components, we may find a certain definite value for each unknown quantity, since we have the same number of equations as of unknown quantities. In this case there is only one value for each c\c 2 . . . c n , p, and T at which the system may exist in equilibrium. When n components are present in n + 2 phases we have equilibrium only for a certain temperature, a certain pres- sure^ and a certain ratio oj concentration oj the single phases, i.e., n + 2 phases of n components can only exist at a certain point (transition point) in a system of coordi- nates. If one of these values is changed, then one phase Iio ELEMENTS OF PHYSICAL CHEMISTRY. disappears entirely, and we have n+i phases of n com- ponents. In this case there will be n + 2 variables (un- known quantities) in n + i equations; hence it is nol possible to find an absolute value for any one. We find, however, relative values of the n + 2 variables c\c^ . . . c n . p, T, i.e., for each value of T we have only a certain value of p and of each of the terms ciC 2 . . . c n at which equilibrium can exist. It is necessary to have at least n components present in order that a system containing n + i phases may exist as a univariant equilibrium. Where we have n + i phases of n components, and the conditions are altered, one phase disappears and we have n phases (equations) and n + 2 variables (unknown quantities), i.e., the composition of the phases is uncertain and we have a divariant equilibrium. Such a one is a mixture of water and alcohol, i.e., two components in two phases, liquid and gas. At a low temperature we shall have a univariant equilibrium, for ice is formed and two components are present in three phases. 31. The equilibrium of water in its phases. Fig. 5 gives the curves for water when plotted in a system of coordinates of which the abscissae are temperatures and the ordinates are pressures in other words, the pressure curves for water and ice, and the p.t curve for water-vapor. Along the curve WV liquid and gas form a univariant equilibrium. The curve IV is made up of the values at which gas and solid can exist in uni- THE PHASE RULE. ill variant equilibrium, i.e., it is the vapor-pressure curve for ice. The curve IW represents the conditions for the coexistence of water and ice. As the freezing-point is but slightly influenced by pressure, this curve forms wv FIG. 5. only a slight angle to the axis of pressures (i atmos. = depression of 0.00752). The point in which the three curves intersect is the transition- point, i.e., the triple point in which all three phases can exist in equilibrium. This is at o the pres- sure being 4.6 mm. of Hg. If we start with the system containing water and vapor (i.e., down the curve WV) in a closed vessel from which heat can be absorbed the point of intersection will be reached, part of the water will freeze, and we shall have three phases of one substance, i.e., the transition- point. If the temperature is still decreased either the liquid or the gaseous phase will disappear. Which of these depends upon the ratio of volume. If the volume of vapor is great enough, all the liquid will freeze and H2 ELEMENTS OF PHYSICAL CHEMISTRY. the curve IV will be followed. On the other hand, if the volume of liquid is large all gas will disappear and the curve IW will be produced, for the amount of ice which separates will increase the pressure and thus cause all the gas to condense. In a system of coordinates, then, a divariant equilib- rium is represented by a surface, i.e., W, V, and I; a univariant equilibrium by a line, as WV, IW, and IV; and an invariant equilibrium by a point, as O. 32. The phase rule and the mdentification of basic salts. An exceedingly clever and most important appli- cation of the phase rule to inorganic chemistry has been made by Miller and Kenrick.* In the identification of basic salts we have a problem which up to the present has seemed utterly impossible of accurate solution. By the application by Miller and Kenrick of the phase rule, however, a great field is opened up and there seems to be no reason now why this subject should not be re- moved entirely from the position of mystery which it has occupied in the past. Naturally analysis alone can- not give much aid in an investigation, for by it it is not possible to tell whether a chemical individual is present or a mixture in varying amounts of two or more chemi- cal individuals. Applying the phase rule to the formation of basic salts * See Jour. Phys. Chem., 7, 259, 1903, of which I have made very free use. THE PHASE RULE. 113 by the action of water on chloride of antimony or on the nitrate of bismuth, the temperature being low and con- stant and the pressure being atmospheric, we know: (i) that if the system which consists of three components has arrived at equilibrium not more than three phases can coexist. Of these the solution forms one, and the pre- cipitate must be either one single homogeneous substance (one phase), or a mixture of two phases, for instance, of two basic salts, or of one basic salt with the oxide. (2) That if the observed difference in composition between two precipitates by the action of different quantities of water on the same salt is due to their being mixtures of the same pair of basic salts in different proportions, the composition of the mother liquors will be the same in the two cases. (For the solution must be in equilibrium with both phases, whatever their actual amount.) The possible cases are thus divided into three groups: 1. The solutions are identical in composition in different experiments, while the composition of the precipitate varies. The precipitate is a mixture of two phases. 2. The solutions differ in composition, but the pre- cipitates have the same composition. The precipitate is a single chemical compound. 3. Both solutions and precipitates vary. The pre- cipitate is a single phase of variable composition, a " solid solution." If it were possible to represent the compositions of the H4 ELEMENTS OF PHYSICAL CHEMISTRY. solution by abscissae, and those of the precipitates b} ordinates, the results of a series of experiments could b( represented by a curve; perpendicular lines would ther correspond to case i, horizontal lines to case 2, anc slanting lines to case 3. In a three-component systerr this is not possible in general. In many cases, however a pair or pairs of components may be found whose ratic in the solution or precipitate changes whenever the composition of the solution or precipitate changes, and only then; and since for the interpretation of the results it is only necessary to know whether the composition of solutions and precipitates remains constant or changes from experiment to experiment, it is sufficient to plot these ratios instead of the compositions themselves. Since in the case of the action of water on nitrate oi bismuth the precipitates are all of the general formula wBi2Os, N2O 5 , nH.2O, the compositions of the solu- tions, also, may be expressed in terms of Bi 2 O 3 , N 2 O5, and H 2 O, which are the components of the system. In the curve below the abscissae give the ratios between N 2 C>5 and H 2 O in the solutions, the ordinates those between Bi 2 O3 and N 2 Os in the precipitates.* The only basic nitrates found are the two noticed in the curve and Bi 2 O3-N 2 O5-2H 2 O, and all the others which have been supposed to exist were found to be mixtures of these in differing proportions. The vertical line corre- * See Allen, Am. Chem. Jour., 25, 307, 1901. fHE PHASE RULS. spends to mixtures of precipitates, and the horizontal ones to pure chemical individuals. It will be seen from this brief sketch that the phase rule itself is not at all difficult to understand, but that Bi,0, ^ToT 2Bi 2 3 .y 2 5 +H 2 5 -HH, H 2 005 010 015 FIG. 6 025 030 its applications may give trouble, owing to the difficulty of choosing the components and other details. For further applications to metallurgy and to chemistry the reader must be referred elsewhere, for here it is only a question of the general principles involved in the law itself. CHAPTER VI. SOLUTIONS. 33. Definition of a solution. A solution is a homo- geneous mixture which cannot be separated into its constituents by mechanical means. The power to form solutions varies with the state of aggregation. Thus for gases it is unconditioned, while for solids, the opposite extreme, it is small, though still present. 34. Gases in liquids. A true solution here is one in which there is no chemical reaction between the liquid and the gas, i.e., the gas may be expelled by heat. Henry's law, which was verified by Bunsen, enables us to find the amount of gas which will be absorbed. When a gas is absorbed in a liquid the weight dissolved is proportional to the pressure of the gas. Since, however, pressure and volume (at constant temperature) are inversely pro- portional, the law may also be expressed as follows: A given amount oj liquid absorbs at any pressure the same volume oj gas. If now we introduce the idea of concen- tration (i.e., the amount of substance divided by the total volume), instead of the pressure, we obtain still 116 SOLUTIONS. II? another form for the law: The amount of gas dissolved by a liquid is such that there is always a constant ratio between the concentration oj the gas in the liquid and in the space above it. This ratio does not depend upon the actual concentrations, being the same for all concentra- tions within wide limits, and its size is dependent only upon the nature of the liquid and its temperature. If, instead of a single gas, a mixture of gases is dis- solved, this law also holds, in that each constituent is absorbed from the mixture to the same extent as if it were present alone at a pressure equal to its partial pressure in the mixture. Among the characteristics of a gas absorbed in a liquid are the following: There is less absorption in a solution than in a pure solvent, and addition of a substance to a liquid already containing gas forces this out. A liquid is easily supersaturated with gas, i.e., reduction of the pres- sure does not cause the gas to evolve immediately. The presence of dust or another gas, porous solids enclosing air and free from the gas in question, all cause the gas to be evolved. When a gas is absorbed in a liquid there is always a change of volume in the latter. As a general rule the less compressible a gas is, the greater is the increase of volume caused by it. The increase of volume caused by the solution of a gas has been found to be approx- imately equal to the value b in the equations of Budde ii 8 ELEMENTS OF PHYSICAL CHEMISTRY. and Van der Waals (pp. 34, 35), as is shown for severa] gases in Table II. TABLE II. Gas. Av b O 0.00115 0.000890 N 0.00145 0.001359 H 0.00106 0.000887 35. Liquids in liquids. We have in this case three possibilities : 1. No solubility, i.e., the formation of two layers and no homogeneous mixture. 2. A mutual solubility, but not in all proportions (as water in ether = 3% and ether in water = 10%). 3. The same relation as with gases, i.e., a solubility in all proportions (as water and ethyl alcohol). The volume of liquid mixtures is not an additive pro- perty. The volume is never equal to the sum of those of the constituents. Almost all mixtures possess a volume smaller, but a few larger, than the sum of the partial volumes. This difference is probably due in some to a mutual change in the molecular weights, and in others to the change in the molecular weight of one of the liquids by the other. The most important point for us to consider here con- cerning these mixtures is the manner in which they distil. We shall consider first (i) cases where two completely immiscible liquids are present; next (2) those which have a mutual solubility; and finally (3) those for which the mixture is homogeneous. SOLUTIONS. 119 i. A liquid which consists of two immiscible constitu- ents, present in two layers, has a vapor-pressure equal to the sum of those of the constituents. The boiling-point of such a system, then, is that temperature at which the vapor-pressure becomes equal to the pressure of the atmosphere; it lies still lower, therefore, than the boiling- point of the lower boiling constituent. This is difficult to observe experimentally by direct heating, owing to the bumping. If, however, the vapor of another boiling liquid is passed through the mixture the whole is heated evenly and to the same degree, and the mixture distils below the temperature at which either constituent alone will do so. Assume the vapor-pressures of the constituents to be pi and p2', the volume of each in the vapor which distils over will be proportional to these pressures, and the total vapor-pressure will be equal to the sum of those of the constituents in the pure state. The weight of the vapor of each is equal to its density multiplied by its volume, or, by what is proportional to it, its vapor-pressure. We have, then, the weights (#1 and q 2 ) of the constituents in the vapor from the proportion qi:q 2 : :pidi'.p 2 d 2 . If the vapor-pressures of the two constituents and their weights hi the -vapor are known it is possible to determine the vapor-density of the one in terms of the T20 ELEMENTS OF PHYSICAL CHEMISTRY. other. Naumann was the first to do this, using the equation , , ?i ?2 dii d 2 : : ' . Pi p2 2. In the case of a mixture of partially miscible liquids, i.e., where each is mutually soluble in the other, we have also a constant boiling-point and a constant dis- tillate so long as two layers remain. The boiling-point here may be higher, equal to, or, as for class i, lowei than that of the lower boiling liquid. It cannot, how- ever, be higher than that of the higher boiling constituent. With regard to the vapor-pressures of the two layers formed when the two constituents are to a slight degree mutually soluble, Konowalow has shown that the liquid B when saturated with A has the same vapor- pressure as the liquid A when saturated with B, both being at the same temperature. Since the same vapor is given off by both layers, the only effect of the distillation is tc change the absolute quantities of these, for the portion lost by one layer will be absorbed from the other. When but one layer remains the liquid is identical with those in class 3. 3. When the liquids mix in all proportions, or in general when there is only one layer of liquid, then it is possible to make a complete separation of the con- stituents by a fractional distillation, provided the vapor- pressures of the two differ. The vapor given off by a SOLUTIONS. 121 homogeneous mixture has a different composition from that of the liquid, and its vapor pressure is no longer equal to the sum of those of the constituents; it depends upon the action of the liquids upon one another and upon the vapors given off. If we use a system of coordinates in which the vapor- pressures are laid out upon the axis of ordinates, and the percentage composition of the liquid upon the axis of abscissae, we find that the vapor-pressure never reaches the sum of those of the constituents, but at times goes even 'below the smaller of these. Fig. 7 gives three FIG. 7. types of curves thus obtained, which will illustrate the principle involved. We find in I a maximum of the vapor-pressure corresponding to a certain strength of solution. In II this maximum has disappeared, while in III the opposite extreme is observed, i.e., a minimum of the vapor-pressure which corresponds to a certain composition. I. Mixtures which correspond to this type possess a maximum of the vapor-pressure for a certain com- position, i.e., the lowest boiling liquid is a solution of 122 ELEMENTS OF PHYSICAL CHEMISTRY. a certain strength. An example of this is a 75% solu- tion of propyl alcohol in water. This 75% solution will distil at the lower temperature from any propyl- alcohol solution until one of the constituents, i.e., either the water or the propyl alcohol, has disappeared, when the other will remain behind in the almost pure state. Thus if we start with a 50% solution of propyl alcohol a 75% solution will be distilled until all the alcohol is removed and water is left in the flask. If we start with a 90% solution of the alcohol the 75% solution will be given off until all the water has been used up and pure propyl alcohol remains behind. II. For mixtures of this type we can make a com- plete separation by distillation. The vapor-pressure of all mixtures lies between those of the two constitu- ents; consequently the one with the higher vapor-pres- sure will be given off in the almost pure state, leaving the other behind. Solutions of ethyl and methyl alcohol in water belong to this class. The method of calculat- ing the value of the vapor-pressure in such a case will be given later. . III. Here we have a minimum of the vapor-pressure corresponding to a certain composition. A solution of this composition will, then, always be the last to be distilled, since its boiling-point is the highest of all pos- sible mixtures. This is just the opposite of case I, where the solution of a certain composition is given off 123 first. All solutions, for example, of formic acid and water will distil off an amount of one constituent until _ the solution remaining contains 70% of formic acid. A 90% solution will give off almost pure formic acid until just enough remains to form a 70% solution. A 50% solution will give off water until the 70% solution is left behind. An interesting technical application of these principles and facts has been made by Young (Trans. Chem. Soc., 81, 707, 1902) in his method of preparing absolute alcohol from strong spirit. As is well known ethyl alcohol and water form a constant boiling liquid (B.P. =78.i5) of a composition of 95.57% of alcohol and 4.43% of water, which by distillation alone cannot be further separated. Instead of employing a dehydrating agent and distilling the alcohol alone, as has been the common practice, he mixes with the solution benzene, w-hexane, or a similar liquid. The behavior of this system on distilling can be seen from the table below: Liquid of Cconstant B.P. B.P. Percentage Composition. Alcohol. Water. Benzene. i. Alcohol, water and benzene. . . 64. 85 68. 25 6o.2 5 78- 15 78.. 3 80. 2 100. 18.5 -32.41 95-57 TOO.O 7-4 8.83 4-43 74-1 67.59 91.17 3. Water and benzene 4. Alcohol and water 5. Alcohol 6 Benzene 100 7 Water 100 124 ELEMENTS OF PHYSICAL CHEMISTRY. The lowest boiling-liquid is the ternary system W.A.B., and, unless one liquid is present in a relatively very small quantity, it will distil first. If there is more than sufficient benzene to carry over the whole of the water, and if alcohol is present in excess, the ternary mixture will be followed by the binary mixture (A.B.), and the last substance to come over will be alcohol. Owing to the fact that there is but 3.4 difference between the boiling- points of the ternary mixture (A.B.W.) and the binary one (A.B.), however, it is impossible to get rid of all the water in one distillation. By redistilling the partially dehydrated alcohol once or twice with a further quantity of benzene complete separation is effected. As the alcohol itself need not be distilled, but can be poured directly from the still, and as the benzene can be recovered without difficulty, this method has many advantages. 36. Solids in liquids. When a soluble solid comes in contact with a liquid it dissolves. When no more substance is taken up by the liquid at any one tempera- ture the solution is said to be saturated at that temperature. There are two general methods of making a saturated solution; but in either it is always to be remembered that in order that a solution be saturated the solid sub- stance must be in con'tact with it. I. The substance, in granular form, is brought into the liquid, and the system agitated at a certain tempera- ture until no more is dissolved; or, SOLUTIONS. 125 II. The solution is made as above, but at a higher temperature than the desired one, to which it is reduced later. Both these methods give the same result if properly carried out, although the latter is quicker and so to be preferred. The importance of the state of the solid from which the solution is made has recently been pointed out by Hulett (J. Am. Chem. Soc., 27, 49, 1905). I quote from him: "The explanation of the phenomenon" (of the greater speed of solution and greater solubility of very small particles of solid) "is based on the following considerations: The boundary between a solid and a liquid is the seat of a certain amount of energy due to the surface-energy of the liquid; if this surface is increased by powdering the solid, the total surf ace- energy is cor- respondingly increased. Further, it is a generally ob- served fact that the form of a substance which has the greater free energy is the more soluble, has the greater vapor-pressure, and is the least stable form, e.g., alkr tropic modifications of substances have different solu- bilities, and the unstable form is always the more soluble. This phenomenon is hardly analogous to the well-known behavior of liquid drops of different sizes. Small drops in the vicinity of large ones grow small and disappear, while the larger ones grow larger, and the reason is quite clear. It is known that the curved surface of a liquid 126 ELEMENTS OF PHYSICAL CHEMISTRY. has a greater vapor-pressure than a plane or less curved surface, therefore, a distillation takes place. The similarity between the vapor-pressure of liquids and the solution-pressure of solids has suggested to some the analogy between the facts just mentioned and the be- havior of solid particles of different sizes in contact with the solution. But we cannot assume that the surface oi the particles of a powder are curved, or, if that is granted, we do not know that a curved surface of a solid or a sharp edge has a greater solution-pressure than a plane surface of the same substance." Hulett has found that a solution, of gypsum saturated at 25, containing 2.080 grs. CaSO 4 per liter (i.e., 0.01530 mole) will increase its concentration rapidly to a maximum when shaken with powdered gypsum and then will decrease to the original value again. In one experiment the con- tent of gypsum reached 2.542 grs. CaSO 4 in a liter in a minute. The fine powder used for this purpose was found, after the concentration had reached its original point, to have increased in the size of its grain. The smallest particles thus go into solution, produce the supersaturation and are then deposited upon the others, so that all increase in size. It has been observed that crystals dissolve more rapidly in some directions than in others, and it is possible thus to cause a circular disk of gypsum (cut parallel to oio) to become elliptical by the action of dilute hydrochloric SOLUTIONS. 127 acid. Hulett calls attention, however, to the fact that solubility has nothing to do with rate of solubility, for solubility is measured by the concentration of the solution which is in final equilibrium with the solid. He found that the plate of gypsum is equally soluble in all directions > and a solution saturated with planes cut in one direction is in equilibrium with those cut in other directions. For these reasons Hulett advocates the preparation of saturated solutions from large particles, as there is then no danger of supersaturation, and the liquid need not be filtered. In case small particles are used and a supersaturation feared, contact with the large particles or crystals will cause the equilibrium to be attained from the supersaturated side. It is often possible to cool a saturated solution (which has been separated from its solid phase) to such an extent that it is supersaturated. Some solutions of this sort can be kept indefinitely, provided germs of the solid phase are excluded. Such solutions are called metastable. On the other hand, there are some solutions in -which the solid phase appears even when its germs are excluded. Such solutions are called labile. The distinction between these two kinds of solution is not so sharp as one would think, however, for by overcooling sufficiently the meta- stable solution may pass over into the labile state. In general the metastable solution is smaller in concentration than is that which is labile. Increase of the concentration 128 ELEMENTS OF PHYSICAL CHEMISTRY. causes the metastable state to be transformed into the labile; the concentration at which this occurs is called the metastable limit. 37. Osmotic pressure. If in a tall jar we place a layer of pure water over a colored solution we observe after a time that the whole liquid is colored. The colored substance has diffused through the liquid until a homo- geneous mixture results. This action shows that there must be a certain tendency or pressure which causes the substance to occupy the entire space available. This pressure causing the substance to diffuse into the solvent is called the osmotic pressure oj that solution. If a vessel is provided with a partition of such a nature that it allows free passage to the solvent, but not to the solute, and we have water on one side and a solution on the other, the solute will exert a pressure upon it ; this is the osmotic pressure of that solution. Semipermeable partitions of this sort had been found by Traube, and Pfeffer perfected them in such a way that he was able to actually measure the different osmotic pressures by means of a manometer. An apparatus by which the principle of Pfeffer's measure- ments may be understood is shown in Fig. 8. The cylinder A is made of porous clay, and is designed to support the semipermeable film. To produce this the pores of the cylinder are filled with a solution of copper sulphate. After the excess of this is removed the cup is filled with a 3% solution of potassium ferrocyanide, and SOLUTIONS 129 allowed to stand for a day in a solution of copper sulphate. In this way the pores are filled with a film of copper ferrocyanide, which is permeable to water, but not to a large number of salts. The process of measuring is FIG. 8. as follows: The now semipermeable cylinder is washed out and filled with the solution to be measured. The rubber stopper CC is next inserted in A in such a way that the solution rises a short distance in the tube, i.e^ the cylinder is entirely filled with solution without air, and the tube connected to a manometer. The cork BB is then fastened in the vessel DD, which is filled with water. The liquid rises slowly in the tube until equilibrium is reached, i.e., until the resistance in the manometer is .just equal to the osmotic pressure. The height of the column of mercury thus represents the osmotic pressure of the solution in A. There is a definite attraction, then, or an apparent one, between the substance in solution and the solvent, 130 ELEMENTS OF PHYSICAL CHEMISTRY. and this is of such a nature that without the partition it would cause the system to become homogeneous. Whether this apparent attraction is actually an attraction between solvent and solute, or whether it is a tendency for the solute to become uniform in concentration in the liquid, owing to forces existing in it, is unknown. As under certain conditions a change in the solvent makes no difference in the osmotic pressure observed (and when it does it can be accounted for quite readily) it would seem that the latter supposition is the more correct. However, whether this phenomenon is a pressure or an attraction can have no difference from our standpoint, for what we shall designate as osmotic pressure is an experimentally observed fact which has been given that name. As such, then, we can define osmotic pressure (as we shall use the term) as that pressure which will prevent pure solvent from going through a semiperme- dble partition to dilute the solution contained within it. The nature of the semipermeable partition seems to have no effect upon the value of the osmotic pres- sure. That this must be true can be proven by the following process: Imagine A and B (Fig. 9) to be two unlike semipermeable partitions placed in a cylinder of glass. Assume the cylinder full of solution and lowered in a horizontal position in a vessel of water. If the pressure P is exerted by the partition A } and p SOLUTIONS. I3 1 X^LT'TV-i *'" (a smaller one) by B, then water will go through A until the internal pressure P is reached. Since B allows only the pressure p, however, this pressure P can never be reached, so that we would have a steady current of water going from A to B under the finite pressure P p. This, however, would be a perpetual motion, which is FIG. 9. impossible; hence the partition A must give rise to the same osmotic pressure as the partition B. Osmotic pressure is, then, independent of the nature of the solvent and the partition, the latter being simply a means of making it visible. Pfeffer measured, by means of an apparatus of the type of the one described, the osmotic pressures which exist in a large number of solutions and found them in many cases to be very great. A few results for sugar solutions of varying concentrations at 15 C. are given in Table III. TABLE III. c P p/c c P p/c I 53-8 53-8 4 208.2 52.1 I 53-2 53-2 6 307-5 5i-3 2 101.6 50.8 i 53-5 53-5 2.74 151.8 55-4 132 ELEMENTS OF PHYSICAL CHEMISTRY. where c is the percentage composition of the solution, p is the osmotic pressure, in cms. of mercury, and p/c is the ratio of pressure to concentration. Considering the difficulties in measuring and the imperfections in the semipermeable film, the term p/c is to be considered as constant; hence the osmotic pres- sure is proportional to the concentration of the solute. Further results at different temperatures showed that the osmotic pressure is proportional to the absolute tem- perature. This, however, is only strictly true when the heat of dilution of the solution is zero, i.e., when the solu- tion is dilute. On account of experimental -difficulties the direct effect of this heat of dilution upon the osmotic pressure of concentrated solutions has not as yet been determined. van't Hoff in 1887 from this much knowledge derived, by the aid of thermodynamics, an analogy between the behavior of gases and substances in solution. Here, however, we shall not go into his reasoning, but shall consider how his results may be arrived at in a more simple manner. As we have seen by Pfeffer's results, the term p/c remains constant for any one solution when p is the osmotic pressure and c the concentration, i/c, however, is equal to v, the volume; we have then p/c=pv. SOLUTIONS. 133 Since now the osmotic pressure is proportional to the absolute temperature, as is also the term p/c, we have pv= constant XT. This equation is so much like the one for gases (9) that it immediately suggests that there is some kind of a connection between them and substances in solution. If now the value of the constant is determined, then by comparing it with the gas constant it is possible to find this connection. Since we are to use the molec- ular gas constant in the comparison, it will be neces- sary also to find this constant for i mole. A i% solu- tion of sugar at o is held in equilibrium by a column of mercury 49.3 cms. high and thus gives an osmotic pressure equal to 49.3X13.59=671 grams per sq. cm. Since the molecular weight of sugar (CnH^On) is 342, the volume in which i mole is dissolved (to give a i% solution) is 34200 cc. We have then PoVo 671X34200 constant = ^=- = - = 84200 gr.-cms., l 273 while the molecular gas constant is #=84800 gr.-cms. 134 ELEMENTS OF PHYSICAL CHEMISTRY, The constant is the same, then (within the experi- mental error), for the gaseous state and the state of a substance in solution; consequently for each state, con- sidering i mole, we have pV=RT. van't Hoff summed up these results in the following form: The osmotic pressure of a substance in solution is the same pressure which that substance would exert were it in gaseous form at the same temperature and occupy- ing the same volume. Since i mole of gas at o and 76 cms. of Hg occupies the volume of 22.4 liters, at the volume of i liter it will have the pressure of 22.4 atmospheres (Boyle's law). This pressure, 22.4 atmospheres, is then the osmotic pressure which is exerted by i mole of solute in i liter of solution at o, i.e., by a molar solution. From this os- motic pressure, 22.4 atmospheres, it is very easy to find that for any other concentration given in terms normal. Thus a n/io solution gives an osmotic pressure of 2.24 atmospheres. That within the limits thus far possible the osmotic pressure is approximately equal to that pressure which would be exerted if the substance were in the gaseous state in the same volume, at that temperature, is not a hypothesis, but an experimental fact. It is unfortunate that as yet osmotic pressures cannot be measured accurately above a few atmospheres. From SOLUTIONS. 135 the advance which has been made lately by Morse and Frazer (Am. Chem. J., 25, i, 1903), it is to be hoped that we shall soon be able to test this at present limited law between wide limits. From preliminary results thus far obtained, however, more concentrated solutions seem to give too high a value by experiment, just as is observed with the gas laws themselves. This law of van't HofFs enables us at once to define the molecular weight of a dissolved substance, and the definition naturally reminds one of the definition for substances in the gaseous state. The molecular weight of a substance in solution is the weight in grams which in the volume of approximately 22.4 liters of solvent will give an osmotic pressure of i atmosphere at o C., or a corresponding value at another temperature or volume. If the osmotic pressure of a solution is known it is then possible to find from it the molecular weight. A 2% solution of sugar gives, at o C., an osmotic pres- sure equal to 101.6 cms. of Hg. If this were a molar solution it would give a pressure equal to 22.4X76 = 1702.4 cms. of Hg. The 2% solution contains 20 grams in the liter; hence 20^1:101.6:1702.4; ^ = 335. The determination of the osmotic pressure directly by the use of Pfeffer's apparatus is not very satisfac- tory, owing to the difficulty in preparing the semiper- meable film and to the fact that this latter is easily broken down so that liquid goes through. Morse and Frazer I3 6 ELEMENTS OF PHYSICAL CHEMISTRY. have succeeded in obtaining electrolytic films which have stood high pressures, and have obtained pressure as high as 31.4 atmospheres for molar and 13 to 14 atmos- pheres for half -molar sugar solutions at 25 C. without be- ing able to detect any breaking down of the film. In the absence of any such method, however, it is usual to cal- culate the osmotic pressure from the molar (i.e., moles per liter) concentration of the solution, as observed by other methods given later, i.e., to assume that the general law would hold if the osmotic pressure could be accu- rately determined. It is not only water solutions that have been measured, however, for Raoult (Zeit. f. phys. Chem. 27, 737, 1895) has found that vulcanite allows passage to ether but not to methyl alcohol. Thus with ether on one side and an equal volume mixture of ether and methyl alcohol on the other, he observed an osmotic pressure of 50 atmospheres at 13 C., when the manometer was fractured. In this case the pressure, just as with water solutions, was exerted in the solution, i.e., the pure solvent goes toward the solution and a pressure is necessary to prevent it. There is one characteristic of osmotic phenomena which must be mentioned. It is the slow rise of the pres- sure up to its maximum value. Thus with a half-molar sugar solution Morse found 95 hours necessary for the establishment of the final equilibrium. This, naturally, is to be attributed to the resistance of the cell to diffusion. SOLUTIONS. 137 A simple and quick method of observing the effect of osmotic pressure is given by Tammann. If in a moder- ately strong solution of copper sulphate we place a drop of a strong solution of potassium ferrocyanide it is imme- diately surrounded by a semipermeable film of copper ferrocyanide. Since the ferrocyanide solution is more concentrated than that of the copper sulphate, water goes through the film, from the copper salt to the ferrocyanide, and dark streaks are observed to sink from the bubble. These streaks are of stronger copper-sulphate solution, which is formed from the other by the loss of water, and sink on account of their increased specific gravity. If the ferrocyanide solution is weak and another salt or more ferrocyanide is added to it until no streaks are observed to sink from a bubble, then there must be an equal number of moles per liter in both of the two solu- tions, i.e., in the copper solution and the ferrocyanide plus salt. If the ferrocyanide solution is weaker than the one of copper sulphate (in moles per liter), then water will go out of the bubble, which will decrease in size and light streaks will appear in the copper solution. In this experiment the tendency for the solutions to become of the same molar concentration is particularly striking, and water is always observed to go from the more dilute to the more concentrated, expressed in moles per liter. For later work see Appendix. Consider a cylinder, as shown in Fig. 10, filled with 138 ELEMENTS OF PHYSICAL CHEMISTRY. a solution, and provided with a semipermeable piston a. If by a certain weight this piston is lowered, water will go through it, and we shall have separated an amount Solution Salt a FIG. 10. of solvent from the solution. If the amount of water separated previously contained i mole of solute, and the total volume is very large, the osmotic work is given by the equation pV=RT, where V is the decrease in volume occupied by the solute under the constant osmotic pressure p. It is possible thus to determine the molecular weight of any substance in solution by finding the work necessary for the separa- tion of a certain amount of the solvent. For each amount which has contained one mole of solute the work is (just as for gases) 2.T cals., so that the number of moles in the definite amount of solvent removed may be calculated from the work, and from this the molecular SOLUTIONS. 139 weight of the solute can be found. The molecular weight of a substance in solution by this method would be the weight which removed from a solution against its osmotic pressure would require the work of 2T cals.; or what is the same thing, the weight which has been dissolved in the volume of solvent requiring the work 2 T cals. to re- move reversibly from the main portion of the liquid. 38. Electrolytic dissociation or ionization. The mo- lecular weights of a very large number of organic sub- stances in water are found correctly from the osmotic- pressure, i.e., they correspond to the values in the gaseous state. For other substances, however, varying results are obtained. The osmotic pressure is here always too large, and consequently from our definition the molecular weight is smaller than that observed in the gaseous state. This reminds one immediately of the abnormal densities of gases. In the case of the gas the volume increases and the pressure is constant; in that of the solution, however, the volume remains constant, while the pressure changes. Arrhenius in attempting to find an explanation for the abnormal osmotic pressures found by experiment that those substances, and only those, which give abnormal osmotic pressures in solution are capable of conducting the electric current, and if they are dissolved in other solvents in which they behave normally they lose this power. Arrhenius determined the electrical conductivity of the substance in terms of molecular conductivity. The *4 ELEMENTS OF PHYSICAL CHEMISTRY. molecular conductivity of a solution may be denned as the reciprocal of the resistance (in ohms) of the volume of liquid which contains one formula weight of the substance, the electrodes being i cm. apart and large enough to contain between them the entire amount of solution. This value, naturally, is not found directly, but is calculated from that value found for a centimeter cube of the solution. (See Chapter IX.) In this way, always having i mole between the electrodes, he found that the more dilute the solution the greater is the mo- lecular conductivity. In many cases, indeed, he was able to reach such a dilution that the molecular conduc- tivity attained a maximum value, which is unaffected by further dilution. This molecular conductivity at infinite dilution, as it is called, is designated by the term jw^, that value for any dilution F, being designated by/V From this it is apparent that the solution undergoes some kind of a change as the result of dilution; and "the investigation of such solutions at various dilutions shows, indeed, that the molecular weight (according to definition) also changes with the dilution; the molecular weight decreasing to a minimum constant value, which for binary electrolytes is one-half the formula weight of the substance dissolved. It is not unreasonable, then, to infer that the breaking down of the molecular weight is the factor which causes the conduction of the electric current. SOLUTIONS. 141 These facts formed the starting point of what is known at present as the "theory of electrolytic dissoci- ation." As this theory to-day is much misunderstood by many, and is the subject of much speculation on the part of others, it will be necessary for us to consider carefully just what is fact and what assumption, and to see clearly which portions are hypothetical, and which are destined to remain under any hypothesis or lack of hypothesis; in other words, which are experimental facts. It may be said, however, that that which is hypothesis in this theory is unessential, as far as the use of the data is concerned, and the only hypothesis present, as we shall consider it, is that inherent in the terminology, which is a relic of the atomistic hypothesis and utterly beyond our power either to prove or disprove. The salient facts which have been grouped in this theory, for it is a theory in the sense that it is a law of nature holding between certain limits, although these are not as yet definitely fixed, are as follows: (1) The molecular conductivity of certain substances in water is found to increase up to a maximum, constant value, and this increase is the result of dilution. (2) Those solutions which conduct the current also give abnormal osmotic pressures; in other words the molecular weight (as defined above) decreases with increased dilution and finally reaches a minimum value, 142 ELEMENTS OF PHYSICAL CHEMISTRY. which, for binary electrolytes, is one-half the formula weight of the substance. (3) Those substances which in water conduct the current and give abnormal osmotic pressures, give normal osmotic pressures when dissolved in other solvents in which they do not conduct. (4) The nearer the value of p v is to that of //>, the more abnormal the value of the osmotic pressure (mo- lecular weight) of the solution. And the solution for which /*> is found also gives the maximum osmotic pressure, i.e., the minimum molecular weight. (5) The molecular conductivity of a solution at infinite dilution is an additive value, i.e.> is equal to the sum of the conductivities of the substances of which it is com- posed. The meaning of this is as follows: The molecu- lar conductivity at infinite dilution of, for example, potassium chloride plus that of nitric acid minus that of potassium nitrate is found to be equal to that of hydro- chloric acid. In other words, For this to be true, and it is true in general for all substances, it is necessary that the molecular conduc- tivity of each substance in solution be the sum of two values which are independent each of the other. Chlorine, for example, as the constituent of an electrolyte, at the dilution giving ,,, has the same conducting effect when SOLUTIONS. H3 part of a compound with one element as it has when combined with any other. It is possible, then, to find the value of //> for any binary electrolyte when the values for the elements composing it are known. In other words, the conductivities of the solution as pro- duced by the presence of any element can be calculated; and from these values, by summation, the value of //< for any binary electrolyte can be found. 6. When a solution is electrolyzed, the products of electrolysis appear instantaneously at the electrodes so soon as the circuit is completed. This indicates (since the solvent, water, does not conduct beyond a very small extent) that whatever does carry the current through the liquid is charged with electricity even before the current is applied, for the conduction is due to the dis- solved substance. (See Chap. IX.) Further, it is observed that the same amount of electricity, 96537 coulombs, is necessary for the separation of one equivalent weight (in grams) of any element; in other words that 96537 coulombs of electricity are transported through the liquid with each equivalent weight (in grams) of an element. (Faraday's Law, see Chapter IX.) 7. The properties of electrolytes are found to be the sum of the properties of the products observed during electrolysis. Thus any solution giving off chlorine on electrolysis, excluding secondary reactions, will precipi- tate silver from its solution as the chloride. And if 144 ELEMENTS OF PHYSICAL CHEMISTRY. chlorine cannot be produced in any way by the elec- trolysis, silver will not be precipitated as chloride from its solutions. And, on the other hand, silver is only precipitated by chlorine when contained in a solution from which silver can be deposited by the current by primary action. The catalytic effect of acids on the inversion of sugar as well as on the decomposition of methyl acetate, is found to be proportional to the ratio for the acid; /ioo and when a large amount of a salt of this acid is added to the acid this effect is decreased. But this is only true when the salt added is an electrolyte. All copper solutions, when very dilute, show the same blue color, and this also depends upon the ratio p , , /*00 and can also be changed, as the effect of acids were above, by the addition of a large amount of an elec- trolyte which contains the same acid radical as the cop- per salt in question. 8. Observation shows that when an element is sepa- rated on one electrode, anode or cathode, it is always separated on that one by primary action ; in other words, the sign of the electricity transported by an element is always the same. Unless an element in the pure state, when dissolved in water, reacts with the water it does not conduct the current. This circumstance is assumed to be due to the fact that only one kind of electricity SOLUTIONS. 145 could be carried by the substance, and hence it pro- duces no conduction. The question now arises as to what theory can be found to correlate these facts and observations so that the generalization thus obtained may be employed to foresee other facts, and applied to other observations that they, in their turn, may be elucidated and general- ized. By the word theory, then, we do not mean a hypothesis, in which something not observed is added to the facts to " explain" them, but only a generalization of observed facts. In other words, what law of nature, holding within definite, if small, limits, can be obtained from the above experimental facts when considered to- gether? The generalization which has been made from these facts is known as the theory of electrolytic dissociation, and, considering those portions which are free from hypothesis and fulfil the above conditions, in other words, omitting the hypothetical portions which it has attained since the time of its inception, we find in it, within cer- tain limits, a definite law of nature. The principal points of this theory are summarized below in brief form, and will each be expanded in the later portions of the book. 146 ELEMENTS OF PHYSICAL CHEMISTRY. That a substance in solution, which conducts the electric current, is dissociated or ionized into its con- stituents, and these constituents, when secondary actions are excluded, appear at the electrodes during elec- trolysis. The extent of the ionization or dissociation in any solution being given at the dilution V (number of liters in which i mole is dissolved) by the ratio That the products of ionization or dissociation are charged with electricity, 96537 coulombs being carried by the gram equivalent of any element (see (6) above). A further proof of this charged state of ionized matter is given by the fact that not only is the current carried by a solution dependent upon the number of gram equiva- lents transported, but, as we shall see later, any other means of depositing the constituents of the solution upon the electrodes liberates an amount of electricity which depends also upon the number of gram equiva- lents deposited. And all cells in order to give a current must contain electrolytes, i.e., solutions which are ionized. Since a solution which by conductivity is shown to be completely ionized, or practically so, leads to a molecu- lar weight, by osmotic pressure or analogous methods, of one-half the value expressed by the formula weight, then, from the case of hydrochloric acid in solution, where I HC1=H'+C1', SOLUTIONS. H7 the molecular weight of the hydrogen and chlorine in the ionic state> according to our definition of molecular weight, must be synonymous with the atomic weight. The ionic state, then, is an allotropic form of the ordinary state of the constituents; it differs from that in being charged with electricity, having less energy than when in the gaseous state, and always being trans- formed into the ordinary state on the loss of its charge of electricity. Since the constituents in the case already mentioned, and in general in all cases, show a molecular weight (by the definition) which is the same as the atomic weight, it is possible to determine a, the degree of ionization, by osmotic-pressure measurements, or from the average mo- lecular weight of the substance in solution, as determined by osmotic pressure or any of the other methods mentioned later. If, for example, we start with one formula weight of hydrochloric acid in a solution, and a moles of it are ionized, the total number of moles will consist of (i a) of un -ionized HC1 and a moles each of H' and Cl' (where the dot indicates positive electricity as the charge, and the accent negative). The total number of moles in the volume of the solution will go then from i to (i a) +2a, i.e., 1 4- a, and the ratio of osmotic pressures when entirely un- ionized, and when partially ionized will be the same as this. In other words, if the formula weight in a certain volume should give the osmotic pressure ^, it will give, when 148 CLEMENTS OF PHYSICAL CHEMISTRY. ionized to the extent a, the pressure p = (i+a)p f . Since the number of moles (by definition) shown by the same weight is thus increased, the molecular weight will be smaller, and we have as the relation between the two values of the molecular weight: M(i+a)=M', where the M'- refers to substance if it were un-ionized, i.e., is the formula weight of the substance, and M is the average molecular weight observed. Just as with gaseous dissociation, the ionization of a solution is affected by the presence of one of the products of the dissociation, and later when we consider the quantitative effect of this for gases, we shall also consider the case for solutions. Owing to the fact that the dissociated constituents in a solution were called ions (in the Faraday sense of charged atoms), it has become common to speak of and to represent the ionized state as though it were character- ized by an atomic structure ; in other words, as if it existed as an accumulation of charged particles of infinitesimal size. As this is pure hypothesis, and cannot be either proven or disproven, we shall only use the word ion here in the sense that it is expressive of the relation , i.e., we shall /*00 only use it in the sense oj ionized matter, according to this definition, and as purely expressive of an experimentally determined fact. One fact may be mentioned here which indicates what SOLUTIONS. *49 a very marked difference dilution makes in the behavior of a substance, and which decidedly supports the con- clusions we have just drawn. Although hydrochloric acid is more volatile than hydrocyanic acid, it has been observed that from a mixture of the dilute acids (o.i molar of HC1) it is possible to distil the HCN quantitatively (provided the dilution of the HC1 is retained at about this value by the frequent replacement of the water lost). In the light of the above theory the difference between the two acids in solution is that while - is nearly equal ft* to i for HC1, it is very small for HCN. In other words, HC1 is cofnposed principally of the ionized constituents H* and Cl', which cannot produce HC1 gas without going through the state HC1 in solution, and that is pre- vented by the nearly constant dilution which 'is re- tained during the distillation. Any gaseous substance then, which in solution is largely ionized, is more difficult to distil than an un-ionized or less ionized one. The HCN, being dissolved and retained in this state in solution, can be expelled readily just as any other gas which undergoes no great change in solution. This method, indeed, was discovered as the result of such theoretical reasoning, and it is but one example of the many practical applications of the above generalization. (For details of the separation see Richards and Singer, Am. Chem. J., 27, 205, 1902). 15 ELEMENTS OF PHYSICAL CHEMISTRY. It is not to be imagined that the facts mentioned above are the only ones leading to these conclusions, for later, throughout our work, we shall find occasion to consider other things which will confirm each of the steps leading to the final conclusion. In other words, it is not to be thought that the whole theory has been described in this place, or that, because some of the points mentioned are not clear, the theory itself is to be condemned, for many of the points can only be brought out after considering certain other methods which will enlarge our horizon. It may be said, however, that these further aids but confirm and make more evident the truth of the con- clusions we have arrived at. At the same time we must not forget that we have been speaking of this subject as lying within certain limits, and so cannot expect our con- clusions to hold outside of them, nor to condemn them because they do not. The relation of substances in non- aqueous solvents to a certain extent is different, and .consequently these conclusions could not be expected to hold. As a matter of fact, the conduction relations for these solutions are so utterly different from the aqueous ones that it would be impossible to consider them together at all in the light of our present knowledge. All of these points will be discussed more fully later, however, and the limits stated, within which our conclusions in general will hold. It is to be remembered, however, that simply because our theory does not hold for solutions in certain SOLUTIONS. 151 non-aqueous solvents (solutions which show no similarity in behavior to the aqueous ones, and which may or may not be solutions, as we consider them, but may involve an entire rearrangement of the composition of the solvent, or solute, or both), it should not be considered as false and of little use. For the two kinds of systems are so different that it would be impossible to imagine that both are subject to the same laws. The value for ct, the degree of ionization for a few electrolytes is given below for varying conditions. These are the values as found from the ratio of molecular con- ductivities, since that method is the most delicate one for this purpose which we possess. 2 4 8 16 V 25 V 60 a a 16 0.828 16 0.841 64 0.8 9 9 64 0.909 5" 0.962 5" 0.964 4 HC1 16 0.832 25 a 64 0.904 2 0.876 5" 0.965 16 -955 HBr 25 HI 25 KC1 NaCl V 25 25 NH s a LiCl 25 a a a a a a 0.897 0.895 2 0.737 0.932 0.926 10 0.86 0.842 0.852 0.803 0.950 0-945 loo 0.94 0.937 0.94 0.007 0.965 0.963 looo 0-98 0.982 0.979- 10000 o 993 16667 0.906 . i$ 2 ELEMENTS OF PHYSICAL CHEMISTRY. For further information of this sort see Rudolphi, Zeit. f. phys., Chem., 17, 385 1895. 39. Solution pressure. The pressure which a solid exerts in going into solution is called the solution pressure. This term, naturally, is expressive of the fact that when the substance has dissolved it exerts osmotic pressure. Since the osmotic pressure increases with the amount of substance dissolved, the solution pressure of a substance can be no greater than its osmotic pressure in a saturated solution, for that pressure prevents the substance from dissolving further. A substance is less soluble in its own solution than in pure water, and it seems that it is the osmotic pressure which is due to that substance which effects its solubility principally; at any rate, no general law is known which governs the solubility of a substance in solutions of other substance, i.e., when the solution pressure is counteracted by an osmotic pressure due to another substance. As will be seen later, one of the ionic products of a substance will when in solution depress the solubility of this substance. The behavior of a sub- stance dissolving, then, is very similar to a substance going into the gaseous state, and we have a solution pressure corresponding to the vapor-pressure. The work of separation of i mole of a dissociated substance from the volume of solvent which contained it is equal to iRT, analogous to the work done by expansion of a dissociated gas, where i is the sum of undissociated SOLUTIONS. 153 and dissociated portions, i.e., the total number of moles present, (i a)+na, where n represents the number of moles of ionized matter formed from i mole of the sub- stance dissolved. Since iRT . RT ~*> a, the degree of dissociation, may be readily calculated, for . i i a= . n i Since in the example above the hydrochloric acid was almost completely ionized, while the hydrocyanic acid was not, the work of separation from the solvent, as given by this formula, is obviously greater for the former than the latter, for the value of i is greater. To make such a separation in general, then, it is necessary that the value of iRT exceed that of RT, and that by the dilution the difference is made as marked as possible. 40. Vapor-pressures of solutions. It has been known for many years that the vapor-pressure of a solution is lower than that of the pure solvent. Babo (1848) and Wullner (1856) found, further, that the depression 0} the vapor- pressure is proportional to the amount oj solute present; and jor the same solution the depression for any temperature is the same fraction oj the vapor- pressure oj the pure solvent. If p is the vapor-pressure of the pure solvent, p r that 154 ELEMENTS OF PHYSICAL CHEMISTRY. of the solution, and w the percentage of solute, then Lj where k is the relative depression, , for a i % solution. Raoult in 1887 found that a more general relation is obtained if the relative depression is determined for i mole of substance in a certain amount of solvent. By experiment he found that all solutions containing one mole of substance in a certain weight of solvent possesses the same vapor-pressure; i.e., the MOLECULAR depression of the vapor- pressure is constant for all substances in the same solvent. Further, he found that solutions of equi- molecular amounts of substance in equal weights of different solvents give relative depressions of the vapor- pressure which are proportional to the molecular weights of the solvents. This shows that the vapor-pressure depends upon the relative number of moles of the solute and solvent. Raoult summed up his results as follows: One mole of any substance dissolved in 100 moles of any solvent causes a i% depression oj the vapor- pressure. (Here the number of moles of solvent is calculated from the gaseous molecular weight.) This law may also be expressed in another form: The vapor-pressure oj a solution is to that oj the pure sol- vent as the number of moles oj the solvent is to the iota] SOLUTIONS. 155 number of moles present in the solution, i.e. ,0/ solvent plus solute. For very dilute solutions the number of moles of the solute is so small in comparison with that of the solvent that the ratio becomes one, i.e., the vapor-pressure of the solution is practically the same as that of the solvent. We thus obtain another definition of the molecular weight of a substance in solution, and this as far as we know agrees perfectly with our previous one based upon osmotic pressure. According to it; the molec- ular weight of a substance in solution is that weight which in 100 formula weights in grams of the solvent depresses the vapor-pressure of this one hundredth of its value. Expressing the above laws as equations we obtain or where N is the number of moles of the solvent and n is the number of moles of the solute. This n is not the same as the one used in the formula (i a)+na, but is equal to this whole expression, i.e., to the total num- ber of moles present. As the text will prevent con- fusion in these two values, and as they are usually used, they are retained here. I5 6 ELEMENTS OF PHYSICAL CHEMISTRY. These equations will only hold when the solute is non- volatile, a condition which is practically fulfilled when its boiling-point is 140 above that of the solvent, and when n represents the correct number of moles in solu- tion. For ionized substances, then, n represents the formula weights dissolved, after they are multiplied by the expression (i a)+na, or, in other words, this n is i times the number of formula weights dissolved. The value of N here is the number of formula weights of solvent, i.e., the weight in grams divided by the molecular weight of the solvent in the gaseous state. For the determination of either the molecular weight or the dissociation it is more convenient to, use an altered form of equation (27), i.e., according to which the depression oj the vapor- pressure is to the vapor- pressure oj the solution as the number oj moles o] solute is to the number oj moles 0} solvent. w Since n=, where w is the weight of solute and m W m its molecular weight, and JY=-rr, W and M being these terms for the solvent, we have p tf wM wM p' r 2) " = -- SOLUTIONS. 157 Example. A solution of 2.47 grams of ethyl ben- zoate in 100 grams of benzene has a vapor-pressure of 742.6 mm. of Hg, while pure benzene has one of 751.86 mm., both at 80 C. Find the molecular weight of ethyl benzoate. 2^ = 2.47, M = jS, W = ioo, pr =742.6, # = 751.86, #-^=9.26; hence 01 = 154, while from the chemical formula we find C6H 5 COOC 2 H 5 = 150. Biddle (Am. Chem. J., 29, 341, 1903) gives a dif- ferential method for determining the difference in vapor- pressures of pure solvent and solution (i.e., pfl) for any concentration and suggests calculating this to i mole of substance in 100 grams of solvent, and using it as the molecular depression of the vapor-pressure. In this way, of course, it is possible to determine molec- ular weights of other substances in thai solvent by the use of a simple proportion. . Thus we would find (p pf) for i mole : (p p') found : : i : x, and x would represent w the number of moles, i.e., , contained in 100 grams of the solvent. Knowing the weight w, it is thus easy to calculate m, the molecular weight. For most organic solvents correct molecular weights are obtained by the use of these formulae, for in them the dissociation is so small that it may be neglected. When both solvent and solute are volatile it is still possible to calculate the vapor-pressure of the mixture 158 ELEMENTS OF PHYSICAL CHEMISTRY. under certain conditions. The conditions here are these: That no chemical reaction takes place between the solvent and solute, and that the molecular weight in the liquid state is the same as that in the gaseous state, for both solvent and solute. Since in such a case the vapor-pressure of each is depressed by the other we can find the total vapor-pres- sure by subtracting from the sum of the vapor-pressures (i.e., assuming that neither affects the other) the sum of the effects of each on the other. We have then where the sub-numerals distinguish the two constituents, the expressions - and - - represent the molar fraction of each in the mixture, and the terms within the parenthesis are obtained from (27), considering each in turn as the solvent. By rearrangement this formula becomes \ n\ n 2 (31) x^Pi where the two terms represent the partial pressures of the constituents in the mixture, or, in other terms, pi and p2 f being the partial pressures of each in the mixture, fi and p2 having represented the vapor-pres- SOLUTIONS. 159 sures at that temperature of the pure substances as measured alone. Since by Dalton's law the total pressure is equal to the sum of the partial pressures, and these by definition (Avagadro's law) are dependent upon the number of moles in the gaseous space, we have and since, by (31) and (32) P\ =P\ / I / rl\ ~Tfl 2 ind Calling #2' , - ^2 7- r=2 and ~=c 2 , fii' +1*2 r ' - where c 2 is the molar fraction of the one constituent in the gaseous phase, and c 2 that of the same in the liquid phase, we have / \ / (33) ^=i -- 160 ELEMENTS OF PHYSICAL CHEMISTRY. and or f \ (34) which was first deduced by Nernst (see Gerber, Dissert. Ueber die Zusammensetzung des Dampfes von Fliis- sigkeitsgemischen. University of Jena, 1894). One conclusion which can be drawn from (34) is of great importance, j If the composition of the gaseous mixture distilling from a liquid mixture is the same as that of the latter, the boiling-point and vapor-pressure of the mixture is the same as that of the pure solvent. Here we must not confuse such mixtures with those which have a constant boiling-point due to a maximum or minimum of vapor-pressure, for those do not leave behind the same concentration as that distilling (see pp. 122, 123). Equation (31) only holds strictly for solutions of liquids which act normally. In case they do not, by an appro- priate change in the formula, their behavior may be expressed with some accuracy. Since this change gives no general relation for all substances, but necessitates the knowledge of specific constants for each we shall not SOLUTIONS. 161 consider it here. (For details as to this see von Zawidzki, Zeit. f. phys. Chem., 35, 129, 1900.) An example of the application of the relation expressed by (30), (31) and (32) is given by a mixture of benzene and ethylene chloride at 49. 99 C. The sub-numeral i refers throughout to the benzene. 29.79 moles of ethylene chloride, of which the vapor- pressure (p 2 ) is 236.2 mm. are mixed with 70.21 of benzene, pi = 268 mm. What is the total vapor-pressure of the mixture? By (31) we find TT = 268 Xo.702 1 + 236.2 X 0.2979 = 258.42, the observed value being 259 mm. Equation (34) may also be applied to these data; we have 268 259 0.2979 -2'= 25Q Xo.7021, cj =0.2735, where the observed value is 0.2722, i.e., the distillate from the solution containing 29.79 molar per cent contains but 27.35 m olar per cent of this and 72.65 of the other. 41. The relation between osmotic pressure and the depression of the vapor-pressure. The relation between these two pressures can be shown by the following pro- cess: It is to be remembered, however, that the equation below (41) is not the only way, nor even the most con- 162 ELEMENTS OF PHYSICAL CHEMISTRY* venient one, to % transform osmotic pressures into vapor- pressures, or vice versa. Since both osmotic pressure and vapor-pressure are dependent, according to what was given above, upon the molar concentrations existing in the solutions, the transformation from one to the other is best made by the reduction of one to concentration, and the calculation of this to the other. This process which we are now going to consider, then, has simply the purpose of showing the theoretical relations existing between the two kinds of pressures, and will, perhaps, make them more real to the reader. Imagine an appa- ratus in the form of Fig. n. The tube h, which contains FIG. ii. a solution, has at its lower end a semipermeable partition, which allows passage to the solvent, but not to the solute. This tube is placed in the vessel F, which contains the SOLUTIONS. 163 pure solvent. Assume now that the Whole apparatus is covered with a bell jar from which the air is exhausted. The two liquids will be in equilibrium when the weight exerted by the column hG is equal to the osmotic pressure of the solution. Both liquids will evaporate, one at h and the other at G. The vapor-pressure of the solution at h must then be the same as that of the solvent at the same place. If this were not true liquid would either condense or evaporate at h, and this would disturb the equilibrium between the height of the column and the osmotic pressure in such a way that water would go through the partition. This would cause, however, a continuous cycle, from which we might obtain work without any change in temperature, i.e., a perpetual motion; hence the vapor-pressure of the two liquids must be the same at h. The actual pressure which the pure solvent has at h is equal to its vapor-pressure minus the weight of the column of vapor hG. This is, then, the vapor-pressure of the solution. Let there be N moles of the solvent and n moles of the solute. The osmotic pressure, i.e., the pressure which the substance would exert in gas form, under the like conditions, from (9), for i mole is or for n moles (35) 1 64 ELEMENTS OF PHYSICAL CHEMISTRY. N moles of the solvent, however, will weigh NM grams, where M is the formula weight ; hence where 5 is the specific gravity of the solvent. Substituting this .in (35), we obtain P nRTs = P, the osmotic pressure, however, is also equal to the weight of the column of solution per square centimeter, i.e., (37) P = hs, the specific gravity of the solvent being used because the hydrostatic pressure depends only upon the liquid which can go through the partition. Combining (36) and (37), we find nRT (30J k= ^ , T.T. MN The weight of the column of vapor hG is proportional to the difference between the vapor-pressure of the solvent, />, and that of the solution, p f , i.e., where a is the specific gravity of the vapor. In case SOLUTIONS. 165 the column is of a considerable height its density will depend upon the pressure, i.e., adh=dp, or dp (39) -5fc- The density a, however, is equal to -^-; hence Mp a =T or when combined with (36) dp Mp dh~RT' i.e., dpRT=Mpdh, or and finally by integration between the limits pc=P and h= ^ p (40) 1 66 ELEMENTS OF PHYSICAL CHEMISTRY. Combining this with (38), we find nRT RT p_ MN~ M g " p" or p n But this term, log<, , can be written in the form and this when developed in a series gives P-P 2 - Here, however, the second term may be dropped, since p p' is small, and we find N' or _ p N+n' which is the equation found experimentally by Raoult. n Whenever ]y : we find 7PT* 4) P = hs =~77S log, ^ gr.-cms. (R =84800 gr.-cms.). To obtain the value of P in atmospheres it is only necessary to divide this result by 1033 gr.-cms., the pressure of i atmosphere. We have then RT p RT p-f (41) P = - s \og e .= -- 775 atmos. 1033^ * p Example. 2.47 grams of ethyl benzoate in 100 grams of benzene gives ^ = 742.6, while ^ = 751.86 (at 80) Find the osmotic pressure. ^=84800, 7^ = 273 + 80 = 353, M = 78, 5=0.8149, #-^.=9.26; hence P=3.6 atmospheres at 80 C. This same result could, of course, be obtained by find- ing the number of moles of ethyl benzoate contained in i liter of benzene (i.e., 10 X 0.8149 X~^J and mul- tiplying this by the osmotic pressure due to i mole 1 68 ELEMENTS OF PHYSICAL CHEMISTRY. per liter at 80 C. In this way, however, we obtain the value 3.9 atmospheres, instead of 3.6, owing to the fact that, as shown on p. 157, the vapor-pressure relation leads to a molecular weight of 154, in place of the formula weight of 150, which we have used here. The ratio of the vapor-pressures , gives in general the relation between the number of moles of solute to those of solvent, i.e., -^. If we know the formula weight of the solvent in the gaseous state, it is then possible to rind the number of moles of solute present, for example, in i liter. The vapor-pressure of an aqueous solution at o is 0.95 of that of the pure solvent, what is the osmotic pres- sure of the solution? According to the formula P-P' _n p' N (or either of the other two may be used) we have i .95 n .95 i OOP' ~78~ where n is the number of moles of solute in i liter of water. The osmotic pressure, then, is 22.4X^ = 22.4X2.78=62.27 atmos. SOLUTIONS. 169 42. Increase of the boiling-point. Since the vapor- pressure of a solvent is depressed by the solution of a substance in it, the boiling-point must also increase, so that from this term it is also possible to define the mo- lecular weight of the substance dissolved. We have only to find for this purpose the relation between the number of moles of solute to the corresponding increase in the boiling temperature. This relation can be found by the aid of the second principle of thermodynamics, and this we shall do later. First, however, it will be well to consider the law as found empirically, from the fact that the vapor-pressure of a solution is lower than that of the pure solvent. One mole of any substance dissolved in 100 grams of solvent must always cause a certain definite increase in the boiling point of that solvent, since it causes a definite depression of the vapor-pressure. If A is the increase due to a i% solution, then MA is that due to i mole in 100 grams of solvent. We have, then, K=MA, or where K is the molecular increase oj the boiling-point, i.e., that due to the solution oj i mole oj substance in 17 ELEMENTS OF PHYSICAL CHEMISTRY. TOO grams o] solvent, which must be constant for all sub- stances in the same solvent. If g grams of substance are dissolved in G grams of solvent and the increase of the boiling-point is J, then JG A = , ioog' and we have for the molecular weight This term K can be found for any solvent by ascer- taining the increase in the boiling-point due to a cer- tain amount of a substance whose molecular weight is known, and solving the equation for K. Thus i mole of any substance dissolved in 100 grams of ether increases the boiling-point of the latter 2i.i. This is calculated from the solution of a smaller amount of substance, so that the observed increase is much smaller and much more accurately determined. K may also be found as already mentioned by thermo- dynamical reasoning, by aid of a cyclic process, or simply by aid of the second principle. The general relation is derived as follows: Assume in 100 grams of a solvent whose formula weight is M and whose boiling-point, under atmospheric pressure p, is T, that there are n moles of solute. Under the pressure p the solution V QC THE UNIVERSITY SOLUTIONS. OF NA boils at the temperature T+dT. At the temperature T the vapor-pressure of the solvent is p, and consequently at the temperature T + dT it will be p-\-dp. The differ- ence in vapor-pressure, then, between the solvent and the solution at the temperature T + dT is dp. The variation of the vapor-pressure with the temperature of a liquid has already been given (p. 72) as dp lp or But the relative depression of the vapor-pressure caused by the solution of n moles of solute in 100 grams of solvent is equal to dp dp+p' or, since dp is small as compared to p, dp P' According to Raoult's law (p. 156) this term -J- is iyi p p' n 100 equal to j^, i.e., , =-^, where N =~^'i f or ^ here I7 2 ELEMENTS OF PHYSICAL CHEMISTRY. is equal to the former value p p', and p is the vapor- pressure of the solution, the jf of that formula. We have, then, dp_^_nM^ JdT_ p ~N~Too ~2T2' Since, however, /, the molecular heat of evaporation of the solvent, is equal to Mw, where w is the heat for i gram, it follows that nM _MwdT 100 = 2 T 2 ' or 0.02 T 2 dT = - n, w which for n = i reduces to w i.e., the increase in the boiling-point caused by the ad- dition of i mole of substance to 100 grams of solvent. Some of the values of K, determined in both ways, for they give the same results, are: benzene, 26.70; chloro- form, 36.60; carbon disulphide, 23.70; water, 5.20. The osmotic pressure may also be found directly from the increase in the boiling-point in the following way, since SOLUTIONS. 173 7/7 T* -// by substituting- in place of ^ in (41), we obtain Rswt where w is latent heat for i gram, since l=Mw, and / =dT is the increase in the boiling-point. Or the value of one may be transformed to the number of moles of substance per liter of solvent, and the other calculated from that. The apparatus which is used for the practical deter- mination was devised by Beckmann and is shown in Fig. 12. This form of apparatus has of late been changed in such a way that the heating-bath is eliminated, and tetrahedrous formed by folding platinum foil substi- tuted for the garnets. (For details see Ostwald-Luther, Physiko-chemische Messungen.) About 20 cc. of the solvent is placed in the vapor-bath B, which causes the boiling-tube A to be heated evenly. The boiling- tube has a piece of platinum wire fused in its bottom, and this is covered with small garnets or beads to prevent all bumping and to cause the boiling to take place gently. On top of these beads a weighted amount of the solvent is placed and the thermometer so fixed that the bulb is covered with the liquid. The heating is to be done carefully until the liquid boils, the part evaporating being condensed in the spiral tubes. After a short time the temperature becomes constant and the reading of the thermometer is taken. This gives the boiling-point of 174 ELEMENTS OF PHYSICAL CHEMISTRY. the solvent, the amount of which present is known. A weighed amount of the substance is now introduced into the tube and the temperature of boiling again observed. FIG. 12. This is the boiling-point of the solution. The thermom- eter is one which was invented by Beckmann and is divided into hundredths of a degree. SOLUTIONS. 175 Example. Beckmann found for a solution of 2.0579 grams of iodine in 30.14 grams of ether an increase in the boiling-point of the ether equal to o.566. What is the molecular weight of iodine? = 2.0579, G= 30.14, ^=0.566, # = 21.10; hence Or 2.0579 = 2i.ioXioo -rr~ = 254. 0.566X30.14 2.0579 2 1. 1 : 0.566:: M gr. per 100 gr. : - Xioo, ^ = 254. I 2 corresponds to 254; hence in a solution of iodine in ether the molecular weight is twice the usual formula weight. 0.02 T 2 If in K = - -- (p. 172) we substitute the value of w I Ap 2 T 2 (p. 72), i.e., == We btam According to Innes (Proc. Chem. Soc., 18, 26-28, 1902) this formula gives better values in some cases than the regular one. It is possible to use this in all cases where the latent heat of evaporation is unknown and where the AT ratio -: is. p, here, is the pressure under which the solvent 176 ELEMENTS OF PHYSICAL CHEMISTRY. AT boils, i.e., for the temperature for which -r is determined, and M is the molecular weight of the solvent in the AT gaseous state. Since the value of -7 can be more Ap easily and more accurately determined than the latent heat, this form is exceedingly useful when using little known solvents. // is always to be remembered that these laws only hold when the pure solvent itself separates. In case the solute also is volatile a correction may be employed, however, which will lead to the correct result. (See Nernst, Zeit. f. phys. Chem., 8, 16, 1891). 43. Depression of the freezing-point. More than one hundred years ago Blagden found by experiment that the freezing-point of a solvent is depressed by the addition of any substance to it. Raoult found further (1887) that I mole of any sub- stance dissolved in 100 grams oj any one solvent causes a constant depression oj the freezing- point. If i mole of any substance dissolved in 100 grams of any one sol- vent causes a depression equal to K, then if A is the depression caused by a i% solution MA=K, where M is the molecular weight of the solute, or M = iooK-, SOLUTIONS. 177 If g grams of substance are dissolved in G grams of solvent, and the depression is J, then or where K may be found experimentally, as in the case of the boiling-point, when a substance of known molecular weight is used. Remembering that K is the depression of the freezing- point caused by the addition of i mole of solute to 100 grams of solvent, all results may be calculated by the aid of a simple proportion in place of the formula above. Thus,- as on page 175, we have i mole per 100 gr. :K: : x gr. per 100: depression due to x gr. per 100. It is also possible to find an equation by which the value of K can be determined for any solvent. This equa- tion is derived as follows : Assume a very large amount of a solution, containing P% of solute, in a cylinder which is provided with a semipermeable piston (Fig. 10, p. 138). Let the freezing-point of the pure solvent be r, and its latent heat of solidification, i.e., for i gram, be w; further, assume T 4 to be the freezing-point 17** ELEMENTS OF PHYSICAL CHEMISTRY. of the solution. Cause an amount of the solvent to be separated from the solution at T by a pressure upon the piston. If the amount of solvent separated previously contained i mole of substance in solution, and if the amount of solution is so great that this has not caused an appreciable increase in the concentration, then the osmotic pressure, p, will remain unchanged, and we shall have to do the osmotic work Now allow this volume of solvent to freeze at T. By looMiv this we obtain 5 cais. at T t since the weight of this volume is ~~p~i where M is the molecular weight of the solute. Next reduce the temperature of both ice and solution J, i.e., to T J, and allow the solid to melt in the solution, by which 5 cals. are absorbed at T J. Finally, raise the temperature of the whole system again to T. The two amounts of heat, i.e., the amount liberated by the cooling of the system J, and that absorbed by the heating J, cancel. By this reversible process we have done the work SOLUTIONS. 179 and by it have transferred 5 cals. from T A to T, for heat has been absorbed at T A and liberated at T. By the second principle of thermodynamics (p. 57) we have, then, 2 T J P or 0.02 T 2 AM ~^T~ = ~' AM This term ~p~> however, is the molecular depression of the freezing-point, i.e., that due to a solution of i mole in 100 grams of solvent. If we call this /, then 0.02 r^ i =^v. W This value can also be determined in the following short way although the principle is identical with the above (see Macloskie, Science, N. S., 9, 206-207, 1899). Since i mole of sugar to the liter of water gives an osmotic pressure of 22.4 atmospheres, the separation of the sol- vent from the solute would require the work of 22.4 liter atmospheres. Separating the solvent as ice would involve 1000X80 calories. By the second law of thermo- dynamics we have, then, the relation io ELEMENTS OF PHYSICAL CHEMISTRY. Work done Lowering of temperature Heat during it, in terms of work Temperature i.e., 22.4 lit. at. K' 80000 X. 04 1 lit. at. 273* from which K = ioK' = ioXi> 86 = 18.6 just as above. The value of K varies naturally with the different solvents, but is constant for any one. Some of the observed values are: water, 18.9; acetic acid, 38.8; benzene, 49.0; phenol, 75.0. From the fact that the vapor-pressure of a solution is lower than that of the pure solvent it is necessary that the freezing-point of the solvent be depressed by the addition of any substance. When the ice which is separated is the pure solvent, and the freezing-point is that temperature at which both solid and liquid may exist together in equilibrium in all proportions, the vapor-pressure of ice and liquid must be the same. This has already been proven, and if it were not true a per- petual motion would be possible. In Fig. 13 ww is the vapor-pressure curve for water, ss that for a solu- tion, and ii that for ice. At the point t=o ice and water have the same vapor-pressure, and so are in equilibrium. The solution and ice, however, will only be in equilibrium at the temperature corresponding to the intersection of the two curves i.e., the freezing-point cf the solution SOLUTIONS. 181 must lie below that of pure water. The more substance in the solution, the lower the vapor-pressure will be and the lower the point of intersection will lie, i.e., the lower the freezing-point will be. This is the same as the empirical law already used by which the depression is proportional to the amount of substance dissolved. We can also define the molecular weight of a sub- FIG. 13. stance in solution by the depression it produces in the freezing-point of the solvent, the definition depending upon the nature of the solvent. Thus the molecular weight of any substance in water is that weight in grams which will depress the freezing-point of 100 grams of water i8.9; or will depress it in the same ratio when dissolved in another amount. ,This holds of course for the boil- ing-point as well, when the proper increase is inserted. These laws also hold for metals dissolved in mercury, i.e., for amalgams. Heycock & Neville have shown 182 ELEMENTS OF PHYSICAL CHEMISTRY. that potassium, sodium, thallium, and zinc in the metallic state all have a molecular weight equal to their atomic weights. And Hoitsema has shown that hydrogen in palladium has a molecular weight equal to i. // is to be remembered, however, that these laws only hold when pure solvent solidifies. When water is used as a solvent, and the substance is dissociated in it, the molecular weight found will be smaller than the formula weight. If oc is the true molec- ular weight and tf is that found either by boiling or freezing-point, then i.e., the ratio of the true molecular weight to that found is the same as that of the total number of moles present to the number present provided no dissociation takes place. This follows from the fact that the greater the number of moles present in a certain volume the smaller must be the molecular weight. i, the total number of moles present when we have started with one, is given by the equation or a -!+*, for binary electrolytes, where i-a represents the frac- tion of the solute which is left in the undissociated state, SOLUTIONS. 183 and 20. gives the number of moles of ions formed from each decomposing mole, a being the degree of disso- ciation. The experimental value of i is readily deter- mined from the true molecular depression and that found, i.e., for water molecular depression found This term 18.9 is the value of K for the solvent used. An example of the method as used to determine molec- ular weight and the dissociation will make the calcu- lation clear. A solution of 0.68 1 gram of acetic acid (which is very slightly dissociated) in 100 grams of water causes a depression of o.2i68. What is the molecular weight of acetic acid? # = 18.9, g=o.68i, G = ioo, J=o.2i68; hence 0.681 M = 18.9X100 -=SQ.4. 0.2168X100 Or 18.9: 0.2 i68::Mgr. per loogr. :o.68i gr. per 100, M = 59.4, while CH 3 COOH= 60. A 0.0107 normal solution of KOH gives a depres- sion of o.o388. Find the degree of dissociation. Since i mole in 100 grams of water causes a depres- I4 ELEMENTS OF PHYSICAL CHEMISTRY. sion of 1 8. 9, i mole in 1000 grams would cause one of i.89. Our molecular depression is then 0.0388 . 3.6261 -=3.6261, and ^= = 1.010. 0.0107 1.89 a=i 1 = 1.919-1 =0.919, or the KOH is 91.9% dis- sociated. The apparatus for the determination of the freezing- point was devised by Beckmann and is shown in Fig. 14. In the freezing-tube A the solvent (15-30 cc.) is introduced and a freezing-mixture (ice and salt) is placed in C. The temperature is allowed to fall until the liquid is overcooled i or 2 degrees; then it is stirred, ice forms, and the thermometer rises to the true freezing- point and remains constant. The cause of the rise in temperature is the sudden formation of ice, which gives up heat to the liquid. B is a tube which acts as an air- bath and causes the cooling to take place more evenly. After the freezing-point of the pure solvent is deter- mined the solution is introduced and the process repeated, Or the solution may be made in the tube by dropping a weighed amount of solute into the known amount of the solvent. The result, however, will only be correct when the solid which separates is the pure solvent. One point to be kept in mind is that the freezing- SOLUTIONS. 185 point thus determined is of a solution which is more concentrated than the one started with, for some of the solvent has been removed by the freezing. This FIG. 14. follows from the fact that the freezing-point is that temperature at which ice and solution exist together in equilibrium. The concentration of the solution due to the separation of ice may be calculated from the overcooling (see p. 100), for it has been found for water that for each degree of overcooling 12.5 grams 1 86 ELEMENTS OF PHYSICAL CHEMISTRY. of water separates from each liter. Each degree of over- cooling, then, causes the solution to be 1/80 more con- centrated than the original one. If the solution is dilute, we may find the freezing-point of the original solution by a proportion. Thus if the overcooling is i, the depression of the freezing-point o.i, and v represents the volume of solution, we have where x gives us the depression of the freezing-point caused by the solution of the weighed amount of sub- stance in the volume v i/So, i.e., in the volume of solvent present after the separation of solid. The osmotic pressure may be found directly from the depression of the freezing-point just as from the increase in the boiling-point, by substituting in (41) the value of -77 in terms of depression of the freezing- p-p' dp IdT . point. The expression , =-r=f5 is the general relation between vapor-pressure and temperature and holds for either the depression of the freezing-point or the increase of the boiling-point, except that / is latent heat of fusion in one case and of evaporation in the other. We have, then, from (41) Rswt P = - - atmospheres, I033X2XT SOLUTIONS. 187 where w is the heat of fusion, since ~Tf =w i an d t is the depression of the freezing-point, T being the tempera- ture of the process. Here, also, the transformation of freezing-point depres- sion to osmotic pressure, increase of the boiling-point or depression of the vapor-pressure can be accomplished by first finding from the freezing-point depression the molar concentration of the solution and calculating the other values from this, according to the appropriate law. If we determine the freezing-point of a solution and then add to it a substance which unites with the one already in solution, thus keeping the number of moles the same, the new freezing-point will not differ from the original one. This method can be used to find the number of moles of each substance uniting to form the new one, for we know the origmal number and the new depression will show how many are present after uniting. If the ice which separates contains solid substance, the freezing-point of the solution will be different from that of the solvent itself, this one, of course, being the point at which the solid solution of ice and substance separates. For such a rase, naturally, the law of the depression of the freezing-point does not hold. 44. Division of a substance between two non-miscible solvents. Depressed solubility. If a water solution of i88 ELEMENTS OF PHYSICAL CHEMISTRY. succinic acid is shaken with ether, the acid is divided between the two solvents in such a way that there is always a certain ratio between the two concentrations, independent of the relative amounts of the two solvents. Table IV shows this. TABLE IV. H 2 O(i). Ether (2). In H 2 O (3). In Ether (4). Coefficient (5). 70 cc. 30 cc. 43-4 7-i 6 almost 49 cc. 49 cc. 43 . 8 7.4 6 ' ' 28 cc. 55-5 cc - 47-4 7-9 6 exact Columns 3 and 4 give the number of cc. of a Ba(OH)2 'solution which is necessary to neutralize 100 cc. of the solutions. The coefficient of partition depends in absolute value upon the temperature and the dilution of the water solution. For decreased temperature and concentration the constant decreases. // there are two or more substances in solution the co- efficients of partition are the same as if each substance were present alone. The solvent behaves here with respect to the substance dissolved just as it would to a gas, i.e.,' between the two parts (liquid and solution) there is a certain coefficient absorption the concentration of substance acting as the pressure of the gas, and the temperature having the same influence in both cases. If the relation of the molecular weights remains the same throughout the various concentrations employed, the coefficient of partition will remain constant. Thus SOLUTIONS. 189 />-nitrophenol shows a coefficient of partition [ ) in \c w ' the system water-chloroform,^ which increases with the temperature and concentration and indicates that the />-nitrophenol is normal in molecular weight in water and is polymerized in chloroform solution. In dilute solu- tion the polymerization is very inconsiderable, however. In few words, speaking from the experiments, this means that the greater the amount of />-nitrophenol in the system and the higher the temperature, the greater is the proportion dissolved by the chloroform, and this variation has nothing to do with the solubility variations with the temperature of the solvents when used alone. An interesting application of this was made by Morse (Zeit. f. phys. Chem., 41, 1902) to find the amount of HgCU remaining uncombined in a complex equilibrium. By finding the coefficient of partition of HgCU between water and toluol for known amounts of HgCb and then that for HgCl 2 between the aqueous complex solution and toluol he was able to find the amount of HgCl 2 which was free and capable of dissolving from the complex solution. (See also 60 and 75, Chapter VIII.) If one solvent is soluble in another the amount of the one dissolved will depend upon its state, i.e., whether it is pure or contains a substance in solution. Just as with solids, we have for liquids a certain solution pressure, j.e,, a force which causes them to dissolve. In the same way 190 ELEMENTS OF PHYSICAL CHEMISTRY. that the vapor-pressure is depressed by the addition of a substance, so is the solubility of one liquid in another. This follows from the analogy between gases and sub- stances in solution. If s is the solubility of A, i.e., the amount of A in the unit of volume of B, and s f is* the solubility of the same after an addition of substance to A, then in analogy to _ p' N we have ss' n ~7~ = N' where N is the number of moles of solvent A and n the number of moles of solute in A . This was first deduced by Nernst, who used the formula as a method for the determination of the molecular weight. If an excess of ether is agitated with water and the freezing-point of the mixture determined it is found to be lower than that for pure water. If the ethei* contains an amount of substance which is insoluble in water, then its solubility in water will be decreased, and the freezing- point of the mixture will be higher than before. Calling this difference the molecular increase of the freezing- point, when the dissolved ether itself contains i mole of dissolved substance, we have the simple relation /-/" SOLUTIONS. 191 where m is the molecular weight of the substance dissolved in ether, w is the weight of this in 100 grams of ether, / is the freezing-point of a saturated aqueous solution of ether, /' the freezing-point of the saturated aqueous solution of ether containing w grams per 100 grams, and ^ is this molecular increase of the freezing-point, just defined. The value of ^ for a water-ether system (it varies with each system) is 3.06. Table V gives a few results of this method. The benzene and naphthalene are the substances which in each case are dissolved in the ether TABLE V. Substance. w t-f tn (cal.) m (found) Benzene 2.04 0.080 78 77 5.87 0.219 78 82 Naphthalene 3.42 0.082 128 128 t in all cases is equal to-3.85, i.e., the freezing-point of a saturated solution of ether in water. The solution of substance in ether is here slightly increased in concentra- tion, owing to the separation of the ether with the ice, but, as very strong ethereal solutions are rarely used, it has but little influence. Other substances than ether may also be used with water, the constant V of course being different for each one. Other results and the calculation of the value of V which transforms freezing-point results into those of solubility must be sought in the original paper. (Nernst, Zeit., f. phys. Chem., 6, 16 and 573, 1890). 1 92 ELEMENTS OF PHYSICAL CHEMISTRY. 45. Solid solutions. Solid solutions are homogeneous solid phases which can vary continuously within certain limits. These limits are smaller for solids than they are for liquids, just as the limits for these are much more restricted than for gases. The most important differences between a solid solution and a mixture of solids are as follows: Work is necessary to separate the constituents of the solution, and consequently these do work in uniting to form the solution; a solution has a lower vapor-pres- sure than the pure solvent; and a solution has a smaller solubility in all liquids than the pure solvent. Examples of solid solutions are given by mixtures of isomorphous crystals; zeoliths, i.e., natural water holding silicates, which lose water and yet retain their transparency, although the vapor-pressure is reduced; hydrogen with palladium; hydrogen with iron; and mixtures of two Crystalline substances which have different Torms, but which give a mix-crystal like that of the substance present in excess. The reader must be referred for further in- formation on this subject to one of the books treating it in detail, Findlay's Phase Rule, for example. 46. Colloidal mixtures.* Colloidal mixtures, for con- venience, are divided into two classes. Of these we shall consider colloidal solutions, the type of which is a solu- * For further information the reader is referred to a paper by Noyes (Jour. Am. Chem. Soc., 27, 85, 1905), of which I have made extensive use here. SOLUTIONS. 193 tion of gelatine, and colloidal suspensions, of which arsenious sulphide is a type. A colloidal mixture is a liquid mixture of two or more substances which are neither separated by the long con- tinued action of gravity nor by passage through filter- paper, although they are separated by the passage through animal membranes or close grained porcelain. Colloidal solutions. These as a rule are more or less viscous, gelatinize upon evaporation and go into solution again on the addition of water. They are not coagu- lated by the addition of small amounts of salts, and, in common with colloidal suspensions, influence the vapor- pressure, freezing-point, boiling-point of the solvent very much less than a corresponding amount of a crystalloid. This is also true for the osmotic pressure of the solution formed. It is evident, then, employing the above defini- tions of molecular weight in solution, that either their molecular weights are very high, or the definitions do not hold in such cases. At present the question is still open, although from the enormous results obtained for the molecular weights, it would seem that the latter conclu- sion is the correct one. The osmotic pressure of a 6% glue solution has been found to be but "a third of an at- mosphere; and the rate of diffusion is also found to be much smaller than for crystalloids. The colloids of this group include gelatine, agar-agar, unheated albumen, caramel, starch, dextrine, etc. Gela- 194 ELEMENTS OF PHYSICAL CHEMISTRY. tinized colloids have been found to be permeable to salts in solution, and to offer but an exceedingly slight hind- rance to their passage. The electrical conductivity, as a result of this, of salt solutions is observed to be but a few per cent different from the value found in pure water, a fact of which we shall take advantage in Chapter IX. These colloids seem to be impermeable to one another, however. Colloidal suspensions. These, in contrast to the other class, are coagulated by the presence of even a small quantity of an electrolyte. They do not gelatinize, as do the others, on cooling, and if gelatinized do not redissolve on heating. The presence of a gelatinizing colloid in even a small quantity prevents the coagulation of colloidal suspensions by salts. Thus, in the presence of gelatine silver chloride precipitated in solution remains in a state of colloidal suspension, and this fact has long been turned to account in the preparing of emulsions in photographic work. It is true, however, that a few other substances serve this same purpose, as sugar, glycerol and even ether. These suspensions are not only coagulated by the presence of electrolytes, but are not affected at all by the presence of non-electrolytes. Thus the addition of hydro- gen sulphide water to a solution of mercuric cyanide pro- duces a colloidal suspension, for neither the hydrocyanic acid formed, nor the other two substances are ionized beyond a slight extent, i.e., all show a small value for SOLUTIONS. 195 = . It seems to be necessary to always exceed a Poo certain small minimum quantity of the electrolyte in order to produce coagulation. Metallic colloidal suspensions may be made (Bredig) by producing an arc under pure water, using electrodes of the metal in question. Some colloidal solutions can be proven to consist of small particles by examination under the microscope, although the number is not very large. One unfailing property of all colloidal mixtures is that they revolve the plane of polarized light, and, in fact, in general behave as though they are made up of exceedingly small particles, although it is impossible to prove that this is always the case. The effect of the electric current upon colloidal mix- tures is exceedingly important, for it gives us some insight into their constitution. Certain suspensions are always carried in the direction of the current through the solution, i.e., coagulate around the cathode (ferric hydroxide). Others, and this is by far the larger class, collect around the anode, i.e., go with the negative current. Of the first group we know of the basic hydroxides (Al, Cr, etc.) and certain dyestuffs. All the rest, the ordinary suspen- sions as well as the colloids, go in the direction of the negative current and coagulate around the anode. When a colloid is mixed with another having a differ- ent sign, i.e., when two colloids, which go in opposite *9 6 ELEMENTS OF PHYSICAL CHEMISTRY. directions under the influence of the current, are mixed we observe a coagulation, leaving clear solvent above. This is the case with ferric hydroxide (+) and arsenious sulphide ( ). Non-electrolytes have no effect upon colloidal mixtures, but electrolytes (above a certain minimum concentration) cause a coagulation, and this effect is apparently proportional to the valence of the ion of the opposite sign to the colloid in question. Since the valence determines the amount of electricity carried by ionized matter, it seems but a natural consequence of our conclusions above, that ionized matter, and that only, outside of the current, can cause the coagulation of these mixtures. From the relation with regard to the migration of colloids with the electric current it is evident that the colloidal particles carry charges of electricity, some always of one sign, some of the other, depending upon their nature. And, further, since non- electrolytes fail to act as do electrolytes, with regard to coagulation, these facts seem to confirm our conclu- sions that the constituents of electrolytes are charged with electricity. It is assumed that the colloidal particles are charged with one kind of electricity while the liquid in which they exist has the opposite charge. Anything which can disturb this apparent electrical equilibrium, then, and the disturbance apparently must be electrical itself, causes the colloidal mixture to coagulate. As to how the particles and solvent become differently SOLUTIONS. 197 charged we have only hypothesis as yet, or rather hypo- theses, for there are several, and nothing definite is known. 47. The molecular weight in solution. Our definitions of molecular weight in solution (osmotic pressure, vapor- pressure, freezing-point and boiling-point) in general lead us to the same results when it is possible to equalize the experimental error and employ all methods. We have found, however, that in some cases the molecular weight so determined is very much smaller than the formula weight (dissociation), and in other cases several times larger than the formula weight (association). Further, two associated liquids (according to definition by surface-tension) when mixed often show a mutual dissociation of the associated state. But these cases, excepting the last, are also found in connection with the molecular weight in the gaseous gas, as it is affected by temperature changes, etc. It is not a question here of any difficulty with our definition of molecular weight, for the various methods seem to agree very well, and it is generally conceded that the various results express the molecular weight as it changes from solvent to solvent, and in the same solvent from dilution to dilution. Certain results are found by the freezing-point method for both electrolytes and non-electrolytes, however, which are hard to follow so accurately by other methods, that are difficult to explain other than by assuming that there 198 ELEMENTS OF PHYSICAL CHEMISTRY. exists a hydration of the substance in solution. The freezing-point depression, in other words, is found to be greater than it should be from results calculated from data for more dilute solutions. The effect of this hydra- tion would be small as to the increase in the molecular weight due to the addition of water; it is the solvent which is removed which would exert the greater influence upon the result, especially on the small amount present in concentrated solutions. By plotting actually observed results and assuming water to be removed from the sol- vent and to unite with the substance, a curve can be found which will be in accord with the law. In this way it is proven that ij a hydrate were formed to the requi- site extent, the law would be followed. Later we shall consider this further (Chap. VIII). Outside of the one" experimental fact (see Chap. IX) observed by Morgan and Kanolt (J. Am. Chem. Soc., 26, 635, 1904) the for- mation of compounds of the solute and solvent has not been actually observed. According to the freezing-point method this hydration has been assumed to take place with both the ionized and the un-ionized portion; the experimental proof just mentioned refers only to the ionized matter in a solution of silver nitrate in a mix- ture of water and pyridine. All the methods given above seem to give explicable results for very dilute solutions; the difficulties will be discussed later. For concentrated solutions, however, SOLUTIONS. 199 it is quite evident that the correct relations have not been obtained as yet. Whether the behavior of very concentrated solutions can ever be brought into accord with the more dilute ones is a question which is still open. It is obvious, though, notwithstanding the diffi- culties which have arisen, that the theory of solution is a law of nature holding between certain limits, and that the problems in this connection for the future have' all to do with enlarging these limits. That this can only be done by following the results closely and avoiding all hypotheses is an opinion which is growing not alone in this branch of science, but in all others. To do this, however, it is not necessary to cast aside what we have learned, thinking it hypothesis, but to use the fragment of a great law of nature which we undoubtedly have in the theory of solution as a basis for further development. CHAPTER VII. THERMOCHEMISTRY. 48. Definition. Thermochemistry is the subject which treats o) the connection between chemical and thermal processes. ' Since almost every chemical process is accompanied by temperature changes, and is more or less influenced by them, this subject is one of great importance. If a substance is changed from one state into another in such a way that all difference in energy appears in one form, as heat, for example, then this amount, when determined, is proportional to the difference of chemical energy in the two states. 49. Applications of the principle of the conservation of energy. Hess was the first to apply this principle to thermochemistry. In 1840 he announced the law that the amount oj heat generated by a chemical reaction is the same whether it takes place all at once or in steps. In other words, all transjormations from the same original state to the same final state liberate the same amount oj heat, irrespective oj the process by which the final state 200 THERMOCHEMISTRY. 201 is reached. The heat liberated in a reaction, then, de- pends only upon the final and initial states. This was entirely the result of experiment on the part of Hess, who did not consider it as self-evident. The truth of this principle can be shown, for example, by the equality in the heat of formation of ammonium chloride in water where prepared in two dissimilar ways. We have (NH 3 , HC1) = +42100 cal. (NH 4 Claq.) =- 3900 " NH 3 ,HC1, aq. = +38200 cal. and (NH 3 ,aq.) -+ 8400 cal. (HC1, aq.) = + 17300 " (NH 3 ,aq.,HCl,aq.) = + 12300 " (NH 3 , HC1, aq.) = + 38000 cal. Upon this law the whole subject of thermochemistry is based, for by it it is possible to find indirectly the heat liberated or absorbed by any reaction. This is true even though it is not possible to carry out the reaction in practice. We have simply to consider the process as a part or a sum of others which have been measured; then, by the addition of the equations, all undesired substances may be caused to disappear until finally the desired re- action is found. The chemical symbols, as we have used them up to now, have represented only the molecular weights of the 202 ELEMENTS OF PHYSICAL CHEMISTRY. substances in question. Now, however, where we are combining with the chemical reaction the quantity of energy absorbed or liberated by the reaction, the sym- bols have a further meaning which must not be over- looked. In addition to the weight in grams expressed by the form'ula weights of the substances, the symbols also represent the amount of energy contained in that weight in this state as compared to the energy contained in another, the standard, state. Thus the equation C + 2(9 = CO '2 + 97000 cals. means that the energy contained in 12 grams of solid carbon and 32 grams of gaseous oxygen is greater by 97000 cals. than that contained at the same tempera- ture in 44 grams of gaseous carbon dioxide. The small calorie, cal., as will be seen, is so small that the numbers employed are very large, and, more important still, do not express well the experimental limitations of the determination. For this reason the Ostwald calorie (K) is much better adapted for the purpose. This is the heat which is necessary to raise the temperature of i cc. of H 2 O from o to 100 C., and is related to the large and small calories in approxi- mately the following way : cals. (i.e., i cc. H 2 O i C.); Cal. (i.e., i kg. of H 2 O i C.). THERMOCHEMISTRY. 203 It will be observed here that one unit in terms of the Ostwald calorie expresses about the limit of accuracy of experimental observations, while of the other two, one leads to the impression that the experimental result is more accurate than it is, the other that it is less accurate. In addition to these units we also have others which are still more convenient, for they are based upon the erg. The joule, designated by j, is equal to io 7 ergs, and a unit one thousand times this, i.e., io 10 ergs is desig- nated by J. We have, then, i cal.=4.i83 j and i j =0.2391 cal. =io 7 ergs, or i cal. =0.004183 J and i J = 239.1 cal. = io 10 ergs. In all reactions we shall distinguish the state of aggregation according to the system proposed by Ost- wald, in which ordinary letters are used for liquids, heavy ones for solids, and italics for gases. In considering substances in solution it is necessary to know the heat which is liberated by this process. This is called the heat of solution. For different amounts of water, of course, this will vary, so that for uniformity the heat of solution is always understood to be the heat (positive or negative) which is liberated by the solution of i mole in so much water that an addition of more water will give no additional heat effect. This is desig- nated by the addition of the abbreviation Aq (aqua) 204 ELEMENTS OF PHYSICAL CHEMISTRY. to the symbol of the substance. It is always to be re- membered, however, that an isolated Aq in an equation is not equal to an H 2 O, which refers to 18 grams of water. The following example of the determination, by the process of elimination, of the heat liberated by a reaction is taken from the work of Thomsen. His object was to find the heat of formation of SeO 2 Aq, i.e., the heat generated when SeO 2 is formed from its elements and dissolved in a large amount of water. He found SeO 2 + 2HClAq + 2 NaHSAq 2NaOHAq + 2 HClAq - 2 NaClAq + 2NaOHAq + 2H 2 S = 2NaHS Aq + 2 50^ ; By adding these equations, with the signs changed as indicated, it follows that In the same way we can find the heat of combustion of carbon in oxygen, a value which cannot be directly measured. Two reactions which have been measured are and CO+ Adding these, after changing the signs of the second, we obtain THERMOCHEMISTRY. 205 C+ O=CO + 2 These facts, taken in connection with those mentioned above (pp. 141-145) and the conclusions arrived at there, are not so startling as one might imagine at first glance. Since for the acid and base we have the relation and, since the salt is observed to have a molecular weight (by definition) equal to one-half the formula weight, i.e., is completely ionized according to all the possible methods of measurement, it is quite certain that it is made up of the substances previously composing the acid and base in the same state as that in which they existed in them. In other words, expressing the chem- ical equation in accord with the experimental facts above, we have 2i6 ELEMENTS OF PHYSICAL CHEMISTRY, 9 where n represents the number of moles of water present in the system before the reaction. Since the conductivity shows the constituents of the salt (the ions) to be present in the same form they were in originally, the only portion of the reaction which could possibly involve heat is the formation of water from ionized hydrogen (H") and ionized hydroxyl (OHO- As we know that hydrogen and oxygen in the ionized state can exist together to but an infinitesimal extent (for pure water conducts only very slightly), the follow- ing conclusion is certainly justified. When an acid unites with a base (at any rate in the condition in which we have assumed them) the cause of the reaction is the inability of ionized hydrogen to exist in the presence of ionized hydroxyl beyond an exceedingly small amount; and the heat of the neutralization (for this case) is that heat which is evolved during the formation of water from its ions in this way, i.e., 137^ for each mole of H* and OH' (by definition) forming one mole of H 2 0. By a method which we shall consider later (Chap. VIII) it is possible not only to show the presence of, but to calculate accurately, the heat involved in the dissociation of a substance. When the acid and salt are completely ionized, for example, and the base but slightly, it is possible to show just how much extra heat (either positive or negative) is involved by the further THERMOCHEMISTRY. 217 dissociation of the base. For the partly dissociated base must increase in dissociation as its ionized OH is used up, since the more dilute the solution of the base the greater is its ionization, up to a certain point. If both the acid and base are but partly dissociated the result will differ still more, for heat will be absorbed or evolved by the further dissociation of both of these. In general, we shall have, then, if the salt, also, is not com- pletely dissociated, i.e., if more heat is liberated by its undissociated product being formed, where a\ = dissociation of acid, w\ =its heat of dissociation, 2 = " " base, w 2 =" " " a 3 = " " salt, w 3 =" " " x = heat of association of i mole of H* ions with i mole of OH' ions to form i mole of H 2 O ; i.e., the heat generated by the neutralization of an acid by a base is equal, for each mole oj water formed, to 13?K plus the product oj the heat of dissociation of the salt into the undissociated portion minus the same products for the acid and base. Naturally the negative value of the heat of associa- tion of H' and OH' ions is the heat of dissociation of 2i8 ELEMENTS OF PHYSICAL CHEMISTRY. water, i.e., the heat necessary to form i mole of H' and i mole of OH' ions, from water. Later we shall consider this relation more in detail, i.e., after we have studied the method to be used for the measurement of the heat of dissociation. It is obvious from the above that the thermal prop- erties of electrolytes are additive when they are in such a dilution that they fulfill the condition = i jMoo When not in this condition the change in the thermal effect depends upon the amount of heat involved in causing them to alter their states. When a precipitate is formed in such a solution (i.e., when a chemical reaction takes place, which was excluded above) it is often possible to find its heat of formation just as we found that of water above. An example of this is the following: Ag- Aq + NO' 3 Aq + Na' Aq + Cl'Aq = AgClAq + Na'Aq or Ag' Aq + Cl'Aq = AgClAq + 1 58^ i.e., when i mole of AgCl is formed from the ionized silver and ionized chlorine in a solution 158^" is pro- duced. Conversely if i mole of AgCl were dissolved; THERMOCHEMISTRY. 219 this amount of heat would be absorbed, i.e., the heat of solution of a substance is equal to the negative value of the heat of precipitation. Although this is not always possible, we can find the heat of formation in solution in another way. The princi- ple of this is as follows: By electrical measurements it has been possible to find the amount of heat involved when 2 grams of gaseous hydrogen form 2 grams of ionized hydrogen in solution. This value is approxi- mately equal to 4 J, but since there is some uncertainty about its exact value, it is usual to assume it equal to zero. Later, then, the results based upon this can be readily recalculated. From this value, by dissolving a metal in a completely ionized acid, i.e., by the substi- tution of metal in the ionized state for the hydrogen, which is evolved as a gas from that state, we can observe directly the heat of formation of the ionized metal from massive metal. By then determining the heat of solu- tion of a completely ionized salt of this metal, the heat due to the negative radical can be determined readily, for the heat of solution of the salt is equal to the sum of the heats of ionization of the constituents, of which we assume that of hydrogen to be zero. In this way the table given below has been prepared by Ostwald. In order to find the heat of formation of the salt it is only necessary to form the sum of the heats due to the ions into which it decomposes, taking into 22O ELEMENTS OF PHYSICAL CHEMISTRY. account the valence of the ions as indicated by the dots for the electro-positive and the accents for the electro- negative substances. Cathions J = joules X io 3 Anions of J = joules X io 3 Hydrogen H- + o Hydrochloric acid Cl' + 164 Potassium K- + 2 59 Hypochlorous acid CIO' + 109 Sodium Na- + 240 Chloric acid cio/ + 98 Lithium Li' + 263 Perchloric acid CIO/ - 162 Rubidium Rb- + 262 Hydrobromic Br' + 118 Ammonium NH 4 +137 Bromic acid BrO 3 ' + 47 Hydroxylamine NH >' +157 Hydriodic acid I' + 55 Magnesium Mg ' +456 lodic acid icv + 234 Calcium Ca- + 458(?) Periodic acid io/ + i95 Strontium Sr" + 501 Hydrosulphuric acid S" - 53 Aluminium Al" + 506 HS' + 5 Manganese Iron Mn Fe" + 210 + 93 Thiosulphuric acid Dithionic acid S 2 3 " S 2 " + 581 + 1166 Fe" - 39 Tetrathionic acid S 4 6 " + 1093 Cobalt Co' + 71 Sulphurous acid SO 3 " + 633 Nickel Ni" + 67 Sulphuric acid SO/' + 897 Zinc Zn" + 147 Hydrogen selenide Se" - 149 Cadmium Cd- + 77 Selenious acid SeO 3 " + 501 Copper Cu- - 66 Selenic acid SeO/' + 607 Cu- - 6 7 (?) Hydrogen telluride Te" 146 Mercury Kg' - 85 Tellurous acid TeO 3 " + 323 Silver Ag- -106 Telluric acid TeO/' + 412 Thallium Tl- + 7 Nitrous acid NO 2 ' + H3 Lead Pb- + 2 Nitric acid NO/ + 205 Tin Sn" + 14 Phosphorous acid HPO/ + 603 Phosphoric acid PO/" + 1246 HPO/' + 1277 Arsenic acid AsO/" + 900 Hydroxyl OH' + 228 Carbonic acid HCO/ + 683 CO/' + 674 These numbers hold only for the case that the ions are in very dilute solution, i.e., Aq should be added to the symbol of each. For stronger solutions, in which the ionization is not complete, other amounts of heat are involved which, unless allowed for, will lead to incorrect results. THERMOCHEMISTRY. 221 The equations Na = Na* + 240 J and mean that by the transformation of the formula weight of metallic sodium into the ionized state 240 J are evolved ; and for the change of the formula weight of chlorine gas into two formula weights of ionized chlorine (p. 147) 2X1647 are liberated. . UNIVERSITY j ( F PAL r r v ,- CHAPTER VIII. CHEMICAL CHANGE. A. EQUILIBRIUM. 56. Reversible reactions. If we bring a number of reacting substances together in a chemical system, and leave them for a sufficient length of time, the reaction will reach an end. To represent any chemical reaction we may use the equation Here n\ moles of A\, n 2 of A 2) n 3 of A 3 , etc., unite to form n\ moles of A\' y n 2 ' of A 2 , n 3 of A 3) etc. When all these substances can remain together for an indefinite length of time, without the reaction going in either direc- tion, they are said to exist in chemical equilibrium. Reactions which go partly from left to right when we start with the substances AI, A 2 , etc., and partly from right to left when we start with AI, A 2 etc., are called reversible or reciprocal reactions, provided that in each 222 CHEMICAL CHANGE. 223 case, starting with equivalent amounts, the final equi- librium is the same for both directions. An excellent example of such a reaction is C 2 H 5 OH + CH 3 COOH^CH 3 COOC 2 H5 + H 2 O. Alcohol. Acetic Acid. Ethyl Acetate. Water. If we start from the left side we obtain a certain defi- nite amount of those on right and vice versa. For example, i mole (46 grams) of alcohol plus i mole (60 grams) of acetic acid, or i mole (88 grams) of ethyl acetate plus i mole (18 grams) of water, will always give the same final state, in which we have 1/3 mole alcohol + 1/3 mole acetic acid + 2/3 mole ethyl acetate + 2/3 mole water. 57. The law of mass action. Considering such a reversible reaction as that above, or, for example, the question at once arises in which direction and to what extent will such a reaction go when we statf, for instance, with a certain concentration or pressure of each of the three gaseous constituents, HI, I and H? From the purely chemical point of view the above equation simply provides that if we start with i mole of hydrogen and i mole of iodine, and if these unite 224 ELEMENTS OF PHYSICAL CHEMISTRY. completely, 2 moles of hydriodic acid gas will be formed; or if we start with 2 moles of hydriodic acid gas, and this is completely decomposed we shall obtain i mole each of hydrogen and iodine. As to what portion of the hydrogen and iodine will unite to form hydriodic acid; or what portion of the total original amount of hydriodic acid will decompose to form hydrogen and iodine; or what will take place if all three are mixed together; we are utterly ignorant, failing further infor- mation than that contained in the chemical equation. The answers to these questions can only be obtained by the application of a very general law which was first announced by Guldberg and Waage in 1864. The qualitative form of this law of mass action is as follows: Chemical action, at any stage of the process, is propor- tional to the active masses of the substances present at that time, i.e., to the amounts of each present in the unit af volume. In this form, however, the law of mass action is of but little practical use. It will be necessary, then, for us to derive a quantitative expression of it, and thus to obtain it in such a form that it may be applied to our needs in answering questions such as those alluded to above. Imagine a reaction of the type n\A i + n 2 A $^n\A \ + n 2 'A J CHEMICAL CHANGE. 225 having taken place in a closed vessel and to have attained a state of equilibrium in which we have the partial pres- sures Pi, p 2 , pi and p2 r . Assume, further, that it is possible to insert each of the substances on the left against its gaseous or osmotic pressure p\ t p2, and to remove each of the products, as they are formed, from the gaseous or osmotic pressure pi ', p2 f to the original external pressure po and that this insertion and removal is isothermal and reversible. Since by such a series of operations we would do work on one side ( ), and obtain work (+) from the other, the sum of the two amounts (regarding the signs) would give us an expression for the work ( + or ) which is done by the system itselj during the transformation, at constant temperature, of n\ moles of A\ and n 2 moles of A 2 to HI moles of A\ and w 2 ' moles of A 2 ', the initial and final pressure being the same, viz., p . And this in its turn would lead to the expression of the quan- titative relation existing between the active masses of the constituents at equilibrium, i.e., to the relation we seek. Since the work required to change the osmotic or gaseous pressure of i mole of substance from p Q to p\ is given by the expression ^riog^ ,* that for n\ moles Po po P Po 226 ELEMENTS OF PHYSICAL CHEMISTRY. Pi will be niRTlog e . For n 2 moles of A 2 we have the ' pQ corresponding expression n 2 RT log. . The sum of po these two terms is the work done, i.e., lost, by us in the process. The gain of work for us then for this stage is By the removal, as they are formed, of n\ moles of A \ and n 2 moles of A 2 , the amount of work (a gain for us) is . i In total, then, our gain in work in transforming n\ moles of A i ^Jn 2 ). CHEMICAL CHANGE. 227 But, as we simply wish to get the relation which depends upon the pressures in the reaction at equilibrium, and the pressure po has nothing to do with this, we can assume po to be i, and obtain, since the first term is equal to zero, W=RT(n l f As the processes of insertion and removal are assumed to be isothermal and reversible this work, W, must be the maximum work which can be done by the reaction, and hence must be a constant at any one temperature. We have, then, (42*) W =constant =RT and since if the logarithm is a constant the expression itself must be a constant, and since T and R are also constants, / Wl/ V 2 ' (42) Constant -K - * , or since pressure and concentration are proportional, '**' 2 K - where the values of K and K' may or may not be alike, 228 ELEMENTS OF PHYSICAL CHEMISTRY. according as we have the same number of moles on each side of the chemical equation, or a different number. The constant (K or K') is known as the constant oj equilibrium. We may express the law 0} mass action as follows, then : at equilibrium the product of the pressures (concentra- tions) oj the substances on the right (final ones), each raised to a power equal to the number 0} molecules reacting, divided by the product oj the pressures (concentrations) oj the substances on the left (initial ones), each raised to a corresponding power, is a constant jor any one reaction at any definite temperature. The variation of this constant, K or K' with the tem- perature is to be considered later, after we have studied the application of this most important and general law. 58. Equilibrium in homogeneous gaseous systems. For gases we can most conveniently use the form of the law of mass action which refers to partial pressures, (42). We have, then, for the equilibrium of a gaseous chemical system An example of this is given by the gaseous reaction CHEMICAL CHANGE. 229 If the partial pressure of H is pi, that of Ip 2y and that of Hip, then This case was investigated by Bodenstein, who found by experiment that at a temperature of boiling sulphur (440) - K =0.02012. .If we heat hydriodic acid, then, to this temperature, it is possible to calculate from its amount the amounts of hydrogen and iodine and undecomposed hydriodic acid in the gaseous state present at equilibrium. The total pressure of a mixture of gases is equal to where the terms on the right are the partial pressures. If H and / are present in the free state at the partial pressures a and b and HI to d, and we wish to find in what direction and to -what extent the reaction will go, we proceed as follows: Let x represent the partial pres- sure of H lost; then, according to the reaction 2HI = H 2 + l2, we shall have, at equilibrium, for the partial pressure of HI 230 ELEMENTS OF PHYSICAL CHEMISTRY. for H uncombined PH=&-X, and for / uncombined x must have such a value, then, (i.e., the reaction must go so far) ihd^^equilibrium it will just satisfy the equa- tion of the law (Blmass action, for the reaction at 440. (a- x )(b- x ) A =O.O2OI2 = Knowing a, b, and d, it is possible to solve the equation for x, and to find how, and how far the reaction will go. For example, in this case, if x is positive in value, the reaction will go toward the left, as we have assumed; if negative, in the opposite direction. There' is one thing to be said of the solution of such equations. There are two possible values of x\ which is to be taken? It will be found in this case, as, indeed, in all others, that only one value is in accord with the existing data, so that it alone could be taken. For instance, if the positive value of x is larger than a or b it would lead to an absurdity, for it would show a nega- tive value for H or /, and the other value is the correct one. In cases of equations of a higher degree, where CHEMICAL CHANGE. 231 more than two roots exist, this same rule is to be fol- lowed. A possible case here is to have two values of the same sign, but one smaller than the other. There can be no question in such a case, however, for if the reaction would be in equilibrium after the smaller change had occurred, it could not go out of this state to attain the equilibrium shown by the greater value, hence the lower value is to be taken as the correct one. Here we have used the partial pressure form of the law of mass action; we could use the other just as well, however, for it will be observed that the constant factor which would transform pressures to concentrations c = -- =- is eliminated, since we have the same 22.4X - 273 number of formula weights (2) on each side of the equa- tion For this reason the constant, K, for this reaction, as for all others with the same number of formula weights on both sides, has the same value for concentrations, pressures, volumes under standard conditions, or any other term proportional to concentrations or pressures. A further effect of this condition of equal volume on the two sides is that the progress of the reaction is per- fectly independent of pressure (Le Chatelier's theorem, 232 ELEMENTS OF PHYSICAL CHEMISTRY. p. 33) ; and Lcmoine has shown this to be true for the decomposition of HI for pressures ranging from 0.2 to 4.5 atmospheres. In using concentrations in place of partial pressures it is always to be remembered that the concentration (i.e., moles per liter) is the actual number of moles, present divided by the total volume (see pp. 29-33). An exam- ple will perhaps make this clearer. In the reaction A=2B + D,&t equilibrium, we have o.i mole of A, 0.3 of B and .05 of D in 10 liters at atmospheric pressure and o. Starting with 0.5 mole of A, o.i of B and 0.4 o D, in 22.4 liters, find direction and extent of the reaction. Here we must first find the constant of equilibrium for the data given at equilibrium. Since we have o.i mole of A in 10 liters, the concentration of A, at equilib- rium is , of B , and of D ^ , hence* 10 10 ' 10 /o.3\ 2 /o.o5\ \io/ \ io / K = Assuming that x moles of A are formed by the reac- tion, the final volume will be [(o.$+x) + (o.i 2%) + (0.4 #)] 22.4 liters, the temperature remaining con- stant at o, i.e., (i 2^)22.4 liters. The concentrations at equilibrium, then, will be 7 '- r - moles per liter (i 2^)22.4 CHEMICAL CHANGE. 233 of A, -r - r - of ^> an d 7 - of Z>, hence the '(i -2^)22.4 (1-200)22.4 value of -K, as found above, is to be equated to these values in the following way : W 0.4 x \ .4/ \(l-2X)22.4/ Q.5+* r_ and the sign of x will show the direction of the reaction, and the numerical value its extent (p. 230). Since according to the law of mass action the con- centration is to be raised to a power, it is the whole frac- tion representing it which is to be so treated, i.e., the number of moles per liter. When applied to the equilibrium resulting from a gaseous dissociation the constant of the law of mass action is usually designated as the constant of dissocia- tion. From it, it is possible, just as above, to calculate the degree of dissociation from a certain amount of the dissociating substance, or how much of the products, when present alone, or with the substance, will unite to form the substance itself. And, conversely, we can calculate K for each of the substances for which data was given on pages 27 and 28; and the values will be dependent only upon the temperature, the units employed, i,e., c or f, and nature of the substance. 234 ELEMENTS OF PHYSICAL CHEMISTRY. Where, as is the case here, we know a, the degree of dissociation of the substance, we can proceed as follows: For the reaction for example, we would have for concentrations the formula v A = C PC1 5 Starting with i mole of PC/s, which, if undissociated, would occupy V liters at atmospheric pressure, with a as the degree of dissociation, the concentrations, where V is the final volume, i.e. (i + a)V at the same tempera- ture and pressure, are as follows : For PC/5, at equilibrium, pr-, for PC/3 -y, and for C/ 2 T?, hence A~, which we wish to determine, is to be found from K (i-a)F V' > 'FF' At 250 for PC/5 a =80% (page 27) and, since the pressure is atmospheric, i mole must be present in 22.4 liters; this is equal to V. V, then, is CHEMICAL CHANGE. 23 5 equal to 1+0.8(22.4 j, and we have as the dis- sociation constant for PCI 5 at 250 (i-o.8)(i+o.8)22. 4 273 2 +25< From the value thus obtained we could then calculate the direction and extent of the reaction at 250 when we start with definite amounts of the three constituents, or the value of a for a different V. A physical idea of the dissociation constant, as found for concentrations, can be obtained by aid of the formula a 2 \K = - ( ^y. Assuming that a is equal to 0.5, i.e., that the degree of dissociation is 50%, for a reaction by which / \2 i mole is transformed into 2, we find that K = , \ v , or 2K =77. The dissociation constant of such a reaction, then, when multiplied by 2 is equal to the reciprocal oj the final volume resulting from the dissociation of I mole into 2 to the extent oj 50%. This volume is that in which i mole of the original substance must be placed in order that at that temperature it may dissociate to the extent of 50% into two others. Since 77, the reciprocal of the volume produced by the dissociation of i mole, is equal 236 ELEMENTS OF PHYSICAL CHEMISTRY. to C, the concentration in moles per liter, we also have ?.K = C. An example of the use of this relation is given by the reaction NzO^NOz + NOz, for which a 2 K=~. -- ^ = 0.0138 (calculated from ^ = 182.69 mm., regulate the reaction, however/ this constant value fulfils all our needs. Thus for the reaction calling the pressures due to gaseous CaCOs and CaO 242 ELEMENTS OP PHYSICAL CHEMISTRY. and 7T 2 , and that of the CC>2 p-> we have, according to the law of mass action, but since n\ and 7r 2 will remain constant at any one tem- perature, we may employ the simpler form i.e., equilibrium at any one temperature depends only upon -the pressure of the carbon dioxide gas produced. In Table VI the pressures are given under which equilibrium exists at different temperatures. TABLE VI. EQUILIBRIUM OF CaCO 3 ^CO2+CaO. erature C. Press, mms. of Hg. Temperature C. Pi ess. mms. 547 27 745 289 610 46 810 678 625 56 812 753 740 2 55 865 1333 This means that CaCOs when heated in a vacuum to any temperature gives off CC>2, CaCOs, and CaO until a certain pressure is reached. Thus at 547 a pressure of 27 mms. of Hg is produced. 60. Dissociation of a solid into more than one gas. The vapor of solid NH^HS shows by its density an almost complete dissociation into NH 3 and H 2 S, CHEMICAL CHANGE. H3 At 2 5. i the gaseous pressure is equal to 501 mms. of Hg, i.e., since the partial pressures of the H 2 S and NHa are the same they are each equal to nearly 250.5 mms. of Hg. Equal only nearly, however, because this pressure includes that of the undissociated NH 4 HS gas. But since this is very small and constant at any one temperature it may be neglected. If TT is the partial pressure of the NH 4 HS gas, and p\ and p2 are those of the NH 3 and H^S, then by the law of mass action Since, however, TT is constant at any one temperature, we have The total pressure, P = 5oi mms., is equal, by Dalton's law, to the sum of the partial pressures, and neglecting TT, we have, or and PP P* l2 =--=- or P 2 . =62750. 4 4 244 ELEMENTS Or PHYSICAL CHEMISTRY. This value K = pip 2 may be verified experimentally by observing the effect of the addition to the dissoci- ating system of one of the products of the dissociation, since the product of the partial pressures must always remain constant for constant temperature. Table VII gives the results of experiments carried out for this purpose, the pressures being mms. of Hg. TABLE VII. 2o8 294 61152 138 458 63204 417 146 60882 453 143 64779 Average = 62 504 The average of which agrees quite well with the value previously found. In each case a certain amount of one of the products is added before the solid is sublimed and the total pressure afterward determined. It is quite simple then to find the amount of solid which has dissociated and thus the total amount of each gas present. The case of the dissociation of ammonium carbamate is quite similar. We have .e., But hence CHEMICAL CHANGE. .*rP?l P 3 ~, which has been found to hold true by experiment. The pressure produced by ammonium carbamate when heated to various temperatures is given below. tperature. Gaseous Pressure. Temperature. Gaseous Pressu, -15 2.6 mm. 24 84 . 8 mm. - 5 7-5 30 124 o 12.4 40 248 2 15-7 48 402 4 19.0 5 470 10 29.8 55 600 18 53-7 60 770 Isambert has proven this relation to hold when NH 3 or CO 2 are initially present, i.e., that K remains constant even after addition of one of the products of the reaction. He also proved that the presence of an indifferent gas does not affect the dissociation, for the dissociation- pressure of a substance is not changed by the presence of an indifferent gas, i.e., the partial pressures remain unaltered. Contrary to the case of a homogeneous equi- librium (pp. 238 and 239), an increase of volume has no\ effect upon the degree of dissociation of a non-homogen- ! eous system, so long as the solid (liquid) phase is present, j for the dissociation-pressure is dependent, in such systems, only upon the temperature and nature of the substance. 246 ELEMENTS OF PHYSICAL CHEMISTRY. 61. Equilibrium in liquid systems. The reaction CH 3 COOH + CaHsOH^CHaCOOCaHs + H 2 O, as already observed, reaches the state of equilibrium when we have present 1/3 mole acid + 1/3 mole alcohol + 2/3 mole ester -f 2/3 mole water, provided we start with i mole of each of the two constituents (either acid and alcohol or ester and water). This reaction goes very slowly at ordinary tempera- tures, but when it reaches the above final state it remains in it indefinitely. If we designate by v the volume of the system, and start with i mole of acid, m moles of alcohol, and n moles of ester (or water), then in the state of equi- librium, after x moles of alcohol have been decomposed, we shall have alcohol = moles per liter. acid- x .. .. ester= _ or __ or i-r " " " CHEMICAL CHANGE. 247 hence, applying the law of mass action, we obtain A =' In the special case of equilibrium above, however, = i, n = o, x = 2/$; hence This value of K is one of the few which are practically independent of temperature. At 10 it is found that 65.2% undergoes change, while at 220 the decomposi- tion is but 66.5%. This equation has been tested by experiment with very satisfactory results. // has been found, also, that by using a large amount of acetic acid to a small amount oj alcohol, or vice versa, the formation of ester and water is almost complete. In the same way a large . amount of water upon a small quantity of ester causes the latter to be almost entirely transformed. In the following table some of the experimental results for this reaction are compared with the values calculated by aid of the above formula, and will serve to show how accurate this law is in its application to liquid systems. Obs. Calc. 0.05 0.049 0.078 0.078 0.171 o. 171 0.226 0.232 0.293 0.311 0.414 0.423 0.519 0.528 0.665 0.667 0.819 0.785 0.858 0.845 0.876 0.864 0.966 0.945 248 ELEMENTS OF PHYSICAL CHEMISTRY. Moles of Alcohol Moles of Ester or Water, to i Mole of Acid. 0.05 O.o8 O.l8 0.28 o-33 0.50 0.67 I.OO 1.50 2.OO 2.24 8.00 Using i mole of acid, i mole of alcohol, and various amounts of water (no ester being added) we find the following results. Moles of H 2 O Moles of Ester Formed. to i mole of acid - 1 mok of alcohol. i.o 1-5 2.0 4-o 6.5 .1-5 Amylene in contact with acid forms an ester, accord- ing to the equation \ ^ Obs. Calc. 0.665 0.667 0.614 0.596 o-547 0.542 0.486 o-S 0.458 0.465 0.341 0.368 0.284 0.288 o. 198 O.2I2 CHEMICAL CHANGE. 249 If x is the amount of ester formed when equilibrium is established, v is the volume of the system, and i mole of acid is used for a moles of amylene, then d x = amount of amylene left; = " " acid -= " " ester formed; hence xv The value for K 'in this case has also been determined experimentally. It was found that K _ .001205* the constant in the form A = having the value 0.001205. 250 ELEMENTS OF PHYSICAL CHEMISTRY. The agreement of theory and experiment in this case, the acid being trichloracetic, is shown by the following summary. a. v (liters). ix obs. * calc. 2.15 361 0.762 0.762 4.12 595 0.814 0.821 4.48 638 0.820 0.826 6.63 894 0.838 0.844 6.80 915 0.839 0.845 7.13 954 0.855 0.846 7.67 1018 0.855 0.846 9.12 1190 0-857 0.853 9.51 1237 0.863 0.853 14-15 *7 8 7 0.873 0.861 As will be observed, the volume here, since there are not the same number of formula weights on both sides, must be retained in the formula. Non-electrolytic dissociation in solution. When a solid goes into solution its action is apparently analogous to its transformation into the gaseous state. A saturated solution, thus, in contact with the solid at any tem- perature will still be saturated. We have, then, by the law of mass action, for any one temperature, K'n=c or K = c, where c is the concentration of solid in solution and varies with the temperature. If the solid in going into solution dissociates into other substances, then an addi- tion of one of these will cause less substance to dissolve, CHEMICAL CHANGE. 251 This has been proven by Behrend for a solution of phenan- threne picrate in absolute alcohol, in which a decom- position into phenanthrene and picric acid takes place to a large extent. By the law of mass action where c = undissociated phenanthrene picrate, Ci=free picric acid, and 2 = free phenanthrene all expressed in moles per liter. For any one temperature c must be constant, since the solution is saturated; hence = constant. Coefficient of partition or distribution. We have al- ready considered the distribution of a substance between two non-miscible solvents in the case that the formula weight is the same in each (pp. 187-189) ; now we must find the effect of a difference in the formula weight. If the formula weight in one is twice that in the other, it is A obvious that k* 2= c i 2 , i.e., that -- and ^ are constant, for all dilutions. An illustration of this is given by the distribution of benzoic acid between water (cj and ben- zene (c 2 ), the values of being 0.062, 0.048, and 0.030, while those of ~/=-, for the same dilutions, are 0.0305, 252 ELEMENTS OF PHYSICAL CHEMISTRY. 0.0304, and 0.0293. Benzoic acid, then, has twice the molecular weight in benzene that it has in water, a fact which has been proven by the freezing-point method. 62. The effect of temperature upon an equilibrium. The variation of the constant of equilibrium or dissocia- tion with the temperature. In the case of an invariant equilibrium under constant pressure a change in the temperature causes one of the phases to disappear entirely. In the case of a univariant equilibrium, how- ever, the effect is quite different. A very small change in temperature causes only a very slight change in the equilibrium. This causes a corresponding change in the relative composition of the reacting constituents, in one direction or the other, which just compensates the change which the reaction coefficient has suffered. By differentiating equation (420) (p. 227) by the aid of (42) we obtain log, K + RTd(\og e K). But by the second principle (p. 58) where dW is the former dQ, i.e., the work in terms of heat. Combining these two equations, we get CHEMICAL CHANGE. 253 Since W expresses the work which is done by the trans- fer of the total heat Q, the difference, Q-W, is equal to the heat appearing as heat in the reaction. We have thus by (420) or where the heat of reaction is given in terms of K and T. To integrate this expression it is necessary to assume that q itself is independent of the temperature. This will undoubtedly be practically true for small temperature intervals; for larger ones, however, we must be satisfied to obtain q as the value for the temperature which is the mean of the two extreme temperatures. By integration, under the above assumption, we find (43) Or, using ordinary logarithms and solving for q, we obtain RX 2.306 (log K' -log K)TT' (44) q= T-T ~ 254 ELEMENTS OF PHYSICAL CHEMISTRY. where 2.306 is the reciprocal of the modulus of the system of logarithms. This formula will then enable us to determine the varia- tion 0} the equilibrium constant K with the temperature. Since q = Q W, it only expresses the heat of the re- action as it would be if no external work were done by or upon the reaction. The allowance for this, in comparing the results of (44) with observed results, must be made in each case, as is illustrated in the applications given below. One consequence of this formula is of special impor- tance. If q is zero, the value of K does not change as the result of a change in temperature. Thus the reaction between acid and alcohol, mentioned above (p. 247), the mutual transformation of optical isomers, and a number of others, are found neither to absorb nor gen- erate heat, nor to suffer a displacement of equilibrium, i.e., a change in the value of K, by a change in tempera- ture. We shall now consider the method of applying this equation for various purposes to various equilibria. Vaporization. The condition regulating the equi- librium between a liquid and its vapor (a univariant system) is the pressure or concentration of the latter, and this depends upon the temperature. We have then V RT CHEMICAL CHANGE. 255 If p and p refer to the two temperatures T and T f , then by (43)> we nave > Regnault found for water ^ = 273, ^ = 4-54 mms. of Hg. T' = 273 + 11.54, j/ = 10.02 mms. of Hg. From which by aid of (44) the heat of evaporation of water at constant volume is found for i mole to be 10100 cals. By experiment it is found to be 10854 cals. when /T+T*\ the volume increases. If we subtract from this 2 ( - J = 557 cals., which is the work of expansion, we obtain q= 10297 cals. Dissociation of solids. If a solid dissociates into gases, the equilibrium is conditioned by the concentration of the latter. If in the dissociation ci and c 2 are the concentrations of H 2 S and NH 3 , that of the NH 4 HS being small and remaining constant, then r = -l-A --L-JL. 1 V RT 2 ~V'~RT" 256 ELEMENTS OF PHYSICAL CHEMISTRY. V meaning in each case the volume in which i mole is present. In this case _P_ 2 where P is the total pressure; hence P 2 "- ^1^2 DTI ^Kl If P' is the total pressure at any other temperature T 1 ', then r>/2 JTt^rJfJ^-L .. hence, or At the temperature = 273+ 9.5, P =175 mms. of Hg; 7 '=273 + 25.1, P' = 5oi mms. of Hg; CHEMICAL CHANGE. hence q= -21550 cals. By direct experiment the molecular heat of sublimation is found to be 22800 cals. We must, however, subtract 4 ( -J cals. from this, which amount has been used for the expansion. We have then 22800 1161 = 21639 cals. = q. The general form of the equation for this process, where n\ moles of one gas n^ of another, etc., are given off, is = -- R\T This formula may also be applied to the dissociation of salts containing water of crystallization into gaseous water and the dehydrated or partially dehydrated salt. Solution oj solids. In this case the equilibrium depends only upon the concentration of soKd substance in the solution, and the temperature; we have, then, K=c, where c is the concentration of a saturated solution at the temperature T. If d is the concentration of such 25 8 ELEMENTS OF PHYSICAL CHEMISTRY. a solution at another temperature T', then it is possible for us to calculate by (44) the heat of solution of the solid, the increase of volume being so small that it is practically equal to zero. van't HofT found by experiment with succinic acid in water that for T = 273, c = 2.88 moles per liter; and for ^' = 273 + 8.5, c' -4.22 moles per liter. For i mole, then, by (44), q=- 6900 cals., while Berthelot found 6700 cals. by direct experiment. lonization of solids in solution. If a substance is very slightly soluble, then the solution must contain prin- cipally ions and very little undissociated substance, and the heat of dissociation must be the same as the heat of solution, i.e., equal to the negative value of the heat of precipitation from the free ions. Thus for AgCl we have AgCl=Ag' If the solubility at T = c, and at T' = c? 9 in moles per liter, then, since 2 moles of the ions form from i mole of the salt, we have, as on page 257, CHEMICAL CHANGE. . 259 For 7^ = 273 + 20, c = i.ioXio~ 5 , hence q=i 5900 = 1 59^. For the negative heat of precipitation we found (p. 218) I5&K, which is an excellent agreement. We have assumed the ionization to be complete here and the fact that the heat results agree, cannot but be considered as confirmatory of our assumption. ^^" Dissociation oj gaseous bodies. When a substance A dissociates according to the scheme the equation of equilibrium is where c, c\, c 2 , . . . are the concentrations of A, A\, A 2 , . . . in moles per liter. If the mole occupies the volume V at T and the volume V at T', then for the dissociation of N 2 C>4 we have hence (p. 241) a' 2 a 2 / v , v (i a')V (i a)V 260 ELEMENTS OF PHYSICAL CHEMISTRY. and IOP" IOP = I I lu e /j _ a '\y 1U & however, is equal to (i+a) (i.e., 3.179:^: :i + a : i) ; hence V = T(i+a) and V' = T'(i+a'), and we obtain (i -a' 2 ) ^ T(i -a 2 ) From the results r = 273 + 26.i, ^ = 2.65, a ^ = 273 + 111.3, ^' = 1.65, a' we find q= 12900 cals. a 2 q /i i\ i -a 2 = 2~ (f~ T> ' CHEMICAL CHANGE. 261' After subtracting the heat equivalent to the work of expansion * we obtain from the experimental result q= 12500. By a similar calculation it is also possible to find the heat of dissociation of other gases. B. CHEMICAL KINETICS 63. Application of the law of mass action. Thus far we have only considered the equilibrium which is attained after the reaction has come to rest, i.e., after the two sides bear a constant relajtion to one another. The question now arises as to the progress of a reaction toward this state, and the factors upon which the time necessary to attain it depends. By aid of the law of mass action, it is possible to find an answer to both portions of this question. Since chemical action at any time, according to it (p. 224), is proportional, for constant temperature, to the active masses of the substances present, i.e., to those portions which are free to act, then, when we have two substances reacting, the concentrations being #1 and a 2 moles per liter, * The gas is supposed to increase its volume without doing work. 262 ELEMENTS OF PHYSICAL CHEMISTRY. where x is the fraction of a mole of each which decom- poses in the time /. The term k in this equation is known as the speed constant of the reaction, and is constant at any one temperature for any value of x in the reaction in question. Suppose we have the reversible reaction which after a time attains a state of equilibrium in which all four products are present. The relative amounts of these are dependent upon the value of K for this reaction at this temperature according to the relation If we start with a\ moles of A\ and a 2 moles of A^ then dx But if we start with a\' moles of A\ and aj of AJ, then dx' where rr is the velocity in the opposite direction. CHEMICAL CHANGE. 263 Starting with the substances on either side, then, those on the other will exert an ever-increasing influence upon the velocity due to the initial substances, and this velocity must decrease continually. Finally, however, equilib- rium will be attained and the ratio of the amounts on the two sides will remain constant, i.e., the reaction as a whole ceases, and any motion which exists is so compen- sated by a contrary one that it does not appear. Imagine we start with a^ moles of A\ and a 2 moles of A 2 . The total velocity due to these at any one time will be 7 y and at equilibrium, i.e., where -jr = o, or TT^ A = i.e., the equilibrium constant, K, oj any reversible reaction is equal to the ratio oj the speed constants oj that reaction jor the two directions. This has been proven to be true for a number of cases. For the system acid-alcohol (p. 223) it was found for a certain strength acid that = 0.000238 and ' = 0.000815, 264 ELEMENTS OF PHYSICAL CHEMISTRY. k from which K=-r^ = 2.g2, while direct experiment gave # = 2.84. All this is only true, however, when the reaction takes place isothermally , i.e., when the heat liberated or absorbed is removed or supplied and no change in the temperature results, j or the constants k and k' are dependent, upon the temperature. In general the application of this formula is very much simplified by the fact that most reactions are almost complete in one direction, so that the second term will be so small that it may be neglected. We have then dx In all these cases the values on the right are obtained by subtracting the loss from the concentration of the original substance and having as many such terms as there are moles in the formula. Thus for the reaction A = 2B + D we would write dx =k(c B -x)(c B -x)(c D -x). This is simply custom (see p. 236), for we could also write it - = k'(c B -2x) 2 (c D -x), CHEMICAL CHANGE. 265 and although the k value would be different, it would be constant. 64. Reactions of the first order. For convenience we shall divide all reactions into orders. Thus a reaction of the first order is one in which but one substance suffers a change in concentration. This definition is to be fur- ther restricted, in that for the first order it is necessary that the equation show but i mole of one substance changing its concentration. Cane-sugar in water solution is transformed in the presence of acids almost completely into dextrose and laevulose; it is inverted. The speed of the reaction is very small and increases with the amount of acid added. The progress of the reaction may be observed by aid of the polariscope. The unin verted portion revolves the plane of polarization to the right, while the two products revolve it to the left. If o is the positive angle of revolution at the time /=o, i.e., that which is due to the uninverted sugar alone, the amount of which is a moles, a ' is the negative angle after com- plete inversion, and a: is the angle at the time /, then, since the revolution is proportional to the concentra- tions, x, the amount inverted, is found from x=a For the time /=o, a=a^ i.e., #=o; for the time / - oo , i.e., after complete inversion, a = an or x = a. 266 ELEMENTS OF PHYSICAL CHEMISTRY. This process was first measured by Wilhelmy (1850), and it has played an important role in the history of chemical mechanics. The process follows the scheme Ci 2 H 22 On +H 2 O = 2C 6 Hi 2 O 6 whether acid is used or not, for the concentration of the latter does not change during the reaction. Accord- ing to the law of mass action the speed is proportional to the amounts of sugar and water. The latter, how- ever, is present in such an excess that its action may be regarded as constant. The speed of reaction, then, is proportional to the amount of sugar present and we have a reaction of the first order, i.e., the formula is dx . where, for t=o, #=o, and k is the inversion constant, which depends only upon the temperature. By integra- tion this becomes log e (a x) =kt+ constant, or since, for /=o, #=o, (a) =the constant, i.e. a i CHEMICAL CHANGE. 267 in other words, a constant fraction of the tota. amount of sugar is inverted in each unit of time. The meaning of the constant k in words is as follows: Its reciprocal value multiplied by the natural logarithm of 2 gives the time in minutes which is necessary for the transformation of one half the total amount of substance, provided the products of the reaction are removed as soon as they are formed and the substances replaced as they are used. This is shown by the substitution of for x. Fur- ther, for all reactions of the first order, this constant k is independent of the original concentration of the sub- stance. Table VIII gives the values for k for a 20% sugar solution in presence of 3,0.5 N. solution of lactic acid at 25 C TABLE VTTI. INVERSION OF SUGAR. /(minutes). a. 1435 43i5 7070 "3 60 14170 19815 29925 -100.77 OT J 3i-i 0.2348 2 5 .0 0-2359 20. 16 0.2343 13. 98 0.2310 10. 01 0.2301 7- 57 0.2316 5-o8 0.2291 i.6 S 0.2330 268 ELEMENTS OF PHYSICAL CHEMISTRY. Since only the constancy of the term - log^ is to be shown, the more convenient system of logarithms, i.e., that of Briggs, has been employed above. 65. Catalytic action of hydrogen ions. Catalysis. The inversion of sugar, as well as all other reactions of the first order which are hastened by the action of acids, is apparently due to the ionized hydrogen which is pres- ent. This action is known as a catalytic action of the acid present, which retains throughout the reaction its original concentration. The reason why H* ions should act in this way is unknown; but that they do have an accelerating effect is an established fact. In general a catalytic action is one that hastens the reaction in which it is without at the same time having its concentration changed by the reaction. An example of such an action is given by the increased speed of solu- tion of metallic mercury, silver, or copper in a nitric-acid solution in which one of these has already been dissolved to a slight extent. The catalytic action here is due to the formation of the oxides of nitrogen, these, when passed into the nitric acid from an exterior source, caus- ing the same action as the above. Since the catalysor takes no part in the reaction it does not alter the equilibrium constant, hence it must also has- k ten the reverse action, so that -77 may remain constant. K CHEMICAL CHANGE. 269 As there is no general law known for catalytic action the reader must be referred elsewhere for the vast number of isolated facts. The more concentrated the acid used the more rapidly the sugar is inverted, without, however, any exact propor- tionality. The inverting action increases more rapidly than the concentration. In presence of a neutral salt of the acid its inverting power is increased by about 10% for the stronger acids, but decreased for the weaker ones. Since the acceleration of the speed of the reaction is caused only by acids, and since they are distinguished by the presence of H* ions, the action must be due to these latter. Further than this, a series of acids arranged in the order of inverting power is found to be in the same order in which they stand in reference to ionization. Thus it is plain that the H* ions have this effect, although in concentrated solutions the inverting power is not strictly proportional to the ionization. For example, a 0.5 N. solution of HC1 inverts 6.07 times as rapidly as one of a o.i N. solution, while it contains only 4.64 times as much ionic H*. Arrhenius found the presence of other ions to increase the catalytic action of those of H*. That the speed of reaction increases more rapidly than the concentration of H' ions is assumed to be due to the fact that the negative ions are also present to an equal amount and so increase the inverting action of the H' ions. In the presence of a neutral salt, then, we have two 270 ELEMENTS OF PHYSICAL CHEMISTRY. separate actions upon the acid: the first causes a de- crease in the concentration of H' ions of the acid (see p. 144) ; the second accelerates the action of the H' ions left. The stronger acids are less influenced by the former than the weaker ones, because they are dissociated more nearly to the extent to which their salts are. Consequently the second action predominates, and we have the greatest inverting power for the strong acids when a salt is present. For the weaker acids, however, the second action is the smaller; hence the inverting action is less for weak acids in presence of a salt. Palmaer has found that in dilute solutions, when the influence of the neutral-salt action is avoided, the inver- sion constant is exactly proportional to the concentration of hydrogen ions. Another reaction of the first order which depends for its speed upon the catalytic action of the H' ions is the formation of alcohol and acid from an ester. For example, CH 3 COOC2H 5 +H 2 = CH 3 COOH + C 2 H 5 OH. The equation is the same as before, i.e., follows the law where a is the amount of ester in moles at the time / = o, and x is the amount transformed at the time /, CHEMICAL CHANGE. 271 and is equal also to the amount of acid or alcohol formed by the reaction during the time /. The progress of this reaction is observed by titrating the amount of free acetic acid present. Ostwald measured the value for the reaction with methylacetate. He started with i c.c. of methylacetate and 10 c.c. HC1 (i mole to the liter), diluted to 15 c.c. To neutralize this acid 13.33 c - c - f a Ba(OH) 2 solution would be necessary. The amounts of acetic acid formed are shown by titration after substracting 13.33; these amounts, expressed in c.c. of Ba(OH) 2 solution, for various times are as follows: o 14 34 IQQ 539 QO Minutes 0.92 2.14 8.82 13.09 14.11 c.c. The amount of ester (a) is represented, then, by 14.11 a x being this value minus 0.92, 2.14, etc. The values thus determined for k are 0.00209, 0.00210, 0.00214, and 0.00212. Another reaction which gives a constant when con- sidered as a reaction of the first order is the one which is usually written / 4AsHs = As4 + 6H2. Since at 310 this fulfils the definition of a reaction of the first order, it should be written not as above, as of the fourth order, but as of the first, i.e. 57$ ELEMENTS OF PHYSICAL CHEMISTRY. Here also the velocity is exactly proportional when the action of the neutral salt is avoided. If the amount of neutral salt in the solution is so large that it can be con- sidered constant, then a constant value will be obtained for k, although its absolute value will be different from that for the case that no neutral salt is present. Walker used this method to -determine H' ions present in a solution due to hydrolytic dissociation, which we shall discuss in detail below. 66. Reactions of the second order. Here two mole- cules suffer a change in concentration during the reaction, i.e., the constant depends upon the concentration of two substances. We have then dx = k(a-x)(b-x), or - [log, (b x) log, (a - x)] = kt + constant. For t = o, x = o, and the constant is i.e., , i . (a-x)b & (* K\t ^&e /A CHEMICAL CHANGE. 273 ,- ^ If we use equivalent amounts of the two substances, then a = b, and we have or i x t (ax)a Here k is inversely proportional to the original concentra- tion. An example of a reaction of the second order is CH 3 CO(3C 2 H5 + NaOH = CH 3 COONa + C 2 H 5 OH. Table IX gives the value of this constant as calculated for different lengths of time at the temperature of 10 C. TABLE IX. . , . . N Amount NaOH fc /(minutes).. in c . c . o f acid . * 61.95 489 5 -59 2.36 1037 42.40 2.38 2818 2 9-35 2-33 00 14.92 The question as to how the speed varies with the nature of the base used has been investigated quite thoroughly. Thus Reicher found for strong bases (those which are much dissociated) approximately equal values for k; for weak ones, however, he found very much smaller values. 274 ELEMENTS OF PHYSICAL CHEMISTRY. Ostwald observed that the weak bases ammonia and allylamine fail to give a constant value for k. Thus for ethyl acetate and ammonia he found the values given in Table X. TABLE X. / (minutes). k. O 60 1.64 * 240 I.O4 1470 0.484 This failure to give a constant he recognized, however, as being due to the effect of the neutral salt formed (am- monium acetate) upon the weak base. When a large amount of this is present a constant is obtained, as is shown by Table XI. TABLE XI. t (minutes). . O 994 0.138 6874 o.i 20 15404 0.119 This shows still a slight effect of the neutral salt formed, for it is impossible to keep it perfectly constant. Arrhenius found that the base acts simply in pro- portion to the amount of ionized OH' it contains, i. e., we have C 2 H 3 2 C 2 H 5 + OH' + Na- = C 2 H 3 O 2 ' + Na' + C 2 H 6 OH. CHEMICAL CHANGS. 275 According to this, all bases containing the same amount of OH' ions should give the same constant. We can rearrange our equation, then, to the form where a is the degree of dissociation of the base. Since the strong bases are but slightly influenced by the addi- tion of salts with an ion in common, the constant of KOH can be taken, with very little error, to be that due to a certain concentration of OH' ions. From this it is then possible to calculate the constant for NH 4 OH, with or without an ammonium salt present. A m/^o solu- tion of KOH at 24. 7 C. gives a constant equal to 6.41. Such a solution of NH 4 OH is but 2.69% dissociated, while that of KOH is 97.2% dissociated. The constant, then, for NH 4 OH in the absence of salt is 0.177. 0.972 ' The effect of the neutral salt upon the NH 4 OH can be readily calculated from the dissociation of the two; we have, then, the constant for NH 4 OH in presence of a neutral salt, 276 ELEMENTS OF PHYSICAL CHEMISTRY. where a is the concentration of OH' ions in the NH 4 OH in presence of the amount of the neutral salts, the concentration of which is designated below by s. Table XII compares the values thus obtained with those actually observed by experiment. TABLE XII. SPEED OF REACTION CAUSED BY NH 4 OH. S a k (calculated) k (observei o 2.69% 0.177 0.156 0.00125 I. 21 0.08 0.062 0.005 0.71 0.047 0.039 0.0175 O.IlS 0.0078 0.0081 0.025 O.O82 0.0054 o . 0062 0.05 0.042 0.0028 0.0033 This is a very good agreement, considering the pos- sible experimental errors, so that we may conclude that the velocity oj saponification is proportional to the con- centration oj OH' ions present, and independent of the radical from which they are split off. 67. Reactions of the third order. If the 3 mole- cules are present in equimolecular amounts, we have V or dx dt i x(2a x) k= t 2a 2 (ax) 2 ' CHEMICAL CHANGE. One reaction of this order has been studied by Noyes (Zeit. f. phys. Chem , 16, 546, 1895). T*he reaction is n 2FeCl 3 + SnCl 2 = 2FeCl 2 + SnCl 4 > of The results, starting with a concentration of .025 moles of each, SnCl 2 , FeQ 3 , SnCl 4 and FeCl 2 , were as follows: Time X ax k 2-5 0.00351 0.02149 113 3 0.00388 O.O2II2 107 6 0.00663 0.01837 114 ii 0.00946 0.01554 116 IS o. 01106 0.01394 118 18 0.01187 0.01313 117 30 0.01440 0.01060 122 60 0.01716 0.00784 122 Average = 1 16 For such a reaction oj the jd order k is inversely pro- portional to the square of the original concentration; and k jor a reaction oj the nth order would be inversely propor- tional to the (n i) power oj the original concentration. Up to the present no reactions of an order higher than the third have been found, so that we need not consider them. In a second order reaction the effect of either con- stituent upon the velocity should be the same; in a 278 ELEMENTS OF PHYSICAL CHEMISTRY. reaction of the third order the effect should be different, and for the above Noyes found, indeed, that an excess of ferric chloride has a greater effect in hastening the reac- tion than has stannous chloride. 68. Incomplete reactions. Thus far we have con- sidered the speed of reactions which are complete, i.e., those which only cease when all the original substance has been transformed. In the case where this is not true it is necessary to use the equation in its original form. Such a case is the one already studied, O, which goes with a certain speed until two thirds of the acid and alcohol are decomposed. When the amount of ester is equal to #, then (p. 263) When we start with i mole of each acid and alcohol and have no water or ester present, and k and V are the speed components in the two directions, we have as above (pp. 262-263) CHEMICAL CHANGE. 279 By observing the change for any time we find k From these two values 77 and k tf we can find k. The K reaction so measured, however, does not give a con- stant value for k. This was accounted for by Knob- lauch. When alcohol and acetic acid form ester and water the reaction goes faster than it should, accord- ing to the theory. This is due to the catalytic action of the H' ions present when the reaction goes in the" one direction. If, then, the concentration of H' ions is retained the same throughout the reaction, k should be constant, which has been found by experiment to be the case. This shows the importance of secondary reactions, which may give entirely false results unless accounted for. 69. Reactions between solids and liquids. The solu- tion of a substance in an acid depends for its speed upon the surface of contact between the two and upon the strength of the acid; but many secondary reactions can take place here, so that our results are only approximate. If we assume the surface to be so large that it changes 280 ELEMENTS OF PHYSICAL CHEMISTRY. but slightly during the reaction we may consider it as constant; we have then, if 5 is the surface, where x is the amount dissolved in the time /. By integration we find t The formula in this form was found by Boguski to give constant results, within the experimental error, for the solution of marble in acid. Noyes and Whitney have proved that the speed of solubility of a solid substance in any instant is proportional to the difference between the concentration of saturation and the one at the time of the experiment. 70. Speed of reaction and temperature. Empirically it has been found that the speed of a reaction always in- creases with an increase in the temperature. The increase in many cases is very great. Thus a rise in tempera- ture of 30 causes the speed of sugar inversion to be five times as great as before. An empirical formula was found by van't Hoff and tested by Arrhenius. We have where CQ is the speed constant at the temperature TQ, k. CHEMICAL CHANGE. 281 Ci is the constant at TI, e is the base of the natural logarithms, and A is a numerical constant. Using the* lower of two temperatures as T , we have for the trans- formation of ammonium cyanate into urea 7^0 = 273 + 25, "0 = 0.000227, A = 11700. Tl C(obs.) C (calc.) 273+39 0.00141 0.00133 273+50.1 0.00520 0.00480 273+64.5 0.0228 0.0227 273+74.7 0.062 0.0623 273+80 O.IOO 0.105 C. APPLICATION OF THE LAW OF MASS ACTION TO ELECTROLYTES IONIC EQUILIBRIA. 71. Organic acids and bases. The Ostwald dilution law. The application of the law of mass action to gaseous equilibria, as well as to those existing in solutions of non- electrolytes in short to chemical equilibria has shown that it is a general law of nature, holding between very wide limits. The question which naturally arises here, then, is, Can the law of mass action also be applied to those equilibria existing between 'the ionized and un- ionized portions of a substance in solution? In other words, is the law of mass action the principle govern- ing the amounts of these portions which can exist together hi equilibrium? It is the purpose of this section to 282 ELEMENTS OF PHYSICAL CHEMISTRY. answer this question so far as is possible from our present knowledge of the facts. The conductivity ratios , as well as the methods for jHoo determining the average molecular weight, where these can be carried out with sufficient accuracy, show that a water solution of acetic acid is ionized according to the following scheme: From this equation, applying the law of mass-action, we obtain or (see page 234) which is known as the Ostwald dilution law, for it gives the relation of ionization to dilution. Substituting the ratio for a: at various dilutions, /*00 Ostwald found K to be constant, with a value at 25 of i.SXio^ 5 . From the value of K at any temperature, then, it is possible, by solving for a, to find the degree CHEMICAL CHANGE. 283 of ionization at any dilution or at any concentration at that temperature, for c = . We find > 2 where ^ is the equilibrium, dissociation, or ionization constant, or the so-catted coefficient oj affinity, of the acid. Further than this, when K is known for any tem- perature it is possible to find a, at any dilution, in the presence of an acid or salt with an ion in common (H* or CHsCOO')- And, just as in the case of gaseous dissociation, we always find a smaller dissociation in the presence of the products arising from an exterior source. Naturally the calculation here is similar to the other (p. 238). For example, we have (CH- + x - y where x represents the concentration of H* ions due to other substances, C H - and CcHaCOO' are the concentrations due to the acid in the absence of other substances, and y is the concentration of each (H* and CHaCOO') lost, i.e., uniting to form un-ionized CH 3 COOH, as the result of the presence of x moles of H' ions. The concentrations of H* and CH 3 COO', now arising from the acid, are 284 ELEMENTS OF PHYSICAL CHEMISTRY. equal, then, to (C H - y) = (^cH 3 coo'->'); and the un-ion- ized acid concentration is (^cH 3 cooH+>')- Indeed everything which we found above (pp. 228-241) to hold true for gaseous dissociation, with the one ex- ception mentioned below, also holds true for the relation of ionized and un-ionized portions of a substance, of the type of acetic acid, in solution. And, as a rule, the re- sults are simpler to calculate, for the volume change of the system is so small as to be negligible. We can also define the ionization constant in terms similar to the definition of the gaseous dissociation constant (p. 235). Thus by multiplying K for acetic acid by 2 we obtain 0.000x^36 as the concentration in moles per liter of a solution of acetic acid which would be 50% ionized, i.e., a solution of i mole of acid in 27777.5 liters of water. All organic acids when treated in this way give a con- a 2 slant value jor the expression ^-,. This is the dif- (i a)V ference between gaseous and electrolytic dissociation equilibria, at least when the latter is for an organic acid. The acid, whether it be mono, di, or polybasic, always ionizes as a monobasic acid up to the dilution at which a = 0.5. This means that, assuming the acid to ionize simply into the two products H* and the negative radical (i.e., for the calculation of /*oo )> which may also contain replaceable hydrogen, a constant is obtained so long as the ionization in this way is 50% or less. Beyond that CHEMICAL CHANGE. 285 point the H* ions, due to a breaking down of the negative ion containing replaceable hydrogen, are great enough in concentration to influence the constant, which begins to vary. Above the dilution at which a = 0.5 the second and following replaceable hydrogens begin to appear as ions, and must be taken into account; and at infinite dilu- tion, if it were possible to attain it, the polybasic acid would be composed of all the replaceable hydrogens as ions, together with the negative radical, without replacea- ble hydrogen, as the negative ion. The calculation of the degree of ionization of dibasic acids above the dilution at which a = 0.5 has been worked out by W. A. Smith (Zeit. f. phys. Chem., 25, 144, and 193, 1898). For a tribasic acid, for example, we have the following processes taking place: HA" = H'-{-A'", for which, by application of the law of mass action, we obtain a? a,, 2 286 ELEMENTS OF PHYSICAL CHEMISTRY. Up to the dilution at which 01 = 0.5 the constants k 2 and 3 are so small that in general they have no effect on the concentration of H* ions, which is practically equal to C H -'- Above that point the value of H* at any dilution must be taken as equal to CH-' + CH>"+CK-'". The values of the constants k< 2 and 3, and the con- centrations CH-" and CH-'" can be obtained by observ- ing the catalytic action of the acid salts, for example NaH 2 A and Na2HA, upon sugar at 100 or methyl acetate (p. 265). In these cases, also, we should have the relations given above for k 2 and 3, and experiment has shown this to be the case. Since the mono and disodium salts of a tribasic acid increase in ionization approximately proportional to the volume, the concentration of H* in acid salts is almost independent of the dilution. In , other words, since we have k 2 = , and CH.A' ^H 2 A' is approximately constant in moles, per liter, c" H - must also remain constant. For citric acid (tribasic) the following values at 25 were determined by Smith (I.e.) : k' 2 = 3-2X10-5, & 3 = 0.07 Xio~ 5 . From these values the actual concentration of H' ions can be calculated at any dilution; and it will be found that the second hydrogen, indeed, has little effect CHEMICAL CHANGE. 287 until the amount of H 2 A' is greater than one-half the total concentration. For further details respecting di- basic acids see Wegscheider (Sitzungsber. d. Akad. d. Wissenschaft. in, 441-510, 1902). A few ionization constants are given below in Table XIII; they all refer to the first hydrogen, i.e., are equal to K in the above s.t of formulas. TABLE XIII. IONIZATION CONSTANTS OF ORGANIC ACIDS AT 25 C. KX 10* Malic 39.5 Fumaric 93 Tartaric 97 Salicylic 102 Orthophthalic 121 Monochloracetic 155 Malonic 158 Maleic 1170 Dichloracetic 5140 Oxalic 10060 ()* Trichloracetic 121000 ()* 34 44 45 49 .61 .80 KX Propionic Isobutyric Capronic Butyric Valerianic Acetic Camphoric 2.25 Anisic 3 .20 Phenylacrylic 3.55 Succinic 6. 65 Lactic 13.8 Glycollic 15.2 Formic 21.4 The value of K at 18 for a few other acids, some of which are inorganic, but characterized by their small ionization, and by obeying the law of mass action, are given by Walker and Cormack (Trans. Chem. Soc., 77, 8, 1900), the value of acetic acid being given for comparison. * These two acids are so largely dissociated that a small error in a affects K to a large degree. For a list containing a very large number of other acids see Zeit. f. phys. Chem., 3, 418-20, 1889, or Kohl- rausch and Holbora, Leitvermogen der Elektrolyte, pp. 176-193. 288 ELEMENTS OF PHYSICAL CHEMISTRY. Per Cent Name. X X io Solution. Acetic acid (25), CH 3 COO' H' ............. 1,800,000 i . 3 Carbonic acid, H' HCO 3 ' .................. 3,040 o . 174 H- C0 3 " .................... 0.6* ..... Hydrogen sulphide, H' HS' .................. 570 0.075 Boric acid, H' H 2 BO 3 ' ..................... 17 0.013 Hydrocyanic acid, H' CN' .................. 13 o .01 1 Phenol, H- C 2 H 6 C" ...... . ................. 1.3 0.0037 As the result of his work on organic acids Ostwald formulated, among others, the following law, which will serve as a means of foreseeing the value of the constant of a substituted acid when that of the simple acid is known. The substitution of O, Cl, Br, I, CN, etc. in short, of negative groups in a molecule increases the separa- tion of ions of H in water solution; the addition oj H, NH 2 , etc., i.e., positive groups, decreases it; and the influence is the greater the nearer (in space) the substi- tuted group is to that of carboxyl. Thus for CH 3 COOH 100^ = 0.00180, forCC! 3 COOH 100^=121. In the case of the addition of a salt with an ion in common to an organic acid (p. 283) the following will be seen at once to be true. If the degree of dissociation * McCoy, Am. Chem. Jour., 29, 455, 1903. CHEMICAL CHANGE. 289 of a salt, with an ion in common with an acid, is d, and n is the number of moles of salt which are present, then the equation of equilibrium of the acid will become For very weak acids we can generalize this as follows: a, the degree of dissociation of the acid, is very small in presence of the salt, so that a in comparison to i and nd may be neglected. Since d for salts is almost inde- pendent of the dilution, we have KV - n i.e., the dissociation of a weak acid in presence of one of its salts is approximately inversely proportional to the amount of the salt. The case of the partition of a base between two acids depends to a certain extent upon their ionization con- stants. This partition takes place when there is not enough base present to saturate both acids. The final mixture consists of water, undissociated salt, and the dissociated and undissociated portions of the acids. The equilibrium is the same as that which would be attained by the mixture of the salt of the one acid with the other acid. The affinity of each acid for the base 290 ELEMENTS OF PHYSICAL CHEMISTRY. will depend upon the percentage of free ions of H* which it possesses, i.e., the one containing the larger quantity will unite with the larger amount of the base. For weak acids it is possible to formulate a general law regarding the partition and the ionization constants. In this case a is so small that it may be neglected in i a ; hence we have a = VKV. And for two acids at the same dilution or a 1y The coefficient oj partition oj two acids, then, is propor- tionate to the ratio 0} their degrees of dissociation at the given VK volume, or jor WEAK acids to -7=. v K r This coefficient of partition is independent of the nature of the base and depends only upon the two acids. For the partition of an acid between two bases the co- efficient will depend only upon the two bases. If a is CHEMICAL CHANGE. 291 the degree of dissociation of the one base and a' that of the other, we shall have for them when weak, just as for acids, a VJt and the same generalization holds true. In order that a base may be divided equally between two acids it is necessary that they be isohydric, i.e., it is necessary that An example of isohydric solutions, i.e., two which contain the same concentration of H* ions, is acetic acid at a dilution of 8 liters and hydrochloric acid at one of 667 liters. These two solutions may be mixed in all proportions without any change in the dissociation resulting. When mixed in equal volumes, if treated with a small amount of base, equal amounts of chloride and acetate will be formed. All these conclusions, except that with regard to substitution (p. 288), hold also for the organic bases, as well as some of the inorganic ones. The ionization constants for these are found from the relation ~ A = 29 2 ELEMENTS OF PHYSICAL CHEMISTRY. where M* is the positive and OH' the negative ion. The value of K for a few bases is given below: Urea, (4Q.2) ,. . .K = 0.0037X10-" Acetamide, (4o.2) o.oo33Xio~ u Acetanilide, (4o.2) o .OO44X io~ u Aniline, (25) 5.4 X lo" 10 />-Toluidine, (4o.2) 20.7 Xio" 10 w-Nitraniline,(4o.2) 4 X io~ 12 p- " (4o.2) i Xio- 12 o- " (4Q.2) o.oi Xio- 12 Ammonium hydrate, NH; OH' (25) 23 X 10 Methylamine, (25) 5 X 10 Dimethalamine, (25) o .074 X io~ Trimethylamine, (25) o.oo74X 10 Propylamine, (25) o .047 X io~ Isopropylamine, (25) -S3 X io~ Isobutylamine, (25) 0.034 X io~ The equation on page 283 has also been used to de- termine the concentration of ions in the solution added to a substance obeying the law of mass action, and with very good results. Thus, starting with acetic acid, know- ing the concentration of H" ions, and adding sodium acetate, we can find the concentration of the CH 3 COO' ions in the salt. In this case the new concentration of H* ions can be determined from the speed of inversion of sugar; and, by solving the equation for the amount of CH 3 COO' added, we can find the dissociation of the salt solution added. This same process may also be carried out with other systems, mentioned below, and by it it is possible to determine the concentration of any one kind of ions. As CHEMICAL CHANGE 293 mentioned above, the ionization has been found to be sen- sibly the same, by whatever method it may be determined. 72. Acids, bases, and salts which are ionized to a con- siderable extent. Empirical dilution laws. The applica- tion of the law of mass action (the Ostwald dilution law) to the above so-called strong electrolytes does not lead to a constant value for K when the ionization is taken from the conductivity ratio - . As in general ;j.*> the degree of dissociation is jound to be the same by all of the possible methods, our only conclusion at present is tliat the law oj mass action cannot be applied to the equi- librium oj un-ionized and ionized portions in such solutions as these. Although this is true with regard to very soluble salts, there is a quantitative relation, which we shall develop below, holding for the ionized portions of difficultly soluble substances, when the un-ionized por- tion is retained constant. The only known case of a dissociating and very soluble salt to which the law of mass action may be applied, i.e., the only exception to the above conclusion, is caesium nitrate, when a is determined by the freezing- point method. And this is not true for a as determined by the other methods. In other words, in this case the freezing-point results point to a different degree of ion- ization than any other method. The results for caesium nitrate as determined by Biltz (Zeit. f. phys. Chem., 40, 2 9 4 ELEMENTS OF PHYSICAL CHEMISTRY. 218, 1902) by the freezing-point and conductivity meth- ods are given below. BY DEPRESSION or THE FREEZING-POINT. c (Moles per Liter). V (Liters per Mole). Depres- sion. a(0bs.). if 2 K =-. r- (i-a) 0.00766 l3-7 0.028 0.98 0-33 0.01940 51.6 0.070 0-95 o-35 0.04648 21.6 o. 164 0.907 [0.41] 0.09884 IO.2 0-331 O.SlO o-34 o. 1421 7.09 0.460 0.750 0.32 O . 2 100 4 .8l 0.662 0.704 0-35. 0.2987 3.385 0.907 0.641 0-34 0.3861 2.62 1.125 0-575 [0-30] 0-4339 2 -33 1.267 0.578 0-34 Average K, excluding values in brackets, = o . 34 BY CONDUCTIVITY, /*oo = i47- V (Liters per Mole). 1024 512 256 128 64 32 16 8 4 Vv. 146.4 144.7 141.9 139-3 I38-3 134.2 I28.I I2O-9 III.9 0.14 0.23 0.30 0-37 0.47 0.61 Biltz attributes the failure of the law of mass action, as. applied to strong electrolytes, to a hydration of the substance i.e., to a chemical reaction between the substance and the solvent which would remove active solvent from the solution, so that it would be really more CHEMICAL CHANGE. 295 concentrated than it appears to be. Just why the solution of caesium nitrate in water should give a constant value for K when a is determined by the freezing-point method, while for conductivity it fails to do so, is unknown and thus far nothing but assumption has been possible. It is to be remembered, however, that this value of K, although giving, when solved for a, the ionization accord- ing to the freezing-point method, does not give the value as determined by any other method, so that too much stress is not to be laid upon it, especially in face of the fact that the law cannot be applied to any other strong electrolyte by any method of determining a. It would seem more probable that some secondary action takes place in the freezing which is absent in all other cases, so that in this one case the freezing-point result is in- correct. This, of course, may not be true, but at the present time, in the absence of any indication of such a result for other substances, it is decidedly the most reasonable and logical one. In other words, then, so far as we know at present, in all cases but this the con- ductivity leads to the true value for a, so that it is but reasonable, until further evidence is at hand to assume it be correct here, and attribute the case above to some abnormality not yet encountered with other substances. Although the Ostwald dilution law fails utterly to hold for the equilibrium between the ionized and un-ionized portions of strong electrolytes, certain other empirical 296 ELEMENTS OF PHYSICAL CHEMISTRY. dilution laws have been found which allow us to find the respective amounts at any dilution. Thus Rudolphi (Zeit. f. phys. Chem., 17, 385, 1895) found a dilution law which gives a constant value between certain limits for such solutions as do not follow the Ostwald dilution law. This law is a 2 7 777= = constant, (i-a)v V where the value of the constant is approximately the same for analogous substances, van't Hoff (Zeit. f. phys. Chem., 18, 300, 1895) altered this to the form a* - f ^r, = constant, (i a) 2 V which holds even better than Rudolphi's. Simplified this relation is c? ^ = constant, i.e., the cube o) the concentration of ions divided by the square oj the undissociated portion is a constant. Writing the Ostwald dilution law in this form we obtain (for binary electrolytes) CHEMICAL CHANGE. 297 while this empirical relation (in either form) remains the same jor binary or ternary substances; in other words, is independent 0} the number oj ions formed jrom one mole of the substance. Some values found by use of vant Hoff's equation are given below. The value of k by the Ostwald dilu- tion law will be found to vary considerably, while k^ is quite constant, i.e., within the experimental error. AgN0 3 25. F= 16 0=0.8283 H = I.H = 32 =0.8748 =1.16 = 64 =0.8993 =1.06 = 128 =0.9262 =1.07 = 256 =0.9467 =1.08 = 512 =0.9619 =1.09 Bancroft (Zeit. f. phys. Chem., 31, 188, 1899) pro- poses a dilution law of the form constant = , in which the constant and n are functions of the nature of the electrolyte. Although this relationship is as yet unknown, Bancroft suggests that we may sometime find a relation of the form w=2-/ (constant) 298 ELEMENTS OF PHYSICAL CHEMISTRY. (where / (constant) varies between zero and a value approximating one-half) which will reconcile the Ost- wald dilution law, holding only for organic acids and bases, and in which n=2, with this form in which n varies between the values 1.36 and 1.5. For KC1 at c - 1 ' 36 18 we have the relation - = 2.63, which holds in a Cs very remarkable way between the volumes 0.3 and 10,000 liters, as will be seen by referring to the results in the following table: KC1 AT 1 8. c .l.*9 a Obs. Calc. 0.987 0.987 0.984 0.983 0.978 0.976 0-973 0.970 0.965 0.962 0.950 0.948 0-934 0-934 0.915 0.917 0.902 0.906 0.883 0.892 0-853 0.864 0.821 0.834 0.803 0-813 0.780 0.786 0.748 0-745 0.706 0.700 0.673 0.672 IOOOO 5000 2OOO IOOO 500 200 IOO SO 33-3 20 10 5 3-3 2 I 0-5 C >n C > For other substances the relation ~ is not so satis c* CHEMICAL CHANGE. 299 factory or constant. Noyes (J. Am. Chem. Soc., 26, 1 68, 1904) has determined the degree of ionization, , for the chlorides of potassium and sodium at various r- temperatures and finds a constant value for the ratio j ; that is, the fraction of salt un-ionized is directly proportional to the cube root of the concentration, or the concentration of un-ionized substance, (ia)c, is directly proportional to the 4/3 power of the total concentration, c, oj salt. The degrees of ionization of potassium and sodium chlorides were found to be nearly identical (the extreme variation being 2%) at all temperatures and dilutions. In a o.i molar solution the dissociation has approximately the following values: 18 84% 281 67% 140 79% 3o6 60% 218 74% The values of K' = ^ are as follows: 18 140 218 281 306 NaCl 0.366 0.448 0.573 0.745 0.877 KC1 0.321 0.468 0.577 0.713 0.853 Some of the observed facts as to the ionization of strong electrolytes, as summarized by Noyes (Tech- nology Quarterly, 17, 307, 1904) are as follows: The form of the concentration function is independent 300 ELEMENTS OF PHYSICAL CHEMISTRY, of the number of ions into which the mole of salt disso- ciates. Instead of being proportional for di-ionic, tri- ionic, and tetra-ionic to the square, cube, or fourth power of the concentration of the ions, the undissociated portion is approximately proportional to the 3/2 power of that concentration, whatever may be the type of salt. The conductivity and freezing-point depression of a mixture of salts having an ion in common are those calculated under the assumption that the degree of ionization of each salt is that which it would have if present alone at such an equivalent concentration that the concentration of either of its ions were equal to the sum of the equivalent concentrations of all the positive or negative ions present in the mixture. Suppose that a mixed solution is o.i molar with respect to sodium chloride and 0.2 molar with respect to sodium sulphate, and that it is 0.18 molar with reference to the positive or negative ions of these salts. The principle then requires that the ionization of either of these salts in the mixture be the same as it is in water alone when its ionic concentration is 0.18 molar. This has been proven conclusively for many mixtures. The decrease of ionization with increasing concentra- tion is roughly constant in the case of different salts of the same type. The un-ionized fraction of any definite molal concen- tration is roughly proportional . to the product of. the CHEMICAL CHANGE. 3 O1 valences of the two ions in the case of salts of different types.* From these facts, together, with others, Noyes (l.c.) concludes that the form of union represented by the un-ionized portion of a substance differs essentially from ordinary chemical combination, it being so much less intimate that the ions still exhibit their characteristic properties, in so far as these are not dependent upon their existence as separate aggregates. In other words, the law of mass action is inapplicable to the relation between ionized and un-ionized portions as they exist in strong electrolytes, and hence this is not to be con- sidered as a simple chemical equilibrium, for which, as we know, the law of mass action appears to hold rigidly. I /73. Heat of ionization. By van't Hoff's equation it ns possible to calulate the heat of ionization of a sub- stance, provided we know the degree of dissociation at two different temperatures. This is true not only for those substances which follow the Ostwald dilution law, but for all others as well, according to Arrhenius (Zeit. f. phys. Chem., 4, 96, 1889, and 9, 339, 1892). For binary electrolytes we have then K' a 2 (i -a) q/i i - * Other generalizations of this kind will be found in Chapter IX under Electrical Conductivity. 302 ELEMENTS OF PHYSICAL CHEMISTRY. where the values of K are for the same dilution and represent the change of ionization, even though the val- ues are not the same as other dilutions, and the values V cancel and need not be considered, q is then the heat liberated when a mole of substance is formed in solu- tion by the union of its ions. It will be observed here that these values differ from those calculated from the table given on page 220, for these refer to the heat of formation of the ions from substance already in solu- tion, while those refer to the compound process of solu- tion and ionization, i.e., the difference in energy between the ionized state and the solid or gaseous state. Some of the results as found by Arrhenius are given below, the unit being the small calorie. HEATS OF IONIZATION. { Substance. Temperature. Calories. -3* Propionic acid ................. ! 35 Q ~ 557 Butyric acid ............ { ^ = 935 Phosphoric acid.. ..{3 s -^5 Hydrofluoric acid ............... 33 3549 Potassium chloride ............. 35 362 iodide ............... " - 916 " bromide ............. " 425 Sodium chloride ................ " 454 " hydrate ................ " 1292 ' ' acetate ................. " 391 Hydrochloric acid .......... .... " - 1080 CHEMICAL CHANGE. HEAT NECESSARY TO COMPLETE THE IONIZATION, (i a)w; (i mole in 200 moles of water). Substance. Temperatflre. Calories. Potassium bromide .............. 35 58 " iodide ................ " - 132 chloride .............. " 56 Sodium hydrate ................. " 180 " chloride ................ .' " - 81 Hydrochloric acid ............... " 136 Hydrofluoric acid ................ 33 3304 Phosphoric acid ................. 2i.5 1682 For the temperatures of 35 in the table, 7^=273 + 18, for 2i., r K f the \og e -g- formula. From data such as the above it is possible to calculate the heat of neutralization of an acid by a base. The formula for this (p. 217) is where the figures i, 2, and 3 refer respectively to acid, base, and salt, and x is the heat of formation of i mole of water from H* and OH' ions, i.e., 13,700 cal. In the table below the calculated values of q at two tempera- tures are given, together with the observed values at one of the temperatures. It is obvious from the results above that the value of the heat of neutralization of an acid by a base cannot be considered as indicative of the strength of the acid. The two latter are relatively weak acids and yet they give rise to the greatest amount of heat. 304 ELEMENTS OF PHYSICAL CHEMISTRY. HEAT OF NEUTRALIZATION OF ACIDS WITH NaOH. (i mole of acid + 1 mole of NaOH + 4oo moles of H 2 O.) At 35. At 21.5. Calc. Caic. Obs. HC1 12867 13447 13740 HBr 12945 13525 13750 HNO 3 : 12970 JSSS 13680 CH 3 COOH 13094 13263 13400 C 2 H 6 COOH 13390 13598 13480 CHC1 2 COOH 14491 14930 14830 HgPO, 14720 14959 14830 HF* 16184 16320 16270 The calculation by van't Hoff's formula (pp. 252-261) of the heat of solution, as has already been observed, is satisfactory when the substance can be considered as either completely ionized or completely un-ionized. The next question to be considered is the calculation of the heat of solution of those substances which are but par- tially ionized. Unfortunately no relation has been found which gives results agreeing very closely with those of ex- periment. There are two formulas, however, which may be used together, for the experimental result is usually found to lie midway between them, i.e., one gives values which are too high, the other values which are too low- van't Hoff's formula (p. 253) leads in such a case to * For data as to the ionization of. HF, see Deussen, Zeit. f. anorg. Chem., 44, 408, 1905. CHEMICAL CHANGE. 305 for binary electrolytes. The forms given by Van Laar are two, one based on the Ostwald dilution law, viz., the other on the Rudolphi dilution law (p. 296), As the results by these laws are not in accord with expe- rience, and we could hardly expect otherwise with the present slight knowledge of the equilibrium between the ionized and un-ionized portions, we shall not con- sider them further, for it is quite evident that until we find a dilution law which will hold for all dilutions we cannot hope to follow closely any relation depending, as this does, entirely upon the degree of ionization. 74. Solubility or ionic product. Although, as we have seen, the law of mass action cannot in general be applied to the equilibrium of the ionized and un-ionized portions of a substance in solution (except to organic acid and bases), it can be applied with considerable accuracy to a very large number of saturated solutions. An example of such an equilibrium is a saturated solution of silver chloride, which is found to be practically com- pletely ionized according to the scheme AgCl-Ag' + Cl'. ELEMENTS OF PHYSICAL CHEMISTRY. Applying the law of mass action to this we obtain 2 Kc = c\c^ or K = (i-a)F' when c is the concentration of un-ionized AgCl, c\ that of Ag', and c 2 that of Cl' ions. Since the solution is satu- rated, the value of c at any temperature must remain con- stant, for if the solution were unsaturated, solid would dissolve, if supersaturated, solid would precipitate. We have then, at any one temperature, in a saturated solu- tion of silver chloride, the relation Kc = constant = c\c 2 \ i.e., in a saturated solution of a binary electrolyte (of this kind) the product of the concentrations 0} the ions must remain constant, with unchanged temperature. Expressing this in a more general form, we have for the reaction in a saturated solution, (46) Ci n 'c 2 n2 ^ constant = s, where 5 was called by Ostwald the solubility product of the substance. This solubility product is of para- mount importance in analytical chemistry, jor a precipi- CHEMICAL CHANGE. 3 7 fate (when due to an ionic reaction, and most oj them can be shown to be due to this) is always and only formed when its solubility product is exceeded. This, of course, presupposes that no supersaturation phenomenon is possible; if it is, then the metastable limit (p. 128) must first be exceeded. Just as we found a decrease in the dissociation of a gas or an organic acid, by the addition of one of the products of dissociation from an exterior source, so here the addition of a substance with an ion in common causes the formation and separation in the solid state of the un-ionized substance. In other words, the term 5 still retains its constant value, and consequently the con- stituent ions of the substance unite to form more of the un-ionized portion, which, since the solution is already saturated with it, separates out as solid. This has been found to be true by experiment, but only true for those substances which are difficultly soluble. The effect may be observed most easily by dissolving the difficultly soluble substance in a solution of the salt with an ion in common; but it can also be attained by adding to the saturated water solution of the substance a strong solution of the salt, when a precipitation of the sub- stance, usually in the crystalline state, will be observed. Thus if we add to one portion of a saturated solution of silver acetate a strong solution of sodium acetate contain- ing x moles of CH 3 COO' ions, and the same amount of ELEMENTS OP PHYSICAL CHEMISTRY. a solution of silver nitrate containing x moles of Ag* to the liter to another equal portion, we observe an equal precipitation of solid silver acetate in the two solutions. Although all this is true, as far as the precipitation is concerned for all saturated solutions, it is only for the difficultly soluble substances that the quantitative rela- tions are found to hold. The examples below will serve to show how the solu- bility product of a substance can be found, and how when once found can be employed to foresee the solubility of the substance in a solution already containing a com- mon ion. Silver bromate is soluble at 25 to the extent of 0.0081 moles per liter. Assuming 'the ionization in this state to be practically complete, and it certainly is nearly so, the concentration of the ions Ag* and BrO 3 ' will be the same, and equal each to 0.0081 mole per liter. The solubility product at this temperature, then, will be (0.008 1 ) (0.008 1 ) = ^AgBr0 3 . The solubility in a solution of silver nitrate containing o.i mole of Ag* ions (or in potassium bromate containing o.i mole of BrOs' ions) can be found by aid of the rela- tion (0.0081 ) 2 = (0.0081 +.1 y) (0.008 1 y), CHEMICAL CHANGE. 309 and is equal to (0.0081 y) t for that is the concentration of Ag* and BrOa' ions now existing in the solution, and coming from the salt; the amount o.i of one being -due to the other salt, and y being the AgBrO 3 remaining un- dissolved owing to the presence of this o.i mole of Ag* or Br0 3 '. This is true for all binary salts when they can be as- sumed to be completely ionized, or practically so. Where the substance dissociates into more than two ions and can be assumed to be completely ionized, the relation is quite similar. Suppose the salt to dissociate according to the scheme As solubility product we shall have, if c is the solubility of the completely ionized salt MA 3 , or for we must have three times the number of moles per liter of A' as we have of M'" according to the chemical equation, i.e., C = C M - and <; = c/. 310 ELEMENTS OF PHYSICAL CHEMISTRY. In case of solution in the presence of o.i mole of one of the ions, we have, then, or (CM--*) (A'+o.i-3*) 3 from which it is apparent that the effect of equal ad- dition is not the same for the two ions, i.e., that x and y, the decreases in the solubility, are not equal. In the case the substance is not completely ionized, the solubility product is not so directly related to the solu- bility of the substance as in the above cases, i.e. to the square in one case and twenty-seven times the fourth power in the other. Consider the case of uric acid, which, at 25, is soluble to 0.0001506 mole per liter, and is ionized in that condition to 9.5% into H* and the negative radical which we shall designate as U. The solubility product here is naturally (0.0001506X0.095) (0.0001506X0.095) = SHU =K H u (0.0001506X0.905). The solubility of uric acid in a molar solution of hydro- chloric acid, for which 01 = 0.78 (i.e., H' =0.78, Cl' = o.78), is to be found in the following way : (0.0001506X0.095+0.78 #) (0.0001506X0.09 5 x) (o.oooi 506 X 0.095)2, CHEMICAL CHANGE. 311 where (0.0001506X0.095*) represents the present con- centration of H* and U' from the uric acid, and its total solubility in the hydrochloric acid solution is (0.0001506X0.905) + (0.0001506X0.095 x), i.e., is equal to the sum of that which is un-ionized and that which is ionized. Just as for organic acids in general, an infinite excess of one of the ions will cause the ionization of the sub- stance to become zero. // is to be observed here, however, that this excess will only cause the solubility to become zero in the case that the ionization is compete. In the case of uric acid, an infinite amount of H" or U' at best can only reduce the solubility by 9.5%, the remaining 90.5% being un-ionized and not affected at that temperature by any addition of substance which does not react chem- ically with it. That a substance is always decreased in solubility by the addition of a substance with an ion in common is not true, as the well-known behavior of silver cyanide m potassium cyanide will show. In all such cases, however, the equilibrium which has previously existed is altered in some way, so that the relations are not the same. These cases are usually characterized by the formation of a complex ion , the product of which is exceeded. The removal of the ions to form this complex ion disturbs the equilibrium of the difficultly soluble salt; the un-ionized portion ionizes further, and its loss is replaced by the 312 ELEMENTS OF PHYSICAL CHEMISTRY. solid phase. This process continues, dissolving new salt, until equilibrium is attained, i.e., until the solubility product, whatever it may be, is just satisfied, when solution ceases. The ionization of silver potassium cyanide takes place almost completely according to the scheme but it has been found by Morgan (Zeit. f. phys. Chem., 17, 513-535, 1895) that in a 0.05 molar solution we have Ag* ions to the extent of 3.5Xio~ n and CN' to 2.76Xio~ 3 moles per liter. Knowing, the concentration of the metal ions, for example (which can be determined by methods given in the next chapter), in the complex salt solution and in a water solution of the difficultly soluble salt, we can fore- see the behavior of that salt when in a solution of a salt which might dissolve it to form a complex solution of that strength. In general, i.e., when the concentration oj metal ions in a water solution of salt is greater than that of a water solution of a complex salt, the simple salt will dissolve in any solution which will produce the complex salt in this concentration. If the concentration of metal ions is smaller, the solid will not dissolve to any greater extent than it does in pure water, for the ionic product of the complex ion cannot be exceeded. By this law it is possible to find the relative solubility CHEMICAL CHANGE. 313 of salts of the same metal in water. Thus silver sulphide is the only silver salt which will not dissolve in potassium cyanide solutions; in other words, is the most insoluble salt of silver, and contains fewer ions of Ag* than exist even in a solution of silver potassium cyanide, such as that given above. It is not only for substances in solution that we find this constancy of the product of the concentrations of the ions, for it also exists in our usual solvent, water, where the ionized portion is so small that the un-ionized portion may be considered as constant, i.e., i a = i. Expressing the concentrations of H* and OH' ions in a liter of water by Ci and c 2 , and the un-ionized portion, which is practically i liter, i.e., 0-^55-S moles, by c, we have 5 = *H 2 o = constant. The values of Ci = c 2 = H' ( = OH') ions in water at various temperatures is as follows: Temp. CiXio 7 . Moles per Liter. Temp. C! X 10*. Moles per Liter. o-35 34 1-47 10 0.56 50 2.48 1 8 0.80 85-5 6.20 '5 1.09 100 8.50 The ionic products (we can hardly call them solubility products), then, are as follows: 3H ELEMENTS OF PHYSICAL CHEMISTRY. *o= (0.35X10-7)2, 0-34 =( S 10 = (0.56XIO- 7 ) 2 , 5 50 =(2.48X10-7)2, 5i 8 = (0.80 X 10-7)2, 585.5 = (6.2 X 10-7)2, *25 " (1 .09 X TO-7)2, 5 100 = (8.5 X 10-7)2, where the value of s, in each case, is equal to 55.5 times a 2 the value of K=- ( - y~, V being 0.0018 liter, i.e., the volume occupied by the formula weight, 18 grams, of water. Knowing the solubility products of two substances with an ion in common, it is possible to find how much of each will dissolve when they are exposed together to the action % of a solvent; and this, of course, may be expanded to three or more substances together. Assume we have the two completely ionized, diffi- cultly soluble salts MA and MAi, with the ion M* in common, and that they are dissolved simultaneously in water. Call the amount of MA which dissolves x, and the amount of MA i y. In the solution then we must have x + y moles of M" ions, x of A' and y of A\ \ and, if 5 is the solubility product of MA and s\ that of the relations must be so that by solving the simultaneous equations we can find x and y. CHEMICAL CHANGE. 315 An example of this is given by dissolving thallium chloride and sulphocyanate together. The solubilities in water, each for itself, are TlCl = o.oi6i and T1SCN = 0.0149. Assuming complete ionization, the solu- bility products are respectively (o.oi6i) 2 and (o.oi49) 2 and if x represents the amount of chloride and y that of sulphocyanate dissolving from the mixture, we have TT 9 x=CY, and ;y = SCN', and x(x+y) = (o.oi6i) 2 , y(x+y) = (0.0149)2, from which we find x= 0.0118 and y = o.oioi, while the values #=0.0119 and y = 0.0107 are found by experiment. It will be observed that in the above examples, except the last, we have tacitly assumed that the dissociation of the added salt, with an ion in common, is not influenced by the ions of the difficultly soluble salt. As a rule this is true, for the substances are so insoluble that their effect is infinitesimal; in the last example, this effect has been allowed for, however, and will show the method of treating such cases. In general, then, we can conclude for difficultly soluble salts (and for complex ions] that they are precipitated (formed) when the product of the concentrations of the ions composing them exceeds the solubility (ionic) product. 316 ELEMENTS OF PHYSICAL CHEMISTRY. Although this law holds in general for difficultly soluble salts, isolated cases are to be found where the un-ionized portion does not remain rigidly constant, after the addi- tion of an ion in common; and, to a smaller extent, a slight variation is sometimes observed in the solubility product. Since these cases are very few, and are usually observed for the more soluble salts, it would seem prob- able that they are due to secondary reactions not yet recognized, or to others not properly accounted for. 75. Hydrolytic dissociation. Hydrolysis. Hydrolysis is the process taking place in the water solution of a salt, which causes the solution to appear alkaline or acid, or results in a neutral equilibrium according to the scheme = MOH+HA. If the acid formed is insoluble or un-ionized, the base being ionized, the reaction will be alkaline (OH' ions). When the base is insoluble or un-ionized, and the acid ionized, the reaction is acid (H* ions). And finally, if both acid and base are insoluble or un-ionized, the salt will be completely transformed into base and acid, and, as there will remain no excess of either H* or OH' ions, the reaction will be neutral. In other words, then, hydrolysis is due to the removal of either H* or OH' ions (or both) from the water by the A' or M* ions of the salt, to form un-ionized or insoluble substances, and this continually causes more water to ionize, i.e., to react with the salt, CHEMICAL CHANGE. 317 Examples of this process are most common. For instance, all mercury, copper, zinc, etc., salts are acid, for an un-ionized basic substance is formed by the reac- tion leaving free, ionized acid; and potassium cyanide is alkaline, owing to the formation of un-ionized hydro- cyanic acid and ionized potassium hydrate. The most striking example of this process, perhaps, is the precipi- tation of bismuth oxychloride when water is added to a hydrochloric acid solution of the chloride, but the basic acetate separation of iron (see below) is just as charac- teristic, although apparently not so direct. Since we know the conditions under which insoluble or un-ionized substances will form, i.e., by the exceeding of their solubility products or analogous values, it is possible to find the conditions necessary to produce a hydrolytic dissociation, and also to find the relations governing the equilibrium finally attained as the result of the process. We recognize at once that if the product of the con- centrations of .M" and OH' ions is larger than that which can exist in pure water, un-ionized substance must form. By this formation, however, the equilibrium of H* and OH' ions will be disturbed, and a further ionization of water must take place, until at length the ionic product is just attained. If the H* and A' ions at this point do not unite to form un-ionized acid, the further ionization of water will be unlike what it would be in the absence ELEMENTS OF PHYSICAL CHEMISTRY. of this excess of H* ions, for, since CH-XCQH' must at the same time be equal to s H2 o> we can only have - moles per liter of OH' ions present, when C H - is the total concentration of H* ions at that time. The process due to the formation of un-ionized or insoluble acid, when no un-ionized base is formed, or forms but slightly, is exactly analogous to the above. In both cases water is decomposed, owing to the removal of one of its ions, and the further ionization of water and formation of the insoluble or un-ionized base or acid continues until the equations for equilibrium are fulfilled. For the sake of simplicity we shall consider separately the cases that the reaction is caused by the base, or by the acid. Case I. The process is due only to the formation of base. Here it is obvious that or where the terms c refer to the ionic concentrations. Cal- ling c the original concentration of salt, and d$ the ioni- zation of the salt, the concentration of OH' ions in water at 25 being 1.09 Xio~ 7 moles per liter, we shall have - or After equilibrium has been established, i.e. ; when the CHEMICAL CHANGE. 3*9 degree of hydrolytic dissociation is a, d& being the dis- sociation of the acid formed, we must have and (470) (orig. M' loss of IV 'Lvll 1 A. = ^MOH XMOH formed, or (476) d$c(i-a) where, if the base has a solubility product, it is to be used in place of the terms on the right, and the value SHaO varies with the temperature, having the value (1.09 X io~ 7 ) 2 at 25. Case II. The process is due only to the formation of acid. Here X H- > HA^HA or Just as above, since CH- = i.o9Xio~ 7 at 25, we shall have, when c is the concentration of salt, and d$ its ioni- zation, or 320 ELEMENTS OF PHYSICAL CHEMISTRY. If ^B is the ionization of the base, c the original con- centration of "salt, and a its hydrolytic dissociation, then total H-- total OH" and formed, or (486) dsc(ia t And, here again, the solubility product may be used on the right, if the acid is difficultly soluble. The formulas above, in both cases, may also be written in another form, which, although it does not illustrate so well the principles involved, is more useful in many ways. From (47 ft), by transformation, we obtain Cd s (l -a or fr )v y- (i-a)v as a CHEMICAL CHANGE. 3* 1 and, from (486), in the same way, K\ <* 2 <% '-* hyd - In dilute solutions where d$, d^ and J B rnay be regarded as unity, (490) and (496) are simplified to the form yd. / vr ~ v (i-a)F AHA We have the following law, then, governing hydrolysis. The expression for the hydrolytic dissociation, of a salt a 2 in water, , _ v , is equal, when due to the formation of base (acid), to the ionic product of water multiplied by the degree of ionization of the salt divided by the ioni- zation constant of the base (acid)> multiplied by the de- gree of ionization of the acid (base). This law has recently been much used to determine from the experimentally observed hydrolysis of the salt, the ionization constant of the acid or base formed. The constant of hydrolytic dissociation, in such a case, can also be defined (see pp. 235 and 285), when d, d& or JB are equal to i, as one-half the concentration, in moles per liter, at which the salt is 50% hydrolyzed. And if a is small enough to be neglected in the term a 2 i a, / _\y = K is a l so reduced (p. 290) to 322 ELEMENTS OF PHYSICAL CHEMISTRY. in other words, for the same substance, the hydrolytic dissociation, when small, is proportional to the square- root of the dilution of the salt, i.e., aoc\/F or -y-, where c, the reciprocal of V, is the original concen- tration of the salt dissolved. Knowing the constant for hydrolytic dissociation it is also possible to calculate the degree of hydrolysis at any dilution by the formula -si The following examples will serve to show the use which may be made of the above relations. What is the ionic product for water at 25? A o.i molar solution of sodium acetate is 0.008% hydrolyzed; the sodium acetate to be considered as completely ionized, as is also the sodium hydrate formed, and the ionization constant of acetic acid is 0.000018. Here CH 3 COOH = OH' = 0.00008X0.1 =0.000008 and since CHEMICAL CHANGE. 323 .00001 8 X CcHaCOOH = ^H' X C Ac > , o.ooooi 8 X 0.000008 =i.44Xio~ 8 and Vi. 44 Xio~ 9 X 0.00008 What is the hydrolysis of a o.i molar solution of potas- sium cyanide (assu ming d$ = i ) ? K for HCN = 13X10" 10 and % 2 o = (i.o9Xio~ 7 ) 2 at 25. a 2 _(i.Q9Xio~ 7 ) 2 Ahyd '~(i-a)F : 13 Xio- 10 ' from which, when d%=i and F=io, #=0.967%. In the table below are given the values of _ . for various equilibria in which but i mole of water reacts with the substance. HYDROLYSIS OF HYDROCHLORIDES AT 25. Base. Anih'ne Per Cent Hydrolysis a 2 7 b j (i-a)F' 2 2SXIO- 5 [onization Constant of Free Base. c -j XlO~ 10 0-Toluidine . . . . 7 O I 62X10^"* 7 -I V IO ll nt- ' ' ^ 6 2 O X IQ 10 1>- " i 8 I 05X10* I 13X10- o-Nitroaniline . . 98 6 2 I e 6 Xlo~ 15 m- " .... #- " 26.6 70 6 3-oiXio- 3 o (;8Xio- 2 4.0 Xio~ 12 i 24X10" 13 Aroinoazobenzene . . 18 i I 2!CXlO- 3 o c Xio 10 Urea. 0.781 y-o /v*w i.C Xio- 14 324 ELEMENTS OF PHYSICAL CHEMISTRY. Thus far we have only considered that one mole of water reacts with the salt; in other words, we have only employed salts containing monovalent elements. In case the reaction involves more than one mole of water the treatment is the same as for the law of mass action in general. It must be said, however, that such cases, so far as we know at present, are not at all common; the salt often reacting with but one mole of water . to form a basic salt which still retains some of the original element. Wherever the relation of hydrolysis to con- centration is that given on page 316, this is so. There is one reaction, however, which gives a good constant assuming two moles of water to react, and we shall con- sider it to show how such relations are to be treated. The reaction is A1C1 3 + 2H 2 O = A1(OH) 2 C1 + 2 HC1, which has been investigated by Kullgren (Om metalls- alters hydrolys, page 108. Dissertation. Stockholm, 1904). We have, then, similarly to (476), where c is the molar concentration of AlCls in solution, and a is the fraction of it hydrolyzed, d s being the ionization of Aids, and JA that of the HC1, of which 20.0 is formed, It is impossible to use this formula in calculations, how- CHEMICAL CHANGE 325 ever, for as yet we know nothing of ^Ai(OH) 2 ci- By using the formula in the other form, analogous to (490), we obtain _ J and from this we can calculate the ionization constant of A1(OH) 2 C1, when a is known, or dispense entirely with it, i.e., using the K^A. so determined for the cal- culation of other values. It will be observed here that a, instead of being proportional to \/V as it is for the reac- tion with i mole of water, is proportional to 3/V 2 . The following results will show how well this equilibrium follows the above law, and how it is possible to find the ionization constant by aid of the hydrolytic dissociation, knowing the ionic product for water at that temperature. HYDROLYSIS OF A1C1 8 AT 100 C. A1C1 3 +3H 2 = A1(OH) 2 C1+2H- tr d A Ac* a 8 a X*. o (i-a)T/2 (i-a)V-~ 96 o . 1488 0.966 0.76 420XIO-' 5l6XlO-' 384 0.3629 0.977 0.85 509X10-' 57lXlO-' 1536 0.7142 i 0.91 54IXIO-' 594X io- Average, -Khyd. =- 560 X io~ * The concentrations of base in the three cases are 0.00155, 0.000945 and 0.000465, respectively, the acid concen- trations being twice these values. The average value of 326 ELEMENTS OF PHYSICAL CHEMISTRY. #h y d. in the last column may be used to determine ^Ai(OH) 2 ci> for H2 =4 -yFT 2 . -A We obtain A A1(OH) 2 C1 (1-ajK^ (1$ in this way the value ^ A1( oH) 2 ci =2 -33X IO ~ 19 5 where the ionization, presumably, gives Aid" and 2OH'. The formation of a substance containing OH as well as the original negative element is very common. In the case of the chloride of bismuth mentioned above the substance separating out by hydrolysis is not the pure hydrated oxide, but an oxychloride; but apparently this is not true in the case of the hydrolysis of ferric chloride. In this case the reaction is not FeCl 3 + 3 H 2 - Fe(OH) 3 + 3 H' + 3 C1 , but, according to Goodwin (Zeit. f. phys. Chem., 21, i, 1896, and Phys. Rev., n, 193, 1900), must rather be where the FeOH" is colloidal. In certain other cases it has been found that hydrolytic dissociation takes place in stages, i.e., first i mole of water reacts, then another, etc. It is quite certain, how- ever, that this does not occur at the dilutions above of Aids, for if it did, the formula used would not give a constant value, hence in this one case between these limits of dilution 2 moles of water react with i of salt. T1(NO 3 ) 3 , according to Spencer and Abegg (Zeit. f. CHEMICAL CHANGE 327 anorg. Chem., 44, 397, 1905), however, seems to react directly with 3 moles of water, T1(OH) 3 having a solu- bility equal to jo" 13 - 58 moles per liter, i.e., j=io~ 52 - 896 , but as yet this is the only case known. We must now consider, very briefly, the methods by which the degree of hydrolytic dissociation can be experi- mentally determined. It is obvious from what has already been said, that the determination of the concen- tration of any one of the reacting constituents, together with the chemical reaction, will give us a complete view of the equilibrium. The general methods so far used for this purpose are as follows: (1) The free acid (H") or free base (OHO is measured by observations of the velocity of the inversion of sugar at 100, or the hydrolysis of esters, as was described above (pp. 269, 270). (2) The determination of the ionized portion by means of conductivity observations. For details as to these methods see Ley (Zeit. f. phys. Chem., 30, 193, 1899), Kullgren (1. c.), and Goodwin (1. c.). When one of the active constituents is colored, the reaction may also be followed by aid of spectro-photo- metric observations, as shown by Moore (Phys. Rev., 12, 151-176, 1900), but, naturally, this method is much limited in its applicability. One method, which can be used for salts of weak acids 328 ELEMENTS OF PHYSICAL CHEMISTRY. with strong bases, or salts of weak bases with strong acids, has been suggested by Farmer (Trans. Chem. Soc., 79, 863, 1901, and ibid., 85, 1713, 1904), which, owing to the importance of the principle involved, is briefly considered below. The method is based upon the coefficient of distribution of a substance between water and another solvent, benzene (p. 188). Thus, hydroxyazobenzene has a coefficient of distribution be- tween water and benzene equal to 539, i.e., benzene always takes up 539 times as much hydroxyazobenzene as the water, when the two solvents are present in equal- volumes. If the two solvents are present in unequal quantity, say i liter of water to q liters of benzene, the hydroxyazobenzene in the water will be distributed between them in the ratio 1:539?. By shaking a water solution of the barium salt of hydroxyazobenzene with benzene, then, the free hydroxy- azobenzene, if it be formed by hydrolysis, will be partially extracted by the benzene. Finding the amount of this present in the benzene solution, multiplying it by - , we find what is left in the aqueous solution. The sum of these two quantities, then, is the concentration of free hydroxyazobenzene which has been formed as the result of hydrolysis, and, knowing the amount of salt initially present, the degree of hydrolytic dissociation is easily calculated. By this method, for numerous dilutions of CHEMICAL CHANGE. 3 2 9 a 2 the barium salt, the formula K=-r- r~ was found to give a constant value for K, which at 25 is equal tc 24.3X10-7. This method was also applied to the hydrolysis of the hydrochlorides of weak bases, as aniline, etc. (where the coefficient of distribution of the free base is deter- mined), with very satisfactory results. The values in the table on page 323 were found in this manner. In all such determinations constancy of temperature is o] paramount, importance, for hydrolytic dissociation, as will have been observed from the foregoing, is largely in- fluenced by the temperature. This is due not only to the increased ionization of water (p. 313) with the temper- ature, but also to the decrease in the ionization constants of acids and bases. Thus for acetic acid #i 8 = i8.3X io~ 6 , #ioo=n.4Xio- 6 , ^i 5 6=5.6Xio~ 6 and K 2 ^ = 1.9 X i o~ 6 ,. while for ammonium hydrate ^i8=i7.iX io~ 6 , Kioo=i4Xio~ 6 , and J i 5 6 = 6.6Xio~ 6 . 76. Determination of the ionization constant from ob- servations of increased solubility. A very ingenious and accurate method for the determination of the ioniza- tion constant of an acid or a base, from its -increased solubility in a base or an acid, with a known ionization constant, is given by Lowenherz (Zeit. f. phys. Chem., I 5> 385? 1898). Either the acid or base to be deter- mined must be difficultly soluble in water, i.e., just those 330 ELEMENTS OF PHYSICAL CHEMISTRY. conditions are advantageous which in the usual methods are sources of trouble. The general form of the equilibrium arising by the neutralization of an acid with a base may be written as follows: +iA'+gM'+hOH'. Calling the total amount of acid present 2 A, and the total base IM , we have the following conditions existing at equilibrium: (1) A' = JA-HA-MA. (2) A /1= M'+H' OH'. The sum of the positive ions is equal to the sum of the negative. (3) M' (4) M'=A' + OH'-H\ (5) KacidHA XX . ^ . (6) = OH'" f*\ nTT , K base MOH (8) CHEMICAL CHANGE. 33 l (9) MA = (i - O: M A) total MA. (10) MOH = (i -MOH) total MOH. (ICKZ) MOH, when difficultly soluble is the same as in a saturated water solution. (n) H 2 O = total MA. (12) HA= (i -HA) total HA. (120) HA, when difficultly soluble is the same as in a saturated water solution. The solubility of a difficultly soluble acid, for example, will be greater in a solution of a base than in pure water, for the reaction causes the removal of H* ions from the acid, and this necessitates the further ionization of the un-ionized portion, and the solution of more solid. The difference in solubility in the base and in water gives directly the amount of water formed, for that amount is the cause of the reaction which increases the solubility. The total salt formed is also equal to this difference, but this, unlike the water formed, may be, and usually is, ionized. The sequence in which the equations above may be used in any individual case depends entirely upon the nature of the equilibrium. The first thing in all cases is to find the amount of salt formed and y from its ionization in this dilution^ the concentration o\ its ions. The following 33 2 ELEMENTS OF PHYSICAL CHEMISTRY. very simple examples, however, will probably do more to show the principles involved than pages of explana- tions. Assume we have an acid, soluble to o.ooi mole per liter, the constant of ionization of which is unknown, which is soluble to 0.003 m le per liter in a o.i molar solution of the base MOH. The base is 90% ionized, and the salt which is formed, has a value of a equal to 98% at this dilution. What is the constant of ionization of the acid? Here we must find ^HA^ - Since the dif- ference in the two solubilities is 0.002, the amount of water and the total salt formed are each equal to 0.002 mole per liter. We have, then, MA = (i -a) (total MA) = (1-0.98) (0.002). M* = M' of MOH -loss of M' as MA = 0.9(0.1) (0,002) (i 0.98). OH' = OH' of MOH - loss of OH' as H 2 O = 0.9(0.1) (0.002). H* = SH 2 O OH' (1.09 Xio~ 7 ) 2 0.9 (o.i) (0.002)" = H'+M*-OH' 0.9(0.1) -(0.002.) [0.9 (o.i) (0.002)] CHEMICAL CHANGE. 333 7 +[-9 (o.i)- (0.002) (i -0.98)] (I.Q9XIQ- 7 ) 2 o.c88 (10.98) (0.002). And since we can find its value by substituting the values given above, i.e., H', A', and HA. Proceeding in this way, and only in this way, we can find the concentrations of the various constituents at equilibrium. The sequence of determination is somewhat different when the base is difficultly soluble and K base is unknown, an acid solution with a known degree of ionization serv- ing as solvent. Assume a base, which is soluble to o.ooi mole per liter, to be soluble in an acid solution (01 = 0.90) to the extent 0.003 m l e P er liter, the salt being ionized to 0.98%. What is K = f or the base? \ ^MOH / Here again MA= (o.oo2)(i 0.98), 334 ELEMENTS OF PHYSICAL CHEMISTRY. but now we must first determine A' instead of M* as above, i.e., A' = A' of HA - loss of A as MA = 0.9(0. 1 ) (O.002)(l 0.98). H* = H' of HA - loss of H* as H 2 O = 0.9(0.1) (0.002). OH' = 1 (1.09 Xio~ 7 ) 2 0.9(0.1) (0.002)" A' + OH'-H' [0.9(0.1) -(0.002X1-0.98)]+ (1-09X10-7)2 0.9(0.1) -(0.002) -[0.9(0.1) -(0.002)]. (1.00X10 7 ) 2 MOH=^M-M'-MA = (i.opXio- 7 ) 2 "! .003- [0.00196 + - -~- -- J (0.002) (l0.98) and the values of which are above, viz , M ', OH', and MOH. In case the acid is dibasic, or the metal of the base is divalent, the appropriate changes may be made in the above formulas without difficulty. CHEMICAL CHANGE. 335 In all cases, however, it is necessary to have the solvent (acid or base) considerably more concentrated than the dissolved substance (base or acid). 77. Ionic equilibria. In order that the importance of the things we have just studied may be more clearly realized, we shall now consider very briefly their appli- cation to a few questions of general chemical interest. Since the states of equilibrium which we most often en- counter in our daily experience are those composed of ionized substances, and since those composed of un- ionized substances are comparatively simple and easy to determine, we shall restrict ourselves here to the con- sideration of systems containing ionized substances. The simpler cases of ionic equilibria, i.e., those exist- ing in simple electrolytes, have already been considered above. We found then that the law of mass action enables us to foresee the equilibrium in systems com- posed of organic acids or bases, of substances which are difficultly soluble, or of those of which the ionization is very slight. But for other systems, characterized by a large degree of ionization, the law of mass action is apparently inapplicable, and must be replaced by certain empirical relations. We also found complications to arise in these equi- libria, produced by the further dissociation at higher dilutions of one of the two ions observed at lower dilu- tions. This is the case with the poly-basic organic 336 ELEMENTS OF PHYSICAL CHEMISTRY. acids, which, up to the dilution at which a = 0.5, behave as though monobasic, only showing their real basicity above this dilution. Even in such a case, however, the process can be followed or foreseen by the law of mass action, each form of ionization leading to a constant ratio of the concentrations of the constituents. An an- alogous case with a strong electrolyte is that of sulphuric acid, which ionizes in two stages, as follows: H 2 SO 4 = H'+HSO 4 , HSO 4 '=H'+SO 4 ". Here, however, we cannot follow the process by the law of mass action, and the first stage is only to.be ob- served in concentrated solutions. Indeed, at and above a dilution of i mole in 5 liters no trace of the HSO/ ion can be detected, and a dilution of i mole in 1000- 2000 liters shows practically complete ionization into 2H* and SO 4 ". The behavior of salts of the type of BaCU is similar to this; and in general we may say that the more dilute the solution the smaller the amount of a complex ion present. All these cases of equilibrium are comparatively simple, however, for all the ionized matter present arises from the one original substance with which we start. The cases we are now to consider, i.e., those equilibria result- ing from the reaction of two or more substances, are somewhat more complicated experimentally, for the CHEMICAL CHANGE. 337 ionized matter may be due to several substances, but theoretically they are to be treated just as the others. Before considering the specific cases of equilibrium which have been observed, it will be well to review very briefly the various methods for the determination of ionization and molecular weight in solution, in order that the physical-chemical analysis of such systems may be quite clear. In those electrolytes ionizing into two portions measurements of the electrical conductivity or freezing-point depression give all the necessary informa- tion, i.e., of ionized, as well as of un-ionized matter. This is also true when three ions are formed completely. When one of the three ions is formed to a smaller extent than the others, however, as is the case with the second H* ion of dibasic organic acids, some other method, in addition to the ones above, must be employed in order to show its concentration. As the equilibrium becomes still more complicated, i.e., contains other ionized and un-ionized substances, still other methods must also be employed. In short, the physical-chemical analysis can only be accomplished by a combination of two or more of these methods. . Starting with a given solution containing various ions and un-ionized substances, by the application of all or some of the following methods, we can, of course, ascer- tain the concentration of each of the constituents if this be necessary. As a matter of fact, however, this is usually un- ELEMENTS OF PHYSICAL CHEMISTRY. necessary, for, as will be seen from page 330, knowing the values of certain ones, those of the others may be found by aid of the chemical and physical-chemical relations. The electrical conductivity of a solution gives directly the total concentration o) the ions which it contains; while the freezing-point, boiling-point, vapor pressure, or osmotic pressure give the total number oj moles per liter, i.e., of ionized plus un-ionized matter. These are the general methods applicable to all kinds of ionized or un-ionized matter and their application to an unknown solution is obvious. In addition to these, however, there are certain other special methods, applicable to certain kinds of ions and un-ionized matter, which enable us to make a more detailed analysis of the solution. For example, by the depression of the solubility of a substance with an ion in common, the concentration of that ion in the solution employed as the solvent may be calculated, and with great accuracy. The nature of the ions in general may be determined by migration experiments, i.e., by observing the changes in concentration due to the passage of the electric current; and by aid of electro- motive force measurements the individual concentrations of many kinds of ions may be accurately calculated. For details as to these methods see Chapter IX. Con- centrations of ionized hydrogen may also be found from the speed of the inversion of cane sugar, using a weak, known solution of an acid as a standard; and OH' ions CHEMICAL CHANGE. 339 can be determined from the speed of saponification of an ester, a known and weak solution of a strong base being used as a standard. And, finally, un-ionized ag- gregates can be determined by partition or distribution experiments, i.e., by finding the concentration of the substance in another, immiscible solvent, after this has been shaken with the original solution. And this, it is to be remembered, is not restricted to any one un-ionized substance, for even when several are present together they behave as if they were alone, and so can be determined. By these methods, then, it is possible for us to find the concentrations of the various substances present at equilibrium, both ionized and un-ionized, and thus to define exactly the conditions necessary and sufficient for the retention of that state. As was said above, we are now to consider the equilibria resulting from the reaction of two or more substances, but, since a chemical reaction in ionized systems depends largely upon the formation of un-ionized or difficultly soluble substances, it will be seen at once that the law of mass action is applicable to these systems. For such cases where it is inapplicable we have no guiding principle at present, but these are few, so far as we know, and at any rate are of lesser importance in ordinary work. One of the first questions arising in qualitative analysis is what occurs when the precipitation of Mg(OH)2 by ammonium hydrate is prevented by the presence 340 ELEMENTS OF PHYSICAL CHEMISTRY. of ammonium chloride? The older theory of this influence assumed the formation of a complex salt (MgCl 4 )(NH 4 ) 2 , which is not decomposed by ammonium hydrate; while according to the new theory it is due to the fact that the ionization of the ammonium hydrate is so de- creased by the presence of the NH 4 * ions of the chloride that the solubility product of Mg(OH) 2 cannot be exceeded. This question was first investigated by Loven (Zeit. f. anorg. Chem., 37, 327, 1896). If the action is due to the driving back of the ionization of NH 4 OH by the NH 4 * of NH 4 C1, the reaction would be MgQ 2 + 2NH 4 OHrfVIg(QH) 2 + 2NH 4 C1. And in every case, even when an excess of NH 4 C1 is present, we must have the following relations at equilib- rium: or, by combination, G J i.e., c Mg - X I ^ 3 1 - "TT" = *i a constant. Love*n found the following concentrations at equilib- rium, starting with different salts of magnesium and various concentration of ammonia and ammonium chloride, the temperature being 16-17. CHZMlCAL CHANGE. 34* Mg NH 3 NH 4 k 0.0203 0.0421 O.OIO22 o-34 0.0281 0.02027 O.OO59 o 33 0.03762 0.0189 0.00655 0.31 o. 1084 0.0499 0.0286 o-33 Although in the calculation of the constant k the ionic concentrations of Mg and NH 4 should be used in the above, complete ionization of the Mg and NH 4 salts are assumed, and the total amounts employed. As will be seen, since these concentrations were determined by ordinary analytical methods, the value of k is constant within the experimental limits. Loven's own formula was expressed somewhat differently, but this form is used here so that it may be in accord with other work of the same kind. Muhs (Dissertation, Breslau, 1904) studied this same equilibrium, when attained from the other direction, i.e., according to the reaction Mg(OH) 2 + 2NH 4 Cl<=MgCl 2 + 2 NH 4 OH. Here, just as above, we have and but, since no NH 4 OH was added, we know that hence 2CMg" C NH ; J 342 ELEMENTS OF PHYSICAL CHEMISTRY. or c* s ^ = \ 9 2 = constant, ^NH; 4&i 2 i.e., 1.5 TCiir- = constant. At 29 Muhs found the following results: Mg NH 4 Constant 0.049 0.0771 0.141 0.0638 0.106 0.152 0.089 0.172 0.154 0.108 0.25 0.140 0.156 0.388 - I 59 for ammonium chloride, while for ammonium nitrate he found 0.0495 0.076 0.145 0.0833 0.049 0.131 Here, also, the ionization is assumed to be complete, and it certainly is within a few per cent of being so. The values of the constants are not to be compared here, for unfortunately the temperatures differ widely, which would probably exercise a great influence on the am- monium hydrate solution, as well as upon the solubility, and consequently solubility product, of Mg(OH)2. In addition to these independent proofs that no com- plex is formed in such solutions we also have another based upon entirely different principles (Treadwell, Zeit. f. anorg. Chem., 37, 327, 1903). The molecular weight of a solution containing the chlorides of magne- sium and ammonium in the ratio to form a complex CHEMICAL CHANGE 343 salt, if it did exist, can be found from the following data : 0.1466 gram of MgCl 2 , and 0.1647 gram of NH 4 C1 (i.e., i mole of MgCl 2 to 2 of NH 4 C1) dissolved in 20 grams of water cause a depression of the freezing-point equal to o.96o9. From this the average molecular weight of the substance in solution is found to be 29.97. Assuming the salts to be present as a mixture, and that they are practically completely ionized (for the justification of which see below), the average molecular weight would be '- = 28.9, for a mixture of i mole of MgCl2 and 2 moles of NH 4 C1 would dissociate into seven moles of ions. If a compound were formed there could not be more than three ions formed at most, and the minimum molecular 202.32 weight in solution would be = 67.44. 5 Solutions formed by each salt alone in this dilution give the following results: 0.1466 gram of MgCl 2 in 20 grams of water depresses the freezing-point o.3956, from which ^ = 34.27, instead MgCl 2 95.26 f 2 =~3i.75. And 0.1647 gram of NH 4 C1 in 20 grams of H 2 O gives a freezing-point depression of o.55ii, i.e., ^=27.75, e NH 4 C1 53.52 instead of =^-^-= 26.74. These results show that the ionization of each alone 344 ELEMENTS OF PHYSICAL CHEMISTRY. at this dilution is nearly complete, and certainly the simultaneous solution cannot change the ionization enough to significantly change the result for the average molecular weight. If the two formed a compound, the value for the average molecular weight could not under any circumstances be as low as 29.97, as we find it. This example illustrates very well the difference between a complex (or compound) salt and a simple mixture of two or more salts. In the former case the ionization is changed, i.e., a complex ion is formed; in the latter the ions formed are the same as those which exist when the substances are present alone in the solution. The behavior of a mixture of MnCU and NH 4 C1 can be seen from the following results: 0.0968 gram MnCl2 and 0.0824 gram NH 4 C1 (i.e., i mole to 2) in 20 grams of H2O give a depression of the freezing-point equal to o.4797, i.e., ^1=34.56. Assuming that the com- pound [MnCl4][NH 4 ] 2 is ionized into three ions, it could not be more, the average molecular weight, just as above, 2^2 06 would be - =79.99, while a mixture of the two, the o ionization remaining unchanged, would lead to = 33-28. We can conclude, then, that in such solutions, neither system forms a complex salt, and that the behavior 0} mag- nesium salts with ammonium hydrate simply depends CHEMICAL CHANGE. 345 upon the concentration oj OH' ions present, and upon those ions which can alter the concentration oj these. One of the questions which gave much trouble before the inception of the electrolytic theory of dissociation was that in regard to the sequence of separation, when it is possible for two salts of differing solubility to be formed in a solution. According to the older theory it was assumed that the more insoluble one always formed first, and could be separated from the other by fractional precipitation. This, however, is found not to be in accord with the experimental facts, which, on the other hand, are perfectly represented by the law of mass action. This subject has been investigated by Findlay (Zeit. f. phys. Chem., 34, 409, 1900) in the case of the reversible reaction PbSO 4 -f 2 NaI<=*PbI 2 + Na 2 SO 4 . solid dissolved solid dissolved Applying the law of mass action to this, since at any one temperature the concentrations of the solids in solution are constant, we have = constant, C S0 4 " for the reaction may also be written PbSO 4 + 2l' PbI 2 + SO 4 ". solid solid By aid of analysis, and conductivity and electromotive force observations. Findlay found the ratio ^ to be a 346 ELEMENTS OF PHYSICAL CHEMISTRY. constant for all such systems, and to have a value at 25 lying between 0.25 and 0.3. The outcome of the investigation may be summed up as follows: From a mixed solution of sodium iodide and sodium sulphate, by the addition oj a soluble lead salt, pure lead iodide (the more soluble] can be precipitated if the ratio oj the square of the concentration of iodine ions to the concentration of the sulphate ions is greater than the equilibrium constant. When the ratio becomes equal to this constant, both lead iodide and sulphate are precipitated together, the ratio c\, - remaining constant. And all this is true for the " sulphate when the ratio is smaller than the constant. It will be observed from these examples that by aid of the law of mass action, even though it fails to hold for strong electrolytes, we can forsee and regulate many, if not most, of the reactions with which we come in con- tact. Many other examples could be cited here to illustrate the methods of application, but the few above will suffice to bring out the general principles, and enable the reader to follow work of this sort. Interesting cases of ionic equilibria have also been studied by von Ende (Dissertation Gottingen, 1899), Morse (Zeit. f. phys. Chem., 41, 709, 1902), Sherrill (Zeit. f. phys. Chem., 43, 705, 1903), Sherrill and Skowronski (J. Am. Chem. Soc., 27, 30, 1905), Noyes and Whitcomb (J, CHEMICAL CHANGE. 347 Am. Chem. Soc., 27, 747, 1905), and Abel (Zeit. f. anorg. Chem., 26, 377, 1901). 78. The color of solutions. The color of a solution depends apparently upon the condition of the solute in the solvent. If a substance is not at all ionized, or but slightly so, any color it may possess must be attrib- uted to the un-ionized substance. In case the ionization is practically complete the color of the solution will be the result of the mixture of the colors of the ions; or if only one is colored that color will be the color of the liquid. When partly dissociated, then, the color of a solution will be the result of the mixture of the colors of the ions and the un-ionized portion; or if only one of these is colored that color will be the color of the liquid. The un-ionized portion in cases, however, may also show the color of the ion; see Noyes, Technology Quarterly, 17, 306, 1904. There is always a chance of error here if the color of the solid is assumed to be its color in solution. The color of a crystal, for example, is very often different from that of the substance in the form of powder, and, further, it is possible that a dissociation takes place in the water of crystallization. In this latter case, of course, the solid would exhibit the same color as the colored ion. The only correct way to find the color of the undissociated portion in solution is to use a solvent in which the substance is not dissociated to any extent; 348 ELEMENTS OF PHYSICAL CHEMISTRY. then the color can be directly observed. This is not difficult to carry out, for all solvents have a different dissociating power, and either alcohol, ether, benzene, chloroform, or acetone will be found to serve the purpose. The ions of most acids are colorless ; consequently all salts of a metal in very dilute solutions will have the same color, i.e., the color of the metallic ion. In more concentrated solutions this is not true, for many un- ionized substances are colored and, as they are now present to a greater amount, the color of the solution is the result of the mixture of these and the ions. An example of this is given by solutions of cuprous chloride, where the color of the un-ionized portion is yellow. But the copper ion is blue, hence the color of a solution of cuprous chloride may be either yellow, green, or blue, according as it is undissociated or ionized to a lesser or greater degree. All copper solutions when very dilute, provided the negative ion is colorless, show the same blue color. The formation of a complex ion can be followed very closely when it is composed of a colored and a colorless ion. Thus if a KCN solution is added to a colored copper solution the color instantaneously disappears, due to the formation of the ion CuCN". The formation of a complex ion can be proven -in this way, but its nature can only be shown by migration experiments, as was men- tioned above. (See Chapter IX.) CHEMICAL CH/tNGE. 349 79. The action of indicators. An indicator is a sub- stance which possesses a different color in an alkaline solution from what.it does in acid. The indicators are themselves slightly dissociated acids or bases, and the change in color is due simply to the rise or disappear- ance of the colored ion or un-ionized substance. In- stead of attempting to find a general rule as to their behavior, we shall consider a few typical cases which will make the principle clear. Phenol- phthalein is a very weak acid, i.e., in water it is ionized to an imperceptible extent into H'and the colored negative radical. In the un-ionized state, i.e., in an alcoholic solution, it is colorless, while the negative ions are red. If potassium hydrate is added to a colorless alcoholic water solution, the following reaction takes place: The indicator is but slightly dissociated into H' ions, but when these come into contact with ions of OH' they unite with them to form undissociated water. This is due to the fact that only an infinitesimally small amount of H* and OH' ions can exist together without forming undissociated water. This removal of H* ions, however, destroys the equilibrium which exists between the ions and the undissociated portion of the indicator (by equation 46); consequently more of the indicator dissociates. Again the H* ions are removed, etc., and the process is repeated until we have only ions of K* and the negative colored radical, and the solution is red, 35 ELEMENTS OF PHYSICAL CHEMISTRY. It is a general rule that all weak acids are very much less dissociated than their sodium or potassium salts. In few words, then, the process consists in the removal of the H* ions by those of OH', and the formation of the potassium salt of the indicator, which is so largely dissociated that the red color of the negative ion is visible. If acid is added to this salt, then ions of H* come in contact with those of the negative radical, and since they cannot exist with them to the extent that those of K' can, they unite again to form the almost undisso- ciated indicator, and the color disappears. Ammonium hydrate is a very weak base, i.e., its dis- sociation is very small (about 1.5% in a n/io solution), and the salt formed from it with the very weak acid phenol- phthalein is hydrolytically dissociated to such an extent that it is colorless, until such an excess of ammonia is added that the hydrolytic dissociation is no longer possible. When titrated with phenol-phthalein, all acids give very satisfactory results, because the number of H* ions in the indicator is very small, and therefore easily influenced. In general, then, with phenol-phthalein only the stronger alkalies can be used, but both strong as well as weak acids may be readily determined (contrast with methyl orange). Methyl orange is a medium strong acid, i.e., it is disso- ciated to a larger extent than phenol-phthalein. Its un-jonized portion is red and its negative ion yellow. CHEMICAL CHANGE. 351 The pure water solution of this is so much dissociated that it shows the color which is produced by the mixture of these two colors; for this reason it is usually used in an alcoholic solution. Addition of a strong acid, by the action of its H" ions, causes the dissociation to decrease and consequently the red color of the un-ionized substance appears. If an alcoholic solution which is red is mixed with a base, H' and OH' ions unite to form H20, and a dissociated salt is left, which shows the yellow color of the negative ion. A very weak acid will have no effect upon this indicator, for the number of its H' ions will not be great enough to influence those of the indicator (for example, H^COs). Methyl orange, then, is adapted to all baces, weak as well as strong, but only to strong acids (contrast with phenol-phthalein) . The other acid indicators act on the same principle as these. The alkaline indicators possess OH' ions and form salts with the acids, so that the color of the positive ion appears. With bases the ionization is decreased, hence the color of the un-ionized product is shown. In all cases it is to be remembered that a certain por- tion of the acid or base used is employed in causing the indicator to change, so that the amount of the latter should be as small as possible. The action of acids in each of these cases is to decrease the ionization by its H* ions, or else to remove OH' ions 35* ELEMENTS OF PHYSICAL CHEMISTRY. with them. The bases decrease the concentration of OH' ions, or remove H* ions. This theory is due to Ostwald and apparently repre- sents the facts. Another theory has recently been advanced by Stieglitz (J. Am. Chem. Soc., 25, 1112, 1903), but as neither can be proven absolutely ^ and as the one above accounts for the facts observed more simply than the other, it may be provisionally accepted. 80. General analytical reactions. Here we shall not treat the analytical scheme in detail, but only consider those reactions which, viewed from the standpoint of the theory of dissociation, possess particular interest. METALS OF THE SECOND GROUP. Calcium oxalate is a comparatively insoluble sub- stance ; its solubility is 0.000059 mole per liter at 25, so that s = 3.3Xio~ 9 . Oxalic acid is not dissociated enough, however, to precipitate CaC2O4 completely from the salts of strong acids. This is due to the fact that the reaction as it progresses gives rise to ions of H* and of the negative radical, i.e., to a strong acid, and the H' ions of this drive back the dissociation of the oxalic acid to such an extent that the solubility product of CaC2O4 is no longer exceeded, even with the large excess of Ca" ions present. In contrast to this, all cal- cium salts are precipitated by ammonium oxalate, since that is dissociated to a larger degree, and no acid is formed CHEMICAL CHANGE. 353 to affect its dissociation. If, however, NaCl is added to CaCU, more CaC2C>4 is precipitated by H2C2O4, since the Cr ions of the NaCl drive back the dissociation of the HC1 formed during the reaction, thus decreasing the influence upon the dissociation of the oxalic acid. Ammonia is too weak a base (1.5% ionized in a n/io solution) to form the hydrate of calcium, since the number of its OH' ions is too small to cause the solubility pro- duct of the Ca(OH) 2 to be exceeded, even with the excess of Ca" ions. The hydrates of sodium and potas- sium, however, are dissociated to a greater degree, and the precipitate is formed. Ca(OH)2 is soluble in water to the extent 0.035 mole per liter, but an excess of the precipitant will reduce this so that Ca(OH)2 may be used as a means to detect calcium. Of course the OH' ions here have a greater effect upon the Ca" than those of a divalent element would have, for here in determining the solubility product the square of the concentration of OH' ions is used. We have thus in a saturated so- lution, since Ca" = 0.035 and 0^ = 2X0.035 = 0.07, s = 0.035 X (0.07)2 = 0.031 72. Thus i mole of OH' ions reduces the solubility of Ca(OH)2 to 0.031 72 = .r(#+i) 2 , or since the x in x+i may be neglected as compared with i, #(i) 2 = o.03i72, i.e., the solubility is reduced by a normal solution of OH' ions from 0.035 to 0.03172. Con- sidering that the compound CaY has the same solu- bility as Ca(OH)2 and that Y is divalent, we find 0.035 354 ELEMENTS OF PHYSICAL CHEMISTRY. X 0.035 = 51 = 0.02123. In this case a normal solu- tion of Y ions would depress the solubility from 0.035 only to x = 0.02! 23. Strontium salts may be precipitated as the sulphate, but as this is quite soluble it is necessary to add alcohol and an excess of the precipitant to decrease the disso- ciation and consequently the solubility. Since H2SO4 is ionized to a lesser degree than HC1 and HNO 3 , the sulphate is soluble to a certain extent in them. The process here is as follows: The SrSC>4 gives off ions of Sr" and SO/'; these latter, however, owing to the large concentration of H' ions in the acid, unite to form HSCV or H 2 SO4- This loss of SCV' ions causes more SrSC>4 to dissolve and dissociate, so that the SrSC>4 dis- solves more than it does in pure water. If a large excess of a solution of a soluble carbonate is poured upon dry SrSC>4 the latter is transformed into SrCO 3 . This is due to the fact that SrCO 3 has a much smaller solubility product than SrSC>4. In the solution we have ions of Sr", SCV, 2Na*, and CO 3 "; the Sr* ions unite with those of CO 3 " to form SrCO 3 , which saturates the solution and then separates out as a solid. This causes more SrSC>4 to dissolve and dissociate, which in its turn is transformed into the carbonate, until finally we have solid SrCO 3 in a solution of NaSC>4. The solubility product of SrCO 3 is so much smaller than that of the SrSC>4 that SrSC>4 is transformed com- CHEMICAL CHANGE. 355 pktely by a mixture of equal amounts of carbonate and sulphate for the decrease of the dissociation of the SrSO 4 by the SO 4 " ions is not sufficient to prevent the formation of the SrCO 3 . If the solubility product of a sulphate differs less from that of the carbonate, as in the case of barium, then the sulphate when treated with a mixture of equal amounts of a carbonate and sulphate will not be changed. This is true because the SO 4" ions added decrease the dissociation to such an extent that the solubility product of the carbonate is not reached. We have so much ionized SO/' in the solution that the metal ions with those of the CO 3 " could only reach the solubility product of the carbonate if it were considerably smaller than that of the sulphate. This is a method of separating Sr from Ba, for the SrCO 3 formed is soluble in HC1, while the unchanged BaSO 4 is not. Morgan (J. Am. Chem. Soc., 21, 522, 1899) has shown this relation to be as follows (the solubilities refer to 18): Ba**XCO 3 " = 5i = o.o 3 iXo.o 3 i=o.o 7 i Ba"X SO 4 " = S2 = 0.041X0.041 = 0.091 Sr" X CO 3 " = s 3 = o.o 4 7.Xo.o 4 7 = o.o 8 5 Sr" X SO 4 " = s 4 = 0.0554 X o.o 3 54 = o.o 6 3 To precipitate BaCO 3 from i liter of BaSO4 solution we must have more than 0.041 Xx = 0.071, x= 0.02! of CO 3 " ions. 35<5 ELEMENTS OF PHYSICAL CHEMISTRY. For SrCOa from SrSO4 more than 0.0354 X#=o.o 8 5, #=0.0592 of COa" ions. If we add SO/' ions to the concentration of o.i mole to a BaSC>4 solution then the amount of BaSC>4 remain- ing will be ;yX.i=o.o 9 i, y = o.osi= moles of Ba", of SCV, and of course of BaSC>4, in solution, i.e., the new solubility of BaSC>4. If we now find the concentration of COa" ions neces- sary to reach the solubility product of BaCOa we find it is increased to o.o 8 iX#=o.o 7 i, oc=io moles from 0.^21 moles, i.e., in the case of Ba a o.i molar solution of SCV ions counteracts the effect of 10 moles of. COa" ions. For Sr'we find this to be in the two cases yX. i =0.063, :v= 0.053, i.e., the solubility of SrSCU in presence of .1 SCV ions. From this in order to precipitate SrCOa it is necessary to have only moles. That is, the effect of SCV ions on the formation of SrCOa from SrSC>4 is so small that Sr will be trans- CHEMICAL CHANGE. 357 formed by solutions of COs" ions which have no effect upon BaSO 4 ; and the NaSO4 formed during the reaction increases this difference between Sr and Ba. In the case of the carbonates here the values are very largely affected by the hydrolytic dissociation. This, however, would have the same effect upon both Sr and Ba, so the relation between the two holds just. as well as if this did not take place. Bodlander has lately determined the hydrolytic dissociation of BaCO 3 and CaCOs and finds them equally dissociated, i.e., to 84%. Strontium sulphate is characterized by its extreme slowness of saturation, so that BaSO 4 , which is formed immediately, can be separated from the SrSO 4 , which is formed only after a time. As will be observed this process is similar in principle to the later work of Findlay described on page 346. Magnesium. The hydrate of this metal is slightly soluble (0.00002 moles per liter), but an excess of OH' ions reduces the solubility very largely, so that it may be used a quantitative precipitate. The fact that Mg" and OH' ions can exist to an extent together without uniting explains why Mg(OH) 2 is not precipitated by NH 4 OH, when it is by NaOH and KOH. The NH 4 OH contains just enough OH r ions to -exceed the value of the solubility product of Mg(OH) 2 , but the ammonium salt which is formed by the reaction has a decreasing effect upon the dissociation of the NH 4 OH, so that 35 8 ELEMENTS OF PHYSICAL CHEMISTRY. the product can no longer be exceeded. In this way an ammonium salt added to one of magnesium will prevent the precipitation of the latter as Mg(OH) 2 by NH 4 OH, and also by other hydrates, unless enough is first added to saturate all the NH 4 * ions in the solution. Undissociated NH 4 OH will be formed from the NH 4 ' and OH' ions and an excess of OH' ions can never be present until all NH 4 ' ions, except those few produced from the NH 4 OH, are removed. For the proof of this see above, pages 339~344- METALS OF THE THIRD AND FOURTH GROUPS. These groups of metals are characterized by the fact that their sulphides have such large solubility products that they are soluble in dilute acids. A general law as to the solubility of the sulphides in acids is as follows, its derivation being self-evident: // the solution of H 2 S gas in acid, such as would be formed if the sulphide dis- solved in acid of a certain strength, contains a smaller number of 5" ions than there are in a water solution oj the sulphide, then the sulphide is soluble in acid. If a larger number, then the sulphide is insoluble. If the sul- phide contains in H 2 O a larger number of S" ions than H 2 S in an acid solution, then when it is placed in acid the solubility product of the H 2 S is overstepped and H* and S" ions unite, saturate the solution, and finally escape as H 2 S gas. If the sulphide contains a smaller number CHEMICAL CHANGE. . 359 of S" ions in water, then it will be impossible for the solubility product to be overstepped and no H 2 S gas will be formed. Iron. This metal has two ionic forms one with two charges of electricity, Fe", and the other with three, Fe"". The solubility product of FeS is so large that it is not formed by H 2 S in neutral salt solutions, for the acid formed drives back the dissociation of the H 2 S (which is ionized to a very small extent into 2H* and S"), so that the S" ions become too small in concentration to exceed the solubility product of the FeS. The iron salts of the strong acids are hydrolytically dissociated; in the case of salts of weak acids this is almost complete. On account of this we have difficulty in obtaining clear solutions except in the presence of acids, for the hydrate formed, FeOH", or whatever formula it may possess, is in the colloidal state and separates out readily. The so-called basic-acetate separation of iron is based upon this dissociation. The iron salt is diluted and nearly neutralized with a carbonate, and then acetic acid and an acetate added, and the whole solution boiled and filtered hot. This treatment causes the acetate of iron to be formed, and when hot, since the hydrolytic dissociation increases w r ith the temperature, the hydrate of iron separates out completely. The ace- tate is added of course to depress the dissociation of the small amount of acetic acid present and thus to destroy 360 . ELEMENTS OF PHYSICAL CHEMISTRY. its dissolving power. In this way it is possible to pre- cipitate iron as a hydrate without making the solution alkaline. This hydrolytic dissociation, of course, can always be predicted by the use of the equation of the solubility product, as has already been shown on pages 316-329. The ferro and ferri ions have different colors, the former being pale greenish, the latter pale cream. The brown color is due to the colloidal hydroxide. H 2 S gas reduces the ferri to ferro ions, with the liberation of sulphur. By boiling iron salts with KCN, the ferro- and ferricyanides are formed, each of which dissociates into a complex ion, so that neither gives a reaction for iron. The volumetric method of estimation of iron by per- manganate of potash depends upon the change from ferro to ferri ions. Thus 2KMnO 4 + ioFeSO 4 + 8H 2 SO 4 = K 2 SO 4 + 2MnSO 4 + 5Fe 2 (SO 4 ) 3 + 8H 2 O, or, since it is an ionic reaction, ioFe" + 2MnO 4 ' + i6H' = ioFe'" + 2Mn" + 8H 2 O. The end of the reaction is the point at which the color of the undecomposed KMnO 4 is observed. Aluminium has only trivalent ions, Al"'. All salts react acid, showing hydrolytic dissociation, which with the weaker acids is nearly complete, as in the case for iron. CHEMICAL CHANGE, 361 The behavior of A1(OH)3 with ammonia, caustic soda, and potash is just the opposite to that of Mg(OH) 2 . The two latter dissolve the hydrate, while the former does not. This is due to the fact that with the stronger alkalies, i.e., those which contain a large concentration of OH' ions, the AT" ion takes the place of the hydrogen in the OH', and we have the complex ion AlOa"'. This causes more of the A1(OH) 3 to dissolve and dissociate, etc. With ammonia the concentration of OH' ions is too small to cause the ion to be formed to any extent; hence the A1(OH) 3 is not so soluble. Cobalt and nickel. The sulphides of these two metals show a marked peculiarity. They are not formed in dilute acid solutions, and yet when once formed are not dissolved by acid of this strength. This is probably to be explained by the fact that the sulphides when once formed change their state in some way and thus become insoluble in acids. This seems to be a common process with the sulphides, cadmium sulphides being the only one which is perfectly reversible. This is formed in a solution and then redissolved by removing the H2S gas in a vacuum a process, which is not possible experi- mentally with any other metal, although theoretically possible with all. With all metals of these groups H 2 S under higher pressure will cause the sulphides to be formed in the presence of acid which would prevent the formation under atmospheric pressure. 362 ELEMENTS OF PHYSICAL CHEMISTRY. Zinc forms with the alkalies a complex ion, ZnO 2 ", just as aluminium. The sulphide of this metal is as insoluble as any of this group, for neutral salt solutions are practically (i.e., 90 to 95%) precipitated by H 2 S, notwithstanding the H* ions of the acid formed during the reaction. The acetate in presence of a neutral acetate is completely precipitated by H 2 S. METALS OF THE FIFTH AND SIXTH GROUPS. The sulphides of the metals of these groups possess a concentration of S" ions in a saturated water solution which is smaller than that of H 2 S under the same con- dition, even in the presence of dilute acids. Conse- quently they are precipitated in acid solutions by H 2 S gas. Cadmium. The halogen compounds of this metal are characterized by their slight ionization. The effect of this upon the solubility is marked, for when no dissocia- tion takes place the precipitate is usually soluble in an excess of the precipitant. As a rule, complex ions con- taining Cd are not stable, i.e., their dissociation is com- parable with that of the simple salts. The sulphide of cadmium is less soluble than that of zinc and is formed from the complex salt K 2 CdCN4 by H 2 S gas. Copper can be estimated from the cuprous iodide. The reaction is In order to make the Cul as insoluble as possible it is CHEMICAL CHANGE. 363 necessary to have ions of I' present. An addition of H 2 SO3 serves this purpose, in that it forms HI, from the undissociated I, which is dissociated into H* and I' ions. Copper sulphide is somewhat more soluble than cad- mium sulphide, and too much acid will entirely prevent the precipitation. An excess of water, however, causes it to form. Mercury halides are not largely ionized, and conse- quently many of the reactions take place between the un-ionized aggregates. The cyanide of mercury does not conduct the electric current in solution, i.e., is ionized only to a very slight extent, at most. As hydrolysis take place to a very large extent with the mercury salts of the oxygen acids (i.e., H 2 SO4, HNOs, etc.), an addition of free acid is necessary if a clear solution is to be obtained. Bismuth is hydrolytically dissociated in its solutions to a greater degree than any other metal, and, indeed, this property is taken advantage of in its separation from other substances. If a salt solution in acid is diluted with water it is noticed that a basic salt, probably BiOCl, is immediately precipitated. For further details on this subject the reader is referred to Ostwald's Foundations of Analytical Chemistry or Bottger's Qualitative Analyse. CHAPTER IX. ELECTROCHEMISTRY . A. THE MIGRATION OF THE IONS. 8 1. Electrical units. The unit of the resistance offered to the electric current is the ohm, i.e., the resist- ance at a temperature of o C. of a column of mercury 106.3 cms - l n g> ^h a cross-section equal to i sq. mm. This is equal to io 9 absolute units. The unit of current strength is that strength which will separate 0.001118 gram of silver from a solution in one second, and is called the ampere equal to lo" 1 absolute units. The unit of electromotive force is the volt, or io 8 absolute units, which gives a value for the Daniell cell equal to i.io volts. The unit of the amount of electricity is the coulomb, i.e., amperes per second; i gram of H* ions carries with it 96,537 coulombs. This number is called the Faraday, and is designated by F. 364 ELECTROCHEMISTRY. 3 6 5 The intensity factor of electrical energy is the volt, while the capacity factor is the coulomb, i.e., where s is the amount of current in coulombs, TT is the electromotive force in volts, and E is the electrical work the unit of which is the watt-second, equal to the volt- ampere-second, i.e., equal to io 7 absolute units, i watt- second is the electrical work done when i ampere flows at the potential i volt for i second. The heat equivalent of electrical energy, since the latter is equal to i voltXi coulomb or io 7 absolute units, is i.e., i watt-second = 0.2394 cal. To separate i gram of H* ions, or the equivalent weight in grams of any other element we need then the amount of work rX 96540 X watt-seconds = 96 540X0.2394 XT: = 231107: cals., where n is the electromotive force of the current giving 96,540 amperes per second. If all electrical energy is transformed into heat, then 366 ELEMENTS OF PHYSICAL CHEMISTRY. where A is the amount of heat and k is the heat equivalent of electrical work. 82. Faraday's law. This is the basis of all our work in electrochemistry. The law may be expressed as follows: 1. The amount oj any substance deposited by the current is proportional to the quantity oj electrictiy flowing through the electrolyte. 2. The amounts of different substances deposited by the same quantity of electricity are proportional to their chemi- cal equivalent weights. One gram equivalent of H* ions carries with it 96,540 coulombs of electricity, as has ..been determined by ex- periment. One coulomb, then, will cause 0.041636 gram of hydrogen ions to separate; hence it will cause the separation of 0.041036 X a grams of any other element, where a is the equivalent weight of the element. 83. The migration of the ions. The chemical effect of the passage of an electric current through an electrolyte can be divided into two distinct portions, viz., the con- duction through the electrolyte, and the separation of substance at the electrodes. It is not necessary that the ions which serve for the conduction of the current through the liquid be those which are also separated at the electrode, for secondary reactions may take place there, causing others to appear as the result of the electro- lysis. It is to be remembered, however, that even in such a case Faraday's law still holds, and the substances ELECTROCHEMISTRY 367 separated are chemically equivalent in amount to those which would have been separated in case the secondary action had been avoided. Although the two different effects are observed to- gether in practice, we shall consider them separately; taking up the question of conduction here, and putting off that of separation to a later period. The conduction through the liquid depends upon what we have designated thus far as ionized matter, and varies according to the mobility of this, which, in turn, is de- pendent upon the specific nature of the matter, its con- centration, the temperature, and the nature of the solvent. After the electrolysis of a solution, excluding, or allow- ing jor, any secondary reaction, it is found experimentally that the concentrations around the anode and cathode are not always identical, as they were initially. In some few cases they are, it is true, but in these it can be shown (as will be done later) that the mobility, i.e., velocity through the liquid at a certain voltage, is the same for both the anion, which goes to the anode,, as it is for the cathion, which goes to the cathode. In all other cases the mobility, or speed of migration through the liquid, is different for the two kinds of ionized matter of which the substance is composed. And further than this, the analysis of the anode and cathode liquids after electrolysis, excluding secondary reactions, leads to an expression for the relative mobilities, 3 68 ELEMENTS OF PHYSICAL CHEMISTRY. i.e., the migration ratio of the two ions. Indeed, when the original solution is colored, this difference in con- centration can be observed qualitatively by the eye. The reasons for this change, together with' the principle upon which its quantitative calculation is based, will be made clear by the following considerations: Assume the vessel in Fig. 15 to be divided into three portions, AC, CD, and DB, and filled with a solution A C D B FIG. 15. containing 30 gram equivalents of HC1. We have, then, 10 gram equivalents in each division. If 96,540 cou- lombs of electricity are passed through the cell from A to B, i gram equivalent of H' ions and i gram equiv- alent of Cl' ions will be separated upon the electrodes B and A, if secondary action is excluded. These gases we assume to be removed as they are formed. These 96,540 coulombs passing, as they do, through the whole solution have a certain effect upon the equilibrium of the ions. First we will imagine the H' and Cl' ions to move with the same velocity and then with differing velocities, and find the relation between the change in concentration and the relative ELECTROCHEMISTRY. 3 6 9 It is to be remembered here that two oppositely electrified bodies composing a system will transport a current equal to the sum of the charges carried by the two bodies in the opposite direction, for a negative charge going in one direction is equivalent to an equal and opposite charge going in the contrary direction. In other words, all of the current may be transported by the positive material, or a portion may be carried by each in opposite directions, and in all cases the total current is the sum of those currents going in the opposite directions. I. If the velocity for each ion is the same, then 1/2 gram equivalent of Cl' ions, charged with 48,270 coulombs, will migrate from BD through DC to CA] and 1/2 gram equivalent of H* ions, with the same amount of electricity, will go from AC through CD to DB. Altogether, then, i gram equivalent, charged with 96,540 coulombs, has passed through the section CD. Since i gram equiva- lent of H* ions has been removed by decomposition from BD and 1/2 gram equivalent has migrated to it, we have left 9} gram equivalents of H* ions and 9! gram equivalents of Cl' ions, since but 1/2 gram equivalent of this has migrated from it. Consequently we have in BD 9^ gram equivalents of HC1. In AC we have the same number, since i gram equivalent of Cl' ions has disappeared and 1/2 gram equivalent has migrated to it and 1/2 gram equivalent of H' ions has migrated from 37 ELEMENTS OF PHYSICAL CHEMISTRY. it. In the section DC the concentration is unaltered, i.e., just as much ionized matter has left it as has been carried to it. The concentration at the anode is the same as that at the cathode, then, ajter the electrolysis of a solution com- posed o) ions with the same mobility. II. Assume the velocity of the H' ions to be five times that of those of Cl'. In this case, after i gram equivalent of H and i gram equivalent of Cl have separated in the gaseous state, the whole system will have suffered a change. 5/6 of a gram equivalent of H" ions, charged with 1(96,540) coulombs, will migrate from AC through DC to BD, and 1/6 of a gram equivalent of Cl' ions, with J (96, 540) coulombs, will go from BD through CD to AC. Al- together, as before, i gram equivalent of ions will go through the section CD, carrying with it 96,540 coulombs of electricity. The original composition of the solution in CD is again unchanged. In BD we have lost i gram equivalent of H* ions in the form of gas, and gained 5/6 of a gram equivalent by the migration; consequently we have 9^ gram equivalents of H' ions left. 1/6 of a gram equivalent of Cl' ions has migrated from it, so that in DC we have 9! gram equivalents of HC1. In A C we have lost 5/6 of a gram equivalent of H* 'ons and i gram equivalent of Cl' ions as gas, but have ELECTROCHEMISTRY. 371 gained 1/6 of a gram equivalent of Cl' by the migra- tion; consequently we have 9^ gram equivalents of HC1 left. From these two examples the following law may be deduced: The loss on the cathode (BD) is related to that on the anode (AC) as the mobility 0} the anion (Cl') is to that of the cathion (H'). In this way Hittorf determined the relative mobilities or migration ratios of the various ions. 84. Determination of the migration ratios. The prac- tical determination of the relative mobilities is merely a matter of analysis. The apparatus which is used for this purpose is a decomposition-cell, so arranged that no metal can drop from one electrode to the other; or a U tube may serve the purpose, so long as the two portions of liquid may be removed and analyzed sepa- rately. The apparatus is filled with solution and the current passed through for a certain length of time, the electrodes being of the metal which is contained in the salt or inert, except in cases where certain secondary actions are to be avoided. After a certain time either the anode or cathode portion is withdrawn and analyzed. This analysis will give us the loss of metal on the one electrode, from which that on the other may be cal- culated. If n is the fraction of the cathion which has migrated from the anode to the cathode when one gram equiva- 372 ELEMENTS OF PHYSICAL CHEMISTRY. lent has been separated, then i n is that fraction of the anion which has gone to the anode. These two quantities, n and i n, are called the migration ratios of the cathion and anion. We have then n loss at anode U c in loss at cathode Ud where U c is the mobility of the cathion and U a that of the anion. An example will make the determination of this clear: Hittorf electrolyzed a solution of AgNO 3 until 1.2591 grams of Ag were separated. The volume of liquid at the cathode before the experiment gave 17.46249 grams of AgCl, and after it but 16.6796 grams, i.e., a loss of 0.7828 gram of AgCl or of 0.5893 gram of Ag. If no Ag had come to the cathode by migration, the solution would have lost 1.2591 grams of Ag; it lost, however, only 0.5893 gram; hence 1.25910.5893 = 0.6698 gram Ag has come to it by the migration. If just as much of the Ag had come by migration as had been separated, the migration ratio of the Ag would have been i, i.e., the NOs ions would not have migrated. Only 0.6698 gram of Ag has migrated, however; hence the migration ratio for the Ag' ions in AgNOs can be found by the proportion 1.2591 10.6698: :i ^=0.532; ELECTROCHEMISTR Y. 373 the migration ratio of the NCV ions is 1-0.532 = 0.468. These values are not the same for all dilutions, al- though in general the variation is but slight. A table containing a large number of results from experiments of this sort is given by Kohlrausch and Hol- born, Leitvermogen der ElektrolytCj a few of which will be found in Table XIV. TABLE XIV. HITTORF'S MIGRATION RATIOS FOR THE IONS. Solutions i/io equivalent normal. Substance. i n NH 4 C1 0.508 1/2 BaClj 0.61 1/2 CaCla o . 68 1/2 MgCl 2 0.68 HC1 0.21 KNO, Substance. i - n 1/2 KO 4 0.60 1/2 CuSO 4 , o . 64 I/2H2SO, 0.21 i/ 2 K 2 C0 3 0.37 1/2 NajCOg 0.48 i/ 2 Li 2 C0 3 0.59 KOH 0.74 NaOH 0.84 KC1 0.507 NaCl 0.63 LiCl o . 70 3 0.50 NaN0 3 0.61 AgNO 3 0.526 i/ 2 Ca(NO 3 ) 2 0.61 KCIO 3 0.46 Naturally, inert anodes can also be used for this pur- pose, and in many cases, to avoid secondary reactions, an anode which is different from the metal of the salt may be used. Thus Hittorf (Ostwald's Klassiker der exakten Wissenschaften, Nos. 21 and 23) used cad- mium as an anode in determining the migration ratio 374 ELEMENTS OF PHYSICAL CHEMISTRY. of sodium chloride, to avoid the H* ions due to the reac- tion of chlorine on water. When the substances separated at the electrodes are known, analysis of the liquids at the cathode and anode enable us to find the nature and composition of the ions into which the substance dissociates. In most cases here qualitative observations suffice for the purpose, and often color-changes make it possible to follow the process, even without analysis. Examples of this method are to be found in many investigations, see, for example, Noyes and Blanchard (J. Am. Chem. Soc., 22, 729 and 732, 1900), Peters (Zeit. f. phys. Chem., 26, 229, 1898), Calvert (ibid. , 38, 535, 1901), and Morse (ibid., 41, 709, 1902). The use of agar-agar or gelatine in these cases enables us to obtain the solution in the form of a jelly, its conduction relation remaining almost unchanged (see p. 194). B. THE CONDUCTIVITY OF ELECTROLYTES. 85. The specific conductivity. In measuring the electrical conductivity of a solution we obtain results in two different units: one refers to the same amount of all solutions, the specific conductivity; the other refers to the equivalent or molecular amount, the molecular, or equivalent, conductivity. The unit oj conductivity is the conductivity which a column oj a length of i cm. and a cross-section oj i cm. 2 ELEC TROCHE MIS TRY. 375 possesses when its resistance is one ohm. The best con- ducting aqueous solutions of strong acids have this con- ducjivity at about 40. The conductivities based on this unit are designated by /c. 86. Molecular and equivalent conductivity. Since the conductivity of a solution depends almost exclusively upon the amount of substance dissolved in it, it is more convenient for us to express our results in molecular or equivalent terms. The equivalent (molecular) conductivity oj a substance is the conductivity oj the solution which contains i equiva- lent (i mole) of substance, the electrodes being separated by i cm. and large enough to contain between them the entire solution. This value can be found by dividing K by the number of equivalents (moles) per cubic centi- meter, or by multiplying by the number of cubic centi- meters in which i equivalent (i mole) is dissolved. When V is the volume containing i equivalent (or i mole) in liters, then and where equivalent conductivity is designated by A and molecular conductivity by //, 376 ELEMENTS OF PHYSICAL CHEMISTRY. 87. Determination of electrical conductivity. Since the degree of dissociation of electrolytes can be deter- mined most accurately by aid of the conductivity, .the method is one of great importance. The apparatus used for this purpose was designed by Kohlrausch,* and is similar to that for determining the resistance of metals, except that an alternating current is used in place of the direct, and a telephone receiver instead of the gal- vanometer. The alternating current is used here to prevent actual decomposition of the solution, for that would decrease its concentration continually and cause polarization of the electrodes. In this way all substance which is deposited by the current in one direction is redissolved by it in the opposite direction, and all polari- zation effect is nullified. The form of the electrode is shown in Fig. 16, the connections being made by aid of the mercury in the glass tubes. The electrodes themselves are of Pt, which are coated electrolytically with platinum-black, so as to do away with any difference of potential between them. It is not necessary to have these electrodes just a square centimeter in cross-section or i cm. apart, for we can easily find the factor which will transform results for any cell into terms of specific conductivity. Kohl- rausch has carefully determined the specific conductivity * For details see Ostwald, Handbook of Physicochemical Measure- ments, Macmillan & Co, ELECTROCHEMISTR Y. 377 of a number of standard solutions, so that a determina- tion of any one of these in a larger or smaller cell will give directly the factor necessary to transform re- FIG. 1 6. (Natural size.) suits by that cell into actual specific conductivities, according to the definition. For a 0.02 molar solution of KC1, Kohlrausch found the values K, 30 = 0.002397, J, 8 o= 119.85, 350=0.002768 and A 2S = 138.54. 378 ELEMENTS OF PHYSICAL CHEMISTRY, 88. Ionic conductivities. Since the ions are the car- riers of electricity in solution, the equivalent conductivity at any dilution divided by that at infinite dilution, i.e., when the substance is present only in the form of ions, will give us the degree of dissociation. We have then A v This conductivity at infinite dilution means simply that the equivalent conductivity is not altered by further dilution. This maximum value for the equivalent con- ductivity Kohlrausch found for a binary electrolyte to be equal to the sum of two single values, one of which refers to the anion and the other to the cathion. This law of the independent migration of the ions shows that conductivity is an additive property. The truth of this law is shown in Table XV. TABLE XV. MOLECULAR CONDUCTIVITIES AT INFINITE DILUTION. Ag 109 103 83 The differences of two corresponding sets of numbers jn the vertical rows, and of any two in the horizontal ones, K Na Li NH 4 H Cl . .. 123 103 95 122 353 NO 3 . .. 118 98 35 OH 228 2OI C10 3 - H5 . . . C 2 H 3 2 . . . 94 73 ELECTROCHEMISTRY. 379 are nearly equal, which can only occur when the result is composed of two single and independent values. One kind of ion, then, always carries the same amount of electricity with its own velocity, independent of the nature of its companion ion. The equivalent conductivity at infinite dilution is con- sequently where l c and l a are the equivalent conductivities of the ions of the substance in solution at infinite dilution. We have then (p. 372) lc , la = -r~ and i - = -:-, in other Words, l c = nA x , l a =(i-n)A 00 . Thus the molecular conductivity at infinite dilution fi M (equal here to the equivalent conductivity A w ) of sodium chloride is no, while ^ = 0.38 and 1^ = 0.62, hence * ' X / c = o.38Xno=4i.8(Na-), l a =0.62 Xi 10 = 68.2 (CIO, i.e., i mole of Na' ions possesses a conductivity of 41.8 when between electrodes i cm. apart and large enough to ELEMENTS OF PHYSICAL CHEMISTRY. contain between them the total volume of solution in which the sodium ions exist; and i mole of Cl r ions under the same conditions has a conductivity equal to 68.2. In all solutions in the same solvent, at the same tem- perature, these values remain constant, so that it is possible for us to calculate what the conductivity would be for any substance at infinite dilution. This is of great use in experimental work, for it is npt always possible to actually reach this limiting value with any degree of accuracy. A table of such results, then, enables us to find the limiting value of the conductivity at infinite dilution, and not only this, for i.e., if we know the fraction of a mole 0} each ion present we can find the equivalent conductivity of the solution at any dilution by multiplying the sum of the ionic conduc- tivities by the degree of ionization. Since most neutral salts are very largely ionized the value of the equivalent conductivity can be readily de- termined. By subtracting the value for the metal ion from this result it is possible, then, to find the ionic conductivity of the acid radical which is present as the negative ion. In the table below are given a few ionic conductivities, from which various values may be cal- culated. ELECTROCHEMISTR Y. IONIC CONDUCTIVITIES AT 18 AND INFINITE 381 Li" / 1-2 44 Temp. Coeff . 0.0265 Zn" . . . / 4? 6 Temp. Coeff. o 0251 Na' AT. . re 0.0244 i Mg" 46 o o 02 c.6 K" 64 67 o 02 1 7 \ Ba ' (-6 2 Rb' 67.6 0.0214 i Pb".. 61 c o 024^ Cs' 68 2 O O2 12 SO/' 68 7 NH/ Tl- 64.4 66 0.0222 o 0215 *C0 3 " BrO' . . 70 46 2 0.0270 Ag' F' 54.02 46 . 64 0.0229 0.0238 CIO/ IO/. 64-7 47 7 Cl' 6? .44 0.0216 MnO/ tr-j 4 Br' 67 61 o 02 1 5 CHO ' 46 7 I' 66 40 o 021 3 C H,O ' 7C SCN' ;6 6? O O2 1 1 C,H.O' TT CIO,' ec Q-t O O2I 5 C,H,O/ 27 6 I0 3 V . . . NO ' . jj - w o 33-87 61 78 0.0234 0.0205 C.H.O,'.... C e H,,O, . 25-7 24 3. Et- on' ' 3i3 174 The ionic conductivity of elementary ions is a periodic function of the atomic weight and rises with it in each series of analogous elements; but analogous elements with atomic weights greater than 35 have approximately equal conductivities. Isomeric and metameric anions have the same equiva- lent conductivities. 89. The conductivity of organic acids. The deter- mination of the ionization constant of organic acid, which has been discussed above, is very easily accom- plished by aid of the conductivity. By substituting for a in the Ostwald dilution law we obtain = KV. * See Kohlrausch, Berl. Akademieber. 26, 581, 1902, and Sitzungsber. d. Akad. d. Wiss. zu Berlin, 26, 572, 1902, 382 ELEMENTS OF PHYSICAL CHEMISTRY. For inorganic acids, bases, and salts, for which the Ost- wald dilution law does not hold, we can use either the Rudolphi or the van't Hoff dilution law, which become, by substitution of for a, = const. -i) VF V *J and (-1) 2 F = const. The molecular conductivity, when the acid is very weak and dilute, is found from the experimentally de- termined value of /c, after subtracting from it the value *H 2 o for the water used, for these two terms, under these conditions, become of the same order. 90. The absolute mobility of the ions. Thus far we have considered only the relative mobility of the ions, or else their conductivities ; now, however, we shall consider the actual velocity in centimeters per second with which they go through the solution. ELECTROCHEMISTRY. 383 Imagine two electrodes i cm. apart having a difference of potential equal to i volt, and assume between them a solution containing i equivalent mole of positive and i equivalent mole of negative ions. In i second the amount of electricity e will go over. Each equivalent mole of ions will transport 96,540 coulombs of electricity; hence the ratio - - will give the fraction of the distance, 96540 i cm., which the ions together have traversed in i second, i.e., the sum of their velocities in centimeter-seconds. By Ohm's law where E is the current strength, v is the electromotive force, and r is the resistance. Since v = i in this case, we have But is the conductivity; hence instead of e we may use the equivalent conductivity of the solution, i.e., 96540 96540* This equation can also be obtained directly from the , or 3^4 ELEMENTS OF PHYSICAL CHEMISTRY. The molecular conductivity of a o.oooi normal solution of KC1 at 18 (in reciprocal ohms) is 128.9; the sum of the distances traversed by the two ions is 128.9 =0.001345 cm. per second. This total velocity is made up of the two single velocities. By Hittorf's experiments the relative velocities of K* and Cl' ions in KC1 are in the ratio of 49 to 51. Hence K* has a velocity of 0.00066 cm. per second and Cl' has a velocity of 0.00069 cm - P er second in a o.oooi molar solution at 18, for a potential difference of i volt. Knowing the absolute mobility of the ions we can thus calculate the equivalent conductivity at infi- nite or any other dilution. We have, in the two cases, and 96540 where U a and U c are for infinite dilution, and expressed in centimeters per second at a definite temperature. ELECTROCHEMISTRY. 3 8 5 In the table below are some of the absolute mobilities as given by Kohlrausch. ABSOLUTE VELOCITY OF THE IONS AT 18 IN CMS. PER SECOND. K' =0.00066 NH 4 - =0.00066 Na" =0.00045 Li* =0.00036 Ag' =0.00057 Cr 2 O 7 " = 0.000473 H" =0.00320 Cl' =0.00069 NO 3 ' =0.00064 C1O 3 ' = 0.0005 7 OH' =o. 00181 Cu" =0.00031 It has been possible to prove the correctness of some of these results by experiment, i.e., by actual speed measurements on a color boundary in a tube. Whet- ham's method for this purpose (Phil. Trans., i893A, 337; i895A, 507) is as follows: If we consider the boundary- line of two equally dense solutions which contain a common colorless ion, but which are colored differently, and call the salts AC and BC, then when a current passes through the boundary-line the C ions go in one direction and the A and B ions in the other. If the A and B ions are the cathions, then the color boundary will move in the direction of the current, and its velocity will be equal to the velocity of the ion which causes the change of color. In this way the velocities for Cu", Cr 2 O7", and Cl' were determined in centimeter-seconds. In the table on p. 386 the values found in this way are compared with those calculated by Kohlrausch. For other methods for determining the mobilities of the ions see Masson (Phil. Trans., IQ2A, 331, 1899); 3^6 ELEMENTS OP PHYSICAL CHEMISTRY. Steele (Phil. Trans., 198, 105, 1902); Abegg and Gans (Zeit. f. phys. Chem., 40, 737, 1902), and Denison (Zeit. f. phys. Chem., 44, 575, 1903). COMPARISON OF ABSOLUTE IONIC VELOCITIES. T Whetham, Kohlrausch, by Experiment. Calculated. Cu" 0.00026 0.00031 0.000309 Cl' 0.00057 0.00069 o .00059 0.00047 Cr 2 O 7 " 0.00048 0.000473 0.00046 91. The basicity of an acid. For all binary organic acids the formula holds strictly. For dibasic acids this is also true when the dissociation is not more than 50%, i.e., all dibasic organic acids under these conditions ionize as monobasic acids. In the case of the neutral salts of these acids, however, the dissociation into more than two ions takes place at a much smaller dilution. The difference of conductivity, then, between two dilutions must be greater for the neutral salt of a polybasic acid than for one of a monobasic one. Ostwald has made use of this fact for the determination of the basicity. He found empiri- cally that the Na salt of a monobasic acid gives a differ- 'ELECTROCHEMISTRY. 3 8 7 ence in equivalent conductivity at 7 = 32 liters and 7 = 1024 liters of approximately 10 units, a dibasic one of 20 units, etc. Hence to find the basicity of an acid we have only to find the equivalent conductivity of its Na salt at 32 and 1024 liters dilution; then w, the basicity J of the acid, is given by n = . 10 VALUES OF J AND n FOR SODIUM SALTS OF ORGANIC ACIDS. Acid. J n = IO Formic ............................ 10.3 i Acetic ............................. 9.5 i Propionic .......................... 10.2 i Benzoic ........................... 8.3 i Quininic .......................... 19.8 2 Pyridin-tricarboxylic (i, 2, 3) ........ 31 .o 3 (i, 2, 4) ........ 29.4 3 Pyridin-tetracarboxylic .............. 41 .8 4 Pyridin-pentacarboxylic ............. 50 . i 5 92. The conductivity of neutral salts. Another em- pirical law enables us to find the equivalent conductivity of a neutral salt at one dilution, provided we know it for another, and the salt is largely ionized i.e., when A v is not very different from A*. The relation observed is as follows: or ELEMENTS OF PHYSICAL CHEMISTRY. where n\ and n^ are the valences of the anion and cathion respectively, and c v is a constant for all electrolytes. When c v is known for all dilutions, and also the terms A v , ni, and n%, we can find the value of A^, i.e., the equivalent conductivity at infinite dilution. If we desig- nate (ni-n 2 -c v ) by d m then Table XVI gives the value of d v for different dilutions and values of n\ -nz&t 25- TABLE X Valence, n\ . a I . . ii 8 ^256 6 ^512 4 ^1024 7 2 . . 21 16 12 8 6 3O 23 17 I 2 8 4 * . . . . 42 23 16 10 c 20 21 13 6. . . 60 48 36 2? 16 This behavior may be summed up in words as follows : The decrease 0} equivalent conductivity is roughly Con- stant jor salts oj the same type; and the decrease in equiva- lent conductivity jor salts oj different types is proportional to the valences oj the ions. The relation of ionization to dilution in those cases where the Ostwald dilution law cannot be applied has already been considered (see pp. 293-301 and 305-316). 93. The ionization of water. The ions of water are H* and OH', i.e., to a very slight degree it is a binary ELECTROCHEMISTRY 3 8 9 electrolyte. The specific conductivity K of a specially purified water was found by Kohlrausch to be 0.014 -io~ 6 at o C. o.o4o-io~ 6 at 1 8 0.055 -io~ 6 at 25 0.084 io~ 6 at 34 0.17 -io~ 6 at 50 One mm. of this water at o has a resistance equal to that of a copper wire of the same section and 40,000,000 kilometers long, i.e., which is long enough to go around the earth a thousand times. From this conductivity the degree of dissociation is easily determined. One mole of H' ions has a con- ductivity equal to 318 units (p. 381), while i mole of OH' ions has one equal to 174; hence the maximum molecular conductivity of water should be A M =318 + 1 74 = 492, if completely dissociated. That of a liter be- tween electrodes i cm. apart at 18 is lo 3 times as large as 0.04 Xio~ 6 , i.e., 0.04 X io~ 6 X lo 3 =0.04 X io~ 3 , hence 0.04 X i o~ 3 492 =o.8Xio~ 7 ; 39 ELEMENTS OF PHYSICAL CHEMISTRY. which is the concentration of H' and OH' ions in moles per liter at 18, or there are 17 grams of OH' and i gram of H' ions in 12,000,000 liters of water. It is to be remembered here that the 492 would be the value if i mole of H' ions and i mole of OH' ions were present together between electrodes i cm. apart and of any cross-section, so long as they will contain between them the amount of solution. Since we use water as a solvent, the equivalent conductivity is usually calculated for a liter, and not, as it might be assumed by definition, 1 8 grams. The calculation was similar in determining the ionic product of H* and OH' ions (p. 313), where Sn 2 o 1000^ was equal to z-K, i.e., 555A. Io 94. The temperature coefficient of conductivity. Ac- cording to Kohlrausch, the conductivity varies with the temperature as follows: where ft 'is the temperature coefficient, or the change in conductivity for i C. We can find /? by experiment from the equation Substances with small molecular conductivities usually have large temperature coefficients. Most of the strongly ELEC TROCHE MIS TRY. 391 dissociated salts have a value for /? equal to 0.025, i.e., the equivalent conductivity changes 2\% for each degree. This temperature coefficient depends upon several factors, for the internal friction of the solvent changes, as well as the degree of dissociation and the velocity of migration. It is only possible to obtain a negative coefficient when the dissociation decreases enough to more than compensate for the increase in the velocity due to the diminished friction; consequently all such substances must have negative heats of dissociation. The equivalent conductivity at infinite dilution also increases with the temperature, so it must not be im- agined that the ionization increases in this way. The examples already given (p. 299) will serve to show how a varies with the temperature. 95. The conductivity of difficultly soluble salts. If the saturated solution of any insoluble salt is so dilute that we may assume complete dissociation, A w =A V , we have ^solution ~~ and A v =A go =*XioooF, i.e., 39 2 . ELEMENTS OF PHYSICAL CHEMISTRY. and i *Xio 3 c = yr = -- moles per liter. This method has been used by Hollemann * and Kohlrausch and Rose f with very good results. In this way a number of solubilities have been determined. Thus i part BaSO4 dissolves in 429,700 parts H^O at i6.i C., i part SrSOd dissolves in 10,070 parts H 2 O at 24. 2 C., i part BaCOs dissolves in 45,566 parts H^O at 24.3 C., i part SrCOa dissolves in 91,468 parts H^O at 24.3 C. BaSO4 is of course so insoluble that it is impossible to find its term A^ by experiment. Its value, however, can be found from the general equation where M and Mi represent metals and X and Xi the negative radicals. Thus from 4,, for JBaCl 2 , =115] 4,, forKCl, =122 18 * Zeit. f. phys. Chem., 12, 125, 1893. f Ibid., 234, 1893. ELECTROCHEMISTRY. 393 we have A^, for BaSO4, = II 5 + 128 122 = 121. This value could also be found from the table illus- trating Kohlrausch's law of the independent migration of the ions, of which it is an application, or from that on p. 381. It must be borne in mind here that this method is only applicable when we can safely assume A v for a saturated solution to be the same as A^. And further, it is limited to those substances which are soluble enough to give a value of K which differs appreciably from .that for the water used. Silver iodide solutions in this respect seems to be about on or just beyond this limit. 96. The dielectric constant and dissociating power. Other solvents than water. According to J. J. Thorn sen and Nernst, there is a relation between the dielectric constant of the solvent and its power of dissociating dissolved substances. The dielectric constant is deter- mined by placing the substance between two plates between which there is a certain potential difference, and measuring the loss of potential. Air is used first between the plates, and then the substance to be de- termined. The factor which we measure is the specific inductive capacity of the liquid for electricity, referred to air as a standard. The dielectric constants of a number of solvents are given in the following table. 394 ELEMENTS OF PHYSICAL CHEMISTRY. Benzene 2 . 24 Ether 4.25 Amyl alcohol 16.05 Ethyl alcohol, 99.8% 25 .8 Water 79 . 6 As will be observed, the constant for water is higher than for any of the other solvents, and its dissociating power is also the greatest. The latest theory as to the relation between the dielec- tric constant and dissociating power is by Briihl.* The dielectric, i.e., the solvent, acts just as the glass in a Leyden jar and prevents the ions from neutralizing their charges. The greater the dielectric constant, then, the greater the amount of free ions which may exist in the solution. This connection between the dielectric constant and dissociation, as found by electrical and other means, may be considered as very strongly confirmatory of the theory of dissociation into charged components or ions. The electrical behavior of substances in many non- aqueous solvents is very different from that in water, and it is difficult from our present knowledge to see how they can be governed by the same laws. For examples in some cases the equivalent conductivity does not attain a maximum constant value, but after attaining a maxi- mum decreases with further dilution; and the degree of ionization determined electrically does not in all cases agree with that determined by the boiling-point, for * Zeit. f. phys. Chem., 30, i, 1899. ELECTROCHEMISTRY. 395 instance. Further than this, the conductivity cannot in all cases be said to be the sum of two independent values, i.e., of those of the ions. We cannot but infer, then, from our present knowl- edge that the process of solution is quite different for these than for water. Neither the Ostwald nor any other form of a dilution law seems to hold in these cases. In other words, the electrical behavior of these is in- explicable as yet, and until the processes taking place in such systems is better understood little of our data can be generalized. There is one exception to all this, however. Franklin and Kraus (Am. Chem. J., 23, 277, 1900) have found that dilute binary electrolytes in liquefied ammonia-gas give an approximately constant value for the Ostwald dilution law, at least a value which very much more nearly constant than that ob- tained in a water solution. For concentrated solutions in the same solvent, see Franklin and Kraus (J. Am. Chem. Soc., 27, 191, 1905). The behavior of metallic sodium in liquefied am- monia (Cady, J. Phys. Chem., i, 707, 1897, and Franklin and Kraus, 1. c., p. 306) is very peculiar, for the con- duction is apparently more metallic than electrolytic. No products of decomposition are observed and there is no polarization, while the conductivity is exceedingly high. Among the tables at the end of this book will be found 39 6 ELEMENTS OF PHYSICAL CHEMISTRY. one giving the conductivities of substances in various non-aqueous solvents, from which an idea, at any rate, may be obtained as to their relations toward the electric current. 97. The conductivity of mixtures of substances having an ion in common. The conductivity of a solution con- taining two substances with an ion in common can be readily shown to be given by the formula K = where ^i and V2 are the volumes of the two simple solu- tions which are mixed, i.e., Vi+v 2 is the total volume of solution containing n\+n 2 equivalents of the two substances. a\ and a 2 are the degrees of ionization of the substances in the mixture, A^\ and A^ 2 are the equivalent conductivities of the two substances in a simple solution, and p is the ratio of the volume of the mixture to the sum of the constituent volumes. For the freezing-point of such a mixture, in an analo- gous way, we have A =[MiNi(i +ai(mi - 1)) + ^2^2(1 +a 2 (m 2 - 1))], where m is the number of moles of ions produced from one original mole, M i and M 2 are the molecular depres- ELEC TROCHEMIS TRY. 397 sions (p. 183) in i liter of the simple solutions and N\ and N2 are the number of moles per liter of each. For substances for which the Ostwald dilution law holds there is, naturally, no difficulty in determining the terms a\ and #2 (p- 283) ; and for strong electrolytes it is also possible, although the process is not always so simple. For illustrations of the use of these formulas and the calculation of the values a\ and a 2 the reader must be referred elsewhere,* for it would lead us too far to consider them in detail here. It will be observed, how- ever, that it is assumed in the above formula that A^\ and A M 2 are not altered by the mixing, a supposition which probably holds only for solutions weaker than 0.5 normal. Noyes and Kohr (J. Am. Chem. Soc., 24, 1145-1146, 1902) give an example of the calculation of the degree of ionization of such substances when mixed, which is based upon another principle and somewhat simpler than the other methods. The principle upon which this de- termination is based is as follows: The dissociation of a substance in the presence of another with an ion in common is that which the conductivity measurements show it to have when it is alone present at a concentra- * See MacGregor (Trans. Nova Scotia Inst. of Science, 9, 101, 1896); Barnes (ibid., 10, 124, 1900), and Archibald (ibid., 9, 291, 1898; also Trans. R. S. of Can. (2), 3, 69, 1897). 39 8 ELEMENTS OF PHYSICAL CHEMISTRY. tion of its ions equal to the square-root of the product of the concentration of its ions in the solution of the mixed salts. This principle was demonstrated by solubility experiments (Noyes and Abbott, Zeit. f. phys. Chem., 16, 138, 1895). It was found there that for those substances which do not follow the Ostwald dilution law the concentration of the un-ionized part of a salt has the same value, when the product of the concentrations of its ions has the same value, whatever may be the two separate factors of that product. In a mixture of potassium chloride and potassium hydroxide, for example, where the former is 0.0033 m la r and the latter 0.35, we first assume the ionization to be 85% for each (as an approximation). The product CK- Xc cl /. then has the approximate value, (0.35 + 0.0033) X 0.85X0.0033X0.85. The square root of this quantity (or 0.030) is then 85% of the concen- tration (0.0355) of pure potassium chloride at which the degree of dissociation is the same as it is in the solution of the mixed potassium salts. The degree of dissociation is then to be ascertained from a plot of the molar con- ductivity values. For potassium chloride in this way we find a to be 0.90; and by the same procedure a for the KOH is found to be 84%. In mixtures of substances without an ion in common, the same general formula for the conductivity and freez- ing-point (p. 396) is also to be employed. Here, how- ELECTROCHEMISTRY. 399 ever, there may be a reaction between the substances which would give rise to a new substance. In such a case, this, also, must be considered, and treated just as the others. All these things also hold for mixtures of more than two substances, although in such cases the calcula- tion is considerably more complicated (see MacGregor, Trans. Roy. Soc. Can. (2), 2, 65, 1896). C. ELECTROMOTIVE FORCE. 98. Determination of the electromotive force. In all the cases we are to consider in this section it is neces- sary that the electromotive force be measured by aid of the compensation method, i.e., that as little actual current be taken from the cell as is possible. As the basis of such methods is a known and con- stant electromotive force, the standard cell is an ex- ceedingly important part of the measurement. The standard cells in general use in this work are of two kinds, one of which has a fixed constant value, while the other may be made up in such a way as to give one of several values. The former one is the Clark cell, Fig. 17, which con- sists of amalgamated Zn in a paste of ZnSC>4 and Hg in a paste of Hg2SO4. This cell gives an E.M.F. equal to 1.4328 o.oong(/ 15) 0.000007 (/ i 5) 2 . 4oo ELEMENTS OF PHYSICAL CHEMISTRY. Another standard cell in common use is the cadmium element (Weston). This is made up with electrodes FIG. 17. of mercury and cadmium amalgam according to the following scheme, the Hg being the positive pole: Hg-Hg 2 SO 4 + CdSO 4 paste sat. CdSC>4 + crystals CdSC>4 Cd amalgam. paste The electromotive force of this cell is 1.0186 + 0.00004(20 /) volts. The Helmholtz element, Fig. 18, consists of amalga- mated Zn in ZnCb solution (sp. gr. =1.409 at 15), ELECT ROCHEM1S TRY. 401 calomel, and Hg. This gives an E.M.F. approximately equal to i volt, but its exact value depends upon the sp. gr. of the ZnCl 2 solution so that it must always be measured by comparison with a Weston or Clark ele- Calomel FIG. 18. ment. Its great advantage is its very small tempera- ture coefficient, which at low temperatures is negligible.* 99. Types of cells. Cells are either reversible or non-reversible, according as the process may be reversed or not. A reversible cell can always be put in its orig- inal condition again by sending a current through it in the opposite direction, while with a non- reversible cell this is not possible. A typical reversible cell is the Daniell, where we have Cu in CuSC>4 and Zn in ZnSO4. * For further details of these measurements, cells, etc., see Ostwald's Physico-chemical Measurements, Macmillan. 402 ELEMENTS OP PHYSICAL CHEMISTRY. The Zn dissolves and Cu is precipitated during the action, but by passing a current through it from Cu to Zn the Zn is precipitated, while Cu is dissolved, until the original state is again reached. A non-reversible cell, on the other hand, is any one from which a gas is given off, for the passage of the current cannot cause the gas to recombine. Here we shall consider principally the reversible type. 100. The chemical or thermodynamical theory of the cell. The simplest theory of the action of a reversible cell was advanced by Helmholtz and Thomson. Accord- ing to this, the chemical energy of the process is trans- formed completely into electrical energy. This theory in the form in which it was advanced, however, has proven to be incorrect, for it holds only for the Daniell cell. The explanation of the variation in the case of other cells was offered by Gibbs and Helmholtz independently, who proved it to be due to the temperature. Assume a reversible primary cell in which the amount of heat q is liberate'd or absorbed when one equivalent of the ions has been carried from one side to the other. Imagine this cell in a temperature-bath so that the cell remains at constant temperature, i.e., if heat is ab- sorbed it is replaced, or if heat is liberated it is removed thus preventing q from causing a change in temperature. If TT, the E.M.F. of this cell, is just compensated by TT, the process will be in equilibrium. For two ener- ELECTROCHEMISTRY. 40 3 gies in equilbirum, however, we have found the general equation (p. 6) where c and i are the factors of one kind of energy and c\ and i\ those of the other kind. In our case the two kinds of energy are electrical and thermal, and we have or which is the change in the E.M.F. of a cell caused by the absorption or liberation of q cals. during the reac- tion. If now we pass 96,540 coulombs of electricity through the cell, during which to keep the temperature constant it is necessary to supply q cals., the electrical energy supplied must be equal to the chemical energy plus the electrical energy due to the heat q. We have, then, when all terms are expressed in the same kind of units, 404 ELEMENTS OF PHYSICAL CHEMISTRY. or, since E e = i.e., the actual E.M.F., TT, is only equal to that calcu- lated from the chemical energy when dn=o, i.e., when the E.M.F. does not change with the temperature. Other- wise TT is smaller or greater than - according as T-^ Q dl is negative or positive. The Daniell cell has such a small temperature coeffi- cient that the electrical energy is nearly equal to the chemical energy. For other cells it is possible to deter- mine the temperature coefficient, and thus from the chemical energy to calculate the E.M.F. of the cell. The results by experiment and calculation agree in most cases within the experimental error, showing that this conception of the process is correct. We see, then, that the amount of heat generated by the chemical process is no criterion of the amount of electrical energy involved, for heat which is absorbed from the environment may also be transformed into electrical energy, and in other cases less heat may be transformed into electricity than is given up by the chemical process. An example of the application of this formula is given by the Grove gas cell. Here 71 = 1.062 and = 34200, 34200 dn dn hence 1.062= - - + T~^, i.e., T-^= -0.4187, m 23110 dT dT ELECTROCHEMISTRY. 45 place of 0.416, as found by experiment. The value 23110 here is obtained from 96540X0.2394X^ = 23110/1 cals., and is the work in calories necessary for the deposition of i gram equivalent of any substance a-t it volts. 101. The osmotic theory of a cell. Considering a cell from the standpoint of our conclusions respecting the nature of electrolytes in solution, it is possible to see more clearly into the cause of the rise of a difference in potential between two solutions or a metal and a solution. Assume that we have two solutions in contact, and that they contain the same monovalent ions in different concentrations. The difference of potential existing on their boundary can be calculated by aid of the following process of reasoning: If U a and U c are the mobilities of the respective ions, then, by the passage of 96,540 cou- lombs of electricity, the following changes take place: If the current enters on the concentrated side and passes through both solutions, - ^~ gram equivalents (moles) U c + U a of positive ions will go from the concentrated to the dilute solution, and during the same time Jr " gram equiva- UC~T U a lents (moles) of negative ions will go from the dilute to the concentrated side. Let p be the osmotic pressure of the positive and negative ions in the concentrated solution, 406 ELEMENTS OP PHYSICAL CHEMISTRY. and p' that of those in the dilute solution. The maxi- mum work to be done by the process, then, will be (P- 22 5)> and this must be equal to QX for the process going in this way, i.e., to the electrical work done at the surface of contact of the two solutions. Since R is in calories this value will also be expressed in calories. To trans- form it into electrical units it is only necessary to divide through by 23,110. This relation was derived and tested experimentally by Nernst (Zeit. f. phys. Chem., 4, 1295 1888), who found it to hold true within the experimental error. If U e is greater than U a we could expect, then, that a weaker solution of an acid or base, when superim- posed on a stronger one, would show the polarity of the faster moving ion. In other words in diffusing through the solution the faster ion would impart its polarity to the weaker solution. With acids, indeed, we do observe that the dilute solution is positive, while with bases it is negative ; and this is in accord with our previous observa- tions that H* and OH 7 are the ions possessing the greatest mobility or velocity. This process does not result in a complete separation of the ions nor even in more than ELECTROCHEMISTRY. 4 7 a very slight one, for it must be remembered that the two kinds of electricity attract one another, the conse- quence of which is that the speed of the faster moving ions is decreased, while that of the slower is increased. This same method of reasoning will also enable us to understand the action of the arrangement Concentrated amal- I Water solution of a I Dilute amalgam of the gam of the metal M. I salt of the metal M . \ metal M. In this case the passage of 96,540 coulombs will cause i gram equivalent of the metal M to go from the con- centrated side to the dilute, and, if there are n equiva- lents to the mole, the maximum work per equivalent will be where c\ and 2 are the concentrations of the metal M in the amalgams. Again, here, if this expression is so transformed as to give electrical units instead of calories, by division by 23,110 (i.e., 96,540X0.2394), it will give the value of TT the electromotive force of the cell operating in this way. And the same is true for cells with electrodes of the same soluble metal and two concentrations of a solution of a salt or salts of that metal. Consider, for example, the cell 408 ELEMENTS OF PHYSICAL CHEMISTRY. Cu dilute CuS0 4 concentrated CuSO 4 Cu. By passing a current through this cell in the direction of the arrow the following changes will take place: 1. For each 96,540 coulombs of .electricity i gram equivalent of copper will dissolve from the electrode in the dilute solution, i.e., will be transformed from the metallic to the ionic state. 2. At the boundary of the two solutions the process described above will take place; and 3. One gram equivalent of Cu" ions will be deposited from the concentrated solution upon the electrode. As the result of processes (i) and (3) i gram equiva- lent of Cu" ions will go from the concentrated to the dilute solution. The maximum work of this process, then, for each 96,540 coulombs will be RT where n is the valence of the metal and p\ and p2 are the osmotic pressures of the Cu" ions in the two solutions. By the second process (contrast with direction of cur- rent in the case above) we have as the work n ELECTROCHEMIS TR Y 409 This second value is usually so small, when compared to the other, that it may generally be neglected, and the two differences of potential between the electrodes and the solutions measured separately. Neglecting the difference of potential between the liquids, then, we have RT p, n= log, , or, including the potential difference between the solutions, RT 2Un i.e., the sum of the two, where R must be given in electri- cal units, if T. is to be expressed in volts. TT> rri , By separating the amount of work log, into two n p2 portions, so that each will give the maximum work at an electrode, we can write RT. P RT where P is a constant for any one metal at the same temperature, in the same solvent, and is called the electro- lytic solution pressure. The portions containing the ex- p pressions log, will then give the difference of potential 410 ELEMENTS OF PHYSICAL CHEMISTRY. of copper in dilute copper sulphate and copper in con- centrated copper sulphate. The electrolytic solution pressure assumed to exist in this way is analogous to the ordinary solution pressure which we have already considered, except in this case it is for the solution of a metal or any other element, i.e., is the pressure with which element tends to attain the ionic state. When we consider a stick of metal in a solution of one of its salts, we have, if P is the electrolytic solution pressure and p is the osmotic pressure of the metal ions in the solution, three possibilities: 1. P=p. 2. P>p. 3 . Pp. This is the case with Zn. Ions of Zn" will go from the electrode into the solution, taking with them positive electricity. This will take place only to a very slight degree, for it would increase the positive charge of the solution. These positive ions around the electrode will then be attracted back to the plate, which is negatively charged owing to the loss of the positive ions. We have then, finally, the metal negative against the solu- ELECTROCHEMISTRY. 411 tion, and the electrode surrounded by what is known as a Helmholtz double layer, in this case a layer of positive ions on a negative plate. If positive electricity is given now to the metal, the double layer is broken up and more Zn will go into solution in the ionic form, but as soon as the current is stopped the double layer will again be formed. 3. P p. The noble metals, on the contrary, are positive against their solutions, although in some few cases it is possible to get a solution in which P>p. In general, though, for the noble metals P p. Here, although 414 ELEMENTS OF PHYSICAL CHEMISTRY. P>p, the electrode is positive against the solution, for the negative ions formed from the electrode leave posi- tive electricity behind. In 'general, the electrolytic solution pressure depends upon the temperature, the nature of the solvent, and the concentration of the substance in the electrode (see p. 422). 102. Mathematical expression of the osmotic theory of the cell. It is possible from the above conception of solution pressure to derive the formula giving the value of the E.M.F. of a single electrode of any metal in a solu- tion of one of its salts. This will depend upon the osmotic pressure of the metal ions working in the one direction and upon the electrolytic solution tension working in the other. Since the ions go from one pressure to the other, a certain amount of work is done and this amount must always be the same no matter in what form it appears, i.e., whether electrical or osmotic. When i mole of ions is transferred from the pressure P to the pressure p, the osmotic work done is equal to { J P from which, by integration, we obtain ELECTROCHEMIS TRY. 4*5 The corresponding electrical work, however, is where TT is the difference of potential between the metal and solution and SQ is the amount of electricity carried by i gram equivalent of ions. We have then or ^ P *" log -7' from which we see that if P=p, TT=O. for i mole of a monovalent ion (i.e., for i equiva- lent) is equal to 96,540 coulombs. We must, however, express both sides of the equation in electrical units. Since R=2 cals., the right side of the equation must be divided by 0.2394 in order to obtain our result in electri- cal units. We have then 2 P 96,540 XTr(volts) -T log , 0.4343X0.2394 3 p> or p 7r=o.o575 log volts at 17 C. P 4 1( $ ELEMENTS OF PHYSICAL CHEMISTRY. For a divalent element we must multiply s by 2; hence if n is the valence of the metal, we have p \ WW>^W^ ff ^ _ J (45) ~^~ 1[ %p or for negative ions 0.0002^, P 0.0002 p TT = jf log = T log p volts. If we consider a cell composed of two metals, each present in a solution of one of its salts, the difference of potential may arise (i) at the place of contact of the two metals, (2) at the place of contact of the two solutions, and (3) and (4) at the points of contact of the, two electrodes with their solutions. The first can be entirely neglected provided the temperature remains constant. The second, as was mentioned above, is always small, so that we shall have to consider here only (3) and (4) , assuming, when necessary, that the value has been determined as on page 406. At 17 the E.M.F. of the cell will be, then, (46) ^-* 2 ) the minus sign being used because ions are formed on the one side, and consequently must disappear upon ELECTROCHEMISTRY 4* 7 the other, for there must always be an equal amount of positive and negative ions in the solution. 103. The measurement of the potential difference be- tween a metal and a solution. The method in use for the determination of differences of potential is based upon the use of a normal electrode, i.e., an electrode in a solution of one of its salts, of which the potential dif- ference is known. Making up a cell by combining this electrode with the one to be measured, it is possible, from observations of the resulting electromotive force, to find the value of the unknown potential difference. The original measurement of the potential difference of the normal electrode was made by forming a cell with this and another electrode, having a potential of zero against its solution. The electromotive force of this combination, then, is identical with the potential dif- ference to be found. Such an electrode with zero poten- tial is formed when mercury flows into a solution of an electrolyte (KC1), for any differences of potential existing between the mercury and solution must eventu- ally be eliminated by the continually dropping mercury. The electrodes in general use are made up according to one of the two following schemes: HgCl in molar KCl/Hg, or HgCl in o.i molar KCl/Hg; 418 ELEMENTS OF PHYSICAL CHEMISTRY. the former giving a difference of potential of 0.56 volt, the latter of 0.0613, both at 18. The sign in each case here refers to the solution, i.e., the solution is negative against the metal. Fig. 21 shows the form of the cell. The syphon tube is so arranged that it is always kept full of the KC1 FIG. 21. solution. This dips into the liquid of the other electrode, provided they form no precipitate. In case they do, a molar solution of KNO 3 is used, into which the siphons from the two electrodes dip. The advantage of using KNO 3 as a connector is that the mobilities of the K* and NCV ions are nearly the same; hence the passage of the current produces no change in the concentration at the two sides and consequently gives rise to no difference of potential. ELECTROCHEMISTRY. 4*9 An example of the use of this electrode in measuring the difference of potential of any other is given by the following: For a combination in the form Zn - molar ZnSO 4 - molar KC1 HgCl - Hg, i.e., for Zn in molar ZnSOi, against the normal electrode, we find 7r = i.o8 volts, the current going from Hg to Zn through the wire. The total E.M.F. of the cell, then, is given by the equation RT PI RT P 2 2 Q & e pi ^ e p 2 * where PI and pi refer to the Zn and P 2 and p 2 to the Hg, for that is the only arrangement which would make n positive. We have then RT Pi 2o e pi ~RT P log^ = (1.080.56) =0.52 volt; i.e., Zn is negative against a molar solution of its sulphate and gives an E.M.F. of 0.52 volt. The sign always refers to the E.M.F. of the electrolyte against the elec- 420 ELEMENTS OF PHYSICAL CHEMISTRY. trade. Thus, i means that the solution is negative and the metal positive, and vice versa. In the case of Cu-wCuSO 4 -wKCl HgCl-Hg, we find 7r= 0.025 volt, the current in the wire going from Cu to Hg. We have here, since the current goes from Hg to Cu in the solution, Pi o or RT P l 0.025 0.56--- log,-, RT, PI log, = -0.025 -0.56 = -0.585; i.e., the electrode of Cu is positive, with a difference of potential against the electrolyte equal to 0.585 volt. The process which takes place is shown more clearly by the diagram Cu- molar CuSO 4 HgCl in molar KCl-Hg; s- 0.56 i I ^_> , 1 0.025 ELECTROCHEMISTRY. 42! hence the copper must be more positive than the Hg by 0.585 volt., 104. The heat of ionization. Knowing the difference of potential existing between a metal and a solution, and its rate of change with a variation in the temperature, it is possible from the formula on page 403 to find the heat of ionization of the metal. We found then that or dn The term E c , here, is the heat produced when i gram equivalent of the metal goes from the metallic to the ionic state, and its value is given by the formula But (page 405), SOT: =96540X0.2394X7: volts =231107: cals.; hence the heat of ionization E c , for i gram equivalent can be found from the relation = n- T- 23,110 cals. 422 ELEMENTS OF PHYSICAL CHEMISTRY. For copper in copper acetate (molar) at 17, 7r=o.6, . dn .. dn . and -7^=0.000774, while for copper in copper sul- phate is 0.000757, i.e., an average value of of 0.000766 volts, hence E c , the heat of ionization of copper, is 8736 cals. per gram equivalent, or 17,472 cals. per mole. It was in this way that the value for H was determined for use in the table given on page 220. 105. Concentration-cells. a. In which the electrodes have different concentrations. If the electrodes are made of two amalgams of the same metal in salts of this metal the E.M.F. is given by the formula 0.0002^ PI 0.0002^. PZ . - -Flog- -T log- -volts. n 5 n 8 2 If the two solutions have the same concentration of metal ions, then p\ =p2> and we have 0.0002 PI TT = - T log -77- volts, n r^2 when PI and P 2 are the electrolytic solution pressures of the two electrodes with respect to the dissolved metal. Amalgams when dilute may be considered as solutions, in which the mercury acts as the solvent; consequently the osmotic pressure of the metal in the amalgam will ELECTROCHEMISTRY. 4 2 3 be proportional to the electrolytic solution pressure of the electrode. Since in all solutions the osmotic pressure is proportional to the concentration, we may substitute for the electrolytic solution pressures the proportional terms, the concentrations of the metal in the two elec- trodes. We obtain then 0.0002 c l TT = T log volts. n 6 c 2 This formula was proven experimentally by G. Meyer (Zeit. f. phys. Chem, 7, 477, 1891), some of whose results for zinc amalgams in a ZnSC>4 solution are given in the table below. TABLE XVII. T Cl C2 T (obs.) *(calc.) n.6 0.003366 O.OOOII3O5 0.0419 v. 0.0416 v. 18 0.003366 O.OOOII3O5 0.0433 v - 0.0425 v. I2. 4 0.002280 o . 0000608 0.0474 v. 0.0445 v - 60 0.002280 0.0000608 0.0520 v. 0.0519 v. The results obtained by Meyer also proved the formula to hold copper for amalgams in CuSOi, etc., solutions. We have assumed the metals to dissolve in mercury as monotomic molecules in this case, and the assumption is upheld by our results, not only by this method but also by others. If Zn had been present in the form of diatomic mole- cules in the mercury, our equation would have had a different value. For the movement of the same amount 424 ELEMENTS OF PHYSICAL CHEMISTRY. of ions as before, we should have had, then, the osmotic work equal to 1 RT Cl log , 2 0.4343 & C 2 and since the electrical work would have been the same as before, i.e., 2X96540?:, we should have had I 0.0002 Ci 71 = T log volts ; b i.e., the E.M.F. would have been one-half what we have found it to be. The electromotive force is found, then, to be de- pendent only upon the concentration of the metal in the amalgam and its valence, and not upon the nature of the metal itself. The mercury has no effect so long as the metal dissolved in it gives the larger E.M.F. Another example of the electrodes having different electrolytic solution pressures, owing to differences of concentration, is given by cells of the type of the Grove gas-battery in an altered form. The electrodes are of platinized platinum, in which the gas is absorbed under different pressures, and are placed partly in the liquid ELECTROCHEMISTRY. 425 and partly in gas of corresponding (partial) pressure. Such an electrode is to be considered as a perfectly revers- ible gas electrode,* i.e., one from which ions of the material absorbed as a gas are given up, for the metal acts simply as a conductor, as has been shown experi- mentally by the use of different metals, the same result being always obtained. In this way reversible gaseous electrodes of all kinds can be made. Oxygen as an electrode, however, gives off ions of OH', since those of O" are not known to exist, and absorbs O when the OH' ions give up their charges to it. If we have two electrodes of H, under different pres- sures, in contact with a liquid containing H* ions, we shall obtain a certain E.M.F. This may be calculated in two ways, as we did in the case of amalgams. In the second way, however, the process is slightly different, since one mole of H gas forms two monovalent ions (p. 146). The osmotic Work is equal to RT PI 0.4343 Cg P 2 ' as before. The electrical work, however, which corre- sponds to this is 2o7Tj for H 2 =2H', hence RT 2^0(0.4343) I g ; * The electrolytic solution pressure here of the gas electrode is proportional to the nth root of the gaseous pressure, where n is the num- ber of combining weights in one formula weight of gas. 426 ELEMENTS OF PHYSICAL CHEMISTRY. i.e., we have 2 in the denominator notwithstanding that the gas is monovalent. b. Different ionic concentrations. In this type of cell we assume the electrodes to be of the same metal, but dipping into solutions which possess different ionic concentrations of the metal. An example is given by the arrangement Ag(AgNO 3 cone.) -(AgNO 3 dilute) Ag. From the general equation we have RT , P l RT, P 2 = log, lg,:r- e e Or, since Pi=P 2 , RT where p\ is the osmotic pressure of the Ag' ions on the concentrated side and p2 that on the dilute side. From this we see that the E.M.F. depends only upon the ratio of the concentrations, and not at all upon the actual concentrations, all of which was found to hold experimentally by Nernst (1. c. p. 406). Since the osmotic pressure is proportional to the ^^w-, OF THE ^ \ f UNIVERSITY | ELECTROCHEMISTR /i 42 7 N^^R!j\X concentration, we may substitute the latter "Tor the former; we obtain RT Cl n= loer . HQ *' C 2 The concentration of Ag' ions in a o.oi molar solution of AgNO 3 is 8.71 times as great as that in a solution which is o.ooi molar (not 10 times), hence the E.M.F. (at 1 8) of a cell made up of Ag in these two concen- trations of AgNO 3 should be TT= 0.0002X29 1 log 8.71 =0.054 volt, while 0.055 v to was found by experiment. Since at 17 C. we have -575 , c \ u 7T= -log VoltS, n c 2 a concentration ratio of the ions equal to 10, would give TT =0.0575 volt for a monovalent ion, or 0.02875 volt for a divalent one. This class includes all cells made up of solutions containing different concentrations of metal ions. In case the contact of the two liquids causes the forma- tion of a precipitate, another solution is used between, the two being connected by small siphons. 1 06. Dissociation by aid of the electromotive force. Since for concentration-cells we have the formula it = - log volts, W X 0.4343 C 2 428 ELEMENTS OF PHYSICAL CHEMISTRY. or for monovalent metals we can find the ionic concentration of the metal on the one side provided that on the other is known. An example of this is the cell as given by Morgan: Ag- ^ AgN0 3 -KN0 3 - ^KAgCN.-Ag, which gives an E.M.F. of 0.542 volt, Ag in AgNOs being the positive pole. If, for simplicity, AgNOs o.i molar is considered as completely dissociated, the concentration of Ag' ions will be o.i molar, and, since 7 = 273 + 17=290, we have i.e., in the m/2o KAgCN 2 solution the Ag' ions are present to a concentration of 3.iXio~ n molar. This method was devised by Ostwald, and is one of the most delicate known, for the more dilute the one solution is, i.e., the smaller the concentration of its metal ions, the greater is the E.M.F. of the cell, and the greater the accuracy of the result. ELECTROCHEMISTRY. 4 2 9 Another typical cell of this kind is Ag - ^ AgN0 3 - KN0 3 - ^ KC1 AgCl - Ag, which was used by Goodwin to determine the solubility of AgCl. In a saturated water solution of AgCl we may assume without error that the dissociation is complete. The solubility product will then be Since in a pure water solution of AgCl, Ag'=Cl', the square root of s, s being determined under any conditions at constant temperature, will give directly the concentration of Ag* (or Cl') in the solution. And, if complete ionization may be assumed, this is equal to the solubility of AgCl under these conditions. For example, if we find the E.M.F. of the cell Ag-AgCl KCl-AgN0 3 -Ag, where c c y on the left and c^. on the right are known, we can calculate c^. on the left, and consequently 5 and the solubility of AgCl. Goodwin (Zeit. f. phys. Chem. 13, 577, 1894) found, 43 ELEMENTS OF PHYSICAL CHEMISTRY. when the concentration of AgNOs and KC1 was o.i molar, an average E.M.F. of 0.45 volt at 25. By con- ductivity measurements at 25 we find a (forAgNO 3 w/io)=82% and a (for KClw/io) =85%, 07 . J.gJ-1 VX -j"*'/ J-V-y ->^ J (j CV1XV4. V*. ^iV^A XX.V^JL"*'/ .J-X/ -< so that l = 0.450 ^2 0.0002 X 298* or i.e., Ag* ions are present in a o.i molar KC1 solution of AgCl to a concentration of i.94Xio~ 9 moles per liter. The product s is then found to be equal to 1.94 X io~ 9 Xo.o85 =s = 1.64 X io~ 10 and i.e., AgCl in a saturated water solution at 25 is present to a concentration of 1,28 Xio~ 5 moles per liter. A cell of this sort may be arranged with bromides, iodides, etc., in place of the chlorides, which was also done by Goodwin, and the solubility of AgBr and Agl determined. Another illlustration of this method of determining dissociation is furnished by Ostwald's arrangement for ELECTROCHEMISTRY. 43 ' determining the dissociation constant for water. The scheme is a gas-cell in the form H-Acid ........ Base-H, when the two H electrodes have the same solution pres- sure. The E.M.F. of such a cell, deducting the E.M.F. caused by the contact of the two solutions, is, for molar solutions at 17 C., TT =0.0575 log A Here, contrary to the case for electrodes of different gaseous concentration, the 2 does not appear in the denominator (page 425), for that factor simply refers to the process taking place between the electrode and the gas. In this case the electrodes remain constant and the same, and only the osmotic pressures of the H' ions is to be considered. By experiment 7r=o.8i volt; hence, since 1=0.8 (i.e., the normal acid is 80% dissociated), or c 2 =o.8Xio~ 14 . Since this is the concentration (in terms molar) of the H* ions in the base, and the concentration of OH' ions 43 2 ELEMENTS OF PHYSICAL CHEMISTRY. in contact with them is 0.8, the ionic product for H 2 is o.8Xo.8Xio~ 14 . Since H 2 O dissociates into H* and OH' ions, the con- centration of H' and OH' ions in pure water is the same, and is equal to V (o.8) 2 Xio- 14 =0.8 X io~ 7 =H* and OH' ions in H 2 O. From the conductivity also, at this temperature, Kohlrausch found o.8Xio~ 7 . It would be possible in this case to use oxygen elec- trodes, by which the same final result would be obtained, but platinum black absorbs very little oxygen, so that the results are not as certain as with hydrogen. For the measurement of the concentration of ions of mercury the normal electrode serves as the basis. The concentration of Hg 2 ions in the normal electrode has been found to be 3.5Xio~ 18 moles per liter (Ley and Heimbucher, Zeit. f. Electrochem., 10, 303, 1904). Since Hga - =120, in the presence of metallic mercury, accord- ing to Abel (Zeit. f. anorg. Chem., 26, 377, 1901), it is possible to find the concentration of Hg" in the normal ELECTROCHEMISTR Y. 433 electrode; it is equal to 2.9Xio~ 20 moles per liter. By measurements of the electromotive force of a combination of mercury in a solution containing Hg" or Hgs ions, with the normal electrode we can thus determine the con- centration of these ions. For further details as to this, see Sherrill and Skowrouski, J. Am. Chem. Soc., 27, 30, 1905. 107. Electrolytic solution pressure. It is possible from the potential difference between a metal and its salt solution to calculate the solution pressure in atmos- pheres. The equation is or at if If we use normal solutions, p is equal to 22. atmospheres if completely dissociated. The pressure of the ions, in case the salt is not completely dissociated, is then easily found (i.e., a X 22. 4}^- J atmospheres). Since TT can be determined, the value for P can be found for all metals. In the table on page 434 the values in atmospheres, as determined by Neumann, are given: These values it must be remembered are merely sym- bolical, for the gas laws may not be applied to such an 434 ELEMENTS OP PHYSICAL CHEMISTRY. extent. The relation between these numbers, however, are the experimentally determined relative effects of the metals, as a glance at the formula will show. TABLE XVIII. SOLUTION PRESSURES OF METALS. Zinc 9 . 0X io 18 atmospheres Cadmium 2 . 7X10* Thallium 7.7Xio 2 Iron 1.2X10* Cobalt 1.9X10 Nickel 1.3X10 Lead i.iX io~ 3 Hydrogen 9.9X10"* Copper 4.8Xio~ 20 Mercury i.iX io~ ic Silver 2.3Xio~ 17 Palladium i.Xio- 30 The influence of the dilution on the difference of potential depends upon the magnitude of the electrolytic solution pressure. If the latter is great, dilute solutions are necessary for maximum action; if small, more con- centrated solutions are to be preferred. 108. Cells with inert electrodes. Elements of this type 'have electrodes in solutions which do not contain the ions of the electrodes. Such cells are also known as oxidation and reduction cells. An example is given by the arrangement plat.Pt-FeC! 3 sol SnQ 2 sol. -plat.Pt. ELECTROCHEMISTR Y. 435 The Sn" ions go over into Sn :: ions, taking up the elec- tricity set free by the transformation of Fe"* ions into those of Fe". The Fe side, then, is positive and the Sn side negative. The current in these cells is due to the oxidation on the one side and the reduction on the other. An oxidizing substance (i.e., one which is reduced) is any substance from which negative ions are formed or upon which positive ions give up their charges. A reducing substance is one in which negative ions are given up or positive ions formed. 109. Processes taking place in the cells in common use. All cells give an E.M.F. which is equal to the difference of the differences of potential at the two single electrodes; but in many cases the action is complicated by other processes. In the following the effect of depolarizers is given, as well as the change in the cell during action. Clark cell. This cell, as already mentioned, is used as a standard of E.M.F. The scheme is Hg-Hg 2 SO 4 -ZnSO 4 -Zn. The Hg2SC>4 is not quite insoluble, therefore a small concentration of Hg ions does exist in the solution. The Zn" ions with their high solution pressure are driven from the electrode into the ZnSC>4 solution, and force before them to the electrode the ions of H', since the 436 ELEMENTS OF PHYSICAL CHEMISTRY. solution can contain only the same concentration of posi- tive as negative ions. The H* ions give up their charges to the electrode, which becomes positive, the Zn being negative from loss of positive ions. The current goes, then, externally through the wire from the Hg to the Zn, and internally from 1 the Zn to the Hg. If the resistance placed against the cell is great enough, only an infini- tesimal amount of Hg 2 ions will be driven from the solu- tion, and the cell remains constant. If the current is passed through a small resistance, however, all the Hg 2 ' ions are soon precipitated upon the electrode and the E.M.F. is reduced, i.e., the cell is polarized. If allowed to stand, however, more Hg 2 SO 4 will -dissolve and the original E.M.F. is attained. The Leclanche cell consists of a solution of ammonium chloride, in which we have two electrodes, Zn and C+MnO2. The action of the MnO2 is to prevent polarization, the processes taking place without it and with it being as follows: In the cases without MnO2 the Zn with its high solu- tion pressure goes into solution, driving before it the other positive ions, i.e., those of NH 4 - These ions de- compose on losing their changes, forming NH 3 and H gases. The bubbles of H collect upon the electrode and are absorbed and so given off to the air, but as this process is slow, other ions are prevented from giving up their charges and consequently the E.M.F. decreases. It is to ELECTROCHEMISTRY. 437 get rid of this action of the H gas that the MnO 2 is used. In contact with water we have, then, to a small extent* a solution of MnO2, which dissociates according to thg scheme MnO 2 4- aHaO = Mn : : These tetravalent ions of Mn have the tendency to go into the divalent state by giving up two equivalents of electricity, i.e., to form the ions Mn". In consequence of this the Mn :: ions with those of NHi are driven to the electrode by the Zn" ions, and, since they give up two equivalents of electricity more readily than any other ion gives up its entire charge, the electricity is given up without any substance which might cause polarization. We have, then, MnCl 2 (i.e., Mn" ions) formed in the solution. The process continues so long as there is solid Zn or MnO 2 present. Bichromate cell. This cell is arranged according to the scheme The action of the two substances in solution forms bichromic acid (H 2 Cr 2 O7). This dissociates into 2H' + Cr 2 7 " ELEMENTS OF PHYSICAL CHEMISTRY. to a large extent, and to a smaller degree as follows : H 2 Cr 2 O 7 + sH 2 O = 2Cr : : + 1 2 OH'. These hexavalent Cr ions have the tendency to give up three equivalents of electricity and to go into the trivalent state (i.e., Cr"). Accordingly the Zn" ions which are forced from the electrode drive before them the Cr :: ions, which give up three of their equivalents and become Cr", remaining as such in the solution as ions in equilibrium with SO^' ions (i.e., as Cr 2 (SO4)s). Finally, then, we have a solution of Cr 2 (SO4)3 left in the jar. This change in the number of electrical equiva- lents by a change of valence always takes place more readily than the change from the ionic to the elemental state and is of great value as a means of preventing polarization. Accumulators. The action of the lead accumulator or storage-cell also depends upon a change of valence. Any reversible cell can be recharged after it is used up by the passage of a current through it in the direction opposite to that in which it goes of itself. The lead cell, however, is generally used for the purpose owing to its high E.M.F. Before charging it consists of two ,lead plates, one of which is coated with litharge (PbO) in a 20% solution of sulphuric acid. If the current is passed through these plates (the PbO being positive), ELECTROCHEMISTRY. 439 the PbO is transformed into PbO 2 , lead superoxide (or supersulphate), while spongy lead is deposited upon the other electrode. The flow of current is now stopped, and the cell is charged. The PbO 2 is soluble to a small degree and ionizes as follows : Pb0 2 + 2H 2 0=Pb :: + 4 OH'. These tetravalent Pb ions have the tendency to give up two of their electrical equivalents and to go into the divalent form. Since this is true, the Pb electrode must have the higher solution pressure, and the ions formed from it will drive those of Pb :: to the electrode, where they will lose two charges of electricity and become divalent. This will continue as long as PbO 2 is present, i.e., until it is all transformed into the divalent state, PbSO4. Other theories, by Liebenow and others, have also been advanced to explain this cell, but the reader must be referred elsewhere for them (see Dolezalek and von Ende, The Theory of the Lead Accumulator). Dolezalek has been able to calculate the value of such a cell from the concentration of acid and the vapor- pressure of the solution, and finds an excellent agree- ment between theory and experiment. This proves the process, according to him, to be a primary one such as the theory of Le Blanc and that of Liebenow would make it. 44 ELEMENTS OF PHYSICAL CHEMISTRY. D. ELECTROLYSIS AND POLARIZATION. no. Decomposition values. We must now consider the processes which take place when a current is passed through a solution from electrodes which are not affected by the liquid, i.e., inert ones, as platinum, gold, carbon, etc. The current causes the deposition of the positive and negative ions on the electrodes. If the primary current is disconnected at any time and the poles of the decomposition-cell connected with a galvanometer, we observe a current in the opposite direction which becomes weaker and weaker, until it finally disappears entirely. This is called the current of polarization, and its E.M.F. is the E.M.F. of polarization. This is caused by the tendency of the substance precipitated upon the elec- trode to go back into solution in the dissociated form. It has been found that a certain minimum of E.M.F. is necessary to cause the steady electrolysis of any solu- tion. If the E.M.F. used is smaller than this, the cur- rent goes through for an instant and then ceases; but at this point or above it the process goes steadily. This was found by Le Blanc, who determined the minimum values for a large number of liquids and solutions. The table below gives these values for molar solutions of the salts which separate metals. ELECTROCHEMISTR Y. 441 TABLE XIX. DECOMPOSITION VALUES FOR SALTS. ZnSO 4 = 2.35 v. Cd(N0 3 ) 2 =i. 9 8v. ZnBr 3 = i . 80 v. NiSO 4 = 2.09V. NiCl, =i.8sv. Pb(N0 3 ) 2 =i.52v. Ag(NO) 4 = 0.70V. CdSO 2 =2. 03v. CdCl 2 =i.88v. CoSO 4 = i . 94 v. CoCl, = i . 78 v. The values for Cd(NO 3 )2 and CdSO 4 show that those for the nitrates and sulphates of the same metal are nearly the same. The following tables contain the values for acids and bases. Here there is a certain maximum, which is reached by many and exceeded by none, which is about 1.70 volts. TABLE XX. DECOMPOSITION VALUES FOR THE ACIDS. Dextrotartaric = i . 62 v. Pyrotartaric = i . 5 7 v. Trichloracetic = i . 5 1 v. Hydrochloric = i . 31 v. Oxalic =o.95v. Hydrobromic = o . 94 v. Hydriodic = o . 5 2 v. TABLE XXI. DECOMPOSITION VALUES FOR THE BASES. Sulphuric = .67 v. Nitric = .69 v. Phosphoric = . 7ov. Monochloracetic = .72V. Dichloracetic = .66v. Malonic = [.72 v. Perchloric = ] [ . 65 v. Sodium hydroxide Potassium hydroxide = Ammonium hydroxide = Methylamine n/4 n/2 . 69 v. n/4 .67 v. n/2 -74v. /8 75 .68 74 442 ELEMENTS OF PHYSICAL CHEMISTRY. The alkalies and alkaline earths when combined with the largely dissociated acids, i.e., those with decomposi- tion values of about 1.70 volts, show approximately the same value, viz., 2.20 volts. The chlorides, bromides, and iodides have lower values, which are independent of the alkali metal. It will be observed that all the acids and bases with the maximum value give off hydrogen and oxygen upon electrolysis. Those with the lower values, which in more dilute solutions give off oxygen and hydrogen, also reach this maximum value. Thus for HC1 at different dilutions we have: DECOMPOSITION VALUES FOR HC1. Concentration. n 2 normal =1.26 1/2 " =i-34 1/6 " =1.41 Concentration. n i/ 1 6 normal =1.62 1/32 =1.69 At the dilution n/^2 H and O are given off instead of H and Cl, as at the strength of 2 normal, and the maximum is reached. All the above results are -for platinum electrodes. For gold the values are slightly different, but the rela- tion remains the same. Hi. Theory of polarization. The process taking place upon an electrode during electrolysis has been studied by Le Blanc, who has been able to make the action quite clear. He measured the difference of ELECTROCHEMIS TR Y. 44 3 potential between the cathode and its solution after the passage of the current, varying the E.M.F. from o up to the decomposition value of the solution. At the de- composition value he found the potential difference to be the same as that exhibited by the solution in which is placed a stick of the metal. Thus a molar solution cf CdSC>4 is decomposed steadily at 2.03 volts, at which the difference of potential between the cathode and solution is found to be +0.16 volt. A stick of cad- mium in a normal solution of CdSC>4 gives an E.M.F. equal to -f 0.16 volt, i.e., the metal is negative. For some solutions this difference of potential between electrode and solution is found to be given without the maximum being reached. This is true'for AgNO 3 , which decomposes at 0.70 volt. The reason for this is the negative solution pressure of the Ag, which causes ions to be deposited upon the metal, even without the current. When an indifferent electrode is placed in a salt solu- tion, a small amount of ions must precipitate upon it; otherwise, from the equation TT will be infinite, since P=o; and if that is true it would be possible to arrange a perpetual motion. Metal ions 444 ELEMENTS OF PHYSICAL CHEMISTRY. will then separate upon the electrode until the tendency for them to go into solution in the ionic form just com- pensates the separating force. In this way the electrode will be charged positively, and the solution negatively, so that there will be between them a certain difference of potential. The value of this difference will depend upon the amount of metal upon the electrode, and will not necessarily be as large as that for the massive metal (compare with concentration of H in a platinum-black electrode). If we now connect the cell with a primary cell at a low E.M.F., more metal will be deposited, but this will increase P, so that no more deposition will take place at that E.M.F. If the potential is then increased, P will again increase and further deposition will be prevented. Finally, when the E.M.F. used is such that P receives its maximum value, i.e., that which the massive metal possesses, then the deposition will take place steadily. Here we assume the osmotic pres- sure of the metal ions to remain constant. For strong currents this is not true; it decreases, and hence the deposition becomes more and more difficult and the potential difference at the cathode increases. It is not difficult, however, to keep the value constant for a short time, and thus to find the difference of potential at the cathode. At the same time that this takes place negative ions are separated upon the anode and the process is exactly ELECTROCHEMISTRY. 445 analogous. If oxygen is the gas which separates, its concentration increases until the maximum is reached (=P) and the gas is given off in the air. From the above it is possible to understand why a certain minimum E.M.F. is required to cause a steady electrolysis. The separation takes place only when the concentration of ions around the electrodes reaches a maximum, i.e., not until the osmotic pressure exerted by them is great enough to overcome the solution pres- sure of the metal. From the case of AgNO 3 it would seem that this maximum of concentration need not be reached on both sides at the same time. The nature of the electrodes, so long as they are of an inert sub- stance, has no influence except for gases, and then in the following way: The cell platinized Pt+H gas H2SO4 O gas -f platinized Pt gives an E.M.F. of 1.07 volts. If 1.07 volts are passed through the cell in the opposite direction, then we shall have equilibrium. If smaller than this, H 2 O is formed at the electrodes from the gas in the electrode and the ions in the solu* tion. If larger than 1.07, H and O are given off from the electrodes. We have here the gas-cell which gives the value 1.07 decomposing water, and reversing at i. 08. When polished platinum electrodes are used for the decomposition, however, 1.68 volts are necessary for steady electrolysis, the difference between the 1.68 and the 1.07 being 0.6 volt. The difference in these two 446 ELEMENTS OF PHYSICAL CHEMISTRY. values is assumed to be due to the different ion which is separated in the two cases. Water may be considered as dissociated as a dibasic acid: OH'=H' At the value 1.08 volts the O" ions present separate, causing water to decompose for a time and then as they are used up the process ceases. The 1.07 volts of the gas-cell are due to this ion O", for the platinum black gives it up to the solution in that form after having absorbed oxygen gas. When OH' ions separate they unite to form H^O and O, 4OH'=2H 2 O+O 2 , and the force necessary to do this is 1.68 volts. Where the concentration of OH' ions is great, in bases for ex- ample, the O" ions would be present to a greater extent than in acids, and the decomposition at 1.08 should be more marked than with acids, as it is found to be. HC1 has too few OH' and O" ions to carry any amount of current, so that in strong solutions Cl separates at 1.31 volts, at which point water cannot be decomposed into hydrogen and oxygen. ELECTROCHEMISTRY. 447 If we have in a decomposition vessel one large platin- ized electrode and one very small pointed one and use the latter as cathode, the large one as anode, we get a rapid decomposition of water at i.i volts, H being given off at the point, the oxygen being absorbed in the platinum black. We get then a reversal of the gas- battery. Using the point as anode, we get bubbles of oxygen first at 1.68 volts, H being absorbed by the large electrode. An example of the use of such a process is the Hildburgh cell (J. Am. Chem. Soc., 22, 300, 1900), which rectifies an alternating current. If the value of the alternating current is below 1.68, current can go through when the point is the cathode and is stopped when this is the anode, as it is by the next reversal; so that we get from this a current made up of a series of impulses all in the same direction instead of the alternating current which entered. A current of large voltage can then be rectified by passage through a series of such cells. 112. Primary decomposition of water. The E.M.F. of decomposition of a cell giving off hydrogen and oxygen is dependent upon the concentration of the two ions H* and OH', but independent of the nature of the electroltye. The E.M.F. is the same for acids and bases so long as only H and O are separated. Since by the law of mass action the product of H* and OH' ions in a solution must always be the same, it follows that for all electro- 448 ELEMENTS OF PHYSICAL CHEMISTRY. lytes, since the E.M.F. of the cell must be the sum of those af the two electrodes, the minimum value must be the same for all substances which give off H and O. With the exception, then, of solutions of metallic salts, which are decomposed by H, and the chlorides, bromides, nnd iodides, which are decomposed by O, the ions of water only are the factors in the decomposition of solu- tions, and not those of the dissolved salt. Excluding those solutions mentioned above, then, all solutions when electrolyzed show primary decomposition of water. The current is conducted by all the ions in the solu- tion. At the electrodes, however, that process takes place which requires the smallest expenditure of work, and that is the separation of H and O as gases. In the case of the electrolysis of K 2 SO4 in solution, when the current is not too strong, there is no necessity for assum- ing that the K' and SO' 4 ' ions are separated upon the elec- trode and then redissolved. They simply collect around the electrode, but, as the H and O separate more easily, these are forced out. With a stronger current, of course, it is possible to cause the separation of K and SC>4, for then the H* and OH' ions, being present to but a small extent, cannot carry all the current, and the work of separation of the ions of the salt is smaller than that necessary to remove the very much diminished amount of OH' and H'. The formation and decomposition of H2O are reversible processes, so that there is no loss ELECTROCHEMISTRY 449 of work connected with them, while with the secondary action there would be. The H* and OH' ions are formed as they are used up, i.e., more water dissociates. All bases and acids must have the same E.M.F., for the product of the H* and OH' ions is always the same. For salts we should obtain a higher value, since upon the electrode at which H separates a base is formed, and OH' ions increase in number, driving back those of H', so that the difference of potential on that electrode increases, i.e., P remains the same, while p decreases. On the other electrode H' ions collect and exert the same decreasing effect upon the OH' ions. The smaller the dissociation of the base and acid formed the smaller naturally this influence will be. In the case of HC1 (w/i), Cl gas is given off steadily at a smaller E.M.F. than for the oxygen acids. As the acid becomes more dilute, however, the amount of H" and Cl' ions decreases, until finally there is such a large concentration of OH' ions present, as compared with that of Cl', that O separates the more readily. This explains the results already given for HC1 in different dilutions. 113. Electrolytic separation of metals by graded elec- tromotive forces. As we have seen, the salts of the different metals have different decomposition values. From this fact Freudenberg* showed how it is possible to separate metals quantitatively. It is only necessary * Zeit. f. phys. Chem., 12, 97, 1893. 45 ELEMENTS OF PHYSICAL CHEMISTRY. to find two salts, i.e., one of each metal, which have decomposition values as far apart as possible. If now a certain E.M.F., lying between these limits, is passed through the cell, the metal with the lower decomposition value will separate; after that is separated, the current will cease and it is only necessary to raise the E.M.F. in order to deposit the other. As the concentration of the ions in the salt of the metal separated first decreases, it is necessary to raise the E.M.F. slightly. This amount is small and may be readily calculated from RT Thus if p decreases from o.i normal to o.oooooi normal (the limit by analytical means), x must be increased only 0.3 volt for a monovalent element and but half that amount for a divalent one. An example of this is given in the table below. AgNO 3 =o.7o and Pb(NO 3 ) 2 = i.52. With the two solutions present the Ag will be entirely deposited by an E.M.F. of less than one volt; after this is done the E.M.F. may be raised to 1.52 or more and all the lead deposited.* In this way it is possible to make separations which cannot be made by varying current strength. * The student is referred for further information to the original article. ELECTROCHEMISTRY. 45 1 The following table gives the separation value of a few ions. Here they are based on the value of H* taken as zero. Thus if the value for H' is added to that of OH', we have the decomposition voltage of H2O 1.68. The values are for molar concentration. Ag- = -0.78 r =0.52 Cu' = -o-34 Br' =0.94 H- = 4-Q.o O" = i. 08 (in acid) Pb' = +0.17 Cl' =1.31 Cd' = +0.38 OH' =1.68 (in acid) Zn' = 4-0.74 OH' =0.88 (in base) SO/' =1.9 HSO/ = 2 .6 The values of O" and OH' are true in the presence of a normal solution of H* ions. If we have H* and OH' in a base, the above value of H* becomes 0.8 and the value of OH' and O" is decreased by 0.8. Speketer (Zeit. f. Elektrochem., 4, 539, 1898) has applied the table above to the quantitative determina- tion of Cl, Br, and I in solutions. A platinum cathode and a silver anode are placed in a normal solution of H2SO 4 containing the mixed halides. The minimum electromotive force of decomposition of this system, i.e., solution of Ag and separation of H, is given by the equa- tion i Pl i P2 7T =0.0575 1( >g ^ -0-0575 lQ g fr at 17, where P\ and P^ are the electrolytic solution pressures of the Ag and H, pi and p2 being the osmotic 45 2 ELEMENTS OF PHYSICAL CHEMISTRY. pressures of the ions Ag* and H'. We have then, taking the decomposition value of H* as zero and that of Ag' as 0.78 (since, when pi=i, 0.0575 log P\ = 0.78), P 1 * =-575 log =0.0575 lo g p i --575 log Pi = -0.78-0.0575 log^i. pij however, the concentration of Ag' ions, can only reach a certain value in the presence of Cl, Br, and I, which value is regulated by the solubility products of AgCl, AgBr, and Agl. The solubilities at 25 are respectively i.25Xio~ 5 , 6.6Xio~ 7 , and 0.97 Xio~ 8 . If I is present in the ionic state to the concentration o.i mole per liter, A 1 =L_^Z L } from which 7r 1 = +o.ooF, o.i i.e., the system acts as a cell and gives a current, Ag dissolving and H being evolved. For Br, x= o.i2F; and for Cl, TTJ= o.2F. If I' ions are reduced in con- centration to o.oooi mole per liter, n becomes 0.084. In this way, by varying the E.M.F., the iodide and then the bromide are transformed into the silver salt, and the difference in weight of the anode before and after the experiment gives the weight of the one which has been separated. Cl is usually determined by titra- tion after the Br and I have been removed. CHAPTER X. PROBLEMS. I. (Sec Chaplei I ) 1. How many ergs are equivalent to 3.52 (18) calories? Ans. 147,241,600 ergs. 2. What velocity will a force of 2.12 dynes acting for 2 seconds upon a body weighing 3.62 grams give the body? Ans. 1.17 cm. per second. 3. The velocity of a body weighing 1.23 grams is 4.3 cm. per second, what force has acted upon it for one second? Ans. 5.29 dynes. 4. What is the pressure in dynes per sq. cm. of 721 nim. of Hg? Ans. 961,000. II. (See Chapter II.) 5. An open vessel is heated to 819 C. What por- tion of the air which the vessel contained at o remains in it? Ans. 0.25. 6. An open vessel is heated until one-half of the gas 453 454 ELEMENTS OF PHYSICAL CHEMISTRY. contained at 15 is driven out. What is the temperature of the vessel ? Ans. 303 C. 7. A volume of gas, measured at 15, is 50 cc. At what temperature would its volume become 44 cc. ? Ans. ig.6 C. 8. A volume of gas at 766 mm. pressure is 137 cc. What would it be at 757 mm.? Ans. 138.7 cc. 9. What volume does i mole of gas occupy at 50, the pressure being 760 mm. ? At 100, p being 900 mm. ? Ans. 50 = 26.5, at 100 = 25.8 liters. 10. A volume of air in a bell jar over water measures 975 cc. The water in the jar is 68 mm. above the water in the trough, and the barometer stands at 756 mm. What would the volume be if exposed to standard pres- sure, the specific gravity of Hg being 13.6 ? Ans. 963.4. 11. At 14 C. and 742 mm. pressure a volume of gas measures 18 cc. What will be its volume at o and 760 mm. pressure ? Ans. 16.72. 12. A volume of H at a temperature of 15 measures 2.7 liters, with the barometer at 752 mm. What would have been its volume had the temperature been 9, and the pressure 762 mm. ? Ans. 2.6 liters. 13. What volume is occupied by 44 grams of oxygen at 70 cm. Hg pressure and 35 C. ? Ans. 37.7 liters. 14. J mole of H, \ mole of O, and \ mole of N are mixed in a volume of 10 liters at o C. and 760 mm. What are the partial pressures of H, O, and N? PROBLEMS. 455 Am. 11 = 1156.96, = 1156.96, and N = 77i.65 grams per sq. cm. 15. What would these pressures (14) be in atmos- pheres at 10 C. ? Ans. H = 1. 164, O = 1.164, and N =0.774. 1 6. i liter of N weighs 1.2579 grams at o and 760 mm. Calculate the specific gas constant, r. Ans. 3307 gr. cm. 17. The specific gas constant, r, for N was found above (16). What it is for H? Atomic weight of N is 14.073, and of H is 1.032. Ans. 41,010 gr. cm. 18. How much will 100 liters of chlorine at 74 cm. Hg pressure and 30 C. weigh? Ans. 278.7 grams. 19. A solid gives off a gas which is dissociated to 41%, into two products. What is the work done, in calories, gram-centimeters, and liter-atmospheres, when i mole of solid goes into the gaseous state, the tempera- ture of dissociation being 55 C. ? Ans. 925 cals.,39,4io,coo gr cm , 37.96 L. A. 20. How much work will be done by i kg. of CC>2 when heated 200? Ans. 373.1 L. A., 9088 cals. 21. The time necessary for CC>2 to flow through a small opening is 26.5 minutes. Under the same con- ditions the time for H is 5.6 minutes. Find the density of CO 2 . Ans. 22.3. 22. H is at the partial pressure of 2.136 atmospheres in a space of 10 liters. How many moles per liter are 45<> ELEMENTS OF PHYSICAL CHEMISTRY. there, the temperature being o and the pressure 76cm.? Ans. c= 0.0954. 23. Starting with i mole of A in 22.4 liters (at o, 760 mm. of Hg), assume the dissociation according to the scheme A = 2B + $D (here A, B, and D represent moles) to be 23%. What will be the final volume, where pressure and temperature remain unchanged? Ans. 43 liters. 24. What is the final concentration of A, B, and D in the above ? Ans. A =0.0179, =0.0107, and D = 0.01604 moles per liter. 25. The molecular weights of the above are M A =170, MB =2 5) and M D =40. How many grams per liter are there of each at equilibrium? Ans. ^4=3.04, B= 0.268, and D = 0.642. 26. Assume 17 grams of A (Af = 170) in 2.24 liters (o, 760 mm. Hg). Find concentrations, partial pressures and grams per liter of A, B, and D where the dissociation of A is 20% (M B = 2 5 > M#=4.o), and the volume and temperature remain constant. What is the total pres- sure of the system? Ans. A =0.036, 5 = o.oi8, 17 = 0.027 moles per liter. ^4=6.o6, 5=0.447, Z)=i.7 gr. per liter. A =0.8, B =0.4, D =0.6 atmospheres. Total pressure = i .80 atmospheres. 27. The time of outflow of a gas is 21.4 minutes, H PROBLEMS. 457 being as in (21) 5.6 minutes. Find molecular weight of the gas. Ans. 29.2 28. When heated PC^ dissociates into PCla and C\2- The molecular weight of PC1 5 is 208.28. At 182 the density is 73.5, and at 230 it is 62. Find the degree of dissociation at 182 and 230. Ans. a l82 o= 4 i.7%, a 230 o=68%. 29. Sulphur molecules, Sg, dissociate under certain conditions into those of the form 82. If this dissocia- tion were complete, what would be the density of the gas formed of the 82 molecules? Ans. 32. 30. The ratio of the specific heat for constant pres- sure to that for constant volume is equal to 1.67 for helium. How many atomic weights are there in the molecular weight? 31. How many atomic weights are there in the molec- ular weight of argon, the specific heat for constant volume being 0.075 ? The density gives the molecular weight of 40. Ans. i. 32. What is the specific heat of CO 2 gas at constant volume, i.e., c v ? The molecular weight is 44 and the temperature is 50. Ans. 0.164. 33. The time of outflow through a small opening is found for O as 22.3 minutes; for H it is 5.6. How many atomic weights are there in one molecular weight of oxygen ? Ans. 2. 34. The specific heat for constant pressure, c p , of 45^ ELEMENTS OF PHYSICAL CHEMISTRY. benzene is 0.295, what is the specific heat for constant volume, c v ? Ans. 0.27. 35. The specific heat, c v , of CO 2 is 0.2094; what is the ratio of that for constant pressure to that at con- stant volume ? Ans. =# = 1.22. V 36. CC>2 gas, for which k was found in (35), is com- pressed in a flask to the pressure 1.5 atmospheres, and this pressure allowed to equalize rapidly against the atmospheric pressure, equal to 760 mm. What is the temperature produced? The original temperature was o C. Ans. -i9.2C. 37. Air ( = 1.4) compressed to pressure equal to 50 atmosphere^, is expanded rapidly to double its initial volume against the pressure of the atmosphere. What temperature is produced by the expansion? The origi- nal temperature was o C. Ans. -66.i C. What effect has the original pressure here? 38. Two gases, #1 = 1.4, #2 = 1-22, are compressed rapidly, each to T V of its original volume. What tem- peratures are produced if o is the initial one? Ans. 4 1 =4i2.8, / fe2 =i8o.i C. 39. Find the mechanical equivalent of heat by Mayer's method, using O. c v for O is 0.2751; i gram of O has the volume of 699.25 cc. at o and 760 mm. pressure. PROBLEMS. 459 III. (See Chapter III.) 40. A volume of 50 liters of air in passing through a liquid at 22 causes the evaporation of 5 grams of substance, the molecular weight of which is 100. What is the vapor-pressure of the liquid in grams per square centimeter? Ans. 25. 41. The vapor-pressure of a substance is 44 mm. at 20 C., the molecular weight is 46. How many grams of liquid will evaporate when 100 liters of air pass through ? Ans. 11.08. 42. w-hexane (^=85.82) has a molecular volume in the gaseous state equal to 34,500 c.c., that in the liquid state being 137.96; both being measured at 60 C. What is the latent heat of evaporation of i gram of n- hexane? Ans. 85.7 cals. 43. For CC>2 at o we have the following values: density in liquid state 0.905, density in gaseous state 0.099. What is the latent heat of evaporation of i gram of liquid CC>2? Ans. 54.9. Observed is 57.48. 44. The latent heat of evaporation of chloroform at 61 is 58.49 cals., the boiling-point is 61, and the molec- ular weight in gaseous form is 119.1. What is the for- mula of the substance in the liquid state ? 45. The latent heat of methyl alcohol at its boiling- 460 ELEMENTS OF PHYSICAL CHEMISTRY. point, 66, is 267.48 cals., the molecular weight in gaseous form being 32. Find molecular weight in the liquid form. What conclusion is to be drawn from this result ? 46. The heat of evaporation of ether at 34. 5 is 88.39 cals. This is for one gram. Find the change in boiling- point due to a change from 760 to 634.8 mm. Ans. -4.76. dp 47. Find heat of evaporation of ethyl formate. -7^ = 27 mm., the boiling-point is 54- 3 at 760 mm., and the molecu- lar weight is 74. Ans. 102.8. 48. Find , for ethyl propionate. The heat of evaporation is 77 cals., the density in liquid form is 0.7958, and in the gaseous state 0.0033, both at the boiling-point, 99. Ans. 21.5 mm. 49. The refractive index of liquid N is 1.2062, and for liquid air is 1.2053. Find that for liquid O. Air is composed by weight of 23.01% of O and 76.99% of N, d o = 1.124, ^=0.9673, dw =0.885. Ans. 1.16. For further examples on refractive index see May- bery and Shepherd. (Am. Chem. Jour. 29, 278, 1904.) 50. What is the surface tension in dynes, of C 6 H 6 ? The radius of the capillary tube is 0.01843 cm., at 46 the height of the liquid is 3.213 cm., and the density is 0.85. Ans. # = 24. 7 dynes. 51. Find height to which CS 2 will rise in a capillary PROBLEMS. 461 tube at 46, r being equal to 0.01708 cm., x (the sur- face tension) being 27.68 dynes, and 5 (the density) = 1.224. Ans. 2.7 cm. 52. Find the molecular weight of C 6 H 6 in the liquid state, # = 24.71 at 46, critical temperature is 288. 5, M in gaseous form= 78, k= 2.12, and density=o.85. 53. Find molecular weight in liquid form of CS 2 . #(46) = 27.68, M in gaseous form = 76, critical tempera- ture is 28o.3, and density =1.224. 54. At i4.8>cetyl chloride (density = 1.079) ascends to a height of 3.28 cm. in a capillary tube of which r = 0.0142 5. At 46.2, in the same tube, A = 2.85 and the density is 1.064. Find the critical temperature of acetyl chloride, ^=78.5. Ans. 2^ .2 C. 55. What is the heat of evaporation of ether at 80 if the external pressure is increased so as to increase the boiling-point from 35 to 80 ? Specific heat of liquid at 80 is 0.690, molecular weight is 74, and heat of evaporation at 35 is 88.39 cals. (see data in text). Ans. 77.14 cals. HT 56. Find -j- for ether at 80 as given in (55); the vapor-pressure of ether at 80 is 299.14 cm. Ans. 0.0146 per mm. 57. What external pressure is necessary to change the boiling-point of ether in (55) from 35 to 80 ? Ans. 4.06 atmos. 462 ELEMENTS OF PHYSICAL CHEMISTRY. 58. Find molecular weight of methyl formate. Critical temperature is 214 C., critical pressure 59.25 atmos- pheres, and critical density 0.3494. Ans. M=6$.2. 59. Find molecular weight of ethyl formate. Criti- cal temperature is 235.2 C., critical pressure 46.83 atmospheres, and critical destiny 0.3232. Ans. If = 77.2. 60. What is the change in vapor pressure for ether (56) per degree ? Ans. 68.5 mm. IV. (See Chapter IV.) . 61. The specific heat of Ni is 0.1092. What is the atomic weight of Ni? Ans. 58. . 62. The specific heat of Fe is 0.112. Find atomic weight of Fe. Ans. 56.6. 63. The specific gravity of solid phenol is 1.072, and in the liquid state it is 1.0561; the latent heat of fusion jrr< is 24.93, and the melting-point is 34. Find j-. Ans. 0.00421 per atmos. dT 64. Acetic acid melts at i6.6 C., j~ =o.o242 per at- mos., the heat of fusion is 46.42 cals. Find change in volume by the liquefaction of i gm. Ans. 0.00016 liter. , 65. The specific heat of silver sulphide is 0.0746, why is the formula regarded as Ag2S? 66. Specific heat of solid acetic acid is 0.4599 fr m io 15, of fluid is 0.5026. Find heat of fusion at 50 from that at 5.6, which is 44.34 cals. Ans. 46.24. PROBLEMS. 463 67. Naphthalene in solid state has specific heat =0.356, in liquid state =0.396, both at 80, and the heat of fusion is ^'=35.5 cals. What is w at 30? Ans. 33.5. 68. Assuming the data given in 67, find the fraction of naphthalene which would separate as the result of an overcooling of 10. Ans. 0.112. 69. The vapor-pressure of both ice and water at o is 4.62 mm., the heat of fusion of ice is 80 cals. What is the difference between the temperature coefficients of vapor-pressure for the two states? Ap AP Ans. 0.0446. Observed is -7^ --77^ =0.3852 -0.3399 = 0.0453 mm. 70. The heat of fusion of ice is 80 cals. and the heat of evaporation of water is 597 all at o. What is the heat of sublimation of ice at o? Ans. 677 cals. 71. How could the heat of sublimation of ice at o (see above) be determined in another way? What value would it give? The specific volume of saturated water-vapor at o is 204,680 (i.e. p =4.619 mm. Hg), that of ice is 1.99 cc. and -j^ for ice is 0.3852 mm. Ans. 687 cals. 72. The specific gravity of water at o is i, the heat of solidification is 80. What change in the freezing- point of water results from an increase of pressure of i atmosphere? (For data see above.) Ans. 0.00748. 464 ELEMENTS OF PHYSICAL CHEMISTRY. 73. The specific in the liquid state is 0.43, in the solid state 0.352. An overcooling of i7.8 causes 0.25 of the liquid to separate as solid. What is the latent heat of fusion at the freezing-point? What is it at 10 below the freezing-point? Ans. 30.6 and 29.84 cals. 74. The specific heat of phosphorus in the liquid state at 44 is 0.204, the latent heat is 5 cals., and at 39. 5 is 4.86 cals. What is the specific heat in the solid state? Ans. 0.173. 75. At 29 the latent heat of fusion of CaCl2-6H2O is 40.7, and at 160 is o. What is the specific heat in the liquid state, that in the solid state being 0.345? Ans. 0.560. V. (See Chapter V.) . 76. One liter of H 2 O absorbs i liter of CO 2 at o, 760 mm. How many grams of CO2 gas are contained in a bottle of carbonic water holding 350 cc. of solution, the pressure being 5 atmospheres? Ans. 3.44 grams. 77. Air is composed of 20.9 volumes of O and 79.1 of N. At 15 water absorbs 0.0299 volumes of O and 0.0148 of N, the pressure of each being that of the at- mosphere. What is the composition of air absorbed in H2O ? Ans. 34.8% by volume of O and 65.2 of N. 78. Find molecular weight of naphthalene; 49.4 grams of H2O and 8.9 grams of naphthalene go over when dis- tilled at 98. 2 and 733 mm. The vapor-pressure of H 2 O at 98.2 is 712.4. Ans. 112. PROBLEMS. 465 79. Supposing that the air dissolved in (77) were, collected and redissolved in water, what would be the composition of the air dissolved? Ans. 51.9% by volume of o and 48.1 of N. 80. What is the surface energy involved on a cubic centimeter of gypsum powder (2.5 grams) when it is re- duced to cubical particles with an edge equal to o.oooi mm. and added to water? The surface tension of water is 80 dynes. Ans. 4.8 X io 7 ergs. 81. What is the osmotic pressure of a i% solution of glucose (M = 1 80) at o C. ? Ans. 94.6 cm. H. Obs. =94 cm. 82. The osmotic pressure of a solution of cane-sugar at o is 49.3 cm. of Hg. What percentage of sugar (M =342) is contained in it ? Ans. 0.99% ; obs. = 1.0%. 83. The molecular weight of H^O is 18, of nitro- benzene is 121. Vapor- pressure at 99 ' (the boiling- point of the mixture) of pure H^O is 733 mm., and of nitrobenzene is 27 mm. How much nitrobenzene is con- tained in the distillate? Ans. 0.2 of total by weight. * 84. The osmotic pressure of a sugar solution at 32 C. . is 54.4 mm. What is it at i4.2? Ans. 51.2 mm. x 85. The osmotic pressure of solution containing io * grams of sugar to a certain volume is 200 mm. What \ is that for the same volume containing 13.5 grams? Ans. 270 mm. 86. 10.442 grams aniline in 100 grams of ether gives a 466 ELEMENTS OP PHYSICAL CHEMISTRY. vapor-pressure of 210.8 mm. Ether alone (Af = 74) gives 229.6. Find molecular weight of aniline. Ans. 87. 87. Find osmotic pressure at o of aniline in (86) in at- mospheres and gram per square centimeter, s for ether is 0.737. Ans. 19.82 atmos. or 20460 gr. per sq. cm. 88. The osmotic pressure of a substance in water solu- tion is 100 cms. at o C. Find the vapor-pressure of the solution; that of water at o is 4.57 mm. Ans. 4.56 mm. 89. What is the work, in gr.-cms. and calories, neces- sary to separate 200 grams of a substance (M =66) from the solvent at 20 C. ? 10 grams of substance to the liter of solvent. Ans. 1953 cals.; 83,200,000 gr. cms. 90. 64.74 grams of propylene bromide (If = 222) are mixed with 145.93 g rams of ethylene bromide (M = 206) at 85.o5; the vapor-pressure of pure propylene bromide at this temperature is 127.2 mm., and that of ethylene bromide is 172.6 mm. What is the partial vapor-pres- sure of each in the mixture? What is the composition of the distillate? What is the total vapor-pressure of the mixture ? Ans. f (ethylene bromide) =122.3 mm> observed = 121. 4 " p' (propylene bromide) = 37.1 " observed = 37.3 " total vapor-pressure = 1 59.4 ' ' observed = 158. 7 " PROBLEMS. 467 24.6 grams of propylene bromide and 75.4 grams of ethylene bromide in each 100 grams of distillate. Ob- served values are 24.9 and 75.1. 91. The increase in the boiling-point of 54.65 grams of CS2 caused by the addition of 1.4475 g r & m s of P is o.486. What is the molecular weight of P in CS 2 ? 9 Ans. 129.2. What is the formula, the atomic weight being 31 ? 92. Calculate the increase in boiling-point of ether when to 100 grams we add a mole of a substance. The boiling-point of ether is 34.97, the latent heat of evapo- ration is 88.39. Ans. 2i.5. 93. The molecular increase of the boiling-point of H 2 O, as caused by the addition of i mole of substance to 100 grams, is 5.2. Find heat of evaporation of H 2 O. Ans. 535.1 cals. Ap 4 94. At 40 ~j=* for benzene (M = 78) is 0.88 1 for a mean pressure of 22.42 cm. What is the molecular increase of the boiling-point of benzene under this reduced pressure? Ans. i9.85. 95. 10 grams of a substance in 100 grams of a solvent increase the boiling-point by o.87. The molecular weight of the substance is 60. Find the molecular increase of the boiling-point. Ans. 5.22. 96. 0.284 grams of the oxime (CH 3 ) 2 CNOH causes a decrease of o.i55 in the freezjng-point of 100 grams of 468 ELEMENTS OF PHYSICAL CHEMISTRY. glacial acetic acid. K for acetic acid is 38.8. Find molecular weight of the oxime. Ans. 71. . 97. The ionization of a normal solution is 80%, two ions being formed. What will the depression of the freezing-point be> water ( = 18.9) being the solvent? Ans. 3.4. 98. The molecular depression of an aqueous solution containing an ionized substance is 22. Find the degree of ionization of the substance in that volume. Ans. a =16.3%. 99. A normal water solution gives a depression of 2. i, when the overcooling is 2.21. What is the strength of the solution whose freezing-point is observed? Ans. 1.028 N. 100. Find the freezing-point depression of the normal solution itself in (99). Ans. 2.O4. . IQI. Find the heat of fusion of nitrobenzene; its melt- ing-point is 5.3, the molecular depression of the freezing- point is # = 70.7. Ans. 21.9 cals. 102. The specific heat of nitrobenzene is 0.3524, its heat of fusion is 21.9 cals. Find the amount of solid separated by an overcooling of i. Ans. 16.1 grams per 1000. 103. In (91) find the osmotic pressure at 46 of P in the CS 2 solution. s CSa = 1.2224. Ans. 6.84 atmos. 104. Find the vapor-pressure in (91) of P in CS 2 solu- tion at o the vapor-pressure of CS 2 at o is 127.91 mm. Ans. 125.9 mm - PROBLEMS. 469 105. In (96) find the osmotic pressure of the oxime in glacial acetic acid at 17, sp. gr. (acetic acid) =1.056? Ans. i atmos. 106. What is the relation between the osmotic pres- sures of .01 mole of substance in a 1000 grams of water and a 1000 grams of ether (sp. gr. =0.7370), assuming the same molecular state of the solute in each ? Ans. P,=o.737oP w . 107. In (96) find the vapor-pressure of the solution at 40 C., the vapor-pressure of glacial acetic acid at 40 being 34.77 mm. Ans. 34.69 mm. 108. A o.i normal solution of acetic acid in water freezes o.i927 lower than H^O. Find the degree of ionization of the acetic acid. Ans. a = 2%. 109. A .15 normal solution of succinic acid freezes o.2864 lower than H 2 O. Find the' ionization of the acid. Ans. a = i%. no. A saturated aqueous solution of ether freezes at 3.85. When 4.76 grams of I are dissolved in 100 grams ether, a saturated aqueous solution freezes at 3.789. Find molecular weight of I in ether. The molecular increase of the freezing-point of water by the addition of i mole of substance to the ether is 3.o6 over that point for pure ether in water (see pp. 190, 191). Ans. 239. 47 ELEMENTS OF PHYSICAL CHEMISTRY. VI. (See Chapter VI.) in. Find the heat of decomposition of H 2 O2 in water from the following reactions: SnCl 2 H 2 Cl 2 Aq + H 2 O 2 Aq = SnCl 4 Aq + 2 H 2 Ans. 112. The heat of combustion of NH 3 in O at constant volume is 2NH 3 + 30 = 3H 2 O + N 2 + 181 2K. Find heat of formation of NHs from its elements. Ans. io6.2K. 113. How does the heat of combustion of H with O to form 1 8 grams of liquid H 2 O at constant volume vary with the temperature? The heat capacity of 72 grams of H 2 O is 0.72 K, and that of 4 grams of H plus 32 grams of O is 0.2056. Ans. 0.0772^ per degree. 114. C# 4 + 2O 2 =CO 2 + 2H 2 O + 2ii9# at constant pressure and 18 C. Find heat of formation of CH 4 . Ans. 2iS.2K. 115. PROBLEMS. 47 * Find the heat formation of KOHAq from the ele- ments. Ans. n6$K. 116. Zn + 2HCLAq=ZnCl 2 Aq + 2H + 342/ at constant pressure. What is it for constant volume? Tempera- ture is 20. Ans. 347.9^. 117. The following reactions take place: C 6 H 6 + 1 5 =6C0 2 + 3 H 2 +yK and C 6 H 6 + 750 = 6CO 2 At 27, assuming these to be at constant pressure, find values for constant volume. 1 1 8. What is the difference in energy between 18 grams of water at a 100 and water- vapor at the same tem- perature? Between water and ice at o? The latent heat of evaporation is 536.4, of fusion is 80 cals. Ans. 9655 and 1440 cals. 119. Find heat of formation of gaseous hydrobromic acid from the reactions, SO 2 Aq + O =SOsAq-f 63700 cals., 2Br + SO 2 Aq + H 2 O = 2HBrAq + SO 3 Aq + 54000 cals., when, in addition, we know that H 2 + O =H 2 O + 68400 cals., and H Br ^HBrAq + 20000 cals. Ans. H+Br=HBr+93$o cals, 47 2 ELEMENTS OF PHYSICAL CHEMISTRY. 1 20. Find the heat of formation and heat of forma- tion in solution of NHa. + 30 =N 2 + 3H 2 O + 181200 cals. j const. + 30 = 3H 2 O -f 3 X 68400 ] pressure. Ans. N + $H = 12,000 cals. ; (N, 3H", aq) = 20,400 cals. 121. NaOH + 5oH 2 O has a molecular specific heat equal to 885.0 cals. and SO 3 + iooH 2 O that of 1797 cals. At 9.i6 these evolve 32,060 cals. by combining, and at 24^42 they generate 31,650 cals. What is the specific heat (i.e., for i gram) of a solution of Na 2 SO4+ 2ooH 2 O ? Ans. 0.9603 cal. 122. What is the heat of formation of a very dilute solution of magnesium chloride? (See text for data.) Ans. 1875 cals. VII. (See Chapter VIII.) 123. In the volume of i liter there are 0.14 mole of hydrogen and 0.081 mole of iodine. At the tempera- ture of 440 C. #1 Find the amount of hydriodic acid formed. Ans. 0.14855 mole. PROBLEMS. 473 124. The initial pressure of I is 38.2 cm., the fraction uniting with H is 0.8. What is the original pressure of the H? K=o.o2. Ans. 40.35 cm. 125. The coefficient of distribution of acetic acid 0.245 , 0.314 between water and benzene is as - - and at 0.043 0.071 two dilutions. What is the molecular weight in benzene ? In water it is 60. Ans. 2.02X60. 126. Find the constant of equilibrium, for concentra- tions, for the reaction 2NO2 = 2NO -fO 2 at 279 from the following data, du = 1.590 X 2 : t yn 4 a 130 718-5 1 .600 184 754-6 J-55 1 0.050 279 737-2 1-493 0.130 494 742-5 1.240 0-.565 620 760.0 i. 060 I.OOO Ans. ^279 = 0.136 for 2 moles NO 2 , using the volume per cent of each in the formula, i.e., parts of NO 2t NO, and O per 100 volumes, viz. 81.67, 12.21, and 6.1 respec- tively. 127. At 440 in 50 liters we have a mixture of 2.74 moles of HI, 0.5 mole of H and 0.3011 mole of I. K =0.02. In which direction does the reaction go? 128. At 440 (#=0.02) 5.30 cc. of H are mixed with 7.94 cc. of iodine- vapor. How much HI is formed? Ans. 9.475 cc. Observed, 9.52 cc. 129. At 3000 a for CO 2 , according to the formula 2 , is equal to 0.4. What is the constant 474 ELEMENTS OF PHYSICAL CHEMISTRY. of dissociation of i mole of CO 2 calculated from the volume of each in 100 volumes? Ans. 2.72. 130. What is K (129) for 2 moles of CO 2 ? Ans. 7.4 for the concentrations expressed in parts per hundred, i.e., 50 of CO 2, 33 of CO and 17 of O. 131. The constant of equilibrium for the reaction of amylene and acetic acid is 830.1 (p. 249). At the tem- perature at which this is true 2 moles of amylene are mixed with a moles of acid in 2 liters, and 0.5 mole of ester is formed. Find a, the original number of moles of acid. Ans. 0.50241 mole. 132. 6.63 moles of amylene with i mole of acid shows that 0.838 mole are formed in the total volume of 894 liters. How much will be found when we start with 4.48 moles of amylene and i of acid in the volume of 683 liters? Ans. 0.8 1 1 1 moles. 133. Write the application of the law of mass action for the reaction Fe + 4H 2 O = Fe 2 O 4 + 4# 2 , and calculate K for H 2 O=4.6 mm., [ = 25.8, and for H 2 O=io.i and 11 = 5.79. 134. For NH 4 HS=# 2 S + ATr 3 , #=62,400 (for pres- sures in mm.) at 25.! C. In a vacuum at 25.! we intro- duce NH$ and H%S until we have a partial pressure of the former of 300 mm., and of the latter 594 mm. Then the reaction is allowed to take place. How much does each gas lose in pressure? Ans. 157.2 mm. PROBLEMS. 475 135. At i8.4 i mole of BaSC>4 dissolves in 50,055 liters at 37. 7 in 31,282 liters. On the justified supposition that BaSC>4 is completely ionized, find the heat of disso- ciation. Ans. 8511 cals. 136. The salt Na2HPO4.i2H 2 O has a vapor-pressure at 15 of 8.84 mm., at i7.3 ^ = 10.53 mm. Find the heat of vaporization, i.e., the heat change during loss of i mole of water of crystallization by evaporation. Ans. 12,728 cals. 137. A mixture of alcohol and hydrochloric acid is an equilibrium in which a certain amount of H 2 O is formed. At 77, #=0.307, at 99, #=0.177. Find the heat generated by the reaction. Ans. +6527 cals. 138. The speed constant of formation of HI from H and I is 0.00023; K, the equilibrium constant at that temperature, 374, is 0.0157. What is the speed con- stant of decomposition? Ans. 0.0146. Obs. 0.0140. 139. To i liter of a molar solution of a monobasic acid (#=0.000018), a salt with an ion in common, having an ionization at that dilution equal to 100%, is added. How much in moles in the dry state must be dissolved to decrease the concentration of H" ions of the acid to o.i of its previous value? Ans. 0.04211 mole. 140. What difference would be observed if this had been a dibasic acid with the same constant? 141. A small amount of a base is mixed with a large 476 ELEMENTS Oh PHYSICAL CHEMISTRY. amount of a solution containing equi-molecular amounts of acetic and lactic acids. In what proportion will they form salts? Ans. Acetic: lactic: 10.00424 -.0.0117. 142. A small amount of base is mixed with a large amount of an equi-molecular mixture of two acids. The ionization of one is 10%, of the other 90%. How much of each salt will be present in 100 parts of the salt formed ? 143. The constant for the first hydrogen of malonic acid is i58Xio~ 5 , of the second is i.Xio" 6 . What is the* amount of H' ions in a solution of i mole of malonic acid in 2000 liters? Ans. 0.8385, i.e., C HA / =0.7985. 144. The degree of ionization, a for KC1 is 0.75 for V = i, 0.94 for F = io, 0.99 for F = 10,000. Calculate K by the dilution laws of Ostwald, Rudolphi, and van't Hoff. 145. The value of K for a 0.05 molar solution of acetic acid is 0.0000175 at 18, and 0.000016*24 at 52. What is the heat of dissociation of acetic acid ? At what tem- perature would this value be true? Ans. 416 cals. 146. Find the heat of neutralization of i mole of acetic acid (in 200 moles of H^O) with i mole of sodium hydrate (in 200 moles of EbO) at 35; a for acetic acid is 0.009, the heat of ionization is 386 cals.; a for NaOH is 0.861, its heat is 1292 cals., and a for sodium acetate is 0.742, its heat being 391 cals. The heat of ioniza- tion of water at 35 is 12,632 cals. Ans. 13,093 cals. 147. The solubility of barium sulphate at 3 7. 7 is i PROBLEMS. 477 mole in 37,282 liters. What is its solubility product at this temperature? Ans. 7.2Xio~ 10 . 148. PbI 2 is soluble to 0.00158 mole per liter at 25.2, i.e., What is its solubility in presence of a o.i molar solu- tion of I' ions ? Ans. 1.58 X io~ 6 moles per liter. 149. MCN (solubility is 0.02, and is ionized completely) is dissociated hydrolytically in solution. K for HCN = i 3 Xio- 10 , and K for H 2 O (25) =(i.o9Xio- 7 ) 2 . Find amount of M ions from an exterior source necessary to be in solution in order to prevent hydrolysis. Ans. 0.02162 mole per liter. 150. Water under atmospheric pressure at 16 absorbs 0.9753 liters of CO 2 to the liter, i.e., 0.04354 moles per liter. Of this 0.000115 mole is ionized (0.264%) mto H* and HCCV ions. What is the solubility product of H 2 CO 3 ? Ans. 1.32 X io- g . 151. At 25 solubility product of H 2 CC>3 into H" and HCO 3 ' is 1.32 Xio~ 8 . Find concentration of H' and HCO 3 ' ions. 152. The solubility product of the substance AC2 is 0.00621. Find concentration of A and C ions when the substance ionized completely into A" and 2C'. Ans. 0.1157 mole per liter of A" and 0.2314 of C". 153. AgCNO is soluble at 100 to 0.008 mole per ELEMENTS OF PHYSICAL CHEMISTRY. liter. How much would dissolve in a solution contain- ing o.i mole of Ag ions? Ans. =6.4X io~ 4 moles per liter. 154. The solubility of a salt is o.oooi mole per liter. The formula of the base is M(OH) 4 , and its solubility is o.ooooi, ionizing into M :: and 4 4256 = 60.23, 4,72 = 82.2, and ^1024 = 116.9. What is the ionization con- stant? A^ for HC1 is 383.9, A^ for NaCl is in, and A n for the sodium salt of lactic acid is 85.1. Ans. K (average) = 0.000136. 172. In a o.oi molar solution of KNO 3 , NO 3 ' has a mobility of 0.497, and K* one of 0.503. Find the equiv- alent conductivity of NOa' and K* in this solution, K=o.ooi044. Ans. / K - = 52.5, / NO3 -= 51.9. 173. At infinite dilution the equivalent conductivity of a solution is 91. What would it be at a dilution at which it is 50% dissociated into 2 ions? Ans. 45.5. 174. For hydriodic acid we have the following data: ^2=364* ^4=376, 4*=3 8 4> 4e=39 I > 4*2=397* 4*4 = 402, ^123=405, and ^256 =406. Compare the con- stancy of the Ostwald and van't Hoff dilution laws. 175. The velocity of migration of Ag* is 0.00057 cm - per second, as is also that of ClOs'. What is the equiva- lent conductivity of a solution of AgClOa which is infi- nitely dilute ? Ans. no. 176. The equivalent conductivity of the sodium salt of an acid is ^32=89.9, ^54=97.1, ^123 = 104.5, ^255 = in. i, ^512 = 117.2, and ^1024 = 122.7. What is the basicity of the acid ? 177. The equivalent conductivity of a solution of Na2SC>4 in 256 liters at 25 is 141.9. What is it at infinite dilution? Ans. 153.9. 4^2 ELEMENTS. OF PHYSICAL CHEMISTRY. 178. The conductivity of a solution of AgCl saturated at 18 is 2.4X10"; that of the water used is i.i6Xio~ 6 . What is the solubility of AgCl? ^ ooK ci = I 3 I - 2 ^ooA g NO 8 = 116.5, and J 00 KN0 3 ==I26 - 1 - Ans. 1.02 X io~ 5 moles per liter. 179. Find the heat of amalgamation of cadmium at o. TT for a cell made up of a i% Cd amalgam and mercury in a solution of CdSC>4 is 0.06836 volts at o, and 0.0735 at 2 4-45- Ans. 4 gives a differ- ence of potential of 0.51 volt and^y,= -0.00076. What is the heat of ionization of Zn at 17? Ans. 337.4^. 181. Zn + (2lT + 2C1') Aq = (Zn" + 2C1') Aq + H 2 + 342^, and from (180) What is the heat of ionization of H gas? Ans. 2H'= 182. A KCN solution is added to the Cu side of a Daniell cell, and as a result the E.M.F. is zero. What conclusion is to be drawn ? 183. A single electrode in connection with the nor- mal electrode (^=0.56) gives an E.M.F. of 1.02 volts, the normal electrode being positive. What is the differ- PROBLEMS. 483 ence of potential between the single electrode and its solution ? Ans. 0.46 volt, metal- is negative. 184. A cell with electrodes of the same univalent metal gives an E.M.F. at 17 of 0.35 volt. The con- centration of metal ions at the positive electrodes is 0.02 mole per liter; what is the ionic concentration on the other side? Ans. i.637Xio~ 8 moles per liter. 185. With a cell with electrodes of a divalent metal, with the ionic concentrations on the two sides equal to 0.02 and i.62Xio~ 8 moles per liter, what would the E.M.F. at 17 have been? Ans. 0.175 volt. 1 86. In a hydrogen gas cell there is acetic acid on one side and propionic on the other of the same con- centration; what is the E.M.F. of the cell at 17, and which pole is positive? Ans. Acetic acid positive, ^=0.00369 volt. 187. In the above ceU the pressure of hydrogen gas is increased to ten times its value on both electrodes. How does the E.M.F. change? 1 88. What increase in E.M.F. is necessary to separate up to o.oooi mole of a univalent metal over that which has separated o.oi mole per liter, the temperature is 17? Ans. 0.115 volt. 189. Assume in 188 that the metal is divalent. What would be the E.M.F. then? Ans. 0.0575 TABLES. DEGREE OF IONIZATION ACCORDING TO CONDUC- TIVITY MEASUREMENTS. HC1 V ao 15 025 035 2 0.899 0.874 0.876 0.863 8 0.948 0-937 0.942 0.941 i6 o-959 0.946 0.962 0.965 32 0.971 0.980 0.980 0.977 128 o-993 0-999 I .OOO I. OOO 512 1. 000 I .000 1. 000 I. OOO HN0 3 . V 13 025 035 2 0.906 0.883 0.879 0.879 8 0.952 0.951 0.952 0.950 16 0.982 0-975 0.978 0.974 32 0.985 0.988 0.992 128 0.995 o-999 o-993 I .000 512 0.992 o-999 o-995 0.999 1024 1. 000 I .000 I .000 I .000 V ffl5 4 25 35 2 o-555 0-53 2 o-547 0.542 8 0.612 0.602 o-597 0-572 16 0.643 0.631 0.631 0.625 32 0.702 0.685 0.687 0.690 128 0.801 0.782 0.817 0.800 512 0.900 0.879 0.947 0.888 1024 0.950 0.969 0-977 0.984 2048 o-977 0.990 0.989 0.996 4096 0.986 I .000 I .000 0.998 8102 0.990 I .000 -I . ooo I .000 485 486 TABLES. DEGREE OF IONIZATION ACCORDING TO CONDUC- TIVITY MEASUREMENTS Continued. KOH. V ai% 26 35 I 0.847 0.803 0.8l9 0.833 2 0.884 0.860 0.876 0.883 8 0.916 0.907 0.892 0.900 16 0.936 0-935 0.938 959 32 0.952 0.958 0.987 0.996 128 0.998 0-993 I .000 i .000 256 I .000 i .000 I .000 i .000 NaOH. i 0.780 0.782 0.766 0-774 2 0.838 0.867 0.835 0.843 8 0.920 0.904 0.920 o-935 16 0.936 0.925 0-944 0.951 32 0.971 0-957 0.968 I .000 128 0.992 0-995 I .000 0.989 256 I .000 i .000 I .000 0.981 KBr. V 18 25 30 35 2 0.788 0.793 0.789 0.768 0-747 8 0.823 o . 849 o . 840 0.828 0.821 16 0.872 a.88i 0.876 0.867 0.862 32 0.892 0.921 0.900 0.884 0.885 128 0.920 0.961 0.955 0.938 o .926 512 0-949 0.984 0.983 0.987 0.985 1024 I .000 I. 000 I. 000 I .OOO I .000 KI. V o ais 25 35 i 0.767 0-777 0.767 0.760 2 0.800 0.817 0.791 0.792 8 0.850 0.859 0.852 0.854 16 0.905 0.886 0.872 0.881 32 o-935 0.940 0.9H 0.930 128 o-979 0.961 0-959 o . 954 512 I .000 o 983 0-993 0.996 1024 I .000 0.999 i .000 0.996 TABLES. . 487 DEGREE OF IONIZATION ACCORDING TO CONDUC- TIVITY MEASUREMENTS. Continued. 2 0.542 o-539 0.546 0-532 8 0.648 0.647 0.658 0.662 i6 0.709 0.706 0.711 0.714 32 0.766 0.740 0.750 0.771 128 0.863 0.852 0.893 0.893 512 0.924 0.938 0.965 0.949 1024 0.985 0.986 0.994 0.989 2048 I .000 I .000 I .000 1. 000 MOLECULAR SURFACE TENSION. HC1. r x(Mv)$ T x(Mv)$ T x(Mvft 163.1 263-7 175.8 244-8 187.2 229-3 168.5 255-9 180.1 239.0 189.9 223.6 171.8 250.8 183.2 233.6 192.6 221 .0 HBr. 181.9 330-1 188.9 314-6 198.2 294-8 184.8 325-6 193-4 37-3 200.5 292.2 186.1 320.1 195.3 299.6 203.9 283.8 HI. 225-3 367.0 230-9 355-3 235-0 348.0 227.1 362.8 232-9 35 1 - 236-5 344-6 229.3 H2S. 189.1 349-5 197-4 334-1 203.9 324.7 191.3 345-3 199.7 328.3 206.9 316.7 194-6 338-0 201.5 326.6 210.8 308.6 PH 3 . 167. i 287 2 I7C.4 273 4 171. 8 *\J ( + A 279-6 * f 3 ^ / J T" 179-9 265.4 VARIATION OF MOLECULAR SURFACE TENSION WITH THE TEMPERATURE, AND FACTOR OF ASSOCIATION. HCl HBr HI H PH 3 2.03 1.99 1.91 1.70 I.O I.O I.I 1.4 488 - TABLES. DATA FOR DIFFICULTLY (Bottger, Zeit. f. phys. Salt. /. Conductivitv Xio6 less tli at of H 2 O. *a+*c i.e. AOQ. Ionic Concentration Equivalents per Liter. CaS0 4 19.94 19.68 125.9 I.56XIO- 2 AgCl J 9-95 J -33 125-5 i.o6Xio- 5 AgBr 19.96 0.057 127.1 4-5 Xio- 7 AgSCN. . . . 19.96 0.096 116. i 8.2 Xio- 7 AgCN 19.96 0.19 ii5-5 1.6 Xio- 8 AgBr0 3 . . . . 19.94 663.9 105-3 6.3oXio~ 3 AgI0 3 iQ-95 14.05 92-5 1.51X10-* A g2 19.96 29.27 237.2 i.23Xio- 4 A g2 24.96 35-98 259.1 I.38XIQ- 4 A g2 C 2 4 . ... 19.96 28.76 123-7 2-32X ID" 4 Ag 3 P0 4 .... 19.46 6. 10 135-0 4-5 Xio- 5 T1C1 19.96 1 68.0 137-3 1.22X10-^ TIBr 20.06 220.9 138.9 i.59Xio- 3 Til 20 i ^ 26.18 138.0 i. 89X10-* T1SCN 4\j . x^ 10 Q4 140 o j. ^w . \j 127 i .ogX iQ- 2 TIBrO, V * V ' 19.94 J.|,W . V 108.0 ^ / y 117.1 9-22X I0~ 3 T1IO 3 IO O^ I ^4 2 1 04 ^ I 47XIQ- 3 T1 2 C 2 4 .... i y yj 19.96 J O' ' 534-4 vt o 135-5 A * T- / r\ J w 3-94Xio- 2 T1 2 S 10 06 216.0 j.y . y\j f Pb" 0.0168 PbCl 2 ...... 19-95 5354 ,3 PbCl'. 0.0109 Cl' 0.0444 I PbCl 2 0.00127 Pb" 0.0106 PbBr 2 19.96 3692 134-6-j PbBr' o.o 1 10 Br' 0.0322 I PbBr 2 0.00114 PbI 2 2O . IO 778 4 13^7 2 ^ ^X IO ' oo^ t 'f * 00 / Pb" 0.0121 ] Pb(SCN) 2 . . 19.96 2640 123. 6J PbSCN' 0.00727 I CNS' 0.00315 1 I Pb(SCN) s o. 000615] Pb(Br0 3 ) 2 . . 19.94 4635 112. 8 4. II X IQ PbO 19.96 2^. 5 244. 7 i .O4X io~" PbC0 3 V V w 19.96 3 3 i-5 **Tt / 127-5 1.2 Xio- PbC 2 4 . . . . 19.96 1.52 I3I.2 1.16X10- PbS0 4 Pb 3 (P0 4 ) 2 . .. 24-95 T 9-95 40.19 0.14 I5I.2 142.5 2.65X10- 9.8 Xio- TABLES. 489 SOLUBLE SALTS. Chem., 46, 521, 1903.) Solubility Product (Moles). Total Con- centration in Equivalents. a Grams per Liter. i Gram in X CC. X 2.44X10"* 2.99X10- 0.524 2.03 491.1 I.I2XIO- 1 " 1.06X10- 1. 000 1.53X10- 653600 2.0 XlO~ lf 4-5 Xio- 1 .000 0.84X10- 11900000 6.8 Xio- 13 8.2 Xio- 1. 000 1.3 Xio- 7300000 2.6 Xio- 12 1.6 Xio- I.OOO 2.2 XlO~ 4540000 3-97Xio- 5 6.7 Xio- 0-943 I.58XIO- 538.8 2.31X10-" i . 54X IQ 0.994 4.35X10- 22970 1.52X10-" 1.84X10- 0.668 2.I4XIO- 46700 1.93X10-" 2.16X10- 0.643 2.50X10- 40000 6.29X10-" 2.40X10- 0.977 3.65X10- 27390 1.3 Xio- 18 4.6 Xio- 0.98 O.64X iQ 155000 1.50X10-* 1.35X10- 0.90 3.25X10- 307.1 2.53X10-' 1.64X10- 0.968 0.476X10- 2101 3.60X10-' i -92X io~ 0.989 0.636X10- 15720 i. i9X IQ * 1.20X10 0.906 3.15X10- 317 I 8.50X10-* 9.94X10- 0.927 3.46X10- 288.7 2. 19X io~* 1.52X10- 0.971 0.578X10- 1730 3.06X10-* 6.35X10- 0.62 15.77X10- 63.4 o.86X IQ O . 2 I X IO 46^0 0.044 t v**y 6.Q2X IO~ 2 undiss. 9-6iX io~ i I O4. PbClj Awi r 0.05 4 . 54X 10 * undiss. 8.34X10 l IIQ PbBr 2 V 8.ioXio~ 7 2.61X10-' 0.97 0.470XIQ- 1 2127 0.044 2.78X10* undiss. 4.50X10" 1 222. O Pb(SCN) 2 3 .47Xio- 5 5.78X10- 0.72 i3-37Xio- 74.78 i <;^Xio"~ 0.68 i . 7iX IQ 58600 o-36Xio- 10 i . 3^ /\ i\j 1.2 XlO- 0.96 1.6 Xio- 595000 o.338Xio- 10 I.22XIO- o-95 1.80X10- 550000 1.76X10-" 2.88X10- 0.92 4.38X10- 22830 1.2 Xio- 32 i.o X io~ 0.98 1.3 Xio- 74OOOOO 490 TABLES. SOLUBILITY OF SILVER SALTS. Chloracetate.. Borate Acetate. . . 17 25 1 8. 6 Propionate. . .. 18 18 Sulphate . Butyrate. . 25 18 6.4 Xio- 2 6 Xio~ 2 5-9 Xio- 2 4.6 Xio- 2 2. 33 Xio- 2 2.57XIQ- 2 2.23X ~ 2 Valerate 18 Benzoate 14. Salicylate 15 Chromate 18 Carbonate. ... 25 Sulphide 18 9. 5 Xio~ 3 7-7Xio- 3 3.9X10-" i. 7 Xio- 1 1.2X10-" 4 SOLUBILITY PRODUCTS. Sub. / Hg 2 (SCN) 2 25 Hg 2 I 2 25 Hg 2 Br 2 25 Hg 2 Cl 2 . 25 HgCl 2 -. ... 25 HgBr 2 25 HgI 2 25 Hg(OH) 2 17 Zn(OH) 2 17 Ni(OH) 2 17 Co(OH) 2 17 Cd(OH) 2 17 Solubility. i-7Xio- 7 2.5Xio- 10 5-6XIO- 8 o.SXio-- 6 0.262 Cone, of Ions. Hg 2 " Hg" Hg" Hg" = 3Xio- 10 = 7 Xio- 8 = i Xio- 6 = 2 Xio i.SXio- 20 I.2XIO- 28 i. 3 Xio- 21 3-5Xio- 18 2.6Xio- 15 8 Xio- 20 3-2Xio- 29 i.SXio- 26 4-SXio- 1 2.iXio- 2 5.4Xio- 3 2.7X10-2 SOLUBILITIES AT i8 c BaF 2 SrF 2 CaF 2 MgF 2 PbF 2 : BaSO 4 SrSO 4 BaCrO 4 PbCr0 4 BaC 2 O 4 -2H 2 O. .. . SrC 2 4 CaC 2 O 4 - 2 H 2 O. . . . MgC 2 O 4 . 2 H 2 O. . .. ZnC 2 O 4 -2H 2 O 8.0 CdC 2 4 - 3 H 2 27.0 / 18 Xio c 1530 . 172 . 40 . 224 431 2.4 . 127 3.2 o. i 78-3 54 9.6 200 Mg. Eq. per Liter. 18.4 1.87 0.42 2.8 5-2 0.020 1.24 0.03 O.OOI o. 76 0.52 0.087 5.36 0.083 TABLES. 491 HYDROLYTIC DISSOCIATION. NH 4 N0 3 . KCN. IT Per Cent Per Cent ,, Per Cent V at 100. at 85. at 25. 8 o . 0498 0.0321 i < 0.31 16 0.0776 0.0563 4 0.72 CuCl 2 * 10 I. 12 2 0.313 0.252 40 2-34 8 0.407 0.276 Na 2 C0 3 . 32 0.612 0-373 5 2. 12 128 0.888 0.5*5 10 3-17 Cu(N0 3 ) 2 * 20 4-87 8 0.648 0-530 40 7.10 128 1.02 0.660 Potassium Phenate. HgCl 2 * 10 3-5 8 0.6 0.451 50 6.65 32 1.35 1. 06 Borax. 128 2.99 1.29 32 0.92 512 6.91 4.87 ] VaC-jHgO,. PbCl 2 .* 10 0.008 128 0.768 0.495 128 0.742 0.482 * V is here the volume in liters containing i equivalent weight in grams. 492 TABLES. ELECTRICAL CONDUCTIVITY (MOLECULAR) Solvent. Salt. Specific Conduc- tivity of Solvent. Allylalcohol FeCl 3 6.5Xio~ 6 Benzylalcohol FeCl 3 i.8Xio~ 6 Paraldehyde FeClo < 7 4.XlO~ 7 SbCl 3 Salicylaldehyde FeCl 3 6 oXio~ 6 Furfurol FeCl 3 2.4Xio~ 5 M^ethylpropylketone FeCl 3 qXlO~ 7 Acetophenone FeCL i 8Xio~" 7 Ethylmonochloracetate < < FeCl 3 SbCl 3 CuC) 2 \j . / w 81 .38 Oj 4. 71 5-6 45.6 o / " 20 78 80.98 T 1 / x 22 . 2 149 . 2 I 26 . 42 13.64 ^w . /_> 28.25 22.83 3I-07 53-36 36.98 100.71 42.76 23.46 10.28 65-77 n-59 124.91 12.03 292.98 13.08 7.76 12-45 22.09 13-75 92.05 16.38 152-55 17.88 42 ^ O 174. ii .49 o . 20 1 4.4. 73 O 337 ^o 13 32 *-* - 1 - /*T I . 24 T-T- / o '' oo / O * O 15-3 8.88 27.08 9.29 44.64 9-8 185.22 "57 10.62 4.19 28.3 5-29 58.57 6.46 110.97 7.66 29.85 7.00 41.67 7-5 59-4 8.08 97-77 12.8 13.15 5-88 42.29 5-92 94-79 6.25 342.85 7-7 20.4^ i "" m. 28 ^ 1.61 CI? 21 i . 91 V TO 21.34 i-54 38.74 i-74 j / 264. 16 3.00 644.56 3-73 8.5 0.8 34.02 0.96 136.07 1.07 272.14 i . ii 460 3 67 -0 *1 . S I . wy 10.94 o "/ 8-37 25-99 10.74 74.28 13-32 201.43 15-24 3-4 0.056 6.55 0.088 19.82 0.244 39. ic 0-39 10.86 6.86 84.77 12-55 448.14 19.00 814.8 18.2 2. I 3-37 16.3 S- 2 24.1 7.66 44.62 10. 12 6.1 7.96 24.58 6.85 93-7 5-9 159-55 5-57 7-55 24.1 5 x -43 29-5 392-3 40. 16 784.6 45-21 14.64 4.78 38.4 5-42 100.47 6.16 393-o 6.^7 2 I . I 0.8 1 06 j . C7 168.5 32^ o 08 Oo V" 1.16 o ^o e -j c \j . y_> O .OI2 I 3 . I 0.014 3^7 d 0.53 o oj 21.8 93- 6 1.4 o J / T- 2OO . I 2 7 - 3 j 14.64 I ^O ^ . O 3C 3 * i O"T^ * 7 OJ -0 74. 06 O.2 80? 3 I 4.^ / *T * wvy 4.24 0.368 2 o Acetylchloride, CHgCOCl . 15.4 Methyliodide, CH 3 1 7.2 Ethylacetate, CH 3 COOC 2 H 5 5 . 8 Chloroform, CHC1 3 5.2 Ether, (C 2 H 5 ) 2 4.36 Benzene, C 6 H 6 2.29 Toluol, C 6 H 5 CH 3 2.31 Aniline, C 6 H 5 NH 2 7.31 Chinolin, CH 7 N 8.9 Benzylcyanide, C 6 H 5 CH 2 CN 15 Benzonitrile, C 6 H 5 CN 26 Nitrobenzene, C 6 H 6 NO 2 . .. 36.45 TABLES. 495 LOGARITHMS. Proportional Parts. 10 11 12 13 14 15 16 17 18 19 1 > A 0128 0531 0899 1239 1553 1847 2122 2380 2625 2856 A 0170 0569 0934 1271 1584 1875 214* 2405 2648 2878 5 6 7 8 9 123 456 789 oooo 0043 0414 0453 0792 0828 H39 "73 1461 1492 1761 1790 2041 2068 2304 2330 2553 2577 2788 2810 3010 3032 3222 3243 3424 3444 3617 36*6 3802 3820 3979 3997 4150 4166 43144330 4472 4487 4624 4639 477i : 4?86 4914 4928 5051 5065 5185 5198 5315 5328 5441 5453 5563 5575 5682 5694 5798 5809 5911 5922 0086 0492 0864 I 206 1523 1818 2095 2355 2601 28.^ 0212 0607 0969 1303 l6l4 1903 2175 2430 2672 2900 3^18 3324 3522 37 1 1 3892 4065 4232 4393 4548 4698 4843 4983 5H9 5250 5378 5502 5623 5740 5855 5966 6075 6180 6284 6385 6484 6580 6675 6767 6857 6946 0253 0645 1004 1335 1644 1931 2201 2455 2695 29^3 0294 0682; 1038 13671 1673 1959 2227 2480 2718 2945 3160 3365 3560 3747 3927 4099 4265 4425 i579 4728 4871 Son 5145 5276 5403 5527 5647 5763 5877 5988 0334 0719 1072 1399 1703 1987 2253 2504 2742 2967 3181 3385 3579 3766 3945 4116 4281 4440 4594 4742 4886 5024 5159 5289 54i6 5539 5658 5775 5888 5999 0374 0755 1 106 1430 1732 2014 2279 2529 2765 2989 3201 3404 3598 3784 3962 4133 4298 4456 4609 4757 4812 4811; 3 7 10 3 6 10 369 368 35 8 5 7 5 7 4 7 17 21 2 S 15 19 23 14 I? 21 13 1 6 19 12 15 18 ii 14 17 ii 13 16 10 12 15 9 12 14 9 ii 13 29 33 37 26 30 34 24 28 31 23 26 29 21 24 27 2a 22 25 18 21 24 17 20 22 16 19 21 16 18 20 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 3054 3263 3464 3655 3838 4014 4183 4346 4502 4654 4800 4942 5079 5211 5340 5465 5587 5705 5821 5933 3075 3284 3483 3674 3856 4031 42OO 4362 4518 4669 4814 4955 5092 5224 5353 5478 5599 5717 5832 5944 3096 3.304 3502 3692 3874 4048 4216 4378 4533 4683 3139 3345 354i 3729 3909 4082 4249 4409 4^64 4713 4857 4997 5263 5391 5514 5635 5752 5866 5977 6085 6191 6294 6395 6493 6590 6684 6776 6866 6955 4 6 4 J 4 6 4 6 4 5. 3 5 3 5 3 5 3 5 3 4 8 ii 13' 15 17 ig 8 10 I2 ( i4 16 18 8 10 12,14 15 17 7 9 11,13 IS 17 7 9 10 12 14 15 7 8 io!ii 13 15 6 8 911 13 14 6 8 9 ii 12 14 6 7 9 10 12 13 4829 4969 5105 5237 5366 5490 5611 5729 5843 5955 6064 6170 6274 6375 6474 6571 6665 6758 6848 6937 4900 5038 5172 5302 5428 5551 5670 5786 5899 6010 3 4 3 4 3 4 679 6 7 8 678 10 1 1 13 10 II I 2 9 II 12 3 4 4 4 3 3 3 568 5 6 7 5 6 7 5 6 7 5 6 7 457 9 10 1 1 9 10 n 8 10 ii 8 9 10 8 9 10 8 g 10 6021 6031 6128 6138 6232 6243 6335 6 345 6435 6 444 6532 6542 6628 6637 6721 6730 6812 6821 6902^691 1 6042 6149 6253 6355 6454 6& 6739 6830 6920 6053 6160 6263 6365 6464 6561 6656 6749 6839 6928 6096 6201 6304 6405 6503 6599 6693 6785 6875 6964 6107 6212 6314 6415 6513 6609 6702 6794 6884 6972 6117 6222 6325 6425 6522 6618 6712 6803 6893 6981 7067 7152 7235 73i6 739<5 7474 7551 7627 7701 7774 3 I 3 ^ 4 5 6 4 5 6 4 5 6 4 5 6 4 5 6 456 8 9 10 7 8 9 7 8 9 7 8 9 789 7 8 9 3 ! 455 445 445 678 6 7 8 678 50 51 52 53 54 55 56 57 58 59 6990 6998 7076:7084 7160 7168 7243*7251 7324 7332 7404 7412 7482 7490 7559 7566 7634 7642 7709 7716 7007 7093 7177 7259 7340 7419 7497 7574 7649 7723 7016 7101 7i85 7267 7348 7427 7505 7582 7057 7731 7024 7110 7193 7275 7356 7435 7513 7589 7664 7738 7033 7118 7202 7284 7364 7443 7520 7597 7672 7745 7042 7126 7210 7292 7372 7451 7528 7604 7679 7752 7050 7135 7218 7300 738o 7459 75 -< 7612 7686 7760 7059 7143 7226 738 7388 7466 7543 7619 7694 7767 3 2 2 1 i 2 : 345 3 4 3 4 3 4 3 4 3 4 3 4 3 4 3 4 3 4 678 678 6 7 7 667 667 5 6 7 5 \ 7 5 6 7 5 6 7 5 6 7 496 TABLES. LOGARITHMS Continued. Proportional Parts. j 60 61 62 63 64 65 66 67 68 69 7782 7853 7924 7993 8062 8129 8i95 8261 8325 8388 1 2 3 4 .|. 7 8 ! 9 1 1 123 456 789 566 566 566 556 5 5 6 5 5 6 556 5 5 6 456 4 5 6 7789 7860 7931 8000 8069 8136 8202 8267 8331 8395 7796 7868 7938 8007 8075 8142 8209 8274 8338 ^40 1 7803 7875 7945 8014 8082 8149 8215 8280 8344 8407 7810 7882 7952 8021 8089 8156 8222 8287 8351 8414 7818 7825 7832^839 7846 7889 7896 793 7910 7917 7959 79667973 7980 7987 8028 8035 8041 8048 8055 8096 8102 8109 8116 8122 8162 81698176 8182 8189 8228 8235 8241^8248 8254 8293 8299 ( 83o6 8312 8319 8357,8363,8370 83768382 8420^426843284398445 344 344 334 334 334 334 334 334 334 234 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 8451 8513 8573 8633 8692 875i 8808 8865 8921 8976 8457 8519 8579 8639 8698 8756 8814 8871 8927 8982 9036 9090 9143 9196 9248 9299 9350 9400 9450 9499 8463 8525 8585 8645 8704 8762 8820 8876 8932 8987 8470 853T 8591 86<;i 8710 8768 8825 8882 8938 8.993 8476 8S17 8597 8657 8716 8774 8831 8887 8943 8998 8482^488 8494'8soo 8506 8S438S49!8SSS 8561 8567 8603 8609 8615 8621 8627 8663 8669 8675 8681 8686 8722 8727 8733 8739 8745 87798785(8791 8797 8802 8837 8842:884888548859 8893^899 8904 8910 8915 8949^954 8960 8965 8971 9004 9009 9015 9020 9025 3 4 3 4 3 4 3 4 3 4 4 5 6 455 455 455 455 3 3 3 3 3 3 3 3 455 445 445 445 445 445 445 9031 9085 9138 9191 9243 9294 9345 9395 9445 9494 9042 9096 9149 9201 9253 9304 9355 9405 9455 9504 9047 9101 9i54 9206 9258 9309 936o 9410 9460 9509 9557 9605 9652 9699 9745 9791 9836 9881 9926 9969 9053 9106 9159 9212 9263 9315 9365 9415 9465 9513 9562 9609 9657 9703 975 9795 9841 9886 9930 9974 905819063 9H2 | 9ii7 9165 9170 9217 9222 9269 9274 9320^325 9370,9375 9420,9425 9469 9474 95189523 9069 9074 9079 9122 9128 9133 9175 9180 9186 9227 9232 9238 9279 9284 9289 9330:9335 9340 I I 3 3 3 3 3 3 445 4 4 5 445 344 344 344 9430 9435 9440 9479 9484 9489 9528I9533 9538 o 3 3 3 9542 959 9638 9685 9731 9777 9823 9868 9912 9956 9547 9595 9643 9689 9736 9782 9827 9872 9917 996i 9552 9600 9647 9694 974i 9786 9832 9877 992i 9965 9566 9571 9614 9619 9661 9666 9708^713 9754'9759 980019805 9845 9850 9890 9894 9934 9939 99789983 9576 9581 9586 9624 9628 9633 9671 9675 9680 9717 9722 9727 9763 9768 977: 9809 9814 9818 9854 9859 9863 9899,9903 9908 9943 9948 9952 99879991 9996 o o o 3 3 3 3 3 3 3 3 3 3 344 344 344 3 4 4 344 344 344 344 344 APPENDIX. THE OSMOTIC PRESSURES AND FREEZING-POINTS OF SOLUTIONS OF CANE-SUGAR. In a paper which has just appeared (July), Morse and Frazer (Am. Chem. Jour., 34, i, 1905) give their results for the osmotic pressures and freezing-points of cane- sugar solutions of varying concentration. These results show that the van't Hoff law (p. 134) holds throughout for solutions of sugar, ij the volume of ike pure solvent is substituted, in the law, jor that oj the solution. In their words: "When we dissolved a gram-molecular weight of cane-sugar (342.22 grams') in 1000 grams of water, i.e., in that mass of solvent which Jtas the unit volume, i liter, at the temperature of maximum density we found its osmotic pressure, at about 20, in quite close accord with the pressure which a gram-molecular weight oj hydrogen would exert, at the same temperature, ij its volume were reduced to i liter, i.e., to that volume which the unit mass oj solvent has at the temperature oj greatest density. " 497 49^ APPENDIX. For cane-sugar, then, and presumably for all other substances, we can express the law as follows: A sub- stance in solution "exerts an osmotic pressure equal to thai which it would exert if it were gasified at the same temperature and the volume oj the gas were reduced to that oj the solvent in the pure state." Or, in other words, it "exerts an osmotic pressure throughout the larger volume oj the solution equal to that which as a gas it would exert ij confined to the smaller volume oj the pure solvent" It will be observed here that the two forms of the law will only differ for concentiated solutions, and the stronger the solution, the greater will be the difference. So long as the amount of substance dissolved is relatively small, the volume of the solution will be practically equal to that of the pure solvent, and no difference in osmotic pressure will be observed. As the amount increases, however, the difference between the volume of solvent and solution will become greater and greater, and the osmotic pressures, according to the two laws, will diverge more and more. Our definition of molecular weight in solution (p. 135) should then be slightly altered in that the 22.4 liters should be the volume of the pure solvent and not that of the solution. The error in the first portion of the definition, however, will be exceed- ingly slight, but will increase as the volume decreases, so that the second part, when it refers to a smaller APPENDIX. 499 volume than 22.4 liters, can only be used when the volume taken is that of the pure solvent. In the light of this new form of the van't Hoff law the abnormal results, mentioned on page 135, become perfectly normal, as a glance at the following table will show. OSMOTIC PRESSURES AND MOLECULAR WEIGHTS. SUGAR SOLUTIONS AT ABOUT 20. Weight- normal. Moles in 1000 gr. H 2 O (wf. Volume- normal. Moles per Liter. Pressures at Same Temperature. ^- = Gaseous. Osmotic (P). P 0.05 o . 04948 I. 21 1.26 327-5 0.10 0.09794 2.40 2.44 336.9 O.2O o. 19192 4.82 4.78 345-2 0.25 0.23748 6.06 6.05 342-9 0.30 0.28213 7.22 7-23 342.0 O.4O 0.36886 9.68 9.66 343-1 0.50 0.45228 12.07 12.09 341-7 O.6o 0-53252 14.58 14.38 347-1 0.70 0.60981 17.16 17.03 344-8 0.80 o . 68428 19.17 19.38 338.5 0.89101 o . 75000 21.48 21.21 346.5 0.90 0.75610 21.73 21. 8l ' 340.0 I .00 0.82534 24.27 24-49 339-2 Mean. . . 341.2 These are but a few of the many values determined by the authors at these concentrations. The slight varia- tions in the calculated and observed values they assume to be due to experimental errors which will be eliminated in future measurements. The law in this form, then, is to be regarded as holding accurately for solutions between these limits. Whether stronger solutions will Soo APPENDIX. show abnormal pressures, as do gases, is not as yet known, but further work for this purpose will undoubtedly be undertaken in the near future. The authors have also studied the freezing-points of sugar solutions. The laws already mentioned above (pp. 176-187) have been found for some unknown reason to be unsuited to strong sugar solutions, but by the following relation, found by them, the freezing-point may be calculated for all concentrations. Calling IF the concentration of the solution according to the weight- normal standard, i.e., the fraction of a mole of sugar which is dissolved in 1000 grams of water, D the density of the solution at its freezing-point, i.e., the relative weights of equal volumes of solution and pure solvent near the temperature of freezing, and J the depression of the freezing-point, they find J = i.S$WD for all concentrations, where 1.85 is the molecular depression of the freezing- 0.02 T 2 point of water, as calculated from the formula K = (p. 179). For dilute solutions this relation is just what we found it to be above, for the density D in such a case will be practically unity. For strong solutions, however, J, as calculated, will be larger than that by the simpler relation. The agreement between the cal- culated and observed values of J is exceedingly close as may be seen from the results below. APPENDIX. 5 01 FREEZING -POINTS OF SUGAR SOLUTIONS. W. D. A Calc. A Obs. O. 10 O.20 0.30 0.40 0.50 0.6o O.7O 0.80 O.OXD I .00 .0129 o.i87 o.i87 .0257 o.379 o.373 .0380 o.576 o.574 .0497 o-777 o-776 .0611 o.98i o.97o .0717 i.i89 i.i87 .0825 i.4oi i-398 .1016 i-834 i-837 .1110 2. 060 2. 082 1.26806 1-1340 2. 660 2. 660 "The last solution (the 1.26806 weight-normal one) is a volume-normal solution, i.e., a solution containing a gram-molecular weight of sugar in a liter volume. The depression of the freezing-point of this solution has been regarded as abnormal by the difference between i.85 and 2.66, but it will be noticed that it conforms to the rule with the same degree of precision as the o.i weight-normal solution." The deviations for the 0.5 and i.o weight-normal solutions may be due to experi- mental errors, according to the authors, for the concen- trations on either side show close conformity to the rule. The authors also find a relation existing between the freezing-point and the osmotic pressure of a solution of sugar. Calling P the theoretical osmotic pressure at o, D, as before, the density, and J the observed depres- sion of the freezing-point, then -^ =0.082. From this, of course, knowing either A or P we can calculate P or 502 APPENDIX. J. The application of this relation for the two purposes is shown in the tables below. FREEZING-POINTS OF SUGAR SOLUTIONS. Concentra- tion Weight- normal. Osmotic Pressure at o. At- mospheres. Density of Solution at its Freezing- point. Observed Lowering of Freezing- point. Osmotic Pressure at o X Den- sity X 0.082. O.I 2.25 1.0129 o.i87 0.l87 0.2 4-5 1.0257 o.373 o.378 0-3 6.75 1.0380 o-574 o-574 0.4 9.00 1.0497 o. 77 6 o.774 o-5 11.24 I .0611 o. 9 6 9 o. 97 8 0.6 13-49 1.0717 i.i87 i.i8s 0.7 15-74 1.0825 i. 39 8 i-397 0.8 17.99 1.0918 I.6l2 i.6io 0.9 20.24 i . 1016 i.8 3 7 i.8 2 8 I.O 22.49* I . I I 10 2. 082 2. 049 I . 26806 28.76 1.1340 2. 660 2. 660 OSMOTIC PRESSURES OF SUGAR SOLUTIONS. Values of Concentra- tion. Weight- normal. Observed Lowering of Freezing- point. Density of Solution at its Freezing- point. Theoretical Gas or Osmotic Pressure at o. 0.082/7' ' '' Calculated Osmotic Pressure. Atmospheres. 0. I o.i8 7 I .0129 2.25 2.25 0.2 o-373 1.0257 4-50 4-43 0-3 o-574 I . 0380 6-75 6-75 0.4 o . 77 6 1.0497 9.00 9.01 0-5 o.969 I .0611 11.24 11.13 0.6 i.i87 1.0717 13-49 13-5 0.7 i. 39 8 1.0825 15-74 J5-75 0.8 I.6l2 I .0918 17.99 18.00 0.9 i.8 3 7 i . 1016 20.24 20.33 I .0 2. 082 I . IIIO 22.49* 22.85 i . 26806 2. 660 1.1340 28.76 28.82 The pressures marked with an asterisk considered as 22.4 atmospheres, for which (*) are those which we have the authors use 22.488. INDEX. Adiabatic compression and expansion 45-48 Ampere 364 Analytical reactions 352-363 Association, Factor of 81-85 Atomic and combining weights 6-8 Atoms in molecule, Number of 49 Avogadro's law 15 Babo, Law of 153 Basicity of an acid 386-387 Boiling-point 65-68 , Determination of the '174 , Increase of the 169 , Molecular increase of the 169 Boyle's law 9 Calorie, Large 202 , Ostwald 202 , Small 202 Capacity factor of energy 3 Catalytic action of hydrogen ions 268-272 Cell, Bichromate 437~438 , Chemical or thermodynamical theory of the 402-405 , Clark 435-436 , Concentration 422-427 , Daniell 401, 402, 404 , Gas 404, 424, 431 504 INDEX. PAGE Cell, Helmholtz 400 , Leclanche" 436 , One-volt 400 , Osmotic theory of the 404, 414 , Storage 438, 439 , Weston 400 Charles, Law of 9 Chemical change 222 at constant pressure 208 volume 207 kinetics 261 Coefficient of affinity 283 partition or distribution 188-189, 251 Color of solutions 346 Conductivity, Determination of the 376 of mixtures 396 neutral salts 387 organic acids 381 , Solubility by aid of the 39 I- 392 , Temperature coefficient of 390 , The molecular and equivalent 375 , The specific 374 Constant, Dissociation 283, 296 lonization 296 of a reaction of the ist order, The 265 2d " " 272 3d " " 276 hydrolytic dissociation, The 32 1 Coulomb 364 Cycle, The 54 Dalton's law 13 Decomposition of H 2 O, Primary 447 values 440-442 Dielectric constant and dissociating power 393 Dissociation and external pressure 33 , Constant of 233-264, 281-306 , Degree of . 23, 146, 283, 297 , Electrolytic ,> 146 from E.M.F 427 INDEX. 505 PAGE Dissociation, Gaseous 23, 233 , Heat of, of solids 255 gases. 259 , Hydrolytic 316-329 solution, Non-electrolytic 250 of double molecules in solution 251 H 2 313, 388, 430-432 polyatomic molecules 26-28 Dissociating power and dielectric constant 393 Dilution law, Bancroft's 297 Ostwald's 282 Rudolphi's 296 van't Hoff's 296 Distribution of a substance between two solvents 188-189, 251 Dyne 2 Electric conductivity, see Conductivity. Electricity, Mechanical equivalent of 365 , Thermal equivalent of 365 Electrochemistry 364-452 Electrodes, Concentration of 422-426 , Normal 417-418 Electrolysis 440 Electrolytic solution -pressure 409, 433 Electromotive force 399-440 Energy, Electrical 5, 365 , Factors of 3 , Factors of heat 59 , Kinetic 5 , Mechanical 2 , Potential 2 , Volume 4 Entropy 59 Equation of equilibrium 227, 282, 296, 297 state for gases 12 van der Waals 35 Equilibrium 222 and external pressure 33 between gases and solids 241-245 , Constant of 228 506 INDEX. Equilibrium, Effect of temperature upon 252 Erg 2 Evaporation, Heat of 68-76 Faraday, Law of 366 Film, Semipermeable. 128 Force 2 Freezing-point and external pressure 93 -96 vapor-pressure 180-181 , Depression of the ... 176 , Determination of the 184-185 , Molecular depression of the 176 of mixtures 396 Fusion, Latent heat of . . 93-103 Gas constant, Molecular 12 , Specific ii , Equation of state for a 12 laws 9-1 5 Gases and liquids, Distinction between 61 in liquids. . . 116-1.18 , Specific gravity of 15 . Determination of specific gravity of 16-21 Gay-Lussac, Law of 9 Heat, Atomic 9 J ~93 , Capacity for 39 , Determination of specific S-52 , Latent 68-69, 7 2 ~?6, 98-103 of association of the ions of water. 216 dissociation 255-256, 259-261, 421 evaporation 69, 72, 254 formation 2 5> 206 ionization. c 301, 421 neutralization 216, 304 precipitation 218-219 solution 203 vaporization . 68-76, 256-260 , Specific 38, S Hess, Law of 200 INDEX. 507 PAGE van't Hoff, Law of 134 Hydrolytic dissociation, Constant of 321 Indicators, Action of 348 Intensity factor 2 Ionic concentrations^ 426 equilibria 281, 334-348 lonization 139-152, 281-316 and solubility 305-316 constant 283-301 from increased solubility 3 2 9~334 Ions 139-152 and analytical chemistry 352-363 in thermal reactions 214-221 , Migration of the 364-374 Isohydric solutions 291 Joule 203 Kohlrausch, Law of 378 Liquids in liquids . '. 1 18-124 Liter atmosphere , 13, 41 Mariotte's Law 9 Mass action, Law oi. . . .^ 228 Migration of the ions, Absolute velocity of 385, 386 in terms of conductivity 381 Mixtures, Conductivity of 396 , Freezing-point of 396 Mole 12 Molecular weight from depressed solubility. 189-191 depression of freezing-point 176-185 depression of vapor-pressure 154 distribution between two solvents 251 heat of evaporation 69 increase of the boiling-point 118 osmotic pressure 135 surface tension 79~&5 vapor density 16 law of Young & Thomas 70 508 INDEX. PAGE Non-homogeneous systems 241, 279 Ohm 364 Osmotic pressure and molecular weight 135 vapor-pressure 161-168 , Simple demonstration of 137 theory of the cell 404, 414 Partition, Coefficient of 188-189, 251 Perpetual motion of the ist kind 41 2d " 53 Phase rule . . . ; 104-115 Physical chemistry i Point, Transition in , Triple in Polarization, Theory of 442-447 Pressure, Critical. . ,. 63 , Electrolytic solution 409, 433 , Molecular depression of the vapor 154 , Osmotic 128^138 , Relative depression of the vapor 154 , Solution 126, 152 , Vapor 153, 161, 180 Problems 453-483 Raoult, Law of 154 Reactions between solids and liquids 279 , Incomplete , 278 of the ist order. ... ... 265 2d " . = 272 3d " 276 , Reversible 222 Separation of metals by graded E.M.F.'s 449 Solid state, The 91-103 solutions 192 Solids in liquids 124 Solution, Heat of ... 203, 257 INDEX. 59 PAGE Solution pressure of metals, Electrolytic 409, 433 Solutions 116 , Color of 346 , Isohydric 291 , Saturated 125-128 , Solid 192 , Vapor-pressure of 153, 161 Solubility, Depressed 189-190, 308-313 , Increased 311 from conductivity 39i~393 E.M.F 429-430 , Law of 312 of sulfids, Law of 358 Specific gravity in gaseous form 13-21 Speed of reaction and temperature 280 Sublimation, Heat of 78-80, 102 Sugar, Inversion of 265-268 Surface-tension 78-85 and critical temperature 86-96 molecular weight 78-85 Tables 485-496 Temperature, Absolute 10 , Critical 63-65, 70 Thermochemistry 200 , Definition of 20 Thermodynamics, ist principle of 38 2d " " 52-59 Trouton, Law of 69 Units, Electrical 364 Van der Waals, equation of 35 Vapor-densities, Abnormal 21-29 Vapor-pressur?, see Pressure. Volume, Critical 63-65, 70 , Molecular 12, 80 Volt 364 5io INDEX. PAGE Water, lonization of r - - - 313, 388, 43Q-432 Work ^ , Maximum 2 Wiillner, Law of 153 Zero, Absolute 10 UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY Return to desk from which borrowed. This book is DUE on the last date stamped below. APft 8 1! 26Mar'57Kl REC'D LI APR 12 LD 21-100m-9,'47(A5702sl6)476 YB 16909 <} UNIVERSITY OF CALIFORNIA LIBRARY