PHYSICO-CHEMICAL CALCULATIONS BY JOSEPH KNOX, D.Sc. LECTURER IN CHEMISTRY, UNIVERSITY OF ABERDEEN NEW YORK D. VAN NOSTRAND COMPANY 25 PARK PLAGE 1916 Kb PREFACE THIS collection of physico-chemical problems is based on Abegg and Sackur's " Physikalisch- Chemische Rechen-aufgaben " (Sammlung Goschen). The original intention was simply to translate the German book, which consists of a short summary of the laws and formulae used in the problems, and fifty-two typical problems, with full solutions. With the consent of the late Professor Abegg and of Dr. Sackur, however, I decided to arrange the subject-matter in chapters deal- ing with the main subdivisions of physical chemistry, and to write a short introduction to each chapter, deal- ing with the theory involved in the problems. Most of the problems in the " Kechenaufgaben " have been re- tained, a good many additional solved problems have been introduced, and a collection of problems for solution (with answers) has been added at the end of each chapter. The size of the book has thus been more than doubled. Most of the problems have been taken direct, or with slight modification, from the original literature. I shall be grateful to have any errors in the text pointed out to me. I would take this opportunity of expressing my in- debtedness to the late Professor Richard Abegg, and to Dr. Otto Sackur, of the University of Breslau, for their kindness in allowing me to make use of their " Kechen- aufgaben " as the basis of this book. J. K. ABERDEEN, December, 1911. , 362452 CONTENTS PAGE CHAPTER I. Gas Laws Gaseous Dissociation Osmotic Pressure Examples Problems for Solution 1 CHAPTER II. Density and Specific Volume of Solids, Liquids, Liquid Mixtures and Solutions Examples Problems for Solution ... 14 CHAPTER III. Specific and Molecular Refractivity Examples Problems for Solution 19 CHAPTER IV. Molecular Weight from Lowering of Vapour- pressure Influence of Temperature on Vapour-pressure Examples Molecular Weight from Lowering of Freezing-point Molecular Lower- ing of Freezing-point from Latent Heat of Fusion Molecular Weight from Elevation of Boiling-point Molecular Eleva- tion of Boiling-point from Latent Heat of Evaporation Examples Problems for Solution - 24 tJHAPTER V. Surface Tension, Molecular Weight and Degree of Association of Liquids Examples Problems for Solution - 41 CHAPTER VI. Thermochemistry Examples Problems for Solution 45 CHAPTER VII. Velocity of Reaction Monomolecular Reaction Bimolecular Re- action Examples Problems for Solution .... 53 vii viii PHYSICO-CHEMICAL CALCULATIONS CHAPTER VIII. Law of Mass Action Equilibrium-constant Influence of Temperature on Equilibrium-constant Affinity, Change of Free Energy or Maximum Work of a Reaction Partition Law Solubility of Gases Examples Problems for Solution 62 CHAPTER IX. Ohm's Law Heating Effect of Current Faraday's Laws Examples Specific, Equivalent, and Molecular Conductivity of Electrolytes Degree of Dissociation Dissociation-con- stant Examples Transport Numbers Examples Solu- bility-product Examples Hydrolysis Examples Pro- blems for Solution --------- 100 CHAPTER X. Electromotive Force Electrode Potential Normal Potential Concentration Cells Electromotive Force of Galvanic Elements Diffusion Potential Oxidation-reduction Poten- tial Affinity, or Maximum Work of a Reaction in a Galvanic Element Gibbs-Helmholtz Equation Examples Pro- blems for Solution - - - - - - - 149 CHAPTER XL Diffusion Examples Radio-activity Examples - - - 182 INDEX 187 PHYSICO-CHEMICAL CALCULATIONS CHAPTBE I GAS LAWS GASEOUS DISSOCIATION OSMOTIC PKESSUBE Gas Laws THE relation between the pressure, volume and tempera- ture of any given mass of gas is expressed by the equation / v Pf> P'v' ~T = ' "2*" where P and v are the pressure and volume corresponding to the absolute temperature T and P' and v' the pressure and volume corresponding to the absolute temperature T. The absolute temperature is equal to 273 + t, where t is the tem- perature centigrade. The pressure and volume may be ex- pressed in any units. This relation may also be expressed in the form Pv jp = constant, or Pv = constant x T. The value of the constant varies with the units of pressure and volume and with the mass of gas considered. For the gram-molecular quantity of all gases, however, the value of the constant is the same. For a gram-molecule of a gas the equation, therefore, becomes Pv = RT, and for n gram-molecules (2) Pv = nET (Boyle, Gay-Lussac, Avogadro). 1 2 GAS. LAWS OSMOTIC PEESSUEE P is the pressure of the gas, n th'e number of gram -molecules or moles present in the volume v, R the gas -constant and T the absolute temperature. The numerical value of R varies with the units of pressure and volume adopted. If we take the atmosphere as the unit of pressure and the litre as the unit of volume, we obtain the corresponding value of R from the fact that one gram- molecule of any gas (n = 1) occupies 22-4 litres at C. (273 absolute) and under a pressure of one atmosphere. Then, from (2), 09-J. 1 x 22-4 = 1 x E x 273 .-. R =44r = 0-08204. 410 If the pressure is given in millimetres of mercury and the volume in cubic centimetres, the former must be divided by 760 to convert it to atmospheres and the latter by 1000 to convert it to litres before this value of R can be used. If the total pressure P of a mixture of gases is replaced by the partial pressure p of any of its components, the same equations hold for each individual gas in the mixture, of which the total volume is v (Dalton's law). Osmotic Pressure For the osmotic pressure TT of a dilute solution of volume v, equations exactly analogous to the gas-equations hold, namely, f , irV ?rV (3) -y - -7jir> and (4) TTV = nRT. TT is the osmotic pressure, n the number of gram-molecules of dissolved substance in the volume v of solution, T the, absolute temperature and R a constant, the value of which is the same as in the gas-equation and numerically equal to '08204 if TT is measured in atmospheres and v in litres* The last equation may be written in the form : a* where c is the concentration of the dissolved substance, (e.g % gram-molecules per litre, if the above units are adopted). GAS LAWS EXAMPLES 3 Gas Laws Examples PROBLEM 1. At ' = 10C. and under a pressure of P = 650 mm. of mercury a certain mass of hydrogen occupies a volume of v' = 200 c.c. What volume, v, will it occupy at t = C. and P = 760 mm. pressure ? SOLUTION 1. Substituting the numerical values in equa- tion (i) we obtain 760 xv 650 x 200 273 283 650 x 200 x 273 760 x 283 - 165 c.c. PROBLEM 2. What volume will a = 1 gram of oxygen occupy at a temperature t = 100 C. and under a pressure P = 740 mm. of mercury ? SOLUTION 2. If M is the molecular weight of oxygen, the number of gram-molecules present is n = -^. Equation (2), therefore, becomes a aET Pv = RT, and v = 740 P = ^atmospheres, M = 32, E = Q'08204, T = 273 + t = 373. Substituting these values in the above equation we obtain 1 x 0-08204 x 373 x 760 v = ~ 32 x 740 - ' 9822 htres - PROBLEM 3. The pressure of the atmosphere at the surface of the earth is equal to the weight of a column of mercury 76 cms. high. With increasing height the density of the air diminishes according to Boyle's law. What is the density of the air at the heights \ = 1000 metres, and h 2 = 3000 metres, if the density S Q at the surface of the earth = 0-00129? (t = C.). SOLUTION 3. According to Boyle's law the volume of a gas at constant temperature is inversely proportional to the pressure on it ; its density, i.e. the mass contained in unit volume, is, therefore, directly proportional to the pressure. Let he pressure of the air at the height h metres = P,, and its 4 GAS LAWS EXAMPLES density = s. Then, since s is proportional to the pressure (1) s = kP t . Similarly at the height h + dh, (2) s + ds = kP s+dt . The difference of the pressures P t+df and P g is equal to the weight of a column of air of height dh (the difference in the two heights), and of unit area of cross-section, since the pres- sure of a mass of gas is defined as equal to the weight on unit area. Accordingly we have From (1), (2) and (3) it follows that ds = - k.s. dh, or d log, s = - k . dh, and, after integration, (4) log, s = - kh + constant. At the height h = we have s = s , the density of the air at the surface of the earth, and (4) becomes log, s = constant, and, therefore, (5) log e s = - kh 4- log e s . From (1 ) it follows that k is the density of air under unit pressure. The value of k can be calculated from the data given in the problem, i.e. from the value of the density s at a pressure P of 76 cms. of Hg. Using the cm. and the gram as our units P = 76 x 13-6 grams per sq. cm. (13'6 is the density of Hg), and s = 0-00129 gram per c.c., therefore, (V001 2Q k = 7 g x 136 =0-00000125 =1-25 x 1Q- C . Then from (5) we get, for the heights h t = 1000 m. = 10 5 cms., and h 2 = 3000 m. = 3 x 10 5 cms., log. S L = - 1-25 x 10~ x ftj + log e 0-00129, = - 1-25 x 10 - x 10 5 + log, 0-00129, and, changing to common logarithms, GAS LAWS EXAMPLES 5 log . where d v s i are the densities ' at the height h lt and d^ s the densities at the earth's surface. The carrying-capacity B l at the height 7^, is, therefore, GAS LAWS EXAMPLES and at the height h^ -A) s / (a) For hydrogen . = 0-0695, therefore, since v = s o 1400 x 10 6 c.c. and s l = 1-14 x 10 - 3 (of. Problem 3), for ^ = 1000 m. = 10 5 cm. B! = 1400 x 10 x 1-14 x 10 ~ 3 x 0'93 gm. = 1484 kg., and, similarly, B 2 = 1400 x 10 6 x 0-885 x 10 - 3 x 0'93 gm. = 1152 kg. (b) For coal gas A = 0'43, therefore s o B! = 1400 x 10 6 x 1-14 x 10 ~ 3 x 0*57 gm. = 910 kg., Bo = 1400 x 10 6 x 0-885 x 10 - 3 x 0-57 gm. = 706 kg. (c) Let the volume of the hydrogen balloon be v m then its carrying-capacity B = v x s(l - Y Similarly, for a coal gas balloon of the same carrying-capacit we have d n , d c and s are the densities at any given height of hydrogen, coal gas and air respectively. From these two equations we obtain ~S and, therefore, PROBLEM 5 (cf. preceding problem). If the balloon of 1400 c.m. capacity is filled completely at the earth's surface (a) with hydrogen, (b) with coal gas, how many grams of gas are in each case lost when it ascends to the heights h l = GASEOUS DISSOCIATION EXAMPLES 7 1000 ro. and h> 2 = 3000 m. ? How many kilograms of ballast must be thrown out so that the completely filled balloon may rise from its equilibrium position at the height /^ = 1000 m. to that at \ = 3000 m., (c) when filled with hydro- gen, (d) when filled with coal gas ? SOLUTION 5. The weight of gas lost on ascending to the height /i x is O J = v (d Q - d^ grams, and on ascending to the height h it is G 2 = v (d Q - d 2 ) grams, where d Q9 d l and d are the densities of the gas at the earth's surface, at the height /^ and at the height h 2 respectively. (a) For hydrogen d - d l = 0-0695 (0-00129 - O'OOIH) = 0-0695 x 0-00015 - 1-04 x 10 ~ 5 , and d - d< 2 = 0-0695 (0-00129 - 0-000885) - 0-0695 x 0-000405 ='2-81 x 10 ~ 5 , therefore the loss of weight G x = 1400 x 10 6 x 1-04 x 10 - 6 gm. = 14-56 kg., and G 2 = 1400 x 10 6 x 2-81 x 10 ~ 5 gm. = 39-3 kg. (b) Similarly, for coal gas G x = 1400 x 10 6 x 6-45 x 10 ~ 5 gm. = 90-3 kg., and G 2 = 1400 x 10 6 x T74 x 10 - 5 gm. = 243-5 k g- The weight of ballast which must be thrown out is equal to the difference between the carrying-capacities at the heights hj_ and /& 2 , and is, therefore, (c) for hydrogen, 1484 - 1152 = 332 kg. (d) for coal gas, 910 - 706 = 204 kg. The throwing out of a given weight of ballast has, there- fore, more effect on the coal-gas balloon than on the hydrogen balloon. Gaseous Dissociation Examples PROBLEM 6. At 90 C. the vapour-density of nitrogen per- oxide (N 2 4 ), is 24-8 (referred to H = 1). Calculate the degree of dissociation into N0 2 molecules at this temperature. SOLUTION 6. Let D be the theoretical vapour-density if no dissociation occurred, and d the observed vapour-density. If one molecule of the dissociating substance gives on dissociating n simpler molecules, and if a is the degree of dissociation, then at equilibrium 1 - a undissociated molecules and no. simpler molecules are present for every molecule of undissociated substance initially present. The total number of molecules is, therefore, 1 - a + na = 1 + a(n - 1). The volume 8 GASEOUS DISSOCIATION EXAMPLES occupied by the gram-molecular quantity of undissociated substance is, therefore, 1 + a(n - 1) times as great as it would have been, had no dissociation occurred. But the density of a given mass is inversely proportional to the volume, therefore, D i + a ( n _ i) -j = ~T- and _ (D - d) = d(n - 1) * In the dissociation of N. 2 O 4 according to the equation N 2 4 = 2N0 2 , n = 2, D = N2 4 = ^ = 46, therefore 2> z> (D -d) 46 - 24-8 PROBLEM 7. When a = 5 grams of ammonium carbarn ate NH 4 CO 2 NH 2 , is completely vaporised at t = 200 C., it occu- pies a volume v = 7'66 litres under a pressure P of 740 mm. of mercury. Calculate the degree of dissociation according to the equation NH 4 C0 2 NH 2 =2NH 3 + CO,. SOLUTION 7. If Mis the molecular weight of ammonium carbamate, a grams = gram-molecules. Let a be the de- gree of dissociation. Then since one molecule on dis- sociating gives 2a molecules of NH 3 and a molecules of CO 2 , whilst 1 - a molecules of undissociated substance remain, the total number of molecules at equilibrium derived from 1 molecule of undissociated substance is 1 - a + 3a = 1 + 2a molecules. From molecules of undissociated substance M there are, therefore, derived ~ (1 + 2a) molecules. The equation Pv = nRT, therefore, becomes Pv = l + OSMOTIC PRESSURE EXAMPLE 9 M = 78 and P = - atmos. Substituting the numerical i t>U values we, therefore, obtain x ? . 66 = 5 (1 + 2a) x . 08 204 x 473, 7bO 7o .'.a = 0*999. The compound is, therefore, completely dissociated. Osmotic Pressure Example PROBLEM 8. A solution of glucose containing a = 36 grams per litre gave an osmotic pressure TT = 4' 77 atmo- spheres at t = 25 C. What is the molecular weight M of glucose ? SOLUTION 8. If a is the weight of substance in solution, and M its molecular weight, the number of dissolved mole- cules is n = T,. Equation (4) therefore becomes and aET M irV Substituting the numerical values in this equation we obtain 36 x 0-08204 x (273 + 25) 4-77 x 1 = I8 4'5- Problems for Solution Gas Laws PROBLEM 9. At C. and 760 mm. pressure the volume occupied by 1 gram-molecule of oxygen is 22-4 litres. Cal- culate the numerical value of the gas-constant E when the units of energy are (1) the litre-atmosphere, (2) the gram- centimetre, (3) the erg, (4) the gram-calorie, (5) the joule. Given : density of mercury = 13*59, gravity-constant = 981, 1 gram-calorie = 42720 gram-centimetres = 4-183 joule. Ans. (1) 0-08204, (2) 84720, (3) 0-8312 x 10 8 , (4) 1-984, (5)J^301. ' PROBLEM 10. At 10 C. and under a pressure of 1000 10 GAS LAWS PROBLEMS grams per square centimetre, a certain mass of carbon dioxide occupies a volume of 30 cubic inches. What volume will it occupy at 20 C. and under a pressure of 850 grams per square centimetre? Ans. 36*55 cubic inches. PROBLEM 11. An open vessel at a temperature of 10 C. is heated at constant pressure to 400 C. What fraction of the weight of air originally contained in the vessel is expelled ? Ans. 0-5794. PROBLEM 12. If the vessel in the preceding problem had been closed, and originally contained air at atmospheric pres- sure, what pressure would have been developed on heating to 400 C.? Ans. 2-378 atmos. PROBLEM 13. From a porcelain bulb of volume 197'8 c.c. filled with air, 169'1 c.c. of air, measured at 10 C., were expelled on heating from 12 C. to t C. The pressure throughout the experiment was 747 mm. Neglecting the expansion of the bulb, calculate t. Ans. t -= 1772 G. PROBLEM 14. A bulb of 111*5 c.c. capacity connected with a mercury manometer was heated to 100 C. whilst open to the atmosphere. When the temperature had become con- stant, the atmospheric connexion was closed, and a capsule containing 0'0448 gram of acetone dropped into the heated bulb. The volume was kept constant by increasing the pres- sure by raising the reservoir of the manometer. The final difference of level was 16' 1 cms. Calculate the molecular weight of acetone. Ans. 57-65. PROBLEM 15. Half a gram of an organic compound of empirical formula CH 2 O gave 327 '6 c.c. of vapour at 200 C., and 750 mm. pressure. What is the molecular formula of the compound ? Ans. C 2 H 4 O 2 . PROBLEM 16. At "17 C. 40 grams of electrolytic gas are contained in a 60 litre gas holder. What are the partial pressures of hydrogen and oxygen ? Ans. ^ = 0-8812 atmos., p 0t) = 0-4406 atmos. PROBLEM 17. The composition of air by weight is 23 per GASEOUS DISSOCIATION PEOBLEMS 11 cent, oxygen and 77 per cent, nitrogen. What are the partial pressures of oxygen and nitrogen in a vessel of 1 litre capacity which eontains 2 grams of air at 15 0. ? Ans. p Q ^ = 0-3397 atmos., p^ = 1*299 atmos. PROBLEM 18.- What weights of hydrogen, oxygen, and nitro- gen are contained in 10 litres, measured at 18 C. and 750 mm., of a gaseous mixture, the volumetric composition of which is H 2 = 10 per cent., O 2 = 15 per cent., N 2 = 75 per cent.? Ans. H 2 = 0-0826, O 2 = 1-985, N 2 = 8 -684 grams. Gaseous Dissociation PROBLEM 19. At 26 C. the vapour-density of nitrogen peroxide is 38, referred to hydrogen as unit. Calculate the proportion of N 2 O 4 molecules to NO 2 molecules in the vapour at this temperature. Ans. N 2 O 4 /N0 2 = 1-869. PROBLEM 20. The vapour-density of bromine (at. wt. = 80) at 1000 C. is 76-94 (H = 1). What is the degree of dissociation of diatomic into monatomic molecules at this temperature ? Ans. 4 per cent. PROBLEM 21. At 70 C. and under atmospheric pressure N,0 4 is 65-6 per cent, dissociated into N0 2 . What volume will 10 grams of N 2 4 occupy under these conditions? Ans. 5-065 litres. PROBLEM 22. The vapour-density of nickel carbonyl, Ni (C0) 4 , is 83-3 at 63 C. and 70-8 at 100 C. Calculate the percentage dissociation at these two temperatures (Ni = 58). Ans. 0*7 per cent, and 6'66 per cent. PROBLEM 23. At 200 C. and atmospheric pressure the vapour-density of PC1 5 is 70 (H = 1) Calculate the degree of dissociation of the PC1 5 vapour. What are the partial pressures and concentrations in gram-molecules per litre of PC1 5 , PC1 3 and C1 2 , when a gram-molecule of PC1 5 is heated under atmospheric pressure to 200 C. ? (P = 31, Cl = 35*5). Ans. Dissociation = 48*9 per cent. Partial pressures. PC1 5 = 0-3431 atmos., PC1 3 and C1 2 = 0-3283 atmos. Concentrations. PC1 5 = 0-008841, PC1 3 and C1 2 = 0-008461. PROBLEM 24. When ammonium carbamate, NH^CO^NEL, 12 . OSMOTIC PEESSUEE PEOBLEMS is heated, it dissociates completely into 2NH 3 + CO 2 . What volume will be occupied by the gaseous products from 7'8 grams of ammonium carbamate at 100 C. and 740 mm. pressure ? Ans. 9-428 litres. Osmotic Pressure PEOBLEM 25. At 10 C. the osmotic pressure of a solution of urea is 500 mm. of mercury. If the solution is diluted to 10 times its original volume, what is the osmotic pressure at 15 C. of the diluted solution? Ans. 50-89 mm. PROBLEM 26. At 20 C. the osmotic pressure of a cane sugar solution is 800 mm. of mercury. What will it be at C. ? Ans. 745-4 mm. PROBLEM 27. At 21-8' J C. the osmotic pressure of a cane sugar solution containing 68-4 grams per litre was 4-81 atmos. Calculate the numerical value of the constant E when the units of pressure and volume are the atmosphere and the litre. Ans. E = 0-0816. PROBLEM 28. The osmotic pressure of a solution of 0'184 gram of urea in 100 c.e. water was 56 cms. of mercury at 30 C. Calculate the molecular weight of urea. Ans. 62-03. PROBLEM 29. What is the osmotic pressure in atmospheres at 24-2 C. of a solution of glucose containing 0-5 gram-mole- cule per litre ? . Ans. 12-2. PROBLEM 30. At 24 C. the osmotic pressure of a cane sugar solution is 2-51 atmos. What is the concentration of the solution in gram-molecules per litre ? Ans. 0-103. PROBLEM 31. At 25*1 C. the osmotic pressure of a solution of glucose containing 18 grams per litre was 2-43 atmos. Calculate the numerical value of the constant E when the unit of energy is the gram-centimetre. Ans. E = 84300. 1*^- PROBLEM 32. At 18 C. a 0-5 N - NaCl solution is 74-3 per cent, electrolytically. dissociated. What would be the osmotic pressure of the solution in atmosphereiHit 18 C. ? Ans. 20-79. OSMOTIC PEESSUEE PEOBLEMS . 13 PROBLEM 33. A solution containing 3 gram-molecules of cane sugar per litre was found by the plasmolytic method to be isosmotic with a solution of potassium nitrate containing 1*8 gram -molecules per litre. What is the degree of dis- sociation of the potassium nitrate? Ans. 0-67. PROBLEM 34. A solution containing 1/9 gram-molecules of calcium chloride per litre is isosmotic with a solution of glucose containing 4/05 gram-molecules per litre. What is the degree of dissociation of the calcium chloride ? Ans 0-566. PROBLEM 35. In a solution containing 1 gram-molecule of potassium bromide in 8 litres, the salt is 82 per cent, dis- sociated at 25 C. What is the osmotic pressure of the solution at this temperature ? Ans. 5 '56 atmos. CHAPTBE II DENSITY AND SPECIFIC VOLUME OF SOLIDS, LIQUIDS, LIQUID MIXTUBES AND SOLUTIONS Definitions THE density d of a substance at a given temperature t is the weight of a given volume of it compared with the weight of the same volume of water at 4 C. ; or it is the mass of unit of volume of the substance (grams per c.c.). This is usually written d\. . I The reciprocal of the density- the specific volume or volume of unit mass, is V = 1/d, and in the usual units gives the volume in c.c. occupied by one gram. For many liquid mixtures and solutions the volume of the mixture is equal to, or approximately equal to, the sum of the volumes of the components, so that the specific volume, and, therefore, the density of the mixture may be calculated from those of its components by the mixture-formula. Conversely, the approximate composition of the mixture may be calcu- lated from its known specific volume or density and those of its components. Thus it the specific volume of the substance A is F A and that of the substance B is F B , the specific volume F M of a mixture of A and B containing p per cent, by weight of A is given by the formula (i) 100 F M = pV A + (100 - p) FB. The more closely chemically related the components are, the more accurately, as a rule, does the formula reproduce the actual results. Density and Specific Volume Examples PROBLEM 36. The density of a c = 0'528 molecular N-urea solution is d = 1*0104. How many grams of water per gram of urea does the solution contain ? What is the specific volume F of urea in the solution ? SOLUTION 36. A c molecular N-solution contains cM gms. per litre, if M is the molecular weight of the dissolved sub. 14 DENSITY, SPECIFIC VOLUME EXAMPLES 15 stance ; a litre of the solution weighs 1000 d gms., and there- fore contains 1000 d - cM gms. of water. The c N-urea solution, therefore, contains per gm. of urea 1000 d - cM cM The specific volume V of urea in the solution is the volume in c.c. occupied by 1 gm. Now 1000 c.c. of the solution contain (1000 d cM) gms. of water which occupies (1000 d cM) c.c., if it is assumed that the specific volume of the water is not changed by solution. The volume occupied by cM gms. of urea is, therefore, 1000 - (1000 d - oM) c.c. Hence the specific volume is 1000 - 1000 d + cM cM 1000 - 10104 + 31-7 31 . 7 - = 0-67 c.c. per gm. PROBLEM 37. What is the normality of an a = 4 per cent. (by weight) cane sugar solution, if the specific volume of solid cane sugar is V = 0'615, and if no change of volume accom- panies solution? SOLUTION 37. Let the density of an a per cent, cane sugar solution = d. Then a litre of the solution weighs 1000 d gms. In these 1000 d gms. there are a x - - 10 JLUU ad gms. of cane sugar. If M is the molecular weight of cane sugar the'normality of the solution is c - -^-! M. d is calculated as follows : the volume of 1 litre of solution is equal to the sum of the volumes of the dissolved sugar and of the water used for its solution. A litre of solution contains 10 x ad gms. of sugar, of which the volume is 10 x a^Fc.c., and (1000 d - 10 ad) gms. of water, of which the volume is (1000 d - 10 ad} c.c. Therefore 1000 - lOadV + 1000 d - Wad and 1000 " 10 o(7- 1)4- 1000' 16 DENSITY, SPECIFIC VOLUME EXAMPLES The normality of the solution is, therefore, 1000 g {a(V - 1) + 1W\M' 1000 x 4 = (100 - 4 x 0-385)342 = ' 9 gram-molecule per litre. PROBLEM 38. What is the specific volume of potassium hydroxide in a solution which contains 1 gram-molecule of KOH in 1000 grams of water, and of which the density is d = 1-052, if the density of water is taken = 1-000, and if it is assumed that the specific volume of water is not changed by the process of solution ? SOLUTION 38. If M is the molecular weight of KOH, then 1000 + M grams of the solution occupy a volume of ^ - c.c. The volume of the water used in making the solution is 1000 c.c., therefore the volume of the dissolved KOH is 1QOQ + M _ 1000 c.c. The specific volume of the dissolved KOH, i.e. the volume in c.c. occupied by 1 gram of KOH, is, therefore, 1000 + M - 1000 d = 1000 + 56-15 - 1052 = Md 56-15 x 1'052 gram. The smallness of this value makes it probable that the assumption, that the water does not change its volume in the process of solution, is inadmissible. PROBLEM 39. <^ for carbon disulphide is 1 -264, and for ethyl alcohol 0'7963. What is its value for a mixture of carbon disulphide and alcohol containing 79-82 per cent, of the former? SOLUTION 39. Since V = l/d we may write formula (i) in the form 100 p (100 - p) p = 79-82, d A = 1-264, d B = 0-7963, therefore 100 79-82 20-18 d M ~ 1-264 + 0-7963' and d M = 1-130. (Observed, 1-122.) DENSITY, SPECIFIC VOLUME PROBLEMS 17 Problems for Solution PROBLEM 40. The density of a 2-31 per cent, solution of ammonia is 0*990. What is the concentration of the solution in gram- molecules per litre ? Ans. 1*345. PROBLEM 41. d^ for a 10 per cent. NH 4 C1 solution = 1-029, for solid NH 4 C1 = 1-536, and for water = 0-9974. Calculate the change of volume per 100 grams solution in making a 10 per cent solution of NH 4 C1. What is the specific volume of NH 4 C1 in the solution, if that of the water remains constant ? Ans. Expansion = 0*48 c.c. per 100 grams solution. Sp, vol. NH 4 C1 = 0-699 c.c. per gram. PROBLEM 42. The density of a 4-526 per cent, solution of HgCl 2 in water at 16 C. = 1-038, and that of an 11-88 per cent, solution in alcohol at 16 C. =0-8857. The density of water at 16 C. = 0;9979, and of alcohol = 0*7939. Calcu- late the concentration of each solution, and the specific volume of HgCl 2 in each. Assume no change in the specific volumes of the water and alcohol on solution. Ans. Cone, aqueous solution = 0-1733 gram-molecule per litre. Cone, alcoholic solution = 0-3883 gram-molecule per litre. Sp. vol. in aqueous solution = 0-1575 c.c. per gram, in alcoholic solution = 0-1635 c.c. per gram. PROBLEM 43. For a 5-18 per cent, solution of phenol in water d = 1-0042 ; for water d = 0-9991. What is the concentration of the solution in moles per litre, and in moles phenol per mole water? What is the specific volume of phenol in the solution, if the specific volume of the water is assumed to remain constant ? Ans. Cone. = 0-5534 mole per litre = 0'0104&mole phenol per mole water. Sp. vol. = 0-9363 c.c. per gram. PROBLEM 44. d for ethyl alcohol = 0'7936, for water = 3-9991, and for a 50 per cent, solution of alcohol in water = D-9180. What is the contraction on mixing 50 grams of ilcohol with 50 grams of water at 15 C. ? Assuming the specific volume of the water to remain constant, compare the 2 18 DENSITY, SPECIFIC VOLUME PROBLEMS specific volume of pure alcohol with that of the alcohol in the solution. Ans. Contraction = 4*16 c.c. Sp. vol. of pure alcohol = 1-260, of alcohol in the solution = T177 c.c. per grain. PROBLEM 45. d 1 ^ for ethylene dibromide = 2-183, for propyl alcohol = 0*8066, and for a solution of p per cent, ethylene dibromide in propyl alcohol = 0-8608. Calculate the value of p. Ans. 10-10. (Observed, 10-01.) PROBLEM 46 (cf. preceding problem). Calculate the density at 18-7 C. of a 20-95 per cent, solution of ethylene dibromide in propyl alcohol. Ans. 0-9292. (Observed, 0-9291.) PROBLEM 47. d for carbon disulphide = 1-264, for ethyl alcohol = 0*7963 and for a 50 - 77 per cent, solution of carbon disulphide in alcohol = x. Calculate the value of x. Ans. 0-9804. (Observed, 0-9718.) PROBLEM 48. d* 6 ' 3 for aniline = 1*025, for ethyl alcohol = 0-8081 and for a p per cent, solution of aniline in alcohol = 0-9763. Calculate p. Ans. 81-34. (Observed, 79-24.) PROBLEM 49. d for benzene = 0-8814, for ethyl alcohol = 0-7930. What is its value for a 21-12 per cent, solution of benzene in alcohol ? Ans. 0-8104. (Observed, 0-8106.) PROBLEM 50 (cf. preceding problem). The density of a p per cent, solution of benzene in alcohol at 20 C. is 0-8604. What is the value of p ? Ans. 77-94. (Observed, 79-10.) CHAPTEE III SPECIFIC AND MOLECULAE KEFKACTIVITlT Definitions '"PHE refractive index n of a substance for light of a given J- wave-length, varies with the temperature, but the ex- pressions n -^- (Gladstone and Dale), and n <) " . . a n* + 2 a (Lorentz and Lorenz), where d is the density of the substance at the temperature at which n is measured, are, for a given substance, almost constant and independent of the tempera- ture. This is especially the case with the latter expression, which remains approximately constant even for different states of aggregation of the substance (e.g. liquid and vapour). The value of ?Ll . or of n et ~ . . -= is called the specific d n 2 + 2 d refractive power or specific refractivity of the substance. If the specific refractivity is multiplied by the molecular weight M of the substance, the molecular refractivity (i) ^ ~ ' d or ( 2 ) n n ~ - -r is obtained. The molecular refractivity v } ri* + 2 d of a compound may be calculated from the atomic refrac- tivities of its component elements. For monovalent elements (e.g. H 2 , C1 2 , Br 2 ) the atomic refractivity is practically con- stant and independent of the nature of the other elements with which they are united. For polyvalent elements (e.g. O 2 , C, N 2 ) the value of the atomic refractivity varies with the mode of union to the other elements of the compound, but- is constant for a given mode of union. Thus in the case of sarbon the atomic refractivity of a singly bound (saturated) carbon atom is different from that of a doubly or trebly linked carbon atom. This can be allowed for by assigning a definite ' atomic " refractivity to an ethylene (double) and to an 19 20 REFKACTIVITY EXAMPLES acetylene (triple) bond. The molecular refractivity, as cal- culated from the refractive index, may, therefore, sometimes serve as a guide to the constitution of a compound. The following are the atomic refractivities for the D line (sodium light), calculated according to Lorentz and Lorenz's formula, of the elements required in the following problems. Carbon (singly bound) = 2-501, hydrogen = 1-051, oxygen in a hydroxyl (- OH) group = 1-521, oxygen in ethers = 1'683, oxygen in a carbonyl (>C = O) group = 2-287, chlorine = 5-998, ethylene bond = 1-707, acetylene bond = 2-10. Refractivity of Mixtures The specific refractivity of a homogeneous mixture or solution may be calculated (as a rule with close approxima- tion to the observed value) from the specific refractivities of its components by the mixture-formula (cf. specific volume). Thus if n A is the refractive index of substance A, n B that of substance B and n^ that of a mixture containing p per cent, by weight of A, all for light of the same wave-length, and d A , d s and d m the corresponding densities, we have, using Glad- stone and Dale's formula, n m -I _ n A - I ^_ n B - 1 (100 - p) (3) du d A ' 100 d s 100 or, using Lorentz and Lorenz's formula, n\-l _! = n*_-_l P n\-l (100 - p) (4) n* + 2 * d n\ + 2' 100 d A n\ + 2 ' 100 d B ' Conversely, if the specific refractivites and densities of the mixture and its components are known, the composition of the mixture may be calculated. Refractivity Examples PROBLEM 51. For propionic acid d 4 = 1-0158 and n D , the refractive index for sodium light, = 1-3953. Calculate the molecular refractivity of propionic acid by Lorentz and Lorenz's formula and compare it with the value calculated from the atomic refractivities given above. SOLUTION 51. M, the molecular weight of propionic acid, = 74. Substituting the numerical values in formula (2) we obtain (1'3953)2 - 1 74 (l-3953) 2 -f 2 1-0158 74 for the molecular refractivity. REFRACTIVITY EXAMPLES, PEOBLEMS 21 From the atomic refractivities we obtain for CH 3 CH 2 G{ \OH 3 carbon atoms = 3 x 2'501 = 7*503 6 hydrogen atoms = 6 x 1-051 = 6-306 1 hydroxylic oxygen atom = 1 x 1-521 = 1-521 1 carbonyl oxygen atom = 1 x 2-287 = 2 -287 Molecular refractivity = 17-62. PROBLEM 52. d for ether = 0-7208, for ethyl alcohol = 0'7935 and for a mixture of ether and alcohol containing^ per cent, of alcohol = 0'7389< At 20 C. the refractive indices for sodium light are, for ether 1-3536, for alcohol 1-3619, and for the mixture 1-3572. Calculate the value of p, using the Gladstone and Dale formula. SOLUTION 52. Substituting the numerical values in equa- tion (3) we obtain (1-3572 -1) _ (1-3619 - 1) p. 0-7389 0-7935 100 (1-3536 - 1) (100-_rt 0-7208 100 ' .-. p = 20-81. (Observed, 20-71.) Problems for Solution PROBLEM 53. At 17'4 the refractive index of methyl alcohol for sodium light (W D ) = 1-3297 and its density = 0'7945. Calculate the molecular refractivity according to both (a) Gladstone and Dale's, and (b) Lorentz and Lorenz's formula, and (c) compare the value for the latter with that calculated from the atomic refractivities. Ans. (a) 13-27, (b) 8-21, (c) 8-23. PROBLEM 54. d' 5 for liquid hydroxylamine = 1-205 and W D = 1-4405. (a) Calculate the molecular refractivity (L. and L.). (b) What is the atomic refractivity of nitrogen in the compound ? Ans. (a) 7-23, (b) 2-556. PROBLEM 55. At 20 the density of chloroform = 1-4823 and the refractive index for the D line = 1-4472. Given the atomic refractivities of carbon and hydrogen, calculate that of chlorine (L. and L.). Ans. 5-999. 22 EEFEACTIVITY PEOBLEMS PEOBLEM 56. At 20 the substance C 3 H 6 O has a density of 0-8005 and a refractive index for the D line of 1-3641. Using Lorentz and Lorenz's formula, determine whether it has the constitution (a] CH 2 : CH CH 2 OH, or (b) CH 3 CO CH 3 . Ans. (b). PEOBLEM 57. At 17 '5 the density of ethyl alcohol = 0-8020, and its refractive index for the D line = 1-3619. At 17 '8 the density of normal propyl alcohol = 0*8074, and its refractive index for the D line = 1-3861. From these clata calculate the molecular refractivity of normal butyl alcohol according to Lorentz and Lorenz's formula. Ans. 22-20, (Observed, 22-11). PEOBLEM 58. For ethyl acetate d* 6 ' 4 = 0-9028 and n D = 1*374:2. (a) Calculate the molecular refractivity (L. and L.), and (b) compare it with the value calculated from the atomic refractivities. Ans. (a) 22-27, (b) 22-38. PEOBLEM 59. For cinnamyl alcohol C 6 H 5 . CH : CH . CH 2 OH, dj 6 ' 8 = 1-0555 and n D = 1-5763. (a) Calculate the molecular refractivity (L. and L.), and (b) compare it with the value calculated from the atomic re- fractivities. Ans. (a) 42 03, (b) 41-37. PEOBLEM 60. At the temperature t the density of water is d* 4 and the refractive index for sodium light n D . From these data test the constancy of the specific refractivity as calculated by Gladstone and Dale's, and Lorentz and Lorenz's formula. t d\ n 20 0-99823 1-3330 40 0-99224 1-3307 80 0-97183 1-3230 PEOBLEM 61. The specific refractivity (L. and L.) of glycerine for the C line = 0-2219, and of propyl alcohol = 0-2903. From these data calculate the atomic refractivity of hydroxylic oxygen for the C line. Ans. 1-50. PEOBLEM 62. At 18-07 the density of ethylene dibromide = 2-1830, of propyl alcohol = 0-80659, and of a mixture of the two containing p per cent, by weight of dibromide = 0*86081. At the same temperature the refractive index of EEFEACTIVITY PROBLEMS 23 the dibromide for the D line = 1-5404, of propyl alcohol = 1-3862 and of the mixture =1-3919. Calculate the value of , (a) using the Gladstone and Dale formula, (b) using the orentz and Lorenz formula. Ans. (a) 1017, (b) 10-16. (Observed, lO'Ol.) PROBLEM 63. At 18'07 the density of a mixture of ethylene dibromide and propyl alcohol containing 20*95 per cent, of the former is 0-92908. From the data in the preceding problem calculate its refractive index for the D line (a) according to Gladstone and Dale's formula, (b) according to Lorentz and Lorenz 's formula. Ans. (a) 1-39933, (b) 1 39932. (Observed, 1-39913.) PROBLEM 64. At 20 the density of ether = 0-72078, of benzene = 0*87953, and of a mixture containing p per cent, of benzene = 0*75299. The corresponding refractive indices are (for sodium light), ether = 1-35360, benzene = 1-49996, mixture = I 1 38227. Calculate the value of p by the Gladstone and Dale formula. Ans. 21-96. (Observed, 21-06.) PROBLEM 65. At 18-07 the density of a 5 per cent, sodium chloride solution = 1-0345 and the refractive index (D line) = 1-3423. At the same temperature the density of water = 0-99866 and the refractive index = 1-3335. Calculate the specific refractivity of sodium chloride (Gladstone and Dale). Ans. 0-274. PROBLEM 66. At 18-07 the density of a_p per cent, sodium chloride solution = T0201 and the refractive index = 1-3388. Using the result obtained in the preceding problem, calculate the value of p (G. and D.). Ans. 3-005. (Observed, 3-000). PROBLEM 67. At 21-8 the density ol water = 0-9978 and of a 6-80 per cent, sodium sulphate solution = 1-Q596. At the same temperature the refractive index of water for sodium :.ight = 1-33308 and of the solution = 1-34291. Calculate the specific refractivity of sodium sulphate by the Gladstone and Dale formula. Ans. 0-18382. PROBLEM 68. At 21-8 the density of a p per cent, sodium sulphate solution = 1*0782 and n D = 1-34571. From the data in the preceding problem calculate the value of p. Ans. 8-78. (Observed, 8'80.) CHAPTBE IV MOLECULAR WEIGHT FROM LOWERING OF VAPOUR-PRES- SURE, LOWERING OF FREEZING-POINT AND ELEVATION OF BOILING-POINT. DEGREE OF DISSOCIATION OF ELEC- TROLYTES. MOLECULAR LOWERING OF FREEZING-POINT AND ELEVATION OF BOILING-POINT FROMTLATENT HEATS OF FUSION AND EVAPORATION Molecular Weight from Lowering of Vapour-pressure '"PHE relative lowering of the vapour-pressure of a pure JL solvent by a dissolved substance is equal to the ratio of the number of molecules of dissolved substance to the number of molecules of solvent in the dilute solution (Eaoult), that is, (,\ Po ~ P _ n where p is the vapour-pressure of the pure solvent, p that of the solution at the same temperature, n the number of mole- cules of dissolved substance, and N the number of molecules of solvent in the solution. p Q and p must, of course, be measured in the same unit, but, since we are here dealing with a ratio, the unit chosen is immaterial. This relation may be used as follows for the determination of the molecular weights of dissolved substances. If a is the weight of the dissolved substance, and b that of the solvent in the solution, m the molecular weight of the dissolved substance, and M that of the solvent in the gaseous state, then the number of molecules of dissolved substance = m and the number of molecules of solvent = -. Therefore __ - N ~ b/M ~ bm 24 VAPOUR-PBESSUBE AND TEMPEBATUBE 25 Vapour-pressure and Temperature The relation between the vapour-pressure of a liquid and the temperature is expressed by the equation (Clausius), W dT ~ RT 2 ' p is the vapour-pressure of the liquid at the absolute tempera- ture T, L the molecular heat of evaporation of the liquid, and 'R the gas-constant, the numerical value of which is 1-985 (in round numbers 2) when the gram-calorie is the unit of energy. If we assume that L is independent of the temperature and integrate equation (3) between the temperatures T and T v we obtain (4) ^ a. _(!__ i) _ L ( r ' - T \ where p and p 1 are the vapour-pressures corresponding to the absolute temperatures T and T lt or, converting the natural to common logarithms, (5) log ^= log p - log ^ = For small intervals of temperature the assumption that L is constant will, in most cases, be practically true. For larger intervals of temperature L may be taken as the latent heat T + T at the mean temperature -^-g \ Since in equation (5) we are concerned only with the ratio of the vapour-pressures, p Q and p 1 may be expressed in any unit. Vapour-pressure Examples PROBLEM 69. What is the concentration, in gram-mole- cules per 1000 grams water, of an aqueous solution which at t = 100'42 C. has a vapour -pressure p = 758'2 mm. of mercury? The molecular heat of evaporation of water is L = - 9600 cal. SOLUTION 69. The problem can be solved by equation (i) if the vapour-pressure p Q of water at 100-42 C. can be cal- culated. This can be done by equation (5) as follows : Under a pressure p 1 = 760 mm. of mercury (atmospheric pressure) water boils at t l = 100 C. or T l = 373 absolute, that is, at T^ = 373 the vapour-pressure of water = p = 760 mm. 26 VAPOUE-PEESSUEE EXAMPLES T = t + 273 = 100-42 + 273 = 373'42 and L = - 9600. Substituting these numerical values in (5), we obtain logp, = log 760 + a.SxxsI^ - 2 ' 89 ' ' Po = 771 '2 mm. Accordingly, by (i), p - p 771-2 - 758-2 Q-Q1685 _n ~^~ 771-2 ~1T~ = ~F The solution, therefore, contains 0-01685 gram-molecule of dissolved substance per gram-molecule of water, and, since the gram-molecule of water = 18 grams, it contains 0-01685 x 1000 -^ - = 0*930 gram-molecule of dissolved sub- stance per 1000 grams water. PROBLEM 70. Under a pressure of 760 mm. ether boils at ^ = 35 C. Its molecular heat of evaporation is - 6640 calories, (a) What is the vapour-pressure p of ether at t = 30 C. ? (b) What volume will be occupied by a litre of air at a pressure P = 720 mm. after being bubbled through ether at 30 C., if the total pressure be kept constant at 720 mm. during the process, and (c) how many (x) grams of ether will evaporate during the process ? SOLUTION 70. (a) The vapour-pressure at 30 C. may be calculated from equation (5). T = t + 273 = 303. T l = $! + 273 = 308. p 1 = 760. L = - 6640. Therefore - 6640 x 5 log Po = log 760 + 2-3 x 1-985 x 303 x 308 and p = 635 mm. (b) The volume of air, which has been bubbled through ether at 30 C. under the constant pressure P = 720 mm., increases until the partial pressure of the air becomes p = P - p Qt where _p is the vapour-pressure of ether at 30 C. From Boyle's law we get the final volume v assumed by 1 litre of air after being bubbled through the ether. Thus v_ _ P 720 720 1 ~ P - p ~ 720 - 635 " 85 ' .'. v = 8's litres. (c) These 8*5 litres contain ether vapour at a partial pres- sure p = 635 mm. The weight x of this ether vapour is to be calculated. VAPOUR-PEESSUEE EXAMPLES 27 According to equation (2), p. 1, (see also Problem 2), Po v = ~ET Q1 where M is the molecular weight of ether vapour. Hence If p and v are expressed in atmospheres and litres the value of R is 0-08204:. Therefore 0-08204 x 303 PBOBLEM 71. The vapour-pressure of a solution* con- taining 6-69 grams of Ca(N0 3 ) 2 in 100 grams of water is 746-9 mm. at 100 C. What is the degree of dissociation of the salt ? SOLUTION 71. Let a be the degree of dissociation, then, since Ca(N0 3 ) 2 dissociates into 3 ions, one Ca" and two NO' 3 , one gram-molecule of Ca(N0 3 ) 2 gives on dissociation a gram-mole- cule of Ca" and 2a gram-molecule of N0' 3 , whilst (1 - a) gram-molecule of undissociated salt remains. The total number of molecules derived from one gram-molecule of Ca(N0 3 ) 2 is, therefore, 1 - a + 3a = 1 + 2a. From 6-69 > * A (\ O grams = -virT g ram - m l ecu l es f C a (N0 3 ) 2 there are, there- fore, derived fi'fiQ n = jgjTj (1 + 2a) gram-molecules, 164-1 being the molecular weight of Ca(N0 3 ) 2 . According to (i) p Q - p = n_ m Po ~ N' p = 746-9 mm. and p for pure water at 100 = 760 mm. N = weight of water in the solution divided by the molecular 100 weight of water as vapour = . Hence 18 760 - 746-9 = 6-69 + ^ _ 18 760 164*1 /. a = 0-675. 28 LOWERING OF FREEZING-POINT Lowering of Freezing-point If t is the freezing-point of a pure solvent, and t^ that of a solution containing c gram- molecules of dissolved substance per 1000 grams of solvent, the lowering of the freezing- point A = t - tj, is proportional to c, that is, the lowering of the freezing-point of a pure solvent by a dissolved substance is proportional to the concentration of the latter, or (6) A = Kc. The value of the constant K depends only on the solvent, and on the unit of concentration chosen. We shall take as our unit concentration, one gram-molecule of solute in 1000 grams of solvent. If in (6) we put c = 1, we obtain A = K. K is, therefore, the lowering of the freezing-point of the solvent caused by dissolving 1 gram-molecule of a normal (non-dissociating and non-associating) solute in 1000 grams of solvent. It is called the molecular lowering of the freezing- point. (Two other units of concentration are often used, namely, (1) one gram-molecule of solute in 100 grams of solvent, and (2) one gram-molecule of solute in 1 gram of solvent. If K' is the molecular lowering in the first case, and K" that in the second case, the relation between K, K and K' is K' = 10 K, K" = 1000 K.) If the solution contains a grams of solute in b grams of solvent, and if M is the molecular weight of the solute, the CL concentration of the solute is -r? gram-molecules in b grams of solvent, or ^ gram-molecules per 1000 grams solvent. Therefore, from (6), , N ^ 1000 a (7) A-*-E5- By various transformations of this equation any of the quantities may be calculated if the others are known. In (?)> is the weight of solute in 1000 grams of solvent. If we put this = W, the equation becomes W A W (8) A = tf-^or-^ =^, a form which is particularly clear and easy of application. FREEZING-POINTEXAMPLES 29 The constant K for a given solvent may be found empiric- ally by (6), (7) or (8), by observing the depressions of the freezing-point of the pure solvent caused by known concen- trations of substances of which the molecular weights are known, and taking the mean value ; or it may be calculated from the latent heat of fusion of the solvent by means of the relation (van't Hoff) T ( = 273 + t} is the freezing-point of the pure solvent on the absolute scale, I the latent heat of fusion of 1 gram of the solvent in calories and R the gas-constant (1*985, or, in round numbers, 2). (If the constants K' or K" as denned above are used, then RT* RT* Elevation of Boiling-point For the elevation qf the boiling-point of a pure solvent by a dissolved substance equations exactly analogous to (6), (7), (8) and (9) hold. In this case, however, A is the elevation of the boiling-point and is equal to ^ - , where t is the boiling-point of the pure solvent and t l that of a solution con- taining c gram-molecules of solute of molecular weight M per 1000 grams of solvent, or a grams of solute in b grams of solvent, or W grams of solute in 1000 grams of solvent. K is the molecular elevation of the boiling-point, caused by dissolving one gram-molecule of a normal solute in 1000 grams of solvent. In equation (9) T (= 273 + t) is the boiling-point of the pure solvent on the absolute scale and I is the latent heat of evaporation of one gram of solvent in calories. (As in the case of the freezing-point, K' an.d K" are the molecular elevations of the boiling-point when either of the other common units of concentration are used. Their relations to K and I are thesame as those given under the freezing-point.) Freezing-point Examples PROBLEM 72. The freezing-point of a solution of 0-684 gram of cane sugar in 100 grams of water is - 0'037 C.,. 28 LOWERING OF FREEZING-POINT Lowering of Freezing-point If t is the freezing-point of a pure solvent, and t^ that of a solution containing c gram- molecules of dissolved substance per 1000 grams of solvent, the lowering of the freezing- point A = t x is proportional to c, that is, the lowering of the freezing-point of a pure solvent by a dissolved substance is proportional to the concentration of the latter, or (6) A = Kc. The value of the constant K depends only on the solvent, and on the unit of concentration chosen. We shall take as our unit concentration, one gram -molecule of solute in 1000 grams of solvent. If in (6) we put c = 1, we obtain A = K. K is, therefore, the lowering of the freezing-point of the solvent caused by dissolving 1 gram-molecule of a normal (non-dissociating and non-associating) solute in 1000 grams of solvent. It is called the molecular lowering of the freezing- point. (Two other units of concentration are often used, namely, (1) one gram- molecule of solute in 100 grams of solvent, and (2) one gram-molecule of solute in 1 gram of solvent. If K is the molecular lowering in the first case, and K" that in the second case, the relation between K, K' and K" is K' = 10 K, K" = 1000 K.) If the solution contains a grams of solute in b grams of solvent, and if M is the molecular weight of the solute, the concentration of the solute is -^ gram-molecules in b grams of solvent, or ^, gram-molecules per 1000 grams solvent. Therefore, from (6), 1000 a By various transformations of this equation any of the quantities may be calculated if the others are known. In (?)> is tne weight of solute in 1000 grams of solvent. If we put this = W, the equation becomes (J~A-*-<>'t-ir. 1 a form which is particularly clear and easy of application. FKEEZING-POINT EXAMPLES 29 The constant K for a given solvent may be found empiric- ally by (6), (7) or (8), by observing the depressions of the freezing-point of the pure solvent caused by known concen- trations of substances of which the molecular weights are known, and taking the mean value ; or it may be calculated from the latent heat of fusion of the solvent by means of the relation (van't Hoff) T ( = 273 + f) is the freezing-point of the pure solvent on the absolute scale, I the latent heat of fusion of 1 gram of the solvent in calories and R the gas-constant (1*985, or, in round numbers, 2). (If the constants K or K" as denned above are used, then RT* RT* Elevation of Boiling-point For the elevation Oyf the boiling-point of a pure solvent by a dissolved substance equations exactly analogous to (6), (7), (8) and (p) hold. In this case, however, A is the elevation of the boiling-point and is equal to ^ - t, where t is the boiling-point of the pure solvent and ^ that of a solution con- taining c gram-molecules of solute of molecular weight M per 1000 grams of solvent, or a grams of solute in b grams of solvent, or W grams of solute in 1000 grams of solvent. K is the molecular elevation of the boiling-point, caused by dissolving one gram-molecule of a normal solute in 1000 grams of solvent. In equation (p) T (= 273 + t) is the boiling-point of the pure solvent on the absolute scale and I is the latent heat of evaporation of one gram of solvent in calories. (As in the case of the freezing-point, K an4 K" are the molecular elevations of the boiling-point when either of the other common units of concentration are used. Their relations to K and / are thesame as those given under the freezing-point.) Freezing-point Examples PROBLEM 72. The freezing-point of a solution of 0-684 gram of cane sugar in 100 grams of water is - 0'037 C.,. 32 VAPOUE-PEESSUEE PEOBLEMS 128 x 2-1 _ 128 ~ 2l ' Substituting this value of K in (1) we obtain 1-985 x (308) 2 1 - 1000 x 2-1 = 8 <>'7 calories - Since the molecular weight of ether is 74, the molecular heat of evaporation is L = 89-7 x 74 = 6638 calories. PBOBLEM 75. A solution of barium nitrate containing a = 11-07 grams in 100 grams of water boils at 100-466 C. What is the degree of dissociation of the salt ? K for water = 0-52. SOLUTION 75. The solution contains 10 a grams barium nitrate per 1000 grams water. If M ( = 261-5) is the mole- cular weight of barium nitrate, it would, therefore, contain -jg- gram-molecules in 1000 grams water if the salt were undissociated. Let a be the degree of dissociation. Since Ba(NO 3 ) 2 dissociates into 3 ions we obtain, exactly as in Problem 71, for the total number of gram-molecules in 1000 grams of water 10 a , C = -^-(1 + 2 a). Putting this value of c into equation (6) we obtain A -*.^-(l + 2 a), and, substituting the numerical values, ..._ 0-52 x 10 x 11-07 261-5 ' ( 1 + 2a ) /. a = 0-558. Problems for Solution Vapour-pressure PROBLEM 76. The vapour-pressure of ether at 20 C. is 442 mm. and that of a solution of 6'1 grams of benzpic acid in 50 grams of ether is 410 mm. at the same temperature. Calculate the molecular weight of benzoic acid in ether. Ans. 124. VAPOUR-PRESSUREPROBLEMS 33 PROBLEM 77. At 10 C. the vapour-pressure of ether is , 291-8 mm. and that of a solution containing 5'3 grams ot / benzaldehyde in 50 grams of ether is 71 '8 mm. What is^ the molecular weight of benzaldehyde ? Ans. 1144. PROBLEM 78. The vapour-pressure of alcohol at 70 C. is ^ 54:0-9 mm., and at 80 C. it is 811-8 mm. Calculate the latent heat of evaporation per gram of alcohol. \J Ans. 212 calories. PROBLEM 79. Under a pressure of 760 mm. ether boils / at 35 C. A solution of 10'44 grams of aniline in 100 grams / of ether has a vapour-pressure of 333 mm. at 15'3 C. ( The latent heat of evaporation of ether is 89'73 calories per gram. Calculate the molecular weight of aniline in the solution. Ana. 97-6. PROBLEM 80. At C. the vapour- pressure of water is 4-620 mm. and of a solution of 8'49 grams of NaN0 3 in 100 grams of water 4*483 mm. Calculate the degree of dissocia-\/ tion of the NaNO 3 . Ans. 0-649. PROBLEM 81. At 25 C. the vapour-pressure of water is 23-55 mm. What is the vapour-pressure of a solution con- taining 6 grams of urea in 100 grams of water atf'the same temperature ? Ans. 23-13 mm. / PROBLEM 82. At C. the vapour-pressure of Water = 4"620 mm. and that of a solution containing 21'24ygrams of glycerol in 100 grams of water = 4-432 mm. ^Calculate the molecular weight of glycerol in the solution. Ans. 93-9. PROBLEM 83. The vapour-pressure of a solution of 8-89 grams of dextrose in 100 grams of water is 4-532 mm. at C., whilst that of pure water at the same /temperature is 4-620 mm. Calculate the molecular weigfit of dextrose. Ans. 194-5. PROBLEM 84. A solution containing 9'21 grams of mercuric cyanide in 100 grams of water has a vapour-pressure of 755-2 mm. at 100 C. What is the molecular weight of the 34 VAPOUR-PRESSUREPROBLEMS 7 salt? What conclusion may be drawias/to the dissociation of mercuric cyanide in aqueous solutions Ans. 262-5. PROBLEM 85. The vapour-pressure of a solution con- taining 11-94 grams of glycocoll (C.,H 5 NO 2 ) iti lOCXgrams of water is 740-9 mm. at 100 C. What is the mole/ular weight of the solute ? Ans. 85-5. / PROBLEM 86. Under what pressure does /water boil at a temperature of 95 C. ? The latent heat oft evaporation of water is - 536 calories per gram. Ans. 636-5 mm. PROBLEM 87. At C. the vapour-pressure/of water = 4-620 mm. and that of a solution of 2-21 grams of CaCl 2 in 100 grams of water = 4'585 mm. CaWla^e the apparent molecular weight and the degree of dissockCtion of the CaCl 2 . Ans. M = 52-5, a = 0-557. PROBLEM 88. A current of dry air was bubbled through a bulb containing a solution of 13-33 grams of urea in 100 grams of water, then through a bulb, at the same temperature, con- taining pure water and finally through & tmje containing pumice moistened with strong sulphuric acid. The loss of weight of the water bulb = 0'0870 grams and the gain of weight of the sulphuric acid tube = 2-036 grams. Calculate the molecular weight of urea in the solution. Ans. 56-2. PROBLEM 89. A current of dry air was passed first through a series of bulbs containing a solution of 8-914 grams of nitrobenzene in 100 grams of alcohol, and therr through a series of bulbs containing pure alcohol. Th temperature was 11 C. After the passage of the air the decrease in the weight of the bulbs containing the solution was 2^340 grams and of the bulbs containing the pure solvent 0-0685 gram. Cal- culate the molecular weight of nitrobenzene in the solution. Ans. 125-8. PROBLEM 90 (of. preceding problem). In/a similar ex- periment with ethyl benzoate the solution contained 7'394 grams of ethyl benzoate in 100 grams of "aj/ohol. The loss of weight of the solution was 2-1585 grams and of the pure FKEEZING-POINT PKOBLEMS 35 solvent Q'0515 gram. Calculate the molecular weight of ethyl benzoate. Ans. 146. PROBLEM 91. The vapour-pressure of boron trichloride is 562-9 mm. at 10 C. and 807'5 mm. at 20 C. )Vhat is the molecular heat of evaporation of boron trichloride and its boiling-point under atmospheric pressure ? Ans. L = - 5936 cals., B.P.= 184 C. Freezing-point PROBLEM 92. The freezing-point of pure benzene = 5-440 and that of a solution containing 2-093 grams of benzalde- hyde in 100 grams of benzene = 4-440. Calculate the mole- cular weight of benzaldehyde in the solution. K for benzene = 5. Ans. 104-6. PROBLEM 93. A solution of 0'502 gram of acetone in 100 grams of glacial acetic acid gave a depression of the freezing-point of 0-339. Calculate the molecular depression for glacial acetic acid. Ans. K = 3-9. PROBLEM 94. 17*79 grams of an aqueous solution contain- ing 0-1834 gram of hydrogen peroxide gave a freezing-point of - 0'571. What is the molecular weight of hydrogen peroxide in the solution ? K for water = 1-86. Ans. 33-9. PROBLEM 95. By dissolving 0'0821 gram of w-hydroxy- benzaldehyde (C 7 H 6 O 2 ) in 20 grams of naphthalene (melting- point 80- 1) the freezing- point is lowered by 0'232. Assum- ing that the molecular weight of the solute is normal in the solution, calculate the molecular depression for naphtha- lene and the latent heat of fusion per gram. Ans. K = 6-896, I = 36-2 cals. PROBLEM 96. A solution of 1 gram of silver nitrate in 50 grams of water freezes at - 0-348 C. Calculate to what extent the salt is ionised in the solution. K for water = T86. Ans. a = 0'59. v PROBLEM 97. The weights a of methyl alcohol dissolved 36 FBEEZING-POINT PKOBLEMS in 15 grams of benzene gave the depressions of the freezing- point A. a 0-0478 0-0988 0-2700 0-4291 0*6636 1-093 A 0-360 0-612 1-265 1-610 1-978 2-475. The molecular lowering of the freezing-point for benzene is K = 5-0. What conclusions as to the molecular condition of methyl alcohol in benzene solution may be drawn from these figures ? PROBLEM 98. An aqueous solution of ethyl alcohol con- taining 8-74 grams alcohol per 1000 grams water gave a freezing-point of -0'354. Find the molecular weight of alcohol in this solution. K for water = 1*86. Ans. 45-9. PROBLEM 99. A solution of NaCl containing 3-668 grams per 1000 grams water freezes at - 0-2207. Calculate the degree of dissociation of the salt. (K 1*86.) Ans. a = 0-892. PROBLEM 100. 0-2274 gram of naphthalene in 10 grams of _p-toluidine (melting-point = 42'1) caused a lowering of the freezing-point = 0'940. Calculate the freezing-point con- stant and latent heat of fusion of j>toluidine. Ans. K = 5-29, I = 37'5 cals. PROBLEM 101. The lowering of the freezing-point A was caused by dissolving a grams of formanilide in 10 grams of benzene. (K = 5'0.) a 0-1255 0-3815 0*7439 1-197 A 0-420 0-9'20 1-360 1-747 What conclusions as to the state of the dissolved substance and its variation with concentration may be drawn from these figures ? PROBLEM 102. A solution of 0-4180 gram KOH in 1000 grams water gave a freezing-point of - 0-0275, and a solution of 1*780 gram in 1000 grams water a freezing-point of - 0-1147. Calculate the degree of dissociation of the KOH in these solutions, taking K = 1-86. Ans. 98-1 per cent, and 94 per cent. PROBLEM 103. The melting-point of phenol is 40 C. A solution containing 0-172 gram acetanilide (C 8 H 9 ON) in 12*54 grams phenol freezes at 39'25 C. Assuming that FREEZING-POINTPROBLEMS 37 acetanilide has its normal molecular weight in phenol, cal- culate the freezing-point constant and the latent heat of fusion of phenol. Ans. K = 7-38, I = 26'5 cals. PROBLEM 104. A solution of 8*535 grams NaNO 3 in 100 grams water freezes at - 3 '04: C. Calculate the degree of dissociation of the NaN0 3 . K = 1-86. Ans. 0-629. PROBLEM 105. The freezing-point of a solution containing 0-510 gram-molecule of strontium formate in 1000 grams water is 2-390. Calculate the degree of dissociation of the salt. K = 1-86. Ans. 0-76. PROBLEM 106. The freezing-point of a solution of barium hydroxide containing 1 gram-molecule in 64 litres is - 0'0833. What is the concentration of hydroxyl-ions in the solution ? Take K = 1*89 for concentrations in gram-molecules per litre. Ans. 0-0284 gram-ion per litre. PROBLEM 107. A solution containing 2-423 grams sulphur in 100 grams naphthalene (melting-point = 80-1) gave a lowering of the freezing-point of ! 641, and a solution con- taining 2 - 192 grams iodine in 100 grams naphthalene a depres- sion of 0*595. The latent heat of fusion of naphthalene is 35-7 calories per gram. What is the molecular formula of sulphur and iodine respectively in naphthalene solution ? Ans. M for S = 264, .-. S 8 , M for I PROBLEM 108. A solution containing 0'063 of calcium formate in 1000 grams water fr< What is the degree of dissociation ? K = 1- Ans. 0-85. PROBLEM 109. A solution containing J0K)2 'gram-molecule of zinc chloride per litre freezes at -\0>tu35. Calculate the degree of dissociation. K = 1-89. Ans. 0-87. PROBLEM 110. A solution of zinc nitrate containing 0-065 gram-molecule per litre freezes at - 0-322. What is the concentration of the zinc ions in the solution? K = 1-89. Ans. 0-0526 gram-ion per litre. 38 FEBBZING-POINT PROBLEMS PROBLEM 111. The melting-point of pure copper is 1084 C. A solution of Cu 2 O in copper, containing 1-16 per cent, by weight of Cu 2 O freezes at 1076 C. Assuming that the molecular weight of Cu 2 O in the solution corresponds to its formula, calculate the latent heat of fusion of copper per gram. Ans. 38 cals. PROBLEM 112. A solution of lithium chloride containing 4-13 grams per litre freezes at - 0-343. What is the degree of dissociation? K = 1-89. Ans. 0-865. PROBLEM 113. The melting-point of tin = 231-611 C. and its latent heat of fusion = 14*25 calories per gram. The freezing-point of a solution containing 1*5463 grams copper in 440 grams tin is 229*692. Calculate the molecular weight of copper in the solution. Ans. 65-4. PROBLEM 114. The freezing-point of a solution containing 0*0199 gram-molecule SrCl 2 per litre is - 0-1015. What is the degree of dissociation of the salt? K = 1*89. Ans. 0*85. PROBLEM 115. A solution containing 0*834 gram Na 2 S0 4 per 1000 grams water freezes at - 0'0280. Assuming dissocia- tion into 3 ions, calculate the degree of dissociation and the concentrations of the Na- and SO 4 " ions. K = 1*86. Ans. a = 0-782 ; cone. Na- = 0*0918 gram-ion per litre ; cone. S0 4 " = 0*0459 gram-ion per litre. PROBLEM 116. The vapour-pressure of water at C. is 4*620 mm., and the depression of the vapour-pressure caused by dissolving 5*64 grams NaCl in 100 grams water is 0*142 mm. What is the freezing-point of this solution? /K = 1*86. Ans. - 3-177. / PROBLEM 117. The vapour-pressure of a solution con- taining 5*85 grams NaCl in 100 grams water /s 4-460 mm. at C., and that of pure water is 4*620 mmXyThe freezing- point of the solution is - 3*424. Compare the degrees of dissociation obtained (1) by the vapour-pressure method, and (2) by the freezing-point method. K = 1-86. Ans. (1) 0*924, (2) 0-841. BOILING-POINT PROBLEMS 39 PROBLEM 118. When mercuric cyanide, Hg(CN) 2 , is dissolved in potassium cyanide solutions, the complex anion Hg(CN) m + 2 is formed according to the equation Hg(GN) 2 + mCN' = Hg(CN) m + 2 . The freezing-point of a solution containing 0*1965 gram- molecule of KCN per litre is - 0-704, and that of the same solution, after the addition of 0'095 gram-molecule of Hg(CN) 2 per litre, is - 0'530. What is the value of m? K = 1-86. As m must be a whole number, give the nearest whole number. Ans. m = 1-985, .-. 2. PROBLEM 119. The freezing-point of a solution containing 0-01052 gram-molecule Na 2 Si0 3 in 1000 grams water is - 0'0676. Show that the salt is largely hydrolysed accord- ing to the equation Na 2 Si0 3 + H 2 O = 2 NaOH + SiO 2 (colloidal). PROBLEM 120. The freezing-point of a 0*25 N-KCN solu- tion is - 0-860. The freezing-point of the same solution with the addition of 0-25 gram-molecule of AgCN per litre is - 0-830. The solution of AgCN in KCN takes place according to the equation m ON' + AgCN = (AgCN) (CN') m . What is the value of m (nearest whole number) ? K = 1-86. Ans. m = 1-06, /. 1. Boiling-point PROBLEM 121. A solution containing 0-5042 gram of a substance dissolved in 42-02 grams of benzene boils at 80-175. Find the molecular weight of the s6lute, having given that the boiling-point of benzene is 80.000, and its/ latent heat of evaporation of 94 calories per gram. Ans. 181-9. PROBLEM 122. A solution containing 0-7269 gram cam- phor (mol. wt. = 152) in 32-08 ram< of acetone (boiling- point = 56-30 C.) boiled at 56-55TT What is the molecular elevation for acetone and the latent heat of evaporation? Ans. K = 1-674, I = 129*5 cals. per gram. PROBLEM 123. The latent heat of evaporation of carbon disulphide (boiling-point = 46- 20 C.) is 85*9 cals. per gram. 40 BOILING-POINT PROBLEMS The weights a of benzole acid dissolved in 50-09 grams of CS 2 gave the elevations A of the boiling-point of CS 2 . What is the molecular condition of benzoic acid in CS 2 , and how does it vary with concentration ? a 0-9378 1-6429 2-5792 4-5519 grafns. A 0187 0-319 0-479 0-789. / PBOBLEM 124. A solution of 9*472 gram/ CdI 2 in 44-69 grams water boiled at 100'303. The latent heat of evapora- tion of water is 536 cals. per gram. What is the molecular weight of CdI 2 in the solution ? ^(haj/conclusion as to the state of CdI 2 in solution may be drawn from the result ? Ans. M = 363-2. PBOBLEM 125. The boiling-point of a solution of 0-4388 gram NaCl in 100 grams water is 100-074 C. /Calculate the apparent molecular weight of the NaCl and its degree of dissociation. K = 0-52. \ / Ans. M = 30-84, a = \g&7. PROBLEM 126. The boiling-point of acetic acid is 118-100 and its latent heat of evaporation 121 cals. per gram. A solution containing 0*4344 gram of anthracene/m 44-16 grams of acetic acid boils at 118 - 240. What is the nlolecular weight of anthracene? \J Ans. 178. PROBLEM 127. By dissolving 3 '614 grams of CuCl 2 in 100 grams of alcohol the boiling-point of the alcohol is raised 0-308. For alcohol K = 1-15. What is the molecular weight of the solute ? Ans. 134-9. . PROBLEM 128. The boiling-point of a solution of 3'40 grams BaCl 2 in 100 grams water ts 100-208. K = 0'52. What is the degree of dis:ociation ofV^BaCl^? Ans. 0-725. PROBLEM 129. At 100 the vapour pressure of a solution of 6-48 grams NH 4 C1 in 100 grams water is 731-4 mm. K = 0*52. What is the boiling-point of the solution? Ans. 101-086. .?/ - CHAPTER V SUKFACE TENSION, MOLECULAR WEIGHT AND DEGBEE OF ASSOCIATION OF LIQUIDS Definitions Example PROBLEM 130. In a capillary tube of radius r = 01425 JL cm., pure liquid formic acid rises to a height of h^ = 4-442 cms. at ^ = 16-8, h z = 4-205 cms. at t 2 = 46-4 and h s - 3-90 cms. at t z = 79'8. The density -of formic acid is d l = 1-207 at 16-8, d 2 = 1-170 at 46-4 and d 3 = 1-129 at 79*8. What is the molecular weight of formic acid in the liquid state, and how does it vary with temperature? (g = 981-1 cms./sec. 2 ). SOLUTION 130. The surface tension of a liquid of density d, as determined by the height h to which it rises in a capillary tube of radius r, is (i) 7 = I grhd, where g is the value of gravity (gravitation constant). If the quantities on the right-hand side of the equation are ex- pressed in C.G.S. units, g in cms./SQc. 2 , r and h in cms. and d in grams per c.c., we obtain y in dynes per cm. If d is the density of the liquid under investigation, and M its molecular weight, its specific volume is v = -3 and its molecular volume is Mv. The molecular surface of the liquid is, therefore, proportional to (Mv)*l*. ' The product y (Mv) 2 l* is called the molecular surface energy, and, if we use the above units, is expressed in ergs. The molecular surface energy of any liquid diminishes with rise of tempera- ture, and Ramsay and Shields have shown that the tempera- ture coefficient of the molecular surface energy is the same for all liquids, and that its numerical value is 2-121 when the surface energy is expressed in ergs. Hence 41 42 SUEPACE TENSION EXAMPLES where y 15 i^ are the surface tension and specific volume at the lower temperature ^ and y v v 2 the corresponding values at the higher temperature 2 . The molecular weight of the substance in the liquid state is not, however, necessarily the same as the normal mole- cular weight, or formula weight, M. This is the case with unassociated liquids only. If the molecules in the liquid condition are associated into more or less complex groups, we must, in order to make equation (2) generally applicable, multiply the normal molecular weight M by a factor x, so that xM expresses the mean molecular weight of the substance in the liquid state between the temperatures ^ and t 2 . x is called the factor of association, and in the case of unassociated liquids its value is unity/ Equation (2), therefore, becomes For formic acid at ^ = 16-8 we have 7i = i fl^Mi = ' 5 x 981 ' 1 x ' 01425 x 4'442 x 1-207 = 37'47 dynes per cm., at 2 = 464 y 2 = 4 9 rh A = ' 5 x 981<1 x 0*01425 x 4-205 x 1-170 = 34'42 dynes per cm., at 3 = 79-8 7s = i grhA = ' 5 x 9 81 ' 1 x 0-01425 x 3-90 x 1-129 = 30*80 dynes per cm. Therefore, y^Mvrfl* = 37'47(46/l-207) 2/8 = 424-4 ergs 2/3 = 34-42(46/l-170) 2/3 = 397'7 = 30-80(46/1-129) 2/3 = 364-6 Putting these values into equation (3) we obtain, between ^ and J 2 ^(424-4 - 397-7) 46-4 - 16-8 21> 2-121 x 29-6 26^7 /. x - (2-352) 3 /2 = 3-60. ,, - x - 26^7 = ' SUEFACE TENSION^-PKOBLEMS 43 The mean molecular weight of formic acid between 16'8 and 46-4 is, therefore, 3'60 x 46 = 166. Similarly, between t iJL and t 3 , 0^(397-7-364-6) 79-8 - 46-4 "' .-. x = 3-13, and the mean molecular weight of formic acid between 46-4 and 79-8 is 3-13 x 46 = 144. The molecules of formic acid are, therefore, associated into complex groups in the liquid state, and the degree of associa- tion, and, therefore, the molecular weight of the complexes dim- inish with increase of temperature. Problems for Solution PROBLEM 131. What is the factor of association of carbon disulphide, for which at 19-4 the surface tension is 33 '58 dynes per cm., and the density 1*264, and at 46 '1 the surface tension is 29-41 dynes per cm., and the density 1'223? Ans. 1-07. PROBLEM 132. Liquid nitrogen peroxide rises to a height of 3*14 cms. at 1-6 in a capillary tube of 0*0129 cm. radius, and to a height of 2-905 cms. in the same tube at 19*8 0. The density of the peroxide is 1-486 at 1-6, and 1-444 at 19'8. From these data determine whether liquid nitrogen peroxide is associated at these temperatures, and if so, to what extent. Ans. x for N 2 4 = 1*01 /. formula is N 2 O 4 . PROBLEM 133. Normal butyl alcohol at the temperature t rose to a height h in a capillary tube of radius 0*01425 cms. The density of the alcohol at each temperature was d. Calculate the factor of association and the mean molecular weight between each pair of temperatures. t 17-4 45-7 77-9 0. h 4-305 4-005 3-63 cms. d 0-8115 0-7907 0-7634. Ans. x 1-94 1-72 xM 143-5 127-3 PROBLEM 134. In a capillary tube of 0-03686 cms. di- ameter water rose at C. to a height of 8-10 cms., and at 44 SUEFACE TENSION PEOBLEMS 10 C. to a height of 7 '96 cms. The density of water at = 0-9999 and at 10 = 0-9997. What is the factor of as- sociation and the mean molecular weight of liquid water between and 10 ? Ans. x = 3-81, xM = 68-6. PROBLEM 135. In a capillary tube of radius 0'0129 cms. ethyl iodide rose to the height h at the temperature t, at which the density is d, t h d 19-1 2-445 cms. 1'937 46-2 2-22 cms. 1-875. What is the factor of association and mean molecular weight of ethyl iodide between these two temperatures ? Ans. x = I'Ol. Molecular weight .-. normal. CHAPTER VI THEBMOCHEMISTEY Mess's Law r I "HE heat effect of a chemical reaction, for a given quantity J- of the reacting substances and for a definite temperature, is dependent only on the initial and final states of the reacting system. It is the same whether the reaction takes place in one or in several stages, provided that the initial and final states are the same (Hess's Law). By means of this law we are enabled to calculate the heat effect of a reaction which, for various reasons, cannot be measured directly. The re- action in question is considered as the sum or difference of others which have be"en measured. By suitable manipulation of the thermochemical equations all undesired substances are eliminated, and only the equation for the required reaction is left. By the union of 12 grams of solid carbon with 32 grams of gaseous oxygen at 18 C. to form 44 grams of gaseous carbon dioxide, there is a heat evolution of 96960 calories, or, the energy-content of the system in its initial state (12 grams carbon + 32 grams oxygen) is greater than the energy-content of the system in its final state (44 grams carbon dioxide) by 96960 calories. This statement is conveniently expressed by the equation C + 2 = CO 2 + 96960 cals. The symbols in such thermochemical equations stand for definite, though unknown, amounts of energy associated with the formula-weight in grams of each substance. A knowledge of the absolute amount of energy associated with each gram- molecule is, however, unnecessary, since it is only with differences in energy that we are concerned. 45 46 THEKMOCHEMISTEY Heat of Reaction at Constant Volume and Constant Pressure The difference in energy-content between the initial and final states of the system is equal to the heat-evolution or absorption only when no external work is done during the reaction. This is the case when only solids and liquids are involved in the reaction, since in such cases the volume changes during the reaction are negligible. When gases are involved in the reaction, however, the heat of the reaction at constant pressure may differ considerably from that at con- stant volume, since, in the formation or absorption of one gram- molecule of a gas at T absolute, external work equiva- lent to ET = %T calories is done by or on the system, and the total energy- difference between the initial and final states of the system is equal to the heat evolved + the external work done by the system. If Q v is the heat of the reaction in calories at constant volume (no external work condition), and Q p that at constant pressure (external work condition), then Q v = Q p + nET = Q p + ZnT calories, where n is the number of gas-molecules in the final state (on the right-hand side of the equation representing the reaction) in excess of the number in the initial state (on the left-hand side of the equation), and Tthe absolute temperature at which the reaction takes place. The heat of formation of a compound is the heat evolved in the formation of a gram-molecule of the compound from its component elements. As may be readily seen from Hess's law, the heat evolved in a chemical reaction is equal to the sum of the heats of formation of the final substances minus the sum of the heats of formation of the original substances, the heats of formation of the elements themselves being taken as zero. In the following problems the symbol Aq denotes a large quantity of water. For example, NaCl + Aq = NaCl Aq means the solution of 1 gram-molecule of NaCl in much water to form a dilute solution. I . t THBBMOGHBMISTBY EXAMPLES 47 Thermochemistry Examples PEOBLEM 136. The heat of combustion of ethyl alcohol is Q l = 341800 cals. ; the heats of formation of carbon di- oxide and water are Q> 2 = 96000 cals. and Q 3 = 68000 cals. respectively, all at constant pressure. What is the heat of formation Q x of ethyl alcohol ? SOLUTION 136. The combustion of ethyl alcohol takes place according to the equation (1) C 2 H 5 OH + 30 2 = 2C0 2 + 3H,0 + Q l cals. The combustion can be regarded as taking place in 3 stages, (a) in the decomposition of the alcohol into its elements ac- cording to the equation (2) C 2 H 5 OH = 20 + 3H 2 + O - Q x cals., and (b) and (c) in the combustion of 20 and 3H 2 according to the equations (3) 2C + 20 2 = 2C0 2 + 2g 2 and (4) 3H 2 + liO 2 = 3H 2 + 3g 3 . By adding (2), (3) and (4) we obtain C 2 H 6 OH + 30 2 - 2C0 2 + 3H 2 Q x + 2Q 2 + SQ y - From a comparison of this equation with (1) it follows that Q l = - Q x + 2Q 2 + 3g 3 , and, therefore, Q x = 2 log In the case of a monomolecular reaction the numerical value of k is independent of the unit chosen to express the concen- trations a and x. Thus if a unit n times less than that used in equation (2) is chosen, we obtain log - = log __ = *. * w(a - #) a - a; Bimolecular Reaction At constant temperature the velocity at any instant of a bimolecular reaction which goes to practical completion is given by the equation (4) *? - k(a - x)(b - x). a and b are the initial concentrations of the reacting sub- stances, x the diminution in concentration of the reacting substances after the time-interval t from the commencement of the reaction, due to their transformation into the products of the reaction, and k the velocity-constant. Equation (4) gives, on integration, * * (a - b)t 10ge (6 - x)a , _ 2-302 . (a - x)b ~ (a - b)t g (6 - x)a When the initial concentrations of the reacting substances are the same, i.e. when a = 6, equation (4) becomes which gives, on integration, (6)*-l * t a(a - x) For a bimolecular reaction the numerical value of k depends on the unit of concentration chosen for a, b and x. Thus if VELOCITY OF EEACTION EXAMPLES 55 a unit I/nth of that used in equations (5) and (6) is employed, these equations become 2-302 , n(a - x)bn 2-302 , (a - x}b f\ ]f' __ lOg = 1O2 J ' w n(a - b)t & n(b - x)an n(a - b)t (b - x)a and(8)it' = i. m *-' t na(a - x)n t na(a - x)' By dividing (7) by (5) or (8) by (6) we obtain Therefore, when using a particular value of k for a given reaction, the concentrations should be expressed in the same unit as was employed in obtaining that value of k. Velocity of Reaction Examples PROBLEM 156. When a solution of dibromsuccinic acid is heated the acid decomposes into brom-maleic acid and hydro- bromic acid according to the equation CHBr . COOH CH . COOH | =|| + HBr. CHBr . COOH CBr . COOH. At 50 the initial titre of a definite volume of the solution was T = 10-095 c.c. of standard alkali. After t minutes the titre of the same volume of solution was T t c.c. of standard alkali. * 214 380 T t 10-095 (T ) 10-37 10-57 (a) Calculate the velocity-constant of the reaction, (b) After what time is ^ of the dibromsuccinic acid decomposed ? SOLUTION 156. (a) The initial equivalent concentration a of the dibromsuccinic acid is proportional to T . Since 1 molecule of dibasic dibromsuccinic acid gives. 1 molecule of dibasic brom-maleic acid and 1 molecule of monobasic hydro- bromic acid, the increase in the titre (T t - T ) after time t is proportional to the equivalent concentration of the hydro- bromic acid at time t. But the formation of 1 equivalent of hydro bromic acid means the disappearance of 2 equivalents of dibromsuccinic acid, therefore x, the decrease in the equi- valent concentration of the dibromsuccinic acid after time t, is proportional to 2 (T t - T ). Therefore, from (2), 56 VELOCITY OF EEACTION EXAMPLES , 2-302 . a 2-302 ,_ T k = _ log - i For t = 214 minutes *!* and for * 380 minutes . x UH x 10*7 The mean velocity-constant is, therefore, k = 0*000260. (b) When ^ of the acid is decomposed the concentration of the remaining acid is f of the initial concentration a ; there- fore (a - x) = fa and See also problem 234, p. 108. PROBLEM 157. A b = O'Ol N-solution of ethyl acetate is saponified to the extent of c = 10 per cent, in ^ = 23 minutes by an a = 0*002 N-solution of sodium hydroxide. In how many, ( 2 ), minutes would it be saponified to the same extent by an a' = 0'005 N-solution of potassium hydroxide ? SOLUTION 157. The saponification of ethyl acetate takes place according to the equation CH 3 . COOC 2 H 5 + H. 2 = CH 3 . COOH + C 2 H 5 OH. In alkaline solutions free acetic acid is not formed, but the alkali salt, which is almost completely dissociated into its ions. The equation for the reaction may, therefore, be written in the ionic form CH 3 . COOC 2 H 5 + OH' = CH 3 . COO' + C 2 H 5 OH. If x is the quantity of ethyl acetate saponified at the time t, the velocity of saponification is given by the equation . COOC 2 H 5 ], k(a - x)(b - x), VELOCITY OF EEACTION EXAMPLES 57 if the NaOH is regarded as completely dissociated. The square brackets denote the concentration of the enclosed sub- stance. The integration of this differential equation is carried out as follows : 7 - m - , - ** (a - x)(b - x) (a - 0) \o - x a - x and, on integration, (1) - -1 [ lo g (b ~ x ) ~ lo ge( - &)] = kt + constant. At the time t = 0, x = 0, and, therefore, (2) [log e & - log e a] By subtracting (2) from (1) we obtain (2) [log e & - log e a] = constant. and, converting to common logarithms, (3) *- ?P. 2 , .log*jl*J. t(a - b) * a(b - x) From the data gjven in tLs problem, at the time ^ = 23 minutes x = -^-6 = b = O'l b. The velocity- constant k may, therefore, be calculated from equation (3), since all the other quantities are known. For this particular case we have k = ' ^(a - b) a(b - O'l ) 2-302 l 0-01 x 0-001 23 x - 0-008 0-002 x 0-009 = - 12-51 log 0-556 = - 12-51 x - 0-255 = 3-19. It is now possible to calculate the time t% in which the same ester solution would be saponified to the same extent by a' = 0-005 N-KOH solution. Since KOH, like NaOH, may be regarded as completely dissociated in dilute solution, we obtain again = k(a' - x)(b - x), 58 VELOCITY OF EEACTION EXAMPLES or, integrated, Tc(a' - b) e a'(b - x)' 2-302 Jo 0-01 x 0-004 3-19 x - 0-005 6 0-005 x 0-009' = - 144 log 0-89 = 144 x 0'051 = 7-34 minutes. PROBLEM 158 (cf. preceding problem). What do the times $! and 2 in the preceding problem become (1) if all the concentrations a, b and a' are diminished to - = of their n 10 values there, (2) if the temperature is raised by 15 C.. if it is assumed that the velocity-constant for any temperature is doubled by a rise of 10 G. ? SOLUTION 158. (1) If all the concentrations in the preced- ing problem are diminished to - ~ TQ of their values there, the new values are a = 0-0002, b = 0-001, a' = 0-0005, x = 0-0001. As in the preceding problem we obtain t, = 2 ' 302 log b ( a ~ *)' 1 k(a - b) g a(b =!3f 2-302 ^ 0-001 x 0-0001 . 3-19 x - 0-0008 & 0-0002 x 0-0009' = - 900 log 0-556 = 230 minutes, and similarly, tj = - 1440 log 0-89 = 73-4 minutes. If, therefore, all the concentrations are diminished to - of n their previous values, the time required for the reaction to pro- ceed to the same extent is increased to n times its previous value. (2) If a rise of temperature of 10 C. doubles the velocity- constant k, i.e. if ^t + 10 = 2fc t o, the relation between k and t may be deduced as follows : !og & + 10 = lo g 2 + log &>. log k t0 + 10 - log k t0 = log 2 *- a - 0-0301, VELOCITY OF BEACTION EXAMPLES 59 and, on integration, (1) log k = 0-0301* + constant. For t 1% the temperature corresponding to the values in the preceding problem, A? x = 3 '19, and, therefore, (2) log k l * 0-0301*! + constant. By subtracting (2) from (1) log | = 0-0301(* - f j). For f - t\ = 15" C. we therefore obtain log k = log fcj + 15 x 0-0301 = 0-504 + 0-452 - 0'956, .-. k = 9-04. If t and *j are the times required for the reaction to proceed to the same extent at the temperatures t and t ^ and k and k l are the velocity-constants at these temperatures, then if a, b and, therefore, since the reaction proceeds to the same ex- tent, x are the same in the two cases, it is evident from the equations _ k(a - b) e a(b - x) and . _1 __ I bfa - x) 1 \(a - b) ^ e a(b-x) that * k The times required for the reaction to proceed to the same extent at the two temperatures, are, therefore, ceteris paribus, inversely proportional to the velocity-constants. For 0-002 N-NaOH, where ^ = 23, ^ = 3-19 and k = 9-06, we obtain q.i q t = r_r x 23 -= 8' i minutes, and for 0*005 N-KOH, where t* = 7'34, ^ = 319 and k = 9-06, q.-i Q t M x 7-34 = 2*6 minutes. 9' 60 VELOCITY OF EEACTION rKOBLEMS Problems for Solution PBOBLEM 159. From the following data show that the decomposition of H^ in aqueous solution is a monomolecular reaction : Time in minutes, 10 20 n 22-8 13-8 8-25 c.c. n is the number of c.c. of KMnO 4 required to decompose a definite volume of the H 2 O 2 solution. PROBLEM 160. The decomposition of AsH 3 into solid arsenic and hydrogen may be followed by measuring the pressure at constant volume from time to time. In an ex- periment at 310 the pressures p in mm. of Hg were obtained after the times t hours. Show from these figures that the reaction is of the first order. 5-5 6-5 8 hours. p 733-32 305-78 818-11 835-34 mm. PROBLEM 161. The conversion of acetchloranilide into 2?-chloracetanilide according to the equation C 6 H 5 . NC1(COCH 3 ) = C 6 H 4 C1 . NH(COCH 3 ) may be followed by removing a measured quantity of the solution from time to time, adding it to a solution of KI, and titrating the liberated iodine with standard thiosulphate. The volume of thiosulphate used is proportional to the con- centration of the acetchloranilide. Thus after t hours from the commencement of the reaction y c.c. of thiosulphate were required. * 1 4 y 35-6 13-8 In what time is the conversion half completed ? Ans. 2-195 hours. PROBLEM 162. The decomposition of diacetonealcohol into acetone according to the equation CH 3 . CO . CH 3 . C(CH 3 ) 2 . OH = 2CH 3 . CO . CH 3 is accompanied by a considerable increase in volume. The reaction is catalytically accelerated by OH'-ions, the velocity- constant being proportional to the concentration of OH'-ions. By allowing the reaction to take place in a dilatometer the expansion may be observed. The quantity of diacetone- e VELOCITY OF EEACTION PEOBLEMS 61 alcohol present at any time is proportional to the expansion from that time to the end of the reaction. The following table gives the dilatometer readings R at the times t, obtained with a mixture of 20 c.c. 01 N-NaOH and 1-0526 grams of diacetonealcohol. Calculate the velocity-constant for 0-1 N-NaOH. t 10 20 30 40 50 60 oc mins. E 60-8 97-7 119-9 133-4 141-4 146'1 153'8 Ans. Mean k = 0-05030. PROBLEM 163. Potassium persulphate and potassium iodide interact with liberation of iodine. 25 c.c. of a solution, which was N/30 with respect to both persulphate and iodide, were titrated from time to time with N/100 Na^SaO^ From the following results show that the reaction is bimolecular. t is the time of titration and x the number of c.c. of thio sulphate used. t 9 16 32 50 x 4-52 7-80 14-19 20-05. PROBLEM 164. In the saponification of ethyl acetate by NaOH at 10, y c.c. of 0-043 N-HC1 were required to neu- tralise 100 c.c. of the reaction mixture t minutes after the commencement of the reaction. t 4-89 10-37 2818 oc y 61-95 50-59 42-40 29-35 14-92. Calculate the velocity-constant when the concentrations are expressed in gram-molecules per litre. Ans. Mean k = 2-38. PROBLEM 165 (cf. preceding problem). 1 litre of N/20 ethyl acetate is mixed at 10 with (a) 1 litre of N/20 NaOH, (6) 1 litre N/10 NaOH, (c) 1 litre N/25 NaOH. In what time is half the ester saponified in each case ? Ans. (a) 16-8 mins., (b) 6'81 mins., (c) 24'2 mins. See also problems 319, 320. CHAPTER VIII LAW OF MASS-ACTION. EQUILIBRIUM-CONSTANT. INFLU- ENCE OF TEMPERATURE ON EQUILIBRIUM-CONSTANT. AFFINITY, CHANGE OF FREE ENERGY OR MAXIMUM WORK OF A REACTION. PARTITION LAW. SOLUBILITY OF GASES. Law of Mass-action Equilibrium-constant IF a number of substances, for example, the four substances A, B, C and D, react according to the equation mA + nB = pC + qD, then, at equilibrium, the following relation exists between the concentrations (number of units of mass in unit volume) of these four substances (Guldberg and Waage), The square brackets denote the concentrations of the en- closed substances, m, n, p and q are the number of mole- cules of A, B, C and D which take part in the reaction. K c is called the constant of the law of mass-action, or the equilibrium-constant for the reaction. For a given reaction, the value of K c depends on the temperature, being constant for a given temperature, and on the units of mass and volume used to express the concentrations. If v is the number of units of volume occupied by the re- action mixture at equilibrium, and a, 6, c and d the number of units of mass of A, B, C and D respectively present in this volume at equilibrium, the concentrations of A, B, C, and D are [A]--*, [B] = -*,[C] = <;,[D]=4 62 c LAW OF MASS-ACTION 63 and, from (i), If in + n = p + q, that is, if the total number of molecules on each side of the equation is the same, equation (2) be- comes > [A] m [B] n that is, for a given temperature and unit of mass, the value of the constant K c is independent of the unit of volume used in expressing the concentrations. In all other cases, as can be readily seen from (2), the value of K c is dependent on the unit of volume. Thus if K c is the value of the equilibrium-constant for a given unit of volume, then for a unit of volume x times smaller, the con- stant is K e X a .i + n-P-q | f or W = , [B] = ^ [C] = , [D] = A and (2) becomes Thus if Z c is the value of the constant when the concentra- tions are expressed in gram-molecules per litre, the value of the constant is K c x 1000 m + n ~ p ~ q when the concentrations are expressed in gram -molecules per c.c. The unit of concentration usually employed, and the one which will be used in the following examples, is the gram- molecule per litre. That the value of K c depends on the unit of mass used in expressing the concentrations may be shown by expressing the concentrations in grams per litre instead of gram-mole- cules per litre. (2) then becomes c_c\p (dM v > \ v ~ c 64 LAW OF MASS-ACTION where M A , M B , M c and M D are the molecular weights of A, B, C and D respectively. In any particular case the unit of concentration employed must be that used in obtaining the value of K involved. From (3) it is evident that for given amounts of the re- acting substances the position of equilibrium, or the quantities of the reacting substances present at equilibrium, is inde- pendent of the volume of the reaction-mixture when the same number of molecules occurs on each side of the reaction- equation. In all other cases we see from (2) that the position of equilibrium is dependent on the volume of the reaction-mixture. In the case of gases, their partial pressures at equilibrium may be substituted for their concentrations in equations (i), (2) and (3). The numerical value of the equilibrium-constant will, of course, depend on whether we use partial pressures or concentrations. The relation between the different values may be found as follows. If we assume that A, B, C, and D are all gases, and that T is the absolute temperature of the equilibrium-mixture, then for concentrations in gram-mole- cules per litre equation (2) holds. If the partial pressures in atmospheres of the four substances are p A , p R , p c , p D , then But according to (2), page 1, if a, b, c and d are the numbers of gram-molecules of A, B, C and D present in v litres at equilibrium, aRT __ bRT _ cRT _ dRT PA i PH ~^~> PC ~ ~ > PD > 7T x ' u * * P / ~ ~DIT\\ in f }\T) r n (aBT\ m (bIlT\*> (a\ m (b\ \ v' "/ V v ) w \9/ Only when p + q = m + n is K p K c . If the equilibrium-constant used in a particular case is that obtained by using concentrations of the reacting substances in gram-molecules per litre, then the active masses of the react- ing substances must always be expressed in gram-molecules per litre when using this value of the equilibrium-constant. EQUILIBEIUM-CONSTANT AND TEMPEEATUEE 65 Similarly, if the equilibrium-constant used was obtained by employing partial pressures instead of concentrations, the active masses of the reacting substances must always be expressed in partial pressures when using this value of the equilibrium-constant. If one or more of the reacting substances are present in the solid state, their active masses, and, therefore, their con- centrations at equilibrium, are constant, and may be omitted from the equation. If, for example, A and C in the reaction mA + nB = pC + qD are solid substances, the law of mass-action requires (s) [ -5L q = K 15; [B] n Equilibrium-constant and Temperature The equilibrium- const ant K c or K p is dependent on the temperature according to the equations (van't Hoff) (6) < - ~ and d\og e K p _ - Q p ~~ ' ~ ' ~dT Q v and Q p are the heats evolved in the reaction from left to right at constant volume and constant pressure respectively ; in the reaction mA -f nB = pC + qD, for example, in the transformation of m gram-molecules of A and n gram-molecules of B into p gram-molecules of C and q gram- molecules of D. Q v and Q p are assumed to be inde- pendent of the temperature (see p. 25). B is the gas-constant (1-985 or, approximately, 2 for calories) and T the absolute temperature. On integration between the absolute tempera- tures T and T lt for which the respective equilibrium-constants are K c and K cl (for concentrations) and K p and K pl (for partial pressures), we obtain on changing from natural to common logarithms (7) log* - log*. (8) What was said about L in equation (5), p. 25, applies equally to Q,, and Q p in the above equations. 6 66 ' EQUILIBRIUM-CONSTANT AND TEMPEEATUEE The following are some of the applications of these equa- tions. In all cases the heat-effect is for the mean temperature T + T x l . The application to ordinary chemical equilibria is obvious from the explanations given above. The equations apply also, however, to many physical equilibria. As we have already seen (p. 25), in the case of vaporisation (9) logp - log ft = where p and p l are the vapour-pressures of the liquid at T and T l respectively, and L is the latent heat of evaporation per gram-molecule (at constant pressure). For the sublima- tion of a solid the same equation holds, L being the molecular heat of sublimation. For the solubility of solids we have Q (10) logc - where c and c x are the concentrations of the saturated solu- tions at T and T : respectively, and Q is the heat of solution per gram-molecule. In the case of a difficul% soluble strong binary electrolyte, where the dissociation in the saturated solution may be re- garded as practically complete, we have (p. 121) K = c 2 and K^ = c^ (solubility-products), where G and c x are the solubilities at T and 2\ respectively. Hence (11) logo* - log Cl = where Q is the heat of ionisation per gram-molecule. The heat of precipitation of the solid from its ions has the same value as Q, but the opposite sign. For electrolytes which obey the dilution-law we have Q /T, - (12) log - log^ = ^B (-%r where K and Xj are the dissociation-constants at T and T l respectively, and Q the heat of ionisation per gram-molecule. AFFINITY PARTITION LAW 67 Affinity, or Change of Free Energy At a definite temperature T, the affinity, or change of free energy, of a chemical reaction, for example, of the reaction mA + nB = pC + qD is (3) * = or (14) A = RT\og.K,, + RTlog. ^jj^jjf K c is the equilibrium-constant for concentrations in gram- molecules per litre, and K p that for partial pressures in at- mospheres. A is the maximum work which can be performed by the reversible transformation of m gram-molecules of A and n gram-molecules of B at the concentrations [A] and [B] gram -molecules per litre, or at the partial pressures (A) and (B) atmospheres respectively into p gram-molecules of C and q gram-molecules of D at the concentrations [C] and [D] gram-molecules per litre, or at the partial pressures (C) and (D) atmospheres. Partition Law When a soluble substance is distributed between two non- miscible solvents, in each of which it has the same molecular weight, the ratio of the concentrations c x and c. 2 of the solute in the two solvents has a constant value for a given tempera- ture, or (15) ~ = S. C/2 This is Nernst's partition law. K is called the partition- coefficient of the dissolved substance for the -two solvents. If the solute has not the same molecular weight in the two solvents, but has, say, its normal molecular weight iu the first solvent, whilst in the second solvent it is more or less associated to complex molecules according to the equation s* then the ratio - 1 is no longer constant. For each definite kind of molecule S, S, S, . . . S n however, there is a conotant 68 SOLUBILITY OF GASES partition-coefficient for the two solvents, and the ratio of the concentrations found experimentally is influenced by all these. If one particular type of complex, say S n , predominates largely in the second solvent, the concentration of the solute in this solvent will be practically that of the S n molecules. Accord- ing to the law of mass-action, there exists between the simple and the complex molecules in the second solvent the relation [S\f = *[SJ and, according to the partition law, between the simple molecules in the first solvent and those in the second solvent the relation [8], where [S]j and [S] 2 are the concentrations of the simple molecules in the first and second solvent respectively, and K is the partition-coefficient for this type of molecule. From these two equations we obtain S]l - Vfc x K = K,. Since the S M molecules are supposed to largely predominate in the second solvent, the total concentration in this solvent will be practically that of the S n molecules. If, then, Cj and c 2 are the total concentrations of the solute in the first and second solvents respectively we obtain (16) JTL = constant. VC2 c x and c 2 may be expressed in any unit, for example, grams or gram -molecules (of normal molecular weight) per litre. Solubility of Gases The mass of a gas dissolved by a given volume of a liquid at a definite temperature is proportional to the pressure of the gas. This is Henry's law. Since the concentrations of the gas in the liquid and in the space above the liquid are both proportional to the pressure of the gas, Henry's law may be stated in the following form, which corresponds to Nernst's partition law. For a given temperature SOLUBILITY OF GASES 69 the concentration of the gas in the (17) . Uquid t t . J-T- = (- - constant = ,. the concentration ot the gas in the space above the liquid s is called the solubility-coefficient of the gas in the liquid for the given temperature. The solubility-coefficient s of a gas in a liquid at a definite temperature t C. may be defined as the volume of gas, measured at t C. and under a pressure of p atmospheres, which is dissolved by unit volume of the liquid when the pressure of the gas on the liquid at equilibrium is p atmo- spheres. That the solubility-coefficient s so defined is the same as the ratio of the concentrations of the gas in the liquid and in the space above may be shown as follows : At C. and under 1 atmosphere pressure, 1 gram-molecule of a gas occupies 22*4 litres. 1 litre of the liquid dissolves s litres of gas measured at t C. and p atmospheres, 273 p litres measured at C. and 1 atmos. 273 + t 273 g ram - molecules of s in the liquid is g ram - molecules P er litre - 273 + t /. the concentration of the gas in the liquid is 8 X 273 ? + t 1 gram-molecule of gas occupies 22*4 litres at C. and 1 atmos. 1 gram -molecule of gas occupies .*. the concentration of the gas in the space above the liquid is er litre ' and concentration of gas in the liquid concentration of gas in space above liquid s x 273 x p 22-4(273 + t) ^ (273 + Q22-4 > 273 x p 70 MASS-ACTIONEXAMPLES The absorption-coefficient of a gas in a liquid at t C., in terms of which the solubility of gases is often expressed, is the volume of gas, measured at C. and 1 atmosphere pressure, absorbed by unit volume of the liquid when the pressure of the gas above the liquid is 1 atmosphere. From a mixture of gases the quantity of each dissolved is proportional to its partial pressure at equilibrium. Law of Mass-action Equilibrium and Temperature Affinity Examples PROBLEM 166. a = 9*2 grams of nitrogen peroxide occupy at x = 27 C. and under a pressure of P = 1 atmosphere a volume v l 2'95 litres, and at t% = 111 C. and under the same pressure a volume v 2 = 6*07 litres. Calculate the de- grees of dissociation a x and a. 2 , and the dissociation- constants IJi and J5T 2 of nitrogen peroxide at t 1 C. and t\ C., and also its molecular heat of dissociation. SOLUTION 166. According to the gas laws the number n of molecules in the volume v lt at the temperature t lt and under the pressure P, is Pv l = B(273-+ y If M = 92 is the molecular weight of N 2 O 4 and o^ the fraction of the N 2 O 4 molecules which are dissociated according to the equation N 2 O 4 2NO 2 , then a grams of nitrogen peroxide contain (l - i) molecules of N 2 O 4 and molecules of NO 2 . From this it follows that a Pv l and MPv l Similarly, for the temperature t z C., we get yji a 2 ** /TP/O7Q .1- M ~~ A * Substituting the numerical values given in the problem, we find MASS-ACTIONEXAMPLES 71 92 x 1 x 2-95 ttl = 9-2 x 0-082 x 300 " = ' 2 ' . 92 x 1 x 6-07 "2 = w^-o-mz^m ~ = 'M- The dissociation-constant K of the reaction N A 2 2 NO 2 is [NO 2 ] 2 = [NAT if the square brackets denote the concentration in moles per litre of the enclosed molecules. For ^ C. [N 2 4 ] = ^- ( ^^ and [NOJ = |^, therefore X^ttnA 2 X a(l - a,) (1 - a Similarly, for t C., Putting in the numerical values we find 4 x 9-2 x (Q-20) 2 KI = 0-80 x 92 x 2-95 = * 68 ' _ 4 x 9-2 x (Q-93) 2 ^ " 0-07 x 92 x 6-07 ~ O ' Sl6 ' The heat of dissociation Q of a gram-molecule of N 2 O 4 can be calculated from the dissociation-constant by van't Hoff's equation d log e K _ - Q dT " RT*' Integration between the temperatures T x = 273 + ^ and T 2 = 273 + 2 gives, if we assume that Q x does not change in this interval, v 'i Q f 1 1 72 EQUILLBBIUM-CONSTANT EXAMPLES Substituting the numerical values gives 1-985 x 300 x 384 0-816 Q= - - 34- - x 2-3 x log -^o68" = ~ ! 3<>oo cats. NOTE. The equation for the dissociation of N 2 4 may also be written N 2 O 4 = NO 2 + NO 2 and the dissociation-constant [N0 2 ] [NOJ [N.OJ ' where each NO 2 molecule is considered separately. We have then (see above) at t l and, similarly, at TN O 1 [N 2 OJ = M ~~^~ ' Honce aa __ K * = (1 - ojjfq, = ' 2 4 ' For a given temperature either of the expressions K = (1 - a)Mv r K = (1 -^^ may be used. The expression chosen must, however, be retained throughout any series of calculations dealing with the same reaction. It is evident that the value we obtain for Q is not affected by the expression chosen for K, since the ratio KJK^ re- mains constant. PROBLEM 167. If a = 3'6 grams of phosphorus penta- chloride is heated to t = 200 C., it volatilises completely, and the vapour occupies a volume v = 1 litre under a pressure P = 1 atmosphere. At the same time it dissociates partially into phosphorus trichloride and chlorine. Calculate the degree of dissociation a and the dissociation-constant K of phosphorus pentachloride at this temperature. Express the concentrations in gram-molecules per litre. MASS-ACTIONEXAMPLES 73 SOLUTION 167. As in problem 166 the number of mole- cules which occupy the volume v at the temperature T abs. and under the pressure P atmospheres is Pv If M is the molecular weight of PC1 5 and a the degree to which PC1 5 is dissociated according to the equation PC1 5 = PC1 3 + Olj, - then the volume v contains (1 - a) moles of PC1 5 , moles of PC1 3 and ^ moles of C1 2 . Accordingly D-. and _ MPv _ -, 208 x 1 x 1 ~~aRT 3-60 x 0-082 x 473 and the dissociation-constant at 200 C. is afa [PC1 5 ] M(l - a)v (0-49) 2 x 3-60 - 208 x 0-51x1 = ' 0815 ' PEOBLEM 168 (of. preceding problem). What pressure P is developed when a = 3 '6 grams of solid phosphorus penta- chloride in v = 1 litre of chlorine at t' = 18 C. and under a pressure p = 1 atmosphere, is heated at constant volume to t = 200 C. ? SOLUTION 168. According to the law of mass-action the dissociation of the PClg is diminished in presence of the gaseous chlorine. In this case also the equation m [PClJOy . K (1) [PC1J must be satisfied. Let a be the degree of dissociation of the PC1 5 at t C. in presence of the gaseous chlorine ; then the number of mole- cules of .PC1 5 in the volume v is 74 MASS-ACTIONEXAMPLES and the concentration of the PC1 5 is (2) eta If' Similarly, the number of PC1 3 molecules in volume v is < n = and the concentration of PC1 3 ffi The number of C1 2 molecules is the sum of the two quanti- ties n and ri. The n molecules are derived from the dissociation of the PC1 5 and are equal to the number of PC1 3 molecules = ( ^LI The ri molecules are derived from the chlorine M originally present, and at t' C. = 273 -f V = T' abs. and under a pressure of p atmospheres the volume v litres contains ri = , molecules of We get, therefore, From (1), (2), (3) and (4) we obtain da! (da p \ f Oa P MV\MV + 7tT r ) _. \MV g(l - a) I - a Mv In this equation only a is unknown, and can, therefore, be calculated as follows : , tt(a') 2 a'p "~Mv~ + ~BT" , ( p t \ Mv _ MvK /a a ' Mv f p \ iM~v' 2 f p \ 2 MvK 8x_V- -^- ,000815^ " 2x3-6 Vo-082 x 291 + woio ) MASS-ACTIONEXAMPLES 75 7208)' / 1~ \a 208x0-00815 \* 4 x (3-6) 2 VO-082 x 291 3-6 = - 28-85 (0-0419 + 0-00815) + V2 T 08 + 0-471 = - 1-442 4- JM51 = 0-155. The total number of, molecules of PC1 5 + PC1 3 + CLj is therefore, a(l - a') 2aa' pv ~~M~ ~M + BT" 3-6 x 0-845 2 x 3-6 x 0-155 1 ~T~ C\f\O l 208 208 r 0-082 x 291 = 0-0146 + 0-00536 + 0-0419 = 0-0619. T f he total pressure P developed by heating a grams of PC1 6 in v litres of chlorine, at p atmospheres and t m C., to 200 C. is, therefore, NET 0-0619 x 0-082 x 473 P = = I - = 2*40 atmospheres. PROBLEM 169. Nitrogen and oxygen combine at high temperatures to form nitric oxide, according to the equation N 2 + 2 = 2 NO. The equilibrium-constant at T = 2675 abs. is *- 3-5x10-3. What yield of NO (in percentage by volume) is obtained at this temperature and at normal pressure (1) from air, (2) from a mixture of 40 per cent. O 2 and 60 per cent. N 2 by volume, and (3) from a mixture of 80 per cent. O 2 and 20 per cent. N 2 by volume ? SOLUTION 169. Let the yield of NO in percentage by volume be x. Then, since the reaction takes place according to the equation N 2 + O. 2 = 2 NO, from a mixture which originally contained a per cent, of O 2 and b per cent, of N. 2 by volume, we get at equilibrium x per cent, of NO. (a - ^) percent, of O 2 and (b - ^j per cent, of N 2 by volume. 76 MASS-ACTIONEXAMPLES Since the concentrations are proportional to these per- centages by volume, we get from the law of mass-action 3-5x10-*, or, solving the quadratic equation for x, K(a + b) lK 2 (a + fr)' 2 abK and since K is small compared with 4, we may take, with sufficient approximation, x = jKab - ^-j . We get, therefore, (1) for air, where a = 20'8, and b = 79'2, x 1 = 2-4 - 0-09 = 2*3 per cent. (2) for a = 40 per cent., and b = 60 per cent. x 2 =2*8 per cent. (3) for a = 80 per cent., and b = 20 per cent. PROBLEM 170 (cf. preceding problem). What must be the initial composition of the mixture of O 2 and N 2 ^to give the maximum yield of NO ? SOLUTION 170. The yield of NO is /xr-T E(a + ^) x = jKab ^- '<> or, since a + b = 100 in mixtures which consist of oxygen and nitrogen only, x = #a(100 - a) - K x 25. To obtain the x particular concentration a for which a; is a maximum, we must differentiate x with respect to a, and put the differential coefficient = : This expression becomes = when a = 50 per cent. The yield is therefore greatest in a mixture of equal volumes of oxygen and nitrogen. MASS-ACTIONEXAMPLES 77 PROBLEM 171. On rapidly heating solid ammonium iodide to t = 357 C. a vapour-pressure P = 275 mm. is pro- duced, which, owing to the almost complete dissociation of the ammonium iodide, is practically entirely made up of the partial pressures of the dissociation-products HI and NH 3 . On keeping the system at this temperature for some time, the vapour-pressure increases, because the hydriodic acid dissociates according to the equation " 2 HI = H 2 + I 2 . What is the value of the vapour-pressure P' at complete equilibrium, if the dissociation-constant for the reaction 2 HI = H 2 + I is SOLUTION 171. The total vapour-pressure P is equal to the sum of the partial pressures of the individual molecular species. These partial pressures may, for shortness, be ex- pressed by the chemical symbols enclosed in round brackets. On heating rapidly we get then P = (NH 4 I) + (NH 3 ) + (HI), and at complete equilibrium P = (NHJ) + (NH 3 ) + (HI) + (Ha) + (I 2 ). Since the dissociation of NH 4 I into NH 3 and HI is almost complete, (NH 4 I) may be neglected in comparison with (NH 3 ) and (HI), so that we obtain P = (NH 3 ) + (HI) and (NH 3 ) = (HI)=-f. For the dissociation of NH 4 I vapour the mass-action equation must always be satisfied. If NH 4 I is present as a solid phase, then (NH 4 I) is constant for a given temperature, and, therefore, The dissociation of HI into H 2 and I 2 takes place accord- ing to the equation 2 HI = H 2 + I 2 . 78 MASS-ACTIONEXAMPLES At complete equilibrium, reached after heating for some time, the following equations must, therefore, be satisfied : (1) (2) (H 2 ) - (I 2 ) (3) (NH 3 )(HI) (5) P' = (NH 3 ) + (HI) + (H 2 ) + (I 2 ). In these five equations the five magnitudes P', (NH 3 ), (HI), (I 2 ) and (H 2 ) are unknown. All the unknowns have, therefore, to be calculated. By eliminating (I 2 ) by means of equation (2) we obtain (6) from(l): (NH 3 ) = (HI) + (H 2 ), (7) from (4) : (H 2 ) = (HI) JK, (8) from (5) : P' = (NH 3 ) + (HI) + 2(H 2 ). By eliminating (H 2 ) by means of equation (7) we obtain (9) from (6) and (7) : (NH 3 ) = (HI) (1 + >JK), (10) from (8) and (7) : P = (NH 3 ) + (HI) (1+2 */K) (3) divided by (9) gives pa PI 4(1 + v/)' " >/l + From (11) and (9) we obtain P By substituting in equation (10) the values of (HI) and (NH 3 ) obtained in (11) and (12), we obtain With P = 275 mm. Hg. P' - 137-5 x 2 + 3 X ' 123 id/ o x - ~ = 307 mm. MASS-ACTIONEXAMPLES 79 PROBLEM 172. Iron and water vapour react according to the equation Fe + H 2 = FeO + H 2 until a certain state of equilibrium is reached. At t l = 1025 C., and under a total pressure of 1 atmosphere, the partial pressure of the hydrogen at equilibrium is p H2 = 427 mm. Hg, and the partial pressure of the water vapour p H20 = 333 mm. At t = 900 C. the corresponding partial pressures are p' H2 = 450 mm., and p'u 20 310 mm. What is the value of p 02 , the dissociation-pressure of ferrous oxide pro- duced by its dissociation into metallic iron and oxygen, at the temperature T 3 = 1000 abs., if the percentage dissociation of pure water vapour at this temperature, and under a pres- sure P = O'l atmosphere, is 646 x 10~ 5 ? SOLUTION 172. In the reaction between iron and steam according to the equation Fe + H 2 O = FeO + KJ, we have at equilibrium, according to the law of mass-action, where the value of K depends on the temperature only. From the data given in the problem K can be calculated for the temperatures T-^ = 273 + ^ and T 2 = 273 + t y We must assume that ferrous oxide is dissociated to a definite, though small, extent, according to the equation 2 FeO = 2 Fe + O 2 . the degree of dissociation depending on the temperature. For every temperature, therefore, the partial pressure of the free oxygen p 02 , has a definite value. This free oxygen can react with the free hydrogen to form water vapour according to the equation 2 + 2H 2 = 2H 2 0. For this reaction we have at equilibrium, From (1) and (2) we obtain 1 Po-2 80 EQUILIBRIUM-CONSTANTEXAMPLES For every temperature for which K and K are known we can, therefore, calculate p OT K is known for the temperatures T 1 and T 2 , and K for the temperature T 3 can be calculated from the dissociation of water vapour at that temperature. In order to calculate p . 2 for T 3 , K for T 3 must, therefore, first be found. This can be done by calculating the heat of reaction, Q, for the reaction Fe + H 2 O = FeO + H 2 by means of van't Hofif's equation dT Integrating between the neighbouring temperatures T l and T 2 we obtain KI ( and, therefore, RT,T 2 log e ^ (3) Q = T _ T 2 - For the temperature interval T 3 - T 2 we obtain similarly (4) and from (3) and (4) , - io ge K,) - io ge ^ 2 - io ga z,. -l ~ l Hence log, ^ 3 = lo ge ^ 2 - ^/y 2 ~ y^gog.^ - log. J5g, ^3^2 ~~ X U or, converting to common logarithms, and substituting the numerical values, 450 1298 173 / 427 450 log ^ 3 = log 310 ~ 1000 x 125V lo S333 ~ I = + 0-258. Therefore Ka = r8i. From the fact that water vapour under a pressure P = 0-1 MASS-ACTION AFFINITY EXAMPLES 81 atmosphere is dissociated to the extent of 6-46 x 10 ~ 5 per cent, at T 3 , the equilibrium-constant K' for the dissociation of water vapour at T 3 may be calculated. The degree of dis- sociation a = 6'46 x 10~ 7 , and, since 1 molecule H 2 O gives 1 molecule H 2 and -J- molecule O 2) we have PH Z = aP, p 02 = %aP and p H2O = (1 - a) P. Hence - a) 2 2(1 - a) 2 _J2_ X ^aP ~" a 3 P " a 3 P ' since a is very small, compared with 1. If P is expressed in mm. of Hg as above, we obtain, therefore, = 1 x 10 17 , (6-46) 3 x 10~ 21 x 76 and (1-81) 2 x 10 17 PROBLEM 173 (of. preceding problem). What is the affinity, in calories, at T 3 = 1000 of iron to oxygen under a pressure equal to its partial pressure in the atmosphere, and at what temperature T x would ferrous oxide, heated in contact with air, dissociate into metallic iron and oxygen, if the mole- cular heat of formation of ferrous oxide is 64600 calories, and is assumed to be independent of the temperature ? SOLUTION 173. The affinity of iron to the oxygen of the air is measured by the work which can be gained by the reversible union of one molecule of gaseous oxygen at a pressure of 1/5 atmosphere with metallic iron. For the reaction 2 Fe + O 2 = 2 FeO the equilibrium -constant at T 3 is K = V.Po2 where p 02 is the oxygen dissociation-pressure of ferrous oxide at T y From equation (14), therefore, at T 3 A = RT 3 \og e l/p 02 + ET 3 \og e 1/5 From Problem 172, for T s = 1000, p 02 = 3-1 x 10~ 18 mm. 3>1 X " 18 = 4-1 x 10- 21 atmosphere. 82 EQUILIBRIUM-CONSTANT EXAMPLE Therefore, A = 2-3 x 1-985 x 1000 (log 1/5 - log 41 x 10' 21 ) = 2-3 x 1-985 x 1000 x 19-7 = 90000 calories. The relation between the dissociation-pressure of ferrous oxide and the temperature is given by the equation geff _ -_Q T "' RT*' or, integrated between T z and T M Q tea s 10 p.- B \T, ~ T where p z and p x are the dissociation-pressures at T s and T x respectively. Q is the heat of dissociation of 2 molecules of FeO, i.e. the heat of the reaction which involves the forma- tion of one molecule of oxygen from 2 molecules of FeO, and therefore, = - 129200 calories. If T, is the temperature at which p x = 1/5 atmosphere, then 41 x IP- 2 * _ 129200 /I _! \ 1/5 " 2-3 x 1-98 V1000 ~ T*) 1 and = 0-001 - 0-000696 = 0-000304, .'. T x = 3290 abs. PROBLEM 174 (using the results of the two preceding pro- blems). Iron and carbon dioxide react according to the equation Fe+ CO 2 = FeO + CO. At 1000 abs., and under a total pressure of 1 atmosphere the partial pressure of the carbon dioxide at equilibrium is Pco2 495 mm. Hg, and that of the carbon monoxide p co 265 mm. What is the degree of dissociation of pure carbon dioxide (into carbon monoxide and oxygen) at this tempera- ture and under the pressures P 1 = 01, P 2 = 1 and P 3 = 10 atmospheres ? SOLUTION 174. For the dissociation of carbon dioxide into carbon monoxide and oxygen according to the equation 2C0 2 = 2CO + 2 , MASS-ACTIONEXAMPLES 83 the law of mass-action gives the equilibrium equation P <;o x If the carbon monoxide and carbon dioxide are in equi- librium, not only with the free oxygen but also with metallic iron and solid FeO according to the equation Fe + C0 2 = FeO + CO, the equilibrium equation (2) . = K. must also be satisfied. Finally, the equilibrium-pressure of the free oxygen is equal to the dissociation- pressure of FeO at the temperature in question (of. Problem 173). From (2) and (1) we obtain for the dissociation-constant of carbon dioxide For T = 1000 abs. p 02 = 4-1 x 10" iJ1 atmosphere, and K = ^ = 0-535, 495 therefore, K = 4-1 x 10 ~ 21 x (0-535) 2 = 1-17 x 10 ~ 21 . From this value the degree of dissociation a of carbon dioxide under the pressure P may be calculated, since = (1 - a)P pco = aP and Poi = From (1) we obtain aP x Since the value of K' is very small, a must be small com- pared with 1 for all but very small values of P ; hence 84 MASS-ACTIONEXAMPLES and, for P 1 = O'l atmosphere, tt! = V 20 X 1<17 X 10 "' 21 = 2 ' S6 X I0 " 7 for P 2 = 1 atmosphere, a 2 = /2 x 1-17 x IO- 21 = 1-33 x io- 7 , for P 3 = 10 atmospheres, a 3 = %/200 x 117 x 10 - 24 = 0*616 x io~ 7 . PROBLEM 175 (cf. preceding problem). Carbon monoxide and water vapour react to form water-gas, that is, a mixture of these two gases with their reaction products, hydrogen and carbon dioxide. What is the composition of this mixture at equilibrium at a temperature of 1000 abs. and under a total pressure (1) of P x = O'l atmosphere, and (2) of P 2 = 10 atmospheres, if the carbon monoxide and water vapour from which the equilibrium mixture is produced were originally present in equal volumes ? SOLUTION 175. Water vapour and carbon monoxide react according to the equation H 2 + CO = H.J + C0 2 . The mass-action equation at equilibrium is, therefore, (1) -r "2 ^ c 2 = K, JPn 2 o PCO where p H2 etc. are the partial pressures of the various gases at equilibrium. Since at the same time both water vapour and carbon dioxide are dissoaiated to a certain extent, the former into H 2 and 2 , and the latter into CO and 2 , the equations /n\ pj*2 PO% _ JT- \ ' /2 x ~ 1 P H20 and (3) ^ co ' ^ 2 - ^ P co 2 must also be satisfied. K t and K 2 for 1000 abs. are known from the preceding examples. By dividing (2) by (3) we obtain or K = . S MASS-ACTIONEXAMPLES 85 From problem 172 the value of K is 1 x 10 ~ 17 , if the partial pressures are expressed in mm. Hg, and, from problem 174, K%= 1*17 x 10 ~ 21 , if the partial pressures are expressed in atmospheres. To obtain the value of K l when the partial pressures are also expressed in atmospheres, 1 x 10 ~ 17 must be divided by 760, giving K l = 1-24 x 10 - 20 . For the numerical value of K we therefore obtain 24 x 10- 20 ~ ^r-f. o.n* "\l-17x 10-" =: Vl< The numerical value of K is independent of the arbitrary unit of pressure, because both numerator and denominator of equation (1) are of the same degree with respect to p, or, in other words, because the reaction takes place without change of volume (or pressure). If the fraction x of unit volume of water originally present is transformed into CO 2 and H 2 , which are formed in equal volumes, there will be in the equilibrium-mixture (1 - x) vol. H 2 0, x vol. C0 2 , x vol. H 2 and, if the initial volumes of GO and H 2 were equal, (1 - x) vol. CO ; hence x - J? x (1 - x? ' T~^~x ~ and x = 0*644. From a mixture which originally contained 50 / by volume of H 2 and 50 / o by volume of CO, there is formed, therefore, an equilibrium-mixture containing 32-2 7 C0 2> 32-2 / H 2 , 17-8 / CO and i 7 -87 H 2 by volume. Since K is independent of the pressure, the composition of the equilibrium-mixture is also independent of the pressure. PROBLEM 176 (cf. precediag problem). The molecular heat of combustion of hydrogen is Q l = 58000 cals., and that of carbon monoxide is Q 2 = 68000 cals. What is the com- position at equilibrium of the water-gas formed from equal volumes of water vapour and carbon monoxide (1) at a temperature Tj = 800 abs., and (2) at a temperature T 2 = 1200 abs. ? SOLUTION 176. The reaction H 2 O + CO = H 2 + C0 2 is made up of the two subsidiary reactions (1) H 2 = H 2 + iO s> and (2) O 2 + CO = CO 2 . 86 EQUILIBEIUM-CONSTANT EXAMPLES The heat of reaction (1) is Q l = - 58000 cals., and of reaction (2) Q 2 = 68000 cals. The heat effect of the total reaction H 2 O + CO = H 2 + CO 2 is, therefore, Q = Q l + Q 2 = + 10000 cals. The equilibrium-constant K changes with temperature accord- ing to the equation diog.*_ -Q dT RT* ' Integration between the neighbouring temperatures T and T l gives and similarly between the temperatures T and T 2 From problem 175, the value of K for T = 1000 abs. is 3-26. K! for T t = 800 abs. and K 2 for T 2 = 1200 abs. may, therefore, be calculated. From (3) we obtain . 513 0-4343 x 10000 x 200 1-985 x 1000 x 800 = 0-513 + 0-552 - 1-065, and, therefore, K x = 11*6. Similarly, from (4), = 0-149, and K 2 = 1-41 If x l is the fraction of the water vapour transformed at T-^ according to the equation H 2 + CO = H 2 + CO 2 , and x 2 the corresponding fraction at T 2 , then, as in problem 175, we obtain .-*, -. = 3 ' 40 ' x l = 0-774, and -^L__ = V^ = VTH = 1-19, i - ic 2 a; 2 = 0-542. MASS-ACTIONEXAMPLES 87 The composition of the water-gas mixture is, therefore, at T l = 800 abs. : 38-7 %C0 2 , 38-7 %H 2 , 1 1 -3 7 H 2 0, 1 1 -3 7 CO, and at T' 2 = 1200 abs. : 27-1 % C0 2 , 27-1 / H 2 . 22-9 %H 2 0, 22-9 7 CO. PROBLEM 177. A mixture of a = 10 / by volume of S0 2 and 6 =* 90 / by volume of O 2 is passed at the rate of v = 5 litres per hour (measured at atmospheric pressure and 20 C.) through a tube filled with a catalytic agent and heated in a furnace to 723 C. In the tube combination to SO 3 takes place, and we shall assume that the gases remain long enough in the tube for the equilibrium between the three gases SO 2 , O 2 and SO 3 to be reached, and that the gases leave the tube by a capillary so quickly that the equilibrium is not disturbed. On leaving the capillary the gas passes through a solution of barium chloride, from which it precipitates in t = 2 hours A = 5 '84 grams of BaSO 4 . What is the equilibrium-constant K p of the reaction 2SO 2 + O 2 = 2SO 3 at 723 C. ? SOLUTION 177. Let the fraction x of the SO 2 present in the initial mixture be converted into SO 3 at equilibrium. Then, since^ for every volume of S0 3 formed one volume of SO 2 and half a volume of O 2 disappears, a volumes of SO 2 + b volumes of oxygen give at equilibrium a(l - x) volumes of SO 2 + ax volumes of SO 3 + b - ^ volumes of oxygen. The total 2 volume at equilibrium is, therefore, a + b - 2L. The total 2 volume has, therefore, diminished from a + b = 100 to a + b - ^ = 100 - ^. Since the total pressure remains constant at one atmosphere, the partial pressures of the three gases at equilibrium are (SO,) = -2JL2" (SO,) 100 - f 2 atmospheres, and, therefore, 88 MASS-ACTION EXAMPLES (100 - -y (100 - j) (100 - # 2 (lOO - ~ (1 - .? (6 - f ) a; and, therefore, -ZT P may be calculated as follows : Let n be the number of gram-molecules of a gas which occupy a volume of 1 litre at 20 C. and atmospheric pressure. Then ^ gram- molecules of S0 2 enter the furnace in time t. But the fraction x of the S0 2 is converted into S0 3 , therefore xnva * gram-molecules of S0 3 leave the furnace in time t. 100 These are converted completely into BaS0 4 and weighed as such. If M is the molecular weight of BaS0 4 , there is, therefore, precipitated in time t A = "^p grams BaS0 4 , and (2) z = !5^j_. nvatM The equation PV = W-RIT enables us to calculate n. Here P = 1 atmosphere, F= 1 litre, E = 0*082 and T = (20 + 273) = 293. Therefore, and, substituting the numerical values in (2), 100 x 5-84 0-0416 x 5 x 10 x 2 x 233 " Finally, substituting the numerical values in (1), we obtain (1 - 0-60) 2 ( 90 - 10 X 0'60\ 0'16 X 87 PARTITION-LAW EXAMPLE 89 Partition Law Example PROBLEM 178. The partition-coefficient of iodine for water and carbon disulphide is K = 0-0017. An aqueous solution of iodine containing a = O'l gram of iodine per 100 c.c. is shaken with carbon disulphide. To what value does the con- centration of the aqueous solution sink (1) when a litre of it is shaken with 50 c.c. of carbon disulphide, (2) when a litre of it is shaken successively with five separate quantities of carbon disulphide of 10 c.c. each ? SOLUTION 178. (1) Let the concentration of the aqueous solution, after shaking with 50 c.c. of carbon disulphide, be x grams per 100 c.c. Then 10(a - x) grams of iodine have been extracted by the carbon disulphide. The concentration of the carbon disulphide solution is, therefore, 10(q - *) x 100 _ 50 grams iodine per 100 c.c. Hence x = K 20(a - x) 0-0034 OST = ' 329 gram per 10 c ' c - (2) Let the concentration of the iodine in the water, after the first extraction with 10 c.c. of carbon disulphide, be x } grams per 100 c.c. Then 100(a -xj IQOaA" Xl 1 + 100J5T ' If the concentration of the aqueous solution after the second extraction is x 2 , we obtain similarly - ar s ) = / IQQg 2 Vl + 2 1 + 100# Vl + lOOtf / ' After the fifth extraction the concentration is, therefore, *17\ 5 = 0-1 r.\ = *>'47 x io~ 6 gram iodine per 100 c.c. The extraction has, therefore, been practically complete. 90 SOLUBILITY OF GASES EXAMPLE Solubility of Gases Example PROBLEM 179. The solubility-coefficient of oxygen at is s = 0-04 and of hyrodgen s' = 0-02. What is the per- centage composition by volume of the dissolved gas when a litre of water at is shaken in a closed space with 3 times its volume of electrolytic gas under an initial pressure of 1 atmosphere ? What are the final partial pressures p and p' of the oxygen and hydrogen in the gas above the liquid at equilibrium ? SOLUTION 179. Since J of the volume of the electrolytic gas is oxygen, the initial partial pressure of the oxygen in the mixture is J atmosphere. Let p be the partial pressure of the oxygen at equilibrium. The volume of oxygen ab- sorbed by the litre of water is s = 0-04 litre measured at and p atmos. =. 0-04 x 3p litre J The oxygen left would occupy a volume of 3 - 0-04 x 3p litres at and J atmos., but must fill the volume of 3 litres at and p atmos. As the volume is inversely proportional to the pressure, we obtain 3 - 0-04 x 3p = p 3 3 "J 1 - 0-04p = 3p, .'. p = 0*329 atmos. The volume of oxygen absorbed by the water is, therefore, = 0-04 litre at and 0-329 atmos. = 0-04 x 0-329 litre at and 1 atmos. = 0-0132 1 Similarly for hydrogen, the initial partial pressure is 2/3 atmos., and the final partial pressure p' is obtained from the equation 0-02 x 3p' 3 ~ 2 _ _3p' 3 2/3 ~ = 2 . . p' = 0*662 atmos. The volume of hydrogen absorbed is 0-02 litre at and 0*662 atmos. = 0-02 x 0-662 = 0-0132 litre at and 1 atmos, MASS-ACTION, ETC. PROBLEMS 91 The total volume of gas dissolved is, therefore, 0-0132 + 0-0132 = 0-0264 litre at and 1 atmos., and since it con- sists of equal volumes of hydrogen and oxygen, the per- centage composition is O 2 == 5 P er cent> H 2 = 50 per cent. Problems for Solution Mass-action Equilibrium and Temperature Affinity PROBLEM 180. When 2-94 moles of iodine and 8*10 moles of hydrogen are heated at constant volume at. 444 till equilibrium is established, 5-64 moles of hydriodic acid are formed. If we start with 5 '30 moles of iodine and 7*94 moles of hydrogen, how much hydriodic acid is present at equilibrium at the same temperature ? Ans. 9-49 moles. PROBLEM 181 (cf. preceding problem). What proportion of hydriodic acid is decomposed when 1 mole is heated to 444 till equilibrium is established ? Ans. 0-2197 mole. PROBLEM 182. When 5-71 moles of iodine and 6-22 moles of hydrogen are heated to 357 till equilibrium is estab- lished, 9-55 moles of hydriodic acid are formed. From the equilibrium constants calculated from these data and the data in problem 180, calculate the heat of formation of hy- driodic acid from hydrogen and iodine vapour according to the equation H 2 + I 2 = 2HI. Ans. 3028 cals. PROBLEM 183. If 1 mole of acetic acid and 1 . mole of ethyl alcohol are mixed, the reaction CH 3 . COOH + C 2 H 6 OH = CH 3 . COOC 2 H 5 + H 2 O proceeds till equilibrium is reached, when 1/3 mole acetic acid, 1/3 mole ethyl alcohol, 2/3 mole ethyl acetate, and 2/3 mole water are present. If we start (a) with 1 mole acid + 2 moles alcohol, (b) with 1 mole acid, 1 mole alcohol, and 1 mole water, (c) with 1 mole ester + 3 moles water, how much ester is present in each case at equilibrium? Ans. (a) 0-845 mole, (b) 0'543 mole, (c) 0-465 mole. 92 MASS-ACTION, ETC. PKOBLEMS PROBLEM 184. Above 150 NO 2 begins to dissociate ac- cording to the equation N0 2 = NO + iO a . At 390 the vapour-density of N0 2 is 19'57 (H = 1), and at 490 it is 18-04 Calculate the degree of dissociation accord- ing to the above equation at each of these temperatures, the equilibrium-constants K = - r Jrk I , expressing the con- L iNU 2J centrations in gram-molecules per litre, and the heat of dis- sociation of NO 2 . Ans. a x = 0-35, a 2 = 0-55. ^ = 2-818 x 10 ~ 2 , KZ = 7-716 x lO- 2 . Q = - 9376 cals. PROBLEM 185. At 49-7 C., and under a total pressure of 261-4 mm. of mercury, N 2 O 4 is 63 per cent, dissociated into NO 2 . What would be its degree of dissociation at the same temperature, but under a pressure of 93-8 mm. ? Ans. 80*4 per cent. PROBLEM 186. What is the equilibrium-constant at 49'7 for the above dissociation, (a) for partial pressures in mm., (b) for partial pressures in atmospheres, (c) for concentra- tions in gram-molecules per litre, (d) for concentrations in grams per litre ? Ans. (a) 172, (b) 172/760, (c) 0-00855, (d) 01966, or 4 times these numbers, according to the expression used for the equilibrium-constant. See problem 166. PROBLEM 187. The vapour-pressure of solid NH 4 HS at 25*1 is 50*1 cms. Assuming that the vapour is practically completely dissociated into NH 3 and H 2 S, calculate the total pressure at equilibrium when solid NH 4 HS is allowed to dis- sociate at 25*1 in a vessel containing NH 3 at a pressure of 32 cms. Ans. 59'5 cms. PROBLEM 188. What is the total pressure in the pre- ceding problem when the vessel contains H 2 S at a pressure of 32 cms. instead of NH 3 ? Ans. 59'5 cms. PROBLEM 189. In the reaction 3Fe + 4H 2 O = Fe 3 O 4 + 4H 2 MASS-ACTION, ETC. PKOBLEMS 93 there is equilibrium at 200 when the partial pressure of steam is 4-6 cms., and that of H 2 is 95-9 cms. What is the pressure of H 2 at equilibrium when that of steam is 9*7 cms. ? Ans. 202-2 cms. PROBLEM 190 (cf. preceding problem). Iron is heated at 200 in a closed vessel with steam at an initial pressure of 1 atmosphere till equilibrium is established. What are the partial pressures of steam and hydrogen at equilibrium ? Ans. H 2 O = 0-0458 atmos., H 2 = 0-954 atmos. PROBLEM 191 (cf. preceding problem). If the capacity of the vessel in the preceding problem is 2 litres, what weight of Fe 3 O 4 is formed at equilibrium ? Ans. 2-85 grams. PROBLEM 192. Amylene and trichloracetic acid react to form an ester according to the equation 001,'. COOH + C 5 H 10 = CC1 3 . COOC 5 H n . In an experiment at 100 the equilibrium mixture contained 3*846 gram-molecules amylene per litre, 0'6594 gram-molecule acid per litre, and 2-111 gram-molecules ester per litre. If we start with 1 gram-molecule acid, and 4-48 gram-molecules amylene in 638 c.c. at 100, what is the composition of the mixture at equilibrium ? Ans. 0'174 gram-molecule acid, 3-654 gram -molecules amy- lene, and 0'826 gram-molecule ester in 638 c.c. PROBLEM 193. Solid NH 4 CN has a considerable vapour- pressure at ordinary temperatures, and the vapour is practi- cally completely dissociated into NH 3 and HCN. At 11 C. the total vapour-pressure is 22'7 cms. of mercury, (a) What will be the partial pressure of HCN if solid NH 4 CN is al- lowed to sublime at 11 in a closed vessel filled with NH 3 at a pressure of 32-28 cms. of mercury? (b) What will be the final total pressure ? Ans. (a) 3-595 cms. Hg, (b) 39'47 cms. Hg. PROBLEM 194. If the volume of the vessel in the preced- ing problem is 1 litre, how much NH 4 CN will sublime ? Ans. 0-00203 gram- molecule. PROBLEM 195. At 96 the ammonia dissociation-pressure of the compound LiCl . NH 3 according to the equation LiCl . NH 3 = LiCl + NH 3 94 EQUILIBBIUM-CONSTANT PROBLEMS is 367 mm. of mercury, and at 109-2 it is 646 mm. Cal- culate the heat of dissociation of the compound. Ans. - 12000 cals. PBOBLEM 196. At 2000 C., and under atmospheric pres- sure, carbon dioxide is 1*80 per cent, dissociated according to the equation 2C0 2 = 200 + 2 . Calculate the equilibrium-constant for the above reaction using partial pressures (in atmospheres). Ans. 3 x 10 ~ 6 . PBOBLEM 197. What is the equilibrium -constant in the preceding problem, if the concentrations are expressed in gram-molecules per litre? Ans. 1-61 x 10- 8 . PBOBLEM 198. At 28*85 the vapour-pressure of the system BaCL 2 , 2H 2 O - BaCl 2 , H 2 O is 7'125 mm., and at 31*65 it is 8-945 mm. Calculate the heat of hydration of BaCl 2 , H 2 O to BaCl 2 , 2H 2 O by water vapour. Ans. 14910 cals. PEOBLEM 199. At 30*20 the vapour-pressure of the system CuS0 4 , 5H 2 O - CuS0 4 , 3H 2 O is 10*90 mm., and at 26-30 it is 8-074 mm. Calculate the heat of hydration of CuSO 4 , 3H 2 O to CuSO 4 , 5H 2 O by water vapour per gram- molecule water. Ans. 13940 cals. PBOBLEM 200 (cf. preceding problem). At 30-20 and 26-30 the vapour-pressure of water is 31*93 mm. and 25*43 mm. respectively. Calculate the heat of hydration of CuS0 4 , 3H 2 O to CuSO 4 , 5H 2 O by liquid water per gram- molecule water. Ans. 3382 cals. PBOBLEM 201. The solubility of boric acid in water is 38-45 grams per litre at 13, and 49-09 grams per litre at 20. Calculate the heat of solution of boric acid per gram- molecule. Ans. - 5840 cals. PBOBLEM 202. The solubility of succinic acid at is 2-88 gram-molecules per litre and at 8'5 it is 4-22 gram-molecules per litre. Calculate the heat of solution of succinic acid per gram-molecule. Ans. - 6900 cals. TEMPERATURE AND EQUILIBRIUM PEOBLEMS 95 PROBLEM 203. The dissociation-pressure of CaCO 3 at 810 C. is 678 mm. and at 865 C. it is 1333 mm. Calculate the heat of dissociation of CaCO 3 . Ans. - 30110 cals. PROBLEM 204 (cf . preceding problem). At what tempera- ture does the dissociation-pressure of CaCO 3 become equal to 1 atmosphere. Ans. 818 C. PROBLEM 205 (cf. preceding problem). What is the affinity of CaO to C0 2 at atmospheric pressure at a tempera- ture of 810 C. ? Ans. 246 cals. PROBLEM 206. At 10 the solubility of HgCl 2 is 6-57 grams per 100 c.c. and at 50 11-84 grams per 100 c.c. Calculate the heat of solution per gram-molecule. Ans. - 2669 cals. PROBLEM 207. At 10 the electrolytic dissociation-con- stants of acetic and butyric acids are 1'79 x 10 ~ 5 and 1'66 x 10 ~ 5 respectively, and at 40 they are 1'87 x 10 ~ 5 and 1-62 x 10 ~ 5 . Calculate the heat of ionisation at 25 (a) of acetic, (b) of butyric acid. Ans. (a) - 256-6 cals., (b) 144 cals. PROBLEM 208. At 10 the ionic product of water is 0-314 x 10 ~ 14 and at 34 2-16 x 10 - u . Calculate the heat of formation of H 2 O from H and OH'. Ans. 13950 cals. PROBLEM 209. At 25 the degree of dissociation of o-chlor- benzoic acid at a dilution of 512 litres is 0'557 as measured by the conductivity ; at 40 and at the same dilution it is 0-521. Calculate the heat of dissociation of the acid at 32-5. Ans. 2614 cals. PROBLEM 210. At 20 the solubility of AgBr is 4-5 x 10 - * gram-molecule per litre and at 25 it is 7 '3 x 10 ~ 7 . What is the heat of precipitation of AgBr ? Ans. 16880 cals. PROBLEM 211. At 20 the solubility of AgCl is !! x 10 " 5 gram-molecule per litre and the heat of precipitation of AgCl is 16000 cals. What is the solubility of AgCl at 30 ? Ans. 1-73 x 10 ~ 5 gram-mol./litre .96 PARTITION LAW PROBLEMS PROBLEM 212. At 670 C. the oxygen dissociation-pressure of Ba0 2 is 80 mm. and at 720 C. 210 mm. (a) What is the heat of the reaction 2BaO 2 = 2BaO + O 2 ? (b) At what temperature is the dissociation-pressure equal to the partial pressure of oxygen in the atmosphere (152 mm.) ? Ans. (a) - 36090 cals., (b) 703 C. PEOBLEM 213 (cf. preceding problem). W T hat is the maxi- mum work obtainable at 670 C. by the formation of 2 gram- molecules of Ba0 2 from BaO and oxygen at the pressure at which it occurs in the atmosphere ? Ans. 1203 cals. PROBLEM 214. At 2000 C. the equilibrium-constant for pressures in atmospheres for the reaction CO -I- O 2 = CO 2 is T07 x 10 2 . What is the maximum work obtainable by the formation at 2000 C. of 1 gram-molecule of -GO 2 at atmospheric pressure from 1 gram-molecule of CO and -J gram-molecule of 2 , both at atmospheric pressure ? Ans. 21210 cals. Partition Law PROBLEM 215,.-' At 15 an aqueous solution of succinic acid containing 0-070 gram in 10 c.c. is in equilibrium with an ethereal solution containing 0'013 gram in 10 c.c. Succinic acid has its normal molecular weight in both water and ether. What is the concentration of an ethereal solution which is in / equilibrium with an aqueous solution containing 0*024 gram in 10 c.c. ? ^-. ^Ans. 0-0044 gram in 10 c.c. PROBLE^ 216} At 25 a solution of iodine in water con- taining 0'05l#"gram per litre is in equilibrium with a CC1 4 solution containing 4-412 grams iodine per litre. The solu- bility of iodine in water at 25 is 0-340 gram per litre. What . is the solubility in CC1 4 ? Ans. 29 '07 grams per litre. PROBLEM 217. In the partition of acetic acid between CC1 4 and water, the concentration of the acetic acid in the CC1 4 layer was C gram-mols. per litre and in the correspond- ing water layer W gram-mols. per litre. C 0-292 0-363 0-725 1-07 1-41 W 4-87 5-42 7-98 9-69 10-7 PAETITION LAW PEOBLEMS 97 Acetic acid has its normal molecular weight in aqueous solu- tion. From these figures show that, at these concentrations, the acetic acid in the CC1 4 solution exists as double molecules. PROBLEM 218 In the partition of succinic acid between water and ether the concentrations of the acid in the water and ether layers were c 1 and c 2 respectively. c, 0-121 0-070 0-024 gram in 10 c.c. c 2 0-022 0-013 0-0046 In the distribution of the same substance between water and benzene the concentrations of the acid in the water and benzene layers were c 3 and c 4 respectively. c 3 0-0150 0-0195 0-0289 gram in 10 c.c. c 4 0-242 0-412 0-970 Succinic acid has its normal molecular weight in water. What is its molecular weight (a) in ether, (b) in benzene ? Ans. (a) 118, (b) 236. PROBLEM 219. Phenol has its normal molecular weight in both water and amyl alcohol. At 25 an amyl alcohol solu- tion containing 10'53 grams phenol per litre is in equilibrium with an aqueous solution containing 0*658 gram per litre. What weight of phenol is extracted from 500 c.c. of an aque- ous solution containing 0*4 gram-mols. phenol per litre by shaking it twice with amyl alcohol, using 100 c.c. each time ? Ans. 17'7 grams. PROBLEM 220. The partition-coefficient of iodine for CS 2 /water is 410. A solution of KI containing 8 grams per litre was shaken with iodine and CS 2 till equilibrium was established. The concentration of the iodine in the two* layers was then determined by titration with Na 2 S 2 O 3 . The aqueous layer contained 2-15 grams iodine per litre and the CS 2 layer 35'42 grams per litre. ^Assuming that in the aque- ous solution KI reacts with iodine according to the equation KI + I 2 = KI 3 , calculate the dissociation-constant of the tri- iodide, K = [KI] [I 2 ]/[KI 3 ], expressing the concentrations in gram-mols. per litre. (The concentration of the I 2 in the aqueous layer obtained by titration is the sum of the free I 2 and the I 2 combined with KI in the tri-iodide, KI and KI 3 are assumed insoluble in CS 2 .) Ans. K= 1-68 x 10 ~ 3 . PROBLEM 221. In aqueous solution benzoic acid exists- 7 98 PAETITION LAW PEOBLEMS as single molecules, partly electrolytically dissociated. In benzene solution it exists partly as single and partly as double molecules, the proportions depending on the concen- tration. Taking the partition-coefficient water/benzene for the single molecules = 0-700, calculate from the following data the equilibrium-constant for the dissociation of the double molecules into single molecules in the benzene solu- tion according to the equation (C 6 H 5 COOH) 2 = 2C 6 H 5 . COOH. At 10 the concentration of benzoic acid in the water layer = 0-0429 gram per 200 c.c. and the degree of electrolytic dis- sociation = 0-169. In the benzene layer the concentration of benzoic acid = 0-1449 gram per 200 c.c. Express con- centrations in grams per litre. Ans. K = 0-138. Solubility of Gases PBOBLEM 222. The solubility-coefficient of oxygen in water at is 0-04, and of nitrogen 0'02. If the composition of air by volume is assumed to be 21 per cent./ oxygen and 79 per cent, nitrogen, what is the percenta^e^p^mposition by volume of the gas expelled by boiling from water which has been saturated by free exposure to air at ? Ans. 34-7 per cent. O 2 . 65'3 per cent. N 2 . PROBLEM 223 (cf. preceding problem). If the composition of the air were O 2 = 20-6 per cent., N 2 = 79 per cent., CO 2 = 0-4 per cent, by volume, and the solubility-coefficient of CO 2 *. 1-79 at 0, what would be the percentage/composition of the dissolved gas ? Ans. O 2 = 26-40 per cent., N 2 = 50-64 per cent., CO 2 = 22-95 per cent. PROBLEM 224. 1 litre of oxygen-free water is shaken in a closed space at with 1 litre of oxygen at an initial pres- sure of 1 atmosphere, (a) What volume of oxygen, measured at and 760 mm., dissolves, and (b) what is the pressure of the oxygen over the water when equilibrium is established ? Ans. (a) 38-5 c.c., (b) 0-9615 atmos. PROBLEM 225 (cf. preceding problem). 1 litre of oxygen is shaken with 10 litres of water at in a closed vessel. SOLUBILITY OF GASES PEOBLEMS 99 The initial pressure of the oxygen is 1 atmosphere, (a) What volume of oxygen (at (T and 760 mm.) dissolves, and (b) what is the final pressure of the undissolved oxygen ? Ans. (a) 285-6 c.c., (b) 0'714 atmos. PROBLEM 226. What is the percentage composition by volume of the gas dissolved when a gas mixture containing 21 per cent. O 2 and 79 per cent. N 2 by volume, and under an initial total pressure of 1 atmosphere, is shaken in a closed vessel at (a) with an equal volume of water, (b) with 10 times its volume of water? Ans. (a) 34-18 per cent. O 2 , 65-83 per cent. N 2 . (b) 31-42 per cent. O 2 , 68*58 per cent. N 2 . PROBLEM 227. The solubility-coefficient of CO 2 at is 1-8. What weight of CO 2 under a pressure of 4 atmospheres will dissolve in a litre of water at ? Ans. 14-1 grams. PROBLEM 228. What is the relation between the solu- bility-coefficient s of a gas at t w and the absorption- coefficient a of the gas ? a(273 + t) Ans. s - - V '. V-* >ft/ CHAPTER IX OHM'S LAW. HEATING EFFECT OF CUBEENT. FARADAY'S LAWS. SPECIFIC, EQUIVALENT, AND MOLECULAR CON- DUCTIVITY OF ELECTROLYTES. DEGREE OF DISSOCIA- TION. DISSOCIATION-CONSTANT. TRANSPORT NUMBERS. SOLUBILITY-PRODUCT. HYDROLYSIS Ohm's Law A CCORDING to Ohm's law, the current C produced in a ** conductor of resistance R by the potential difference E between the ends of the conductor is If E is measured in volts, and R in ohms, C is obtained in amperes. Quantity of Electricity If a current of C amperes flows through a conductor, the quantity of electricity which passes any cross-section of the conductor hi t seconds is (2) W = Ct coulombs. Heating Effect of Current The current C amperes flowing through the resistance E ohms for t seconds, under the potential difference E volts, de- velops the quantity of energy ( 3 ) Q = WE = CEt = C 2 Rt volt-coulombs or joules = 0-239 x C 2 Rt calories since 1 joule = 0*239 calorie. Faraday's Laws Faraday's laws state that the quantity of an electrolyte de- composed by an electric current is proportional to the quantity 100 HEATING EFFECT; OF- OTEPNTEMPLES 101 of electricity which passes any cross-section of the electro- lyte, and that the quantities of different electrolytes decom- posed by the same quantity of electricity are proportional to their chemical equivalents. These statements may be summed up as follows. The quantity of electricity, 96540 coulombs, is required for the decomposition of 1 gram-equi- valent of any electrolyte, or for the liberation of 1 gram- equivalent of any cation or anion. The quantity of electricity, 96540 coulombs, is called one faraday, and is usually denoted by the symbol F. Heating Effect Examples PEOBLEM 229. A thermostat of v = 850 litres capacity is kept at a constant temperature of 25 C. by the heat developed by maintaining a current of C = 3'1 amp. through a resist- ance of E = 22 ohms placed in the thermostat. How many degrees does the temperature of the thermostat fall in t = 30 minutes, after switching off the heating current ? SOLUTION 229. If the thermostat maintains a constant temperature, the quantity of heat given up to the surround- ings at any instant by evaporation of water, conduction and radiation of heat, must be equal to the quantity supplied by the heating current. The quantity of heat supplied by the heating current can be calculated from the data given in the problem. According to formula (3), the heat developed by the current in 1 second (t = 1) is 0-239 C' 2 B calories. If the heating current is switched off, the thermostat, therefore, gives up 0-239 C 2 E calories per second, and in t minutes 60 x 0'239 C' z Rt calories. If the thermostat contains v litres of water (specific heat == 1), in t minutes the temperature, therefore, falls _ 60 x 0-23g x C'*Bt _ 60 x Q-39 x (3-1)' 2 x 22 x 30 1000 v 1000 x 850 = 0-107 C. PROBLEM 230. For the production of the heating current in the previous problem only a potential of 220 volts is avail- able. A lamp-resistance is to be used for the necessary current-regulation. How many lamps (in parallel) are re- quired (1) if each lamp has a resistance of JS, = 1000 ohms ? (2) if each lamp has a resistance of J5 2 = 300 ohms ? 102 FABADAY'3 LAWS* -EX AMPLE E SOLUTION 230. According to Ohm's law (i), C = ^>. If, therefore, the only resistance in the circuit were the 220 heating-resistance R = 22 ohms, the current would be -^ty = 10 amps. In order that the current in the circuit may be 220 C = 3*1 amps., the resistance must be increased to-orr = 71 ohms. An additional resistance of 71 - 22 = 49 ohms, in series with the heating-resistance is, therefore, required. This additional resistance is to take the form of glow-lamps arranged in parallel. (1) In the first case the resistance of each lamp is B l ohms. If %! such lamps are arranged in parallel their total resistance T* is - 1 ohms, and this must be = 49 ohms, n i _RI _ looo ' w i - 49 ~ 49 ' (2) In the second case the resistance of each lamp is R% ohms. The total resistance of w 2 such lamps in parallel is ^ ohms, which must also be equal to 49 ohms, the additional resistance required, BZ 300 ' n2 = 49 = 19 ' Since n x and w 2 must be whole numbers, the necessary con- ditions are very approximately fulfilled by H! = 20 and n 2 = 6. In the first case the current is 220 ^i = ctcT~ t 1000 = ^ '06 amps. ** 1 2TT~ and in the second case 220 ^2 = 99 . 300 = 3-06 amps. A* + -*- Faraday's Laws Example PROBLEM 231. A current passed for t = 6 minutes through a voltameter containing dilute H a S0 4 liberated CONDUCTIVITY DEGREE OF DISSOCIATION 103 40 c.c. of electrolytic gas, measured at 15 C. and 748 mm. What was the average value of the current ? SOLUTION 231. If C is the average value of the current in amperes, then during the time t minutes = 60 t seconds W = 60 Ct = 60 x 6 x C coulombs pass through the solu- tion. At C. and 760 mm. the volume occupied by 40 c.c. at 15 C. and 748 mm. is 748 x 40 273 288 x 760 = 37 ' 33 c ' c - But 96540 coulombs liberate 1 gram-equivalent of hydrogen and 1 gram-equivalent of oxygen. At C. and 760 mm. 1 gram-equivalent of hydrogen (1 gram) occupies *o __ 11200 c.c., and 1 gram-equivalent of oxygen (8 grams) occu- mm. occupied by the electrolytic gas liberated by 96540 coulombs, is, therefore, 16800 c.c. Since the amount of electrolyte decomposed is proportional to the quantity of electricity 60 x 6 x C _ 37-33 96540 ~ 16800' 37-33 x 96540 '" C ~ 16800 x 60 x 6 = Specific and Equivalent Conductivity Degree of Dis- sociation The specific resistance of a conductor is the resistance in ohms of a regular cube of the substance of side 1 cm. long. The reciprocal of this quantity is the specific conductivity *. The unit of specific conductivity is, therefore, that of a substance of which the specific resistance is 1 ohm ; this unit is called the reciprocal ohm or mho, for which, in future, we shall use the contraction r.o. According to Ohm's law, the specific conductivity gives the current in amperes produced in a regular cube of 1 cm. side when a potential difference of 1 volt is applied between two opposite faces of the cube. If v is the volume in litres containing 1 gram-equivalent of an electrolyte, the concentration of which is, therefore, c = - gram-equivalents per litre, and of which the specific conductivity is K, the equivalent conductivity of the solu- tion is 104 DISSOCIATION-CONSTANT * (^L.A = K x 1000 v = - recip. ohms. If A^ is the equivalent conductivity at infinite dilution, that is, when the dissociation of the electrolyte is complete, the degree of dis- sociation in the solution of equivalent conductivity A 'is, according to Arrhenius, A Yelocity of Migration Ionic Conductivity If fA. A and fi c are the velocities of migration of auion and cation respectively in cms. per second under a potential gradient of 1 volt per cm., then the equivalent conductivity (6) A = a x 96540 (p A + p c ) (7) - a ft, + l a ), where 1 A = 96540 ^ A and l c = 96540 //, e are the equivalent ionic conductivities at infinite dilution of anion and cation respectively (Kohlrausch). When a 1, that is, when the dissociation is complete, equation (7) becomes (8) A w = 1 A + l m or, in words, the equivalent conductivity at infinite dilution is equal to the sum of the equivalent ionic conductivities. Molecular Conductivity The molecular conductivity ^ of a solution of an electro- lyte is i nnn (9) /A = 1000 KV = - recip. ohms, c where v is the volume in litres containing 1 gram-molecule and c the concentration in gram-molecules per litre. The relation between A and /x is fj, = ah. and /x,^ = flA^, where a is the number of gram-equivalents contained in 1 gram-molecule. Dilution Law Dissociation-constant If a is the degree of dissociation of a binary electrolyte, and c its concentration in gram-equivalents (or gram-molecules) ELECTROLYTIC DISSOCIATION EXAMPLES 105 per litre, the concentration of each of the ions is ac and that of the undissociated electrolyte (1 - a)c. For the dissocia- tion of a binary electrolyte the law of mass-action, therefore, assumes the form (1 - a)c ~ (1 - a ) - (1 - a)v ~ v is the dilution in litres per gram-equivalent (or gram- molecule) and .BTthe dissociation-constant. This is Ostwald's dilution law. It holds only for weak electrolytes, that is, only with such electrolytes does K remain constant with varying concentration. Since a = -r , equation (10) may be written in the form A 2 (")/ ""AX -7: -* -)v A M (A,,, - A)t> From these equations it is evident that the numerical value of K depends on the unit of concentration or dilution chosen. Here we shall work throughout with the units given above. In the case of very weak electrolytes, where a and K are very small, (1 - a) in equation (10) may, without sensible error, be put equal to 1, since a in such cases is very small, compared with 1. We thus obtain from (10) and (n) the simplified formulas n (13) a?C =^ = K, and In a great many cases these simplified formulas can be used with sufficient approximation. Electrolytic Dissociation Examples PROBLEM 232. The specific conductivity of a c l = Ol N- acetic acid solution at 18 is K^ = 0-000471*r.o., and that of a c, - 0-001 N-sodium acetate solution is K . 2 = 0-0000781 r.o. What is the dissociation constant K of acetic acid at 18, if the ionic conductivity of the H'-ion is l c = 318, and that 106 ELECTROLYTIC DISSOCIATION EXAMPLES of the Na'-ion is- l' c = 44-4, and if the sodium acetate solution is regarded as completely dissociated? SOLUTION 232. From (4) we obtain A, the equivalent conductivity of G 1 N -acetic acid, 1000 KI 1000 x 0-000471 A.-j^-- -g^- -=4-71r.o. According to (8), A^, the equivalent conductivity of acetic acid at infinite dilution, is equal to 1 A + l c , where 1 A is the ionic conductivity of the anion of acetic acid, C 2 H 3 0' 2 , and l c that of the IT-ion. l c is given, and 1 A may be calculated from the equivalent conductivity of the c 2 N-sodium acetate solution, which is equal to A'^, the equivalent conductivity of sodium acetate at infinite dilution, as the salt is regarded as completely dissociated. For sodium acetate, therefore, 1000 1000 K 1000 x 0-0000781 ., A o-ooi = 78-1 - 44-4 = 33-7 r.o. For acetic acid, therefore, A^ = 1 A + l c = 33-7 + 318 = 351-7, and, from (5), Since a is known the dissociation-constant of acetic acid may now be calculated ; according to Ostwald's dilution law (10) a 2 Cl (0-0134) 2 x 01 - L - ' " 5 PROBLEM 233. The velocity-constant for the inversion of cane sugar by c = 0*25 N-acetic acid at 25 is k = 0-75 x 10 ~ 3 . If the velocity-constant is assumed to be proportional to the H'-ion concentration, find the value of the constant when the acid solution is also c l = 0*025 N with respect to sodium acetate, being giyen that the dissociation-constant of acetic acid is K == 0-000018, and that the sodium acetate is dis- sociated to the extent of c^ = 86 per cent. ELECTROLYTIC DISSOCIATION EXAMPLES 107 SOLUTION 233. Let a be the degree of dissociation of the pure acetic acid solution, then, according to (10) j~ ^ = K or Q' 25 " 2 = 0-000018 (I- a) (1 - a) .-. a = 0-00845. The concentration of hydrogen ions in the solution is, therefore, [H-] = ac = 0-00845 x 0-25 = 2-11 x 10 ~ 3 N. Let a 2 be the degree of dissociation of the acetic acid in the solution containing sodium acetate. The degree of dis- sociation of the latter is a r The concentration of IT-ions derived from the dissociation of the acetic acid is, therefore, a 2 c, of the C 2 H 3 O' 2 -ions a 2 c, and of the undissociated acetic acid (1 - a 2 )c, whilst the concentration of the C 2 H 3 O' 2 -k>ns derived from the sodium acetate is a 1 c 1 . The total concen- tration of the C 2 H 3 O' 2 -k>ns is, therefore, a%c + a^. Hence, if we use square brackets to denote the concentrations of the enclosed substances, [C 2 H 3 0' 2 ] _ a 2 c(a 2 C + q lCl ) [C 2 H 4 2 ] and, substituting the numerical values, a 2 x (0-25) 2 x (q 2 x 0-25 + 0-86 x 0-025) ~ _ " - oj x 0-25 l00018 ' a 2 = 8-32 x 10 - and [H-jj = a 2 c = 8-32 x 10' 4 x 0-25 = 2-08 x 10~ 4 . If &! is the veloci since the velocity-con centration, we obtain If &! is the velocity-constant for the mixed solution, then since the velocity-constant is proportional to the H'-ion con- [ill _ Q'75 x 10 -a x 2-08 x 10 ~* :] 2-11. x io- 3 = 7-39 x io~ 5 . To illustrate the degree of approximation attained by the use of formula (12) instead of (io), we shall work the problem on the assumption that a may be neglected in comparison with 1. For the degree of dissociation of acetic acid in the pure aqueous solution we obtain from (12) 108 ELECTROLYTIC DISSOCIATION EXAMPLES a 2 c = K, IT 3 ' 48xl ' and, therefore, [H-] = ac = 8-48 x 10- 3 x 0-25 = 2-12 x 1Q- 3 . In the mixed solution we may assume that the acetic acid> in presence of the excess of C 2 H 3 0' 2 -ions from the dissociation of the sodium acetate, is practically undissociated, that is, [C 2 H 4 O 2 ] = c, and that the C 2 H 3 O' 2 -ions are furnished only by' the sodium acetate. Therefore [C 2 H 3 O' 2 ] = a 1 c 1 and [O.H,0-J _ [C.HA] Kc 0-000018 . P*]. - a^; - 0-86 x 0-025 ' 2 ' 09 *. 1( - Since, k-v 0-75 x 10- 3 x 2-09 x 1Q- 4 -.- 2-12 x 10 -~ = r4 X ' ' PROBLEM 234. The angle of rotation of an a = 5 per cent, cane sugar solution in a tube 20 cms. long is A = + 66*7. After complete inversion the angle of rotation is A^ = - 19*7. If the solution contains in addition c x = O'Oi N-HC1, the angle of rotation diminishes by 69 '2 in t = 20 minutes. By how much does the angle of rotation diminish in the same time, when, instead of HC1, the solution contains c 2 = O'l N- lactic acid, if the dissociation- constant of lactic acid is K = 1*4 x 10~ 4 ? Assume the HC1 to be completely dis- sociated. SOLUTION 234. If A denotes the initial angle of rotation, before the inversion begins, and A x the final angle of rota- tion after complete inversion, the total change in the angle of rotation, A - A M , is proportional to the initial amount of cane sugar, a per cent. Similarly, if at the time t, when the solution contains x per cent, of cane sugar, the angle of rota- tion is A x , the change in the angle of rotation, A x - A^, is proportional to x. Therefore, ELECTROLYTIC DISSOCIATION EXAMPLES 109 Since in the presence of a large excess of water the inver- sion proceeds as a monomolecular reaction, the velocity of inversion is given by the equation dx where k is the velocity-constant of the reaction, and x the concentration of the cane sugar at the time t. On integrating we obtain (2) log e x = - kt + constant. When t = 0, x = a, (3) /. log e a = constant. Subtracting (2) from (3) we obtain and, therefore, from (1) or, changing from natural to common logarithms, The inversion is catalytically accelerated by the H'-ions, and the velocity-constant k is proportional to the IT-ion concentration. If we assume the HC1 to be completely dis- sociated, the concentration of the H'-ions is [H*^ = c r The velocity-constant k- for this particular case may be calculated from the data given in the problem by means of equation (4) which in this case becomes In the c 2 . N-lactic acid let the H'-ion concentration be [H*] 2 . If, for brevity, we represent lactic acid by the formula LH, where L is the anion, we obtain for the dissociation of lactic acid the equilibrium-equation 110 ELECTKOLYTIC DISSOCIATION EXAMPLES EiU5i = ^ or, since [I/] = [H-] 2 , [H-V and [H-], = - j If k 2 is the velocity-constant for the inversion of cane sugar by c 2 N-lactic acid, we obtain, since the velocity-constant is proportional to the H'-ion concentration, T * - ** ~ 1 In this equation only & 2 is unknown, and can, therefore, be calculated. The angle of rotation A v in the lactic acid solution after t minutes is obtained from the equation (cf. (4)) Substituting the numerical values in the various equations we obtain k lt & 2 and A y , and, therefore, the diminution of the angle of rotation required, A A y . Thus in the case of the inversion by HC1, since the diminution of the angle of rotation after 20 minutes is 69*2, we obtain A f = 66T - 69'2 = - 2-5, and from equation (5) 2-3 *i = 7- 2-3. 66-7 +19-7 K1 86-4 = 20 l0 ^ - 2-5 + 19-7 = 115 lo 1T2 = 0-115 x 0-703 = 0-0808. From (6) fc, 0-1 ELECTROLYTIC DISSOCIATION EXAMPLES 111 = 8-08(VO-49 x 10- 8 + 1-4 x 10~ 6 - 0'7 x 10"*) = 8-08(3-74 x 10- 3 - 0-7 x 10~ 4 ) = 0-0297. From (7) 0-0297 x 20 = log 86-4- c^j = 1-937 - 0-258 = 1-679, .-. A y ~A ca = 47'8' J , and A y = 47'8 + A M = 47'8 - 19-7 = 28-1. The diminution of the angle of rotation after 20 minutes in the lactic acid solution is, therefore, A - A y = 66-7 - 28-1 = 38*6. PROBLEM 235 (cf. preceding problem). What is the dis- sociation-constant of acetic acid if a mixture of c 3 = 0'05 N- lactic acid, and c 4 = 0*38 N-acetic acid has the same inverting action as the c 2 = O'l N-lactic acid alone ? SOLUTION 235. If a solution containing c 3 N-lactic acid and c 4 N-acetic acid has the same inverting action as a c 2 N- lactic acid solution alone, the H*-ion concentrations of these two solutions must be the same, since the inverting action is proportional to the H--ion concentration. If, for brevity, we represent lactic and acetic acids by the formulae LH and AH, where L and A are the respective anions, we obtain for the electrolytic dissociation of lactic acid and of acetic acid where K and K' are the dissociation-constants of lactic acid and acetic acid respectively. As in Solution 234 we have also (3) 112 ELECTROLYTIC DISSOCIATION EXAMPLES and in the mixed solution (4) [L'] + [LH] - c s , (5) [A'] + [AH] - e v and (6) [L'] + [A'] = [H-]. From these six equations the six unknown quantities [L'], [A'], [LH], [AH], [H-] and K may be calculated. From (3) -^T 3 8 + 1-4x10-6- -^ = 3-67 x 10- 3 , and from (4) [L'] = c 3 - [LH] = 0-05 - [LH]. Substituting these values of [IT] and [L'] and the numeri- cal value of K in (1) we obtain (0-05 - [LH]) x 3-67 x 10 ~ 3 == 1-4 x 10 ~ 4 [LH], and, therefore, from (4) [L'] = 0-05 - 0-04818 = 0-00182. From (6) [A'] = [H-] - [L'] = 0-00367 - 0-00182 = 0-00185. From (5) [AH] = c 4 - [A'] = 0-38 - 0-00185 = 0-378, and, finally, from (2) K/ - [AH] _ 000185 x 0-00367 0-378 = i '8 x 10 ~ 5 . PROBLEM 236. In a mineral water with an acid reaction the total concentration of carbonic acid, free and -combined, is a = 0-620 gram CO 2 per litre, the total concentration of sulphuretted hydrogenffree and combined, is b = 0*018 gram H 2 S per litre, and the concentration of sodium oxide is c = ELECTEOLYTIC DISSOCIATION EXAMPLES 113 0-279 gram Na 2 O per litre. What is the concentration of free carbonic acid [H 2 CO 3 ] and of free sulphuretted hydrogen [H 2 S], if the first dissociation-constant of H 2 CO 3 is K^ = 3-04 x 10- 7 , and that of H 2 S is 0'91 x 10 ~ 7 ? SOLUTION 236. In the acid mineral water the carbon dioxide is present as undissociated molecules of carbonic acid H 2 CO 3 and as the anion HCO' 3 , an( ^ ^ ne hydrogen sul- phide is present as undissociated H 2 S and as the anion HS'. If M 1 is the molecular weight of C0 2 and M% that of H 2 S then (1) [H 2 C0 3 ] + [HCO'J = Jl, (2) [H,S] + [HS'] = A, where the square brackets denote, as usual, the concentra- tions of the enclosed substances. From the law of electro-neutrality, according to which the total number of equivalents of cations must be equal to the total number of equivalents of anions in the solution, it follows that (3) [HC0 3 '] + [HS'] = [Na-] = ^, where M 3 denotes the molecular weight of Na 2 O. For the solution of these three equations with the four unknowns [H 2 CO 3 ], [H 2 S], [HCOg'] and [HS'], a fourth equation is required. This is obtained by an application of the law of mass-action. For the ionic dissociation of the two weak acids the fol- lowing equations hold, [HC0 3 '] [H-] = K, [H 2 C0 3 ] and [HS'] [H>] = K, [H 2 S], or, dividing the latter by the former, (i) } [HOOT] The problem is, therefore, capable of solution. From a consideration of the numerical values given in the problem, it follows that [HS'J is small compared with [HCO/J, especially as hydrogen sulphide is the weaker acid. Neglect- 8 114 ELECTROLYTIC DISSOCIATION EXAMPLES ing [HS'] in comparison with [HC0 3 'J, we therefore obtain from (3) [HCO 3 '] = j = - 6 o = O'OO 9 gm.-ion per litre, x 3 Va and from (1) [H 2 CO 3 ] = 4- - 0-009 = 0-0141 - 0-009 = 0-0051 mole i per litre. From (4) we, thus obtain [HS'] _ 0-91 x 10-T 0-009 W [H 2 S]~ 3-04 x 10 - 7 X 0-0051 , and [HS'] = 0-53 [H 2 S]. Substituting this value of [HS'] in equation (2) we have [H 2 S] (1 + 0-53) - jf* - 0-00053, /. [H 2 S] = 0*000346 gram-molecule per litre, and [HS'] = 0-000184 gram-ion per litre. PROBLEM 237. When picric acid distributes itself between water and benzene, only the undissociated molecules of the acid dissolve in the benzene, and the partition-coefficient of the undissociated molecules for benzene/water is K^ = 1/0-0281. The dissociation-constant of picric acid in water is K 2 = 0-164. What is the concentration c of an aqueous solution of picric acid which is in equilibrium with a c = 0-07 N-solution of picric acid in benzene? SOLUTION 237. Picric acid is soluble both in water and in benzene ; if an aqueous solution of picric acid is shaken with benzene, the picric acid distributes itself between the two solvents and the ratio of the concentrations of the undis- sociated picric acid molecules in the two solvents has a con- stant value K v which is independent of the concentration. The aqueous solution conducts the electric current, and, therefore, contains the ions of picric acid besides undissociated molecules ; the benzene solution, on the other hand,contains no ions since it is a non-conductor. Let the total concentration of an aqueous solution, which is in equilibrium with a c N-benzene solution, be c = c l + c 2 , where c 1 is the concentration of the undissociated molecules ELECTEOLYTIG DISSOCIATION EXAMPLES 115 and c 2 that of the ions. At equilibrium the following equations must hold : (1) from the partition law, = K v c i c 2 (2) from Ostwald's dilution law = K 2 . From these two equations and the further equation = c l + c 2 , the three unknowns la ted. By dividing (1) by (2) we obtain c = c l + c 2 , the three unknowns c , c l and c 2 may be cal culated. fOL and c 2 = \j ', A i from (1) therefore c = = 0-07 x 0-0281 + jQWTx 0-0281 x 0-161 = 0*0200 N. PROBLEM 238. A methyl acetate solution is saponified by a c l = O'l N-solution of KCN a = 3'3 times as fast as by a c 2 = 0-01 N-solution. What is the dissociation-constant K of hydrocyanic acid, if the ionic product of water is K w = 0-8 x 10 ~ 14 ? SOLUTION 238. The solution of KCN is alkaline on ac- count of hydrolysis, and in alkaline solution the saponification of the ester takes place according to the equation CH 3 COOCH 3 + OH' CH 3 OH + CH 3 COO'. The velocity of saponification is, therefore, proportional to the product of the concentrations of the ester and of the free hydroxyl ions. In two solutions, in which the concentrations of the ester are equal, the velocities of saponification are, therefore, proportional to the concentrations of the hydroxyl ions [OH']! and [OH'] 2 . In the present case we have, therefore, 116 ELECTROLYTIC DISSOCIATION EXAMPLES [OH'] 2 From the law of mass-action we obtain for the dissociation of hydrocyanic acid [ONI [H-] _ K [HCN] and, since [H-] [OH'] = K w , K The concentration of the undissociated KCN may be neglected, since the salt may be regarded as completely dissociated. The concentration of the K'-ions is, therefore, practically equal to the total concentration c of the KCN. The principle of electro-neutrality requires c = [K-] = [ON'] + [OH'], as the hydrolysis of KCN takes place according to the equation KCN + H 2 O = HCN + KOH, and the KOH like the KCN may be regarded as completely dissociated. We obtain, therefore, (3) o, = [ON'], + [OH'],, and (4) c, - [ONI + [OH 1 ], = [CN'] 2 + [ ^Ji (from (1)). For the solution of the three equations (2), (3) and (4), in which the five quantities #, [HCN], [CN'] 1} [CN'] 2 and [OH'^ are unknown, we require two further equations. These we obtain as follows. The concentration of the OH'-ions pro- duced by the hydrolysis is equal to the concentration of the undissociated HCN produced at the same time according to the equation ON' + H 2 O = OH' + HCN. Therefore, from (2), I /TM''"] T7" [HCN] = [OH'] - LJ x and ELECTKOLYTIC DISSOCIATION EXAMPLES 117 and By dividing (5) by (6) we obtain ESl_ [ONI, and, from (4), (7)C2 From (3) and (7) we obtain a a* a .-.[CN-],-^- and [OH'], = Cl - [ON'], Finally, from (5), K = *:*m _ o(c 1 - c 2 a) (a - 1) ~ (a' 2 c c ) 2 _ 0-8 x 10-" x 3-3 x 2-3 x (0-1 - Q-Q33) {(3-3) 2 x 0-01 - O-l} 2 = 5 x io-". Transport Numbers If /JL A and p, c are the velocities of migration of the anion and cation respectively, the fraction - $ = n is called PA + Po the transport number (migration number) of the anion, and the LL ' ' * fraction - ^ = 1 " n the transport number of the cation PA + PC (Hittorf). Of the total quantity of electricity which passes any cross-section of an electrolyte, the fraction n is carried 118 TKANSPOKT NUMBERS by the anious or negative ions and the fraction 1 - n by the cations or positive ions. From equations (6), (7) and (8) it is evident that n = 7 IT = A^ I A + lc A oo so that from a knowledge of n and A^ we can obtain 1 A and l c . n is determined experimentally by electrolysing a solution of the electrolyte in question, measuring the total quantity of electricity which has passed through the solution (e.g. by a silver voltameter in series with the electrolyte) and the changes of concentration at the electrodes. If the quantity of electricity F is passed through the solution, 1 gram-equiva- lent of anion is deposited at the anode, n gram-equivalent of anion migrates to the anode, and 1 - n gram-equivalent of cation migrates from the anode, that is, at the anode 1 n gram-equivalent of both anion and cation disappears, or 1 - n gram-equivalent is the fall in concentration of the electrolyte round the anode. Similarly at the cathode, 1 gram-equivalent of cation is deposited, 1 - n gram-equivalent of cation migrates to the electrode, and n gram-equivalent of anion migrates from the electrode, that is, at the cathode n gram-equivalent of both cation and anion disappears from the solution round the electrode, or the fall in concentration of the solution round the cathode is n gram-equivalent. If then- we assume that the changes in concentration round the electrodes are due only to discharge and migration of the ions, that no secondary changes, such as solution of the electrodes, etc., take place, and that no diffusion occurs, we have the relation fall of cone, of electrolyte round cathode n fall of cone, of electrolyte round anode ~ 1 - n Since the quantities on the left-hand side of the equation are known, n may be calculated. It is not, however, necessary to determine the change in concentration at both electrodes. If the theoretical change (fall) in concentration in gram-equivalents at one electrode, say the cathode, and the total quantity of electricity in coulombs passed through the solution (e.g. by the silver voltameter) are determined, all the necessary data are obtained. Then TKANSPOET NUMBERS EXAMPLES 119 fall (in gram-equivs.) round cathode x 96540 or total quantity of electricity passed fall (in gram-equivs.) round anode x 96540 total quantity of electricity passed Since, however, the experimental arrangements for the deter- mination of transport numbers vary with the nature of the ions under investigation, and secondary reactions at the elec- trodes, which vary from case to case, have to be taken into account, the calculation for each particular case has to be thought out independently. Transport Numbers Example PEOBLEM 239. A solution containing 0-1605 per cent. NaOH was electrolysed between platinum electrodes. After electrolysis 55-25 grams of the cathode solution contained 0*09473 grams NaOH, whilst the concentration of the middle portion of the electrolyte was unchanged. In a silver volta- meter in series the equivalent of 0*0290 gram NaOH was deposited during electrolysis. Calculate the transport num- bers of the Na* and OH' ions. SOLUTION 239. After electrolysis 55 '25 grams of the cathode solution contained 0*09473 gram NaOH or (55-25 - 0-09473 = ) 55*155 grams of water contained 0-09473 gram NaOH. Before electrolysis 100 grims of the cathode solution con- tained 0*1605 gram NaOH, therefore, 55*155 grams water contained n = 0*08867 gram NaOH., ' The increase in the amount of NaOH round the cathode is therefore, 0-09473 - 0-08867 = 0-00606 gram. At the cathode, however, the secondary reaction Na + H 2 = NaOH + H ' takes place, the discharged Na'-ions reacting with the water to re-form their equivalent of NaOH. Had no secondary reaction taken place, therefore, the cathode solution would have contained 0*0290 gram NaOH less 'than it did, since 0*0290 gram NaOH is the equivalent of the Nations dis- charged, as measured by the silver voltameter. The amount; of NaQH in trie cathode sqlutiqn would, therefore, have been 120 TRANSPORT NUMBERS EXAMPLES 0-09473 - 0-0290 = 0-06573 gram in 55-155 grams water, compared with 0-08867 gram before electrolysis. The theo- retical fall in concentration round the cathode, that is, on the assumption that no secondary reaction had taken place, is, therefore, 0-08867 - 0-06573 = 0-02294 gram NaOH. If M is the equivalent weight of NaOH we obtain _ fall in gram-equivs. round cathode x 96540 total quantity of electricity passed 0-02294 qfi540 ~M- X 9654 _ 0-02294 _ 96540 " ' 0290 = M and i - n = 1 - 0-791 = 0*209. Alternative Method. For the deposition of _ gram- equivalent of silver in the voltameter, the increase in the 0*00606 amount of NaOH round the cathode is - gram - equiv. Therefore, for the passage of 1 faraday (96540 coulombs), corresponding to the deposition of 1 gram-equivatent of silver in the voltameter, the increase in the amount of NaOH round the cathode would have been 0-00606 1 = 0-00606 M = Q-QQ606 gram- M < 0-0290/M M X 0-0290 0290" equiv. But for the passage of 1 faraday n gram- equivalent of OH' migrates from the cathode, 1 - n gram-equivalent of Na- migrates to the cathode and 1 gram-equivalent of Na* is dis- charged at the cathode. The theoretical fall in the amount of electrolyte round the cathode for the passage of 1 faraday is, therefore, n gram-equivalent of OH' and 1 - (1 - n) = n gram-equivalent of Na-, or n gram-equivalent of NaOH. But owing to the secondary reaction of the discharged gram- equivalent of Na- with the water, 1 gram-equivalent of NaOH is re-formed. Instead of a fall of n gram-equivalent there is, therefore, an increase round the cathode of 1 - n gram- equivalent of NaOH. Therefore and n = 1 - 0-209 = 0791, SOLUBILITY-PRODUCT 121 Solubility and Solubility-product From the law of mass-action it follows that for a saturated solution of a difficultly soluble electrolyte A m B M in contact with the solid phase (14) [A-]" [B'] - k [A. BJ = L, where the square brackets denote the concentration of the enclosed ions or molecules, the former in gram-ions per litre, and the latter in gram-molecules per litre. Since the solution is in equilibrium with solid phase the concentration of the undissociated part [A m BJ for a given temperature is constant and independent of any excess of either of the ions A- or B' in the form of another electrolyte with a common ion. The product [A-]- [BT - L is called the solubility-product (or ionic-product) of the elec- trolyte. For binary electrolytes (14) takes the form (5) [A'] [B'] _ L, and since in pure aqueous solution [A*] = [B'] we obtain for that case = [B'j = Further, if the binary electrolyte is very insoluble, and be- longs to the class of strong electrolytes, like AgCl, for example, we may regard it as practically completely dissoci- ated in solution, and the concentration of the undissociated part as negligible compared with that of the ions. In such a case (16) where S is the solubility of the electrolyte in gram-molecules per litre. For a ternary electrolyte, e.g. A 2 B, we obtain similarly (17) [A-]' [B"] = L, and since in pure aqueous solution two ions of A are formed for one of B, the concentration of B" in gram -ions per litre is half that of A-, i.e. [B"] = [A-J/2 or [A-] = 2 [B"J. There- fore, from (17), 122 SOLUBILITY-PEODUCT EXAMPLES or (19) 4[B'? [B"] = 4[B"] = L. Finally, if the ternary electrolyte is a very insoluble, strong electrolyte, so that practically complete dissociation may ba assumed, from (18) or (19) the solubility S of the electrolyte in gram-molecules per litre is since one molecule of the undissociated electrolyte contains one atom of B and 2 atoms of A. Solubility-product Examples PBOBLEM 240. At 20 the specific conductivity of a satu- rated solution of silver bromide was 1'576 x 10 ~ r.o. and that of the water used was 1-519 x 10 ~ 6 r.o. On the assumption that the AgBr is completely dissociated, calculate the solubility and solubility-product of AgBr, given that the ' equivalent conductivities of KBr, KNO 3 and AgNO 3 at infinite dilution are 137*4, 131'3 and 121-0 r.o. respectively. SOLUTION 240. The specific conductivity K due to the AgBr alone is (total specific conductivity - specific conduc- tivity of the water used), .-. K = 1-576 x 10 - 6 - 1-519 x 10 ~ 6 = 0-057 x 10 ~ 6 r.o. Since the AgBr is assumed to be completely dissociated the equivalent conductivity calculated from the specific conduc- tivity according to (4) is equal to the equivalent conductivity at infinite dilution Aoo . A. for AgBr may be obtained from (8) as follows : A, = l Ag . + Z BI , = (l Ag . + Z N03 ') + (*K-+ JBT-) - fa- + W = 121-0 + 137-4 - 131-3 = 127-1 r.o. Therefore, from (4), 1000* A, = _, 127 . 1 ^ 1000 x 0-057 x 10 - 6 SOLUBILITY-PKODUCT EXAMPLES 123 1000 x 0-057 x 10 - 6 - 127-1 where c is the concentration of the saturated solution, or solubility, in gram-equivalents per litre. Since the AgBr is completely dissociated, the concentration of both the Ag - and Br'-ions is 4*49 x 10 ~ 7 gram-ion per litre and the solubility-product is L = [Ag-][Br] = (4-49 x 10 ~ 7 ) 2 = 2-03 x lo' 18 . PROBLEM 241. At a.given temperature a litre of a saturated solution of silver bromate contains S = 0'0081 gram-molecule of salt ; c = 0-0085 gram-molecule of silver nitrate is then added. Calculate the new solubility S' of silver bromate, assuming that both salts are practically completely dissociated in the solution. SOLUTION 241. In the saturated pure aqueous solution the concentration of both the Ag - and BrO 3 '-ions is S gram-ion per litre, since the salt is completely dissociated. The solubility-product of AgBr0 3 is, therefore, L = [Ag -][BrO,'] = (S) 2 = (0-Ocfel) 2 . In any solution saturated with AgBrO 3 the product of the concentrations of the Ag - and Br0 3 '-ions must be L. Let S' be the solubility, in gram-molecules per litre, of AgBrO 3 in the solution containing c N-AgN0 3 . Since both salts are assumed to be completely dissociated, the concentration of the Br0 3 '-ions, derived from the dissociation of the AgBrO 3 , is S', the concentration of the Ag '-ions from the AgBrO 3 is also S', and the concentration of the Ag --ions from the AgNO 3 is c gram -ions per litre. The total concentration of the Ag *-ions is, therefore, (c + S') gram-ions per litre. Therefore L = [Ag -][Br0 3 '] = (c + S')(S'), or, putting in the numerical values, (0-0085 + S')(S') = (0-0081) 2 .'. S' = 0-0049 gram-molecule per litre. PROBLEM 242. The solubility of Ag 2 C 2 O 4 at 25 is S = 1-48 x 10 ~ 4 gram-molecule per litre. A solution of potas- sium oxalate, containing c = 0-2942 gram- molecule per litre, was shaken at 25 with excess of Ag 2 CrO 4 till equilibrium was established according to the equation Ag 2 Cr0 4 + K 2 C 2 4 = Ag 2 C,0 4 + K 2 Cr0 4 . 124 SOLUBILITY-PRODUCT EXAMPLES The solution then contained c 1 = 0-0602 . gram - molecule K 2 CrO 4 per litre. Assuming that the degrees of dissociation of the oxalate and chromate in the solution are equal and that the dissociation of the silver salts is practically complete, calculate the solubility-product L Cr and the solubility S' of Ag 2 Cr0 4 . SOLUTION 242. The solubility-product of Ag 2 C 2 O 4 is ra =[Ag-p[CA"] since, on complete dissociation, S gram-molecule of Ag 2 C 2 O 4 gives 25 gram-ion of Ag and S gram-ion of C 2 4 ", .-. L ox = 4(1-48 x 10 -*) 3 = 1-3 x 10 - 11 . In the mixed solution at equilibrium both Ag 2 C 2 O 4 and Ag 2 Cr0 4 are present as solid phases, therefore the equations (1) c , r = [Ag-H00 4 "] and must be simultaneously satisfied. Since the Ag--ion con- centration is the same in both, namely that in the solution in equilibrium with the solid phases, we obtain, by dividing (1) by (2), [ C 2 O 4"J L OX The concentration of the K 2 CrO 4 at equilibrium is Cj and of the K 2 C 2 O 4 (c - Cj) gram- molecules per litre. If a is the degree of dissociation of both these salts, the concentrations of the CrO 4 "- and C 2 O 4 "-ions are a.c l and a(c c x ) gram-ions per litre respectively. From (3), therefore, [CrO 4 "] = ac x [C 2 4 "] a(c - Cj " and c, x L n9 L Cf 0-0602 x 1-3 x 10 0-2942 - 0-0602 = 3 34 x 10 -12 If S' gram-molecule per litre is the^solubility of Ag 2 CrO 4 in pure aqueous solution, and if the salt is completely dissoci- SOLUBILITY-PRODUCTEXAMPLES 125 ated, the concentration of the Ag '-ions is 2S' and of the CrO 4 "-ions S' gram-ions per litre. Therefore [Ag-p [CKY] = (2S7 (S f ) = 4(57 = L and 10 " 12 - = 0-94 x io~ 4 gram- molecule per litre. PROBLEM 243. The dissociation-constant of ammonium hydroxide is K = 1'8 x 10 ~ 5 , and the solubility-product of magnesium hydroxide is L = 1'22 x 10 ~ n . How many grams of solid ammonium chloride must be added to a mixture of 50 c.c. of N-NH 4 OH solution and 50 c.c. of N-MgCl 2 solution, so that the precipitate of magnesium hydroxide may just disappear ? Assume that the volume of the solution is not changed by dissolving the solid NH 4 C1, and that the dissociation of the neutral salts is complete. SOLUTION 243. In any solution saturated with Mg(OH) 2 the equation must be satisfied. The precipitate of Mg(OH) 2 disappears, therefore, in a solution in which the concentration of the OH'- ions is given by the equation t H 'J VfHfT where [Mg ] is equal to the total amount of magnesium present. The equilibrium equation for the electrolytic dis- sociation of the ammonia in the solution, namely, [NH 4 -] [OH'] = K [NH 4 OH], must also be satisfied. The NH 4 --ions are derived practically entirely from the solution of the solid ammonium chloride, since the NH 4 OH can be regarded as practically undissociated in presence of the NH 4 C1. The concentration of the NH 4 --ions, is, there- fore, [OH'] 126 SOLUBILITY-PRODUCTEXAMPLES In order to produce this concentration of NH 4 '-ions we must dissolve in the 100 c.c. of solution ( = 1/10 litre) x = .[NH 4 OH] x grams of NH 4 C1, where M denotes the molecular weight of NH 4 C1. Substituting the numerical values in this equation we obtain 1-8 x 10- x 0-5 x ' 5 x10 -u = 9-8 grams NH 4 C1, since the total concentration of both NH 4 OH and Mg in the mixed solutions is 0*5 N. PROBLEM 244. The solubility of calcium carbonate in pure water is S = 1-3 x 10 ~ 4 gram-molecules per litre. What is its solubility in water which is saturated with carbon dioxide under a pressure (1) of _pj = ^ atmosphere, and (2) of p 2 atmosphere, and which, therefore, contains carbonic acid, i the concentration of carbonic acid in water is K^ = 0'04354 times the pressure of the carbon dioxide in atmospheres, and the first and second dissociation-constants of carbonic acid in water are K 2 = 3-04 x 10 ~ 7 and K s = 1-3 x 10 ~ 14 respec- tively ? SOLUTION 244. In the case of all solutions which are saturated with CaCO 3 the equation [Ca ] [CO 8 "] = L must be satisfied, where L is the solubility-product of calcium car- bonate. In pure aqueous solutions we may assume that the ionic dissociation is complete, and can, therefore, put [Ca ] = [CO,"] = S. We thus obtain (1) [Ca--][C0 3 "] = L = S*. By the addition of free carbonic acid the solubility of calcium carbonate is increased, i.e. the concentration of the Ca - ions is increased owing to the diminution of the CO 3 "-ion concentration by the reaction CO 3 " + H 2 CO 3 = 2HC0 3 ', or, in other words, owing to the , formation of bicarbonate. The mass-action equation for this reaction is ~ [H.CIOJ SOLUBILITY-PRODUCT EXAMPLES 127 This constant K may be calculated as follows, from the given values of the dissociation-constants of carbonic acid. For the first dissociation of carbonic according to the equation H 2 CO 3 = H- -f HCO 3 ', the law of mass-action gives [H-J [H00.1 _ K -[3J50T and for the second dissociation according to the equation HCO 3 ' = H- + CO 3 " we obtain similarly H-nccv] (3) divided by (4) gives (2), thus [HOO.T _i, [00,"] [ ' By multiplying (1) by (2) we obtain [Oa--][HCO,T , J& [H.OOJ ' K- The law of electro-neutrality requires 2[Ca--] + [H-] = [HC0 8 '] + 2[CO 3 "]. Since carbonic acid is a very weak acid, [H ] is very small even in presence of free H 2 C0 3 , and [C0 3 "] may be neglected in comparison with [HG0 3 ']. We thus obtain 2[Ca ] = [HC0 3 ']. The concentration of Ca --ions, and, therefore, the solu- bility of calcium carbonate' in water containing carbonic acid, may be calculated from equation (5) if the concentration of the free carbonic acid, [H 2 CO 3 ], in the solution is known. This may be obtained with the help of the factor of propor- tionality K lt which is given. If the partial pressure of the CO 2 over the solution is p, then [H. 2 C0 3 ] = K lP . Substituting this value of [H 2 C0 3 ] in equation (5), and bearing in mind that 2[Ca- ] = [HC0 3 '], we obtain from (5), ' and 128 HYDKOLYSIS 3 /(l-3) 2 x 10 - 8 x 4-354 x 10 -2 x 3-04 x 10 - t = V 4 x 1'3 xlO~ n " ^ 0-016 55". ^5 atmosphere [C a ' '] = 0*016 x 0*37 = 0*0059 gram -molecule per litre, and for p 2 = % atmosphere [C a ' '] 0'016 x 0-794 = o'oi27 gram-molecule per litre. Hydrolysis of Salts In an aqueous solution of a salt BA of a weak acid HA and a strong base BOH the salt is more or less hydrolysed according to the equation BA + H 2 O = HA + BOH. Let x be the degree of hydrolysis, that is, of each gram- molecule of salt in solution let the fraction x be hydrolysed according to the above equation ; the fraction (1 - x) of each gram-molecule remains, therefore, unhydrolysed. If G is the total concentration of the salt in gram-molecules per litre, the concentration of HA and of BOH is, therefore, xc, since they are produced in equivalent amounts, and the concentra- tion of the unhydrolysed salt BA is (1 - x)c. We shall assume further that the strong electrolytes, the salt BA and the base BOH, are completely dissociated, whilst the weak electrolyte, the acid HA, is practically undissociated, a condi- tion which will be very approximately fulfilled, since the dis- sociation of the weak acid into H*- and A'-ions, which is feeble even in a pure aqueous solution, will be still further diminished by the presence of the large excess of A'-ions derived from the dissociation of the salt BA. If the concentrations of the various molecules and ions are denoted by square brackets, we may, therefore, put [HA] = xc, [OH'] = xc a-nd [A'] = (1 - x)c, since, in the last case, the concentration of the A'-ions derived from the dissociation of the HA must, for the reasons given abjjre, be practically negligible. The concentration of the Herons, [H*], we obtain from the fact that in any aqueous HYDROLYSIS 129 solution at a given temperature the product of the concen- trations of the IT- and OH'-ions must be constant, or K w is called the ionic product of water. We obtain, there- fore [OH'] xc Applying the law of mass-action to the dissociation of the weak acid HA we obtain -T7- _. [H-][A']_^ - K ~ ^ where K a is the dissociation-constant of the acid. Hence K f! is called the hydrolysis-constant of the salt BA and, as is evident from equation ^21), -g = (cone, of free acid)(conc. of free base) ^ (cone, of unhydrolysed salt) If Kft and, therefore, the degree of hydrolysis x, is very small, that is, if K a is very much greater than K w , we may put (1 - x) = 1 without great" error, and we obtain from (21) the approximation formula (22) K, = x*c. In an exactly similar way it may be shown that for a salt BA of a weak base BOH and a strong acid HA, (**} K - K - x * c (23) K m - ^ - -_, where K H is the hydrolysis-constant of the salt, K b the dissoci- ation-constant of the weak base, x the degree of hydrolysis of the salt and c the total concentration of the salt. If the salt BA of the weak acid HA and the weak base BOH is hydrolysed according to the equation BA + H 2 O = HA + BOH and if we assume that the dissociation of the unhydrolysed salt (a strong electrolyte) is complete, whilst the weak acid 9 130 HYDROLYSIS and base are practically undissociated in presence of the excess of A and B'-ions furnished by the dissociation of the salt, then if c molecules per litre is the total concentration of BA, and x is the degree of hydrolysis, [A-] = (1 - x)c, [B'J = (1 - x)c, [HA] = xc, [BOH] = xc, where the square brackets denote the concentrations of the enclosed substances in gram-molecules per litre. Therefore " [HA] xc ']_(1 -qQc[OH'] [BOH] xc and Z x K = [H "I XC XC Therefore It (24) x K h _ (cone, of free acid) (cone, of free base) (cone, of unhy Jrolysed salt)' 2 Since the concentration c does not appear in this equation the degree of hydrolysis of a salt of a weak acid and a weak base is independent of the concentration. Hydrolysis Examples PBOBLEM 245. At 25 C. the dissociation-constant of ammonia in aqueous solution is K b = 1-8 x 10 ~ 5 , and at the same temperature the ionic product of water is K w = 0'8 x 10 ~ 14 . The concentration of hydrogen ions in a solution, which is c l = 0-01 N with respect to ammonia and c 2 = 0-02 N with respect to diketotetrahydrothiazole, is a = T7 x 10 ~ 7 gram-ion per litre. What is the dissociation-constant K a of the acid, diketotetrahydrothiazole, and to what degree per cent, is the neutral ammonium salt of the acid hydrolysed (1) inac 3 = 0-001 N-solution, and (2) in a C 4 = O'l N-solution ? SOLUTION 245. The acid, diketotetrahydrothiazole, may, HYDEOLYSIS EXAMPLES 131 for brevity, be represented by the formula AH, in which H is the acid hydrogen or cation, and A the acid radical or anion. According to the law of mass-action, the equilibrium equations for the dissociation of the base, NH 4 OH, and of the acid,,- AH, are [NH 4 -] [OH'] _ * ' [NH 4 OH] 6 . and For the dissociation of water the ionic product is (3) [H-] [OH'] = K w . In an aqueous solution containing both the base and the acid, all three equations must be simultaneously satisfied. In these equations K a and the concentrations [OH'], [NH 4 '], [NH 4 OH], [A'] and [AH] are unknown, whilst [H*] is given = a. For the evaluation of these six unknowns three further equations are, therefore, required. These are (4) [NH 4 -] + [NH 4 OH] = c v (5) [A'] + [AH] = c 2 and (6) [NH 4 -] + [H-] = [A'] + [OH']. The concentration of the undissociated salt molecules, [NH 4 A], may be neglected, since the salt may be regarded as practically completely dissociated. The problem is, there- fore, capable of solution, and its solution is simplified by the fact that [H-] = a is very small compared with c l and c >2 ; the anions A' are, therefore, derived practically entirely from the dissociation of the salt NH 4 A, and the the excess of the acid AH is only slightly dissociated. Equation (6) therefore simplifies to (7) [NH 4 -] = [A']. From (1) and (3) we obtain [NHJ K b K h {R-] _ 1-8 x 10-5 x jL-7 x ip-7 [NH 4 OH] [OH'] K~ 0-8 xTO"-' 1 * 382 "I" [NH 4 OH] may, therefore, be neglected in comparison with [NH 4 -]. We thus obtain 132 HYDEOLYSIS EXAMPLES from (7) and (4) : [NH 4 -] = [A'] = c l = 0-01, from (5) : [AH] = c. 2 - c 1 = 0-01, and from(2) : K a = "Tn^fH = [H-] = a = 1-7 x IO - 7 . In a c 3 N-solution of the neutral ammonium salt let the fraction x be hydrolysed according to the equation NH 4 A + H 2 = AH + NH 4 OH. Then [NH 4 OH] = [AH] = xc z , and [NH 4 -] = [A'J = c 3 - xc,. The ions H- and OH' are present only in very small con- centration, since both the base and the acid are very weak ; and the concentration of the undissociated salt molecules, [NH 4 A], may again be neglected on account of the practically complete dissociation of the salt. From (1) we thus obtain (8) and from (2) (8) x (9) gives n ^2 \ L ~ ^) rmrn nr.n v x z L OH J L H ] = K b and by dividing this result by (3) we obtain and iKJta 1 + /1-8 x 1-7 \ J^ V 0-8 x : - J 1+ x/382 = 0-049. The degree of hydrolysis is, therefore, 4-9 per cent., and is the same for all concentrations of the neutral salt, since x is independent of the total concentration c y The degree of HYDEOLYSIS EXAMPLES 133 hydrolysis of a salt of a weak acid and a weak base is, therefore, independent of the concentration. PROBLEM 246. At 25 the distribution- coefficient of ani- line between benzene and water is 10*1. A litre of 0'03138 N-aniline hydrochloride was shaken with 59 c.c. of benzene. After equilibrium was established 50 c.c. of the benzene was found to contain 0-02916 gram of aniline. Calculate (a) the hydrolysis-constant of aniline hydrochloride, (6) the percent- age hydrolysis in 0*1 N-solution and (c) the dissociation- constant of aniline as a base, given that the ionic-product for water at 25 is 1-2 x 10 - 14 . SOLUTION 246. (a) The molecular weight of aniline is 93, therefore (1) 50 c.c. of benzene contain gram- equivalent of 93 aniline, and the concentration of the aniline in benzene is 2^ x 1000 gram . equivalent per litre . 93 oO Since concentration of aniline in benzene __ JQ.-^ concentration of aniline in water we obtain (2) concentration of free aniline in water layer concentration in benzene _ 0-02916 x 1000 10-1 " 93 x 50 x 10-1 = 0-000621 gram-equiv. per litre. The total concentration of aniline originally present in the water was 0*03138 gram-equiv. per litre, but, from (1), of this amount 59 c.c. of benzene extracted 0*02Qlfi ^Q x _ = 0-00037 gram-equiv. 93 50 The total concentration of aniline (free and combined as hydrochloride) in the water layer at equilibrium is, therefore, 0-03138 - 0-00037 = 0-03101 gram-equiv. per litre. From (2), however, the concentration of the free aniline is 0-000621 gram-equiv. per litre, therefore the concentration of the aniline hydrochloride is (3) 0-03101 - 0-000621 = 0-03039 gram-equiv. per litre. 134 FAEADAY'S LAWS PEOBLEMS Since the benzene extracts only aniline and not hydrochloric acid or aniline hydrochloride from the aqueous layer, the total concentration of HC1 in the aqueous layer at equilibrium is equal to that of the aniline hydrochloride originally present, namely 0*03138 gram equiv. per litre. Of this amount, how- ever, 0-03039 gram-equiv. exists as aniline hydrochloride (from (3)), therefore the concentration of free HC1 at equi- librium is 0-03138 - 0-03039 = 0-00099 gram-equiv. per litre. .From (21) -g _ (cone, of free acid) (cone, of free base) (cone, of unhydrolysed salt) _ (0-00099)(0-000621) . ' (0-03039) (b) If x is the degree of hydrolysis in O'l N-solution, from (21) K H = -*---- or 2-02 x 10- 5 = 1 - x 1 - x .'. x - 1-41 x lO- 2 = 1^41 per cent. (o) From (21) K H = Problems for Solution In the following problems the answers to those marked * have been obtained by using the simplified formula? (12) and (13)1 i-G. by putting (1 - a) = 1, and (22), i.e. by putting (i - *) - 1. Faraday's Laws PROBLEM 247. A current passed through a water volta- meter liberated in 4 minutes 50 c.c. of hydrogen at 17 C. and 750 mm. Calculate the mean current during the whole time. Ans. 1'669 amp. PROBLEM 248. In the preparation of NaOH by the electro- lysis of a sodium chloride solution, 600 c.c. of solution con- taining 40 grams NaOH per litre was obtained after a certain TEANSPORT NUMBERS PROBLEMS 135 time. During the same time 3O4 grams of Cu had been deposited in a copper voltameter in series with the electro- lytic cell. Calculate the percentage of the theoretical yield of NaOH obtained. Ans. 62'8 per cent. PROBLEM 249. What volume of hydrogen at 18 and 737 mm. is liberated by passing a current of 1-54 arnp. through a solution of dilute H 2 SO 4 for 2 minutes ? Ans. 23-56 c.c. PROBLEM 250. A current of 1-5 amp. is passed through a solution of CuCl 2 for 1 hour. What weight of electrolyte is decomposed ? Ans. 3*764 grams. Transport Numbers PROBLEM 251. A solution of AgN0 3 containing 1-139 mg. of silver per gram of solution was electrolysed between silver \ electrodes and the anode liquid after electrolysis contained 39-66 mg. silver in 2O09 grams of solution. In a silver volta- meter in series with the electrolytic cell 32' 10 mg. of silver was deposited. Calculate the transport numbers of Ag- and N0' 3 . Ans. Ag = 0/476, NO' 3 - 0-524. PROBLEM 252. A solution of CuSO 4 containing 1 gram CuSO 4 per 41-59 grams water was electrolysed between a copper anode and a platinum cathode, a silver voltameter being placed in series with the electrolytic cell. After electro- lysis 54-706 grams of the cathode solution gave on analysis 0-5118 gram CuO. During the electrolysis 0*6934 gram of silver was deposited in the voltameter. Calculate the trans- port numbers of S0" 4 and Cu . (Cu = 63-6, S = 32-06, O = 16, Ag = 107-9.) Ans. SO" 4 = 0-5146, Cu = 0-.4854. PROBLEM 253. A solution of NaCl containing 0*01784 per cent, of chlorine was electrolysed between a Cd anode and a Pt cathode, with a silver voltameter in series. After electro- lysis the solution was divided into three portions, cathode, anode, and middle portion. The concentration of the latter was unchanged, whilst 226-99 grams of anode solution con- tained 0-04679 gram chlorine, and 331-49 grams cathode solution contained '05302 gram chlorine. In the silver v/v. E 136 ELECTROLYTIC DISSOCIATION PROBLEMS voltameter the equivalent of 0*01021 gram chlorine was de- posited. Calculate the transport numbers of Cl' and Na\ Ans. Cl' = 0-611, Na- = 0*389. PROBLEM 254. A solution of CdCl 2 containing 0-2016 per cent, of chlorine was electrolysed between a Cd anode and a t cathode. After electrolysis 33-59 grams of the anode quid contained 0-08020 gram Cl and 54*12 grams of the athode liquid contained 0-09662 gram Cl. The concentra- tion of the middle portion was unchanged. In a silver volta- meter in series with the electrolytic cell 0-06662 gram of silver was deposited during the electrolysis. Calculate the transport numbers of Cl' and Cd . Ans. 01' - 0-570, Cd - 430. Electrolytic Dissociation PROBLEM 255. The equiv. conductivity of LiCl at 18 is 101-4 r.o. at infinite dilution and 93-6 r.o. for a O'Ol N-solution. What is the degree of dissociation and the concentration of the Cl' ions in this solution ? Ans. a = 0-923, [01'] = 0-00923 gram-ion/litre. PROBLEM 256. At 18 the equiv. conductivity of HI at infinite dilution is 384 r.o. and the specific conductivity of a 0-405 N-solution is 0-1332 r.o. What is the concentration of the H --ions in this solution ? Ans. [H ] = 0-347 gram-ion/litre. PROBLEM 257. The equiv. conductivity at 25 of acetic acid containing 1 gram-equiv. in 32 litres is 9'2 r.o. The equjv. conductivity at infinite dilution is 389 r.o. Find the dissoci- ation-constant of the acid. Ans. 1-79 x 10- 6 . PROBLEM 258. At 25 the specific conductivity of butyric acid at a dilution of 64 litres is 1-812 x 10 ~ 4 r.o. The equiv. conductivity at infinite dilution is 380 r.o. What is die degree of dissociation and the concentration of H --ions in^he solution, and the dissociation-constant of the acid ? Ans. a = 0-0305, [H ] = 4 -765 x 10 - 4 gram-ions/litre, K= 1-5 x 10 - 5 . PROBLEM 259 (cf. preceding problem). What is the value of the dissociation-constant of butyric acid if the concentra- ELECTROLYTIC DISSOCIATION PROBLEMS 137 tion is measured in gram-equivalents per c.c. instead equivalents per litre ? Ans. 1-5 x 10 ~ 8 . PROBLEM 260. The specific conductivity of a 5 per cent. (by weight) BaCl 2 solution is 389 x 10 ~ 4 r.o. and its density 1-0445 at 18. What is the equivalent conductivity and de- gree of dissociation of the solution, if the equiv. conductivity of BaCl 2 at infinite dilution is 123 r.o. at 18 ? Ans. A = 77-7, a = 0-632. PROBLEM 261 (cf. preceding problem). What is the freez- ing-point of the BaCl 2 solution ? Ans. - 1-064. PROBLEM 262. At 25 the specific conductivity of ammonia solutions, containing c gram-equivs. per litre is * r.o. The ionic conductivity of the NH 4 '-ion at this temperature is 70-4 and of the OH'-ion 200-6. Calculate the equiv. conductivity and degree of dissociation at each concentration and the mean dissociation-constant. c K 0-0109 1-220 x 10-' ^0-0219 1-730 x 10 ~ 4 Ans. A = 11-2 and 7'9 r.o., a = 0-0413 and 0-0231, K = 1-94 x 10~ 5 and 1-92 x 10 ~ 5 , mean = 1-93 x 10 ~ 5 . * PROBLEM 263 (cf. preceding problem). At what con- centration is ammonia 1 per cent, dissociated in solution ? Ans. c = 0-193 N. PROBLEM 264. At 25 the specific conductivity of ethyl- amine at a dilution of 16 litres is 1-312 x 10 ~ 3 r.o., the equiv. conductivity at infinite dilution is 232-6 r.o. Calculate the dissociation-constant. Ans. 5-6 x lO- 4 . PROBLEM 265 (cf. preceding problem). What is the con- centration of OH'-ions in the ethylamine solution ? Ans. 5-64 x 10 ~ 3 gram -ion/litre. PROBLEM 266. At what concentration of ethylamine is the concentration of the OH'-ions 0-01 N ? Ans. c = 0-1886 mole/litre. PROBLEM 267. At 25 the dissociation-constant of mono- 138 ELECTROLYTIC DISSOCIATION PROBLEMS chloracetic acid is T55 x 10 ~ 3 , and its equiv. conductivity at a dilution of 32 litres is 77-2 r.o. What is its equiv. conductivity at infinite dilution ? Ans. 388 r.o. * PROBLEM 268. At 25 the dissociation-constant of lactic acid is 1*4 x 10 ~ 4 and of monochloracetic acid 1-55 x 10 ~ 3 . At what concentrations of lactic and chloracetic acids is the H --ion concentration = O'Ol N ? Ans. For lactic, c = 0-724 N. For chloracetic, c = 0-0745 N. PROBLEM 269. The specific conductivity of KC1 is practi- cally proportional to its concentration in solutions of moderate concentration. The specific conductivity at 18 of a 10 per cent. KC1 solution is 01359 r.o. and of a 15 per cent, solution 0-2020 r.o. What is the percentage concentration of a KC1 solution of which the specific conductivity is 0-1640 r.o. ? Ans. 12-13 per cent. PROBLEM 270. The density at of a Ca(NO 3 ) 2 solution containing 0-208 gram -molecule per litre is 1*010. Its freez- ing-point is 0'910 and its molecular conductivity at is 78-8 r.o. Taking the molecular conductivity at infinite dilu- tion = 129 r.o. at and the freezing-point constant for water = 1-86 (gram -mole in 1000 grams water), calculate the degree of dissociation from the conductivity and from the freezing- point. Ans. From conductivity = 0-611. From freezing-point = 0-648. PROBLEM 271. At 25 the specific conductivities of KC1 and LiCl at a dilution of 32 litres are 4-25 x 10 ~ 3 r.o. and 3-243 x 10 ~ 3 r.o. respectively. Compare the degrees of dissociation of the two salts at this dilution, taking the ionic conductivities of K, Li and Cl at 25 as 75'3, 41 and 76 r.o. respectively. Ans. KC1 = 0-899, LiCl = 0*892. * PROBLEM 272. The dissociation-constant of acetic acid at 18 is 1*8 x 10 ~ 5 . Calculate the degree of dissociation and the H'-ion concentration (1) in a 0-25 N-acetic acid solu- tion, (2) in a 0*25 N-acetic acid solution containing 0'25 N- sodium acetate, if the sodium acetate is assumed to be com- pletely dissociated. ELECTKOLYTIC DISSOCIATION PROBLEMS '139 An3.(l)a = 848 x 10- 3 , [H] = 2'12 x 10 - 3 gram-ion/litre, (2) a = 7 x 10 ~ 4 , [H-] = 1-75 x 10 4 gram-ion/litre. PROBLEM 273. At 18 the specific conductivity of a 5 per cent, solution of Mg(NO 3 ) 2 is 438 x 10 ~ 4 r.o. and its density 1-0378. What is its degree of dissociation if the equivalent conductivity of Mg(NO 3 ) 2 at infinite dilution is 109 '8 r.o. ? ^Ans. 0-57. PROBLEM 274. At 18 the molecular conductivity of LiNO a is 94-46 r.o. at infinite dilution, and 75'01 r.o. in a 0*2 N-solution. What is the concentration of Li '-ions in this solution ? Ans. 0-159 N. PROBLEM 275. The density of a 5 per cent. NaCl solution at 18 is 1-0345 and its specific conductivity 672 x 10 ~ 4 r.o. What is (1) the molecular conductivity, (2) the degree of dissociation of the solution, if the molecular conductivity of NaCl at infinite dilution is 109 r.o. ? (3) What is the vapour- pressure of the solution at 18 if that of pure water is 15*33 mm. ? Ans. (1) 76 r.o., (2) 0-697, (3) 14-91 mm. PROBLEM 276. The velocity of migration of the Ag--ion at 18 is 0-000577 cms. per sec. and of the NO' 3 -km 0-000630 cms. per sec. The specific conductivity of 0-1 N-AgN0 3 at 18 is 0-00947 r.o. What is the degree of dissociation of the AgN0 3 ? Ans. 81-3 per cent. PROBLEM 277. The freezing-point of a 0*1 molecular N - CaCl., solution is - 0-482. (1) Calculate the degree of dis- sociation (freezing-point constant = 1-89 for gram-mol. per litre). (2) Compare the value with that found from the equi- valent conductivity at 18, which is 82'79 r.o., whilst the equivalent conductivity of CaCl 2 at infinite dilution is 115*8 r.o. Ans. (1) 0-775, (2) 0-715. * PROBLEM 278. What is the concentration of H'-ions in a solution containing 1 gram-molecule of acetic acid and 1 gram-molecule of cyanacetic acid per litre ? The dissociation- constant of acetic acid is 18 x 10 ~ 5 and of cyanacetic acid 370 x 10 - 5 . Ans. 6-1 x 10 -*N. 140 ELECTEOLYTIC DISSOCIATION PEOBLEMS PROBLEM 279. The molecular conductivity at 25 of ethyl hydrogen malonate is 356 at infinite dilution, and 21*5, 41*9 and 57*3 at the dilutions 8-57, 34-28 and 68-56 litres respec- tively. Calculate the degree of dissociation and the dissocia- tion-constant at each dilution. Ans. a = 6-04, 11-8 and 16-1 per cent. K = 4-54 x 10 - 4 , 4-58 x 10 ~ 4 and 4-51 x 10 - 4 . PROBLEM 280. At 25 the molecular conductivities of malonic acid at the dilutions 32, 64 and 128 litres are 77' 1, 103-6 and 137'0 r.o. respectively. The molecular conductivity at infinite dilution for dissociation into IT- and C 3 H 3 0' 4 -ions is 382 r.o. Calculate the degree of dissociation and the dis- sociation-constant at each dilution. Wha f ; conclusion may be drawn from these figures as to the manner in which malonic acid dissociates at these dilutions ? Ans. a = 0-202, 271, 0-359 K= 1-59 x 10 ~ 3 , 1-58 x 10 ^ 3 , 1-57 x 10 ~ 3 . PROBLEM 281 (of. preceding problem). For the dilutions 256, 512^ 1024 and 2048 litres the molecular conductivities of malonic acid at 25 are 176-8, 222-6, 269'9, and 313-9 respectively. What conclusions may be drawn from the values of K calculated from Ostwald's dilution law as to the dissociation of the acid at these dilutions ? Ans. K = 1-57, 1-62, 1-68,1-87 x 10 ~ 3 , .-. second dissocia- tion begins to be appreciable after ca. 256 litres. PROBLEM 282. A solution of NaCl containing 0-0585 gram per 100 c.c. freezes at 0'03674. The equiv. con- ductivity of NaCl at infinite dilution is 109 r.o. Assuming the degree of dissociation at 18 to be the same as at 0, cal- culate the specific conductivity of the above solution. Take freezing-point constant = 1-89 for gram-molecule per litre. Ans. 1-029 x 10 ~ 3 r.o. PROBLEM 283. At 18 the molecular conductivity of boric acid at the dilution v litres per gram-molecule is //, r.o. v 11-1 22-2 33-3 I* 0-0474* 0-0670 0-0825 Calculate the degree of dissociation and dissociation-con- stant at each dilution, given that the molecular conductivity at infinite dilution for dissociation into H-- and H 2 BO' 3 -ions is 346 r.o. ELECTROLYTIC DISSOCIATION PROBLEMS 141 Ans. a = 1-37 x 10-*,. 1-94: x 10 - 4 , 2-39 x 10 ~ 4 K = 1-70 x 10 - 9 , 1-69 x 10 - 9 , 1-71 x 10 ~ 9 . PEOBLEM 284. The specific conductivity of a 0*509 N- KNO 3 solution at 18 is 454 x 10 ~ * r.o. The temperature- coefficient of the specific conductivity between 18 and 22 is 0*0208. What is the equivalent conductivity of the solution at 20? Ans. 92-9 r.o. PROBLEM 285. At 25 the specific conductivity of malic acid at a dilution of 32 litres is 1-263 x 10 ~ 3 r.o. The equivalent conductivity at infinite dilution for dissociation into IT- and C 4 H 5 O' -kms i s 380 r.o. Calculate the degree of dissociation and the dissociation-constant. Ans. a = 0-106, K = 3-93 x 10 ~ 4 . r - PROBLEM 286 (cf. preceding problem). At what dilution { the coi 0-02 N? is the concentration of H'-ions in a solution of malic acid Ans. 0-963 litre. PROBLEM 287. At 25 the equivalent conductivities of fumaric and maleic acids at a dilution of 32 litres are 60'1 and 179 r.o. respectively. The equivalent conductivities at infinite dilution for dissociation into two ions are 385*6 and 391 '8 r.o. respectively. Compare their dissociation-constants. Aris. Fumaric 9 x 10 - 4 , maleic 1-2 x 10 - 2 . PROBLEM 288. The molecular conductivity of a Ca (NO S ) 2 solution containing 1 gram-molecule in 23*81 litres is 98 -9 r.o. at 0. The molecular conductivity at infinite dilution is 129-2 r.o. at 0. What is the freezing-point of the solution ? Take freezing-point constant = 1*89 for gram-mole/litre. Ans. - 0-2008. PROBLEM 289. The specific conductivity pf 0*135 N-prop- ionic acid at 18 is 4-79 x 10~ 4 ro. and that of O'OOl N- sodium propionate is 7*54 x 10 ~ 5 r.o. The ionic conductivity of the Na'-ion is 44-4 and of the H--ion 318 r.o. Assuming the sodium propionate to be completely dissociated, calculate the dissociation constant of propionic acid. Ans. 1*41 x 10 - 5 . * PROBLEM 290 (cf. preceding problem). The dissociation- constant of benzoic acid is 6 x 10 ~ 5 . What is the ratio of 142 SOLUBILITY-PRODUCTPROBLEMS the H'-ion concentrations in benzoic and propionic acids at equal dilutions ? Ans. 2-07 : 1. PROBLEM 291 (cf. preceding problem). At what dilution is the concentration of H'-ions in a benzoic acid solution 0-005 N ? Ans. 2-37 litres. PROBLEM 292. The velocity-constant for the inversion of cane sugar by 0*0125 N-HC1 at 54 is 0-00469. The velocity- constant for inversion by 0*25 N-formic acid at the same temperature is 0*00255. Calculate the dissociation-constant of formic acid, if it is assumed that the velocity-constant is proportional to the H'-ion concentration, and that the HC1 is completely dissociated. Ans. K = 1-9 x 10 ~ 4 . PROBLEM 293 (cf. preceding problem). What would be the velocity-constant at 54 for inversion by a 0-25 N-formic acid solution, containing O'l N-sodium formate, if the latter is regarded as completely dissociated ? Ans. 0-000169. Solubility-ProductHydrolysis PROBLEM 294. At 18 the specific conductivity of a saturated solution of silver chloride in water was 2-40 x 10 ~ 6 r.o., and that of the water used was 1-16 x 10 ~ G r.o. Given the equivalent conductivity at infinite dilution of AgNO 3 = 116-5 r.o., of NaCl = 110-3 r.o. and of NaNO 3 = 105*2 r.o., and on the assumption that the AgCl is completely dissociated in solution, calculate the solubility in gram- molecules per litre and the solubility-product of AgCl at 18. Ans. S = 1-018 x 10- 5 gram-mol./litre, = 1-035 x lO' 10 . PROBLEM 295. At 16'3 a saturated solution of barium oxalate has a specific conductivity of 67*7 x 10 ~ 6 r.o. whilst that of the water used was 1-2 x 10 ~ 6 r.o. The ionic con- ductivities for I Ba-- and i C 2 O" 4 at 16'3 are 50'6 and 58-4 r.o. respectively. On the assumption that the dissocia- tion of the salt is complete, calculate the solubility, in gram- molecules per litre, of BaC,O 4 at 16-3. Ans. 3-05 x 10 - * SOLUBILITY-PEODUCT HYDBOLTSIS 143 PKOBLEM 296. At 25 the concentration of a saturated solution of silver acetate is 0-0664 gram-molecule per litre. The molecular conductivity of the solution is 75*2 r.o. and the molecular conductivity of silver acetate at infinite dilution is 101-5 r.o. What is the solubility-product of silver acetate? Ans. 0-00241. PROBLEM 297. The solubility of CaSO 4 at 20 is 2-036 grams per litre. The specific conductivity of the saturated solution at 20 is 1968 x 10 ~ 6 r.o. The ionic conductivity for ^ Ca at 18 is 52 r.o. and the temperature- coefficient of the conductivity is 0-0238. The ionic con- ductivity for SO" 4 at 18 is 68'3 r.o. and the temperature- coefficient 0'0227. Calculate the degree of dissociation of CaSO 4 in the saturated solution and the solubility-product at 20. Ans. a = 52-25 per cent., L = 6-11 x 10 ~ 5 . PROBLEM 298. The solubility of benzoic acid at 25 is 3 '40 grams per litre and its dissociation-constant is 6 x 10 ~ 5 . What would be its solubility in a 01 N -sodium benzoate solution and in a O'Ol N-HC1 solution, if both these sub- stances are regarded as completely dissociated ? Ans. 0*02678 gram-mol. per litre for both. PROBLEM 299. Magnesium carbonate, which is practically insoluble in water, dissolves in water containing C0 2 owing to formation of bicarbonate. In a solution saturated with CO 2 under atmospheric pressure the solubility of MgCO 3 at 25 was 0*325 gram-molecule per litre. The first dissociation- rTT'l TTTPO' 1 constant of carbonic acid is rTT r*r\ T = 3-04 x 10 ~ 7 and [H 2 G(J 3 J [-TT.I rpry i the second dissociation-constant is L rrrnrv i* = 1'295 x 10~ n . L-hLUU 3 J According to Henry's law the concentration of H 2 CO 3 is proportional to the partial pressure of CO n the relation at 25 being given by [H 2 CO 3 ] = 4'92 x 10-*(OO), if [H 2 CO 3 ] is expressed in gram-molecules per litre and (CO 2 ) in atmos- pheres. On the assumption that the whole of the magnesium in the solution is in the form of bicarbonate, and that the bicarbonate is 61 percent, dissociated, calculate the solubility- product of MgCO 3 at 25. (Express all concentrations in gram-molecules or gram-ions per litre.) Ans. 2-7 x 10 ~ 5 . 144 SOLUBILITY-PRODUCT HYDKOLYS1S PROBLEM 300. At 25 the molecular conductivity of aniline hydrochloride at a dilution of 256 litres is 130*5. At the same dilution, but in presence of a sufficient excess of aniline to practically prevent hydrolysis, it is 107' 1. The conduc- tivity of the aniline in presence of its hydrochloride may be neglected. The equivalent conductivity of HC1 at a dilution of 256 litres is 410. Calculate the degree of hydrolysis of aniline hydrochloride at this dilution. Ans. 7*72 per cent. PROBLEM 301. From the degree of hydrolysis obtained in the preceding problem calculats the hydrolysis-constant of aniline hydrochloride, and the dissociation-constant of aniline as a base, given that the ionic product for water at 25 is 1-21 x 10 ~ 14 . Assume the aniline hydrochloride and the HC1 to be completely dissociated, and the aniline to be prac- tically undissociated. Ans. K B = 2-52 x 10 ~ 5 , K b = 4-8 x 10 - 10 . PROBLEM 302. From the result obtained in the preceding problem calculate the degree of hydrolysis of aniline hydro- chloride in a O'Ol N-solution. Ans. 4-88 per cent. PROBLEM 303. At 20 the specific conductivity of a satur- ated solution of T1C1 is 1680 x 10 ~ c r.o. The total concen- tration of T1C1 as determined directly at 20 is 1*36 x 10 ~ 2 gram-molecules per litre. The equivalent conductivity of T1C1 at infinite dilution is 137'3 r.o. What is the degree of dissociation of a saturated solution of T1C1 at 20, and what is the solubility-product of T1C1 ? Ans. a = 90 per cent., L = 1-5 x 10 ~ 4 . PROBLEM 304. The molecular conductivity of carbonic acid at the dilution of 1 gram-molecule in v litres is /x, r.o. v 27'5 55-0 110-0 fi 1-033 1-450 2-040. Calculate the degree of dissociation and the dissociation- constant for each dilution for dissociation into H-- and HCO' 3 - ions, given p for NaHC0 3 = 84-9, forNaCl = 110-3 and for HC1 = 381-9 r.o. Ans. a = 2-9 x 10 ~ 3 , 4-07 x 10 - 3 , 5*72 x 10 ~ 3 K = 3-06 x 10 - 7 , 3-02 x 10 ~ 7 , 2-99 x 10 ~ 7 , mean = 3*02 x 10 ~ 7 . SOLUBILITY-PRODUCT HYDROLYSIS 145 * PROBLEM 305. From the dissociation-constant of carbonic acid obtained in the preceding problem calculate the hydrolysis- constant for the hydrolysis of NaHCO 3 according to the equation NaHC0 3 + H,0 = NaOH + H 2 CO 3 , and the degree of hydrolysis of the salt in O'l N- solution. Assume the strong electrolytes to be completely dissociated, and the weak acid to be practically undissociated, and take the ionic product of water = 1*2 x 10 ~ 14 . Ans. K = 3-97 x 10 ~ 8 , x = 0'063 per cent. PROBLEM 306. At 100 C. 100 c.c. of water dissolve 0'12 gram of AgCNO. How much AgCNO will be dissolved at this temperature by 100 c.c. of a solution containing 1 gram of AgN0 3 ? The silver salts may be regarded as completely ionised. Ans. 0-016 gram. n PROBLEM 307. The solubility of Ag 2 CO 3 in water at 25 is 1 x 10 ~ 4 gram-molecule per litre. Calculate the solu- bility-product, assuming that the dissociation is complete. Ans. [Ag-] 2 [CO" 3 ] = 4 x 10 ~ 12 . PROBLEM 308 (cf. preceding problem). What is the solubility at 25, in gram-molecules per litre, of Ag 2 CO 3 in 01 molecular N-Na 2 CO 3 solution. Assume complete dis- sociation of both salts. Ans. 316 x 10 - . PROBLEM 309. At 25 the solubility of AgCl in water is 1'5 x 10 ~ 5 gram-molecule per litre and that of AgBr 7 x 10 ~ 7 gram-molecule per litre. On the assumption that both salts are practically completely dissociated, calculate the concentrations of Ag -, 01'- and Br'-ions in a solution which at 25 is saturated with both AgCl and AgBr. Ans. [Ag-] = 15-03 x 10 ~ 6 , [CF] = 15 x 10 ~ 6 , [Br'] = 3^26 x 10 - 8 gram-ion/litre. PROBLEM 310. A solution of MgCl 2 was treated with a solution of NH 4 OH and the precipitated Mg(OH) 2 jfhaken with the solution till equilibrium was established 1 : The solution on analysis gave 0-0219 gram-molecule MgCl 2 per litre, 0-0115 gram-molecule NH 4 C1 per litre and 0-0394 gram- molecule of free NH 4 OH per litre. The dissociation-constant of NH 4 OH is 1-8 x 10 " 5 . Assuming the Mg(OH) 2 , MgCl 2 10 146 HYDEOLYSIS PROBLEMS and NH 4 C1 to be completely dissociated and the NH 4 OH in presence of the NH 4 C1 to be practically undissociated, calculate the solubility of Mg(OH) 2 in water in gram- molecules per litre and its solubility-product. Ans. 8 = 2-75 x 10 - 4 gram-mol./litre, L = 8-33 x 10 - ". PROBLEM 311. The solubility of silver hydroxide as determined directly is 2*16 x 10 ~ 4 gram-molecule per litre. The solubility of AgCl as determined by the conductivity method is 1-5 x 10 ~ 5 gram-molecule per litre. A dilute solution of KOH was shaken with an excess of moist silver oxide and silver chloride till equilibrium according to the equation AgCl + KOH = AgOH + KC1 was established. The solution then contained 0-666 milli- gram-molecule of KC1 per litre and 70-7 milligram-molecule of KOH per litre. Assuming that the KC1 is 95 per cent, and the KOH 90 per cent, dissociated, calculate the solubility- product and degree of dissociation of AgOH in a pure aqueous saturated solution. Ans. L = 2-26 x 10 ~ 8 , a = 69-6 per cent. * PROBLEM 312. Calculate the percentage hydrolysis of sodium acetate in 0*1 N-solution at 25 from the following data, assuming that the salt is completely dissociated. Dissociation-constant of acetic acid = 0-000018 Ionic product for water = 1-21 x 10 ~ 14 . Ans. 8-2 x 10 ~ 3 per cent. PROBLEM 313. At 25 the velocity-constant for the saponi- fication of methyl acetate by N-HC1 containing c gram-mole- cule of urea per litre is k. c 0-0 0-5 1-0 2-0 k 0-00315 0-00237 0-00184 0-00114. Assuming that the velocity-constant is proportional to the concentration of free acid, calculate the hydrolysis-constant of urea hydrochloride for each concentration and the mean value. Ans. 0-766, 0-820, 0-772, mean 0-786. p" PROBLEM 314. From the mean hydrolysis-constant of urea hydrochloride found in the preceding example, calculate the dissociation-constant of urea into the ions CON 2 H- 5 and OH'. Take the ionic product of water at 25 = T2 x 10 - 14 HYDROLYSIS PKOBLEMS 147 and assume that the free acid and the salt are completely dissociated, whilst the weak base is practically undissociated. Ans. K b = 1-53 x 10 - 14 . PROBLEM 315 (cf. problem 162). The velocity-constant for the decomposition of diacetonealcohol by 0-1 N-NaHS, under the same conditions as in Problem 162, is 0-000037. Taking the degree of dissociation of 01 N-NaOH = 90 per cent., calculate the degree of hydrolysis of the 0-1 N-NaHS according to the equation NaHS + H 2 O = NaOH + H 2 S. Assume that the degree of dissociation of both NaOH and NaHS in the hydrosulphide solution is 90 per cent. Ans. 0'17 per cent. * PROBLEM 316 (cf. preceding problem). What would be the degree of hydrolysis of NaHS in O'Ol N- solution ? Ans. 0-537 per cent. PROBLEM 317. At 25 the dissociation-constant of aniline is 4-8 x 10 ~ 10 and of acetic acid 1-8 x 10 ~ 5 . The ionic product of water is 1-2 x 10 ~ 14 . What is the degree of hydrolysis of O'Ol and 0*05 N-solutions of aniline acetate, if the unhydrolysed aniline acetate is assumed to be completely dissociated ? Ans. 54-12 per cent, for both concentrations. PROBLEM 318 (cf. preceding problem). What is the H--ion concentration in each of the aniline acetate solutions ? Ans. 2-12 x 10 ~ 5 gram-ion per litre in each. PROBLEM 319. At 100 the following figures were obtained in the catalysis of N/32 methyl acetate solution by N/500 HC1, A being the titre of 10 c.c. of the solution at the time t, t 64 113 152 oo minutes A 1-10 4-15 6-03 7-35 15*70 c.c. N/50 NaOH. The velocity-constant for the catalysis of N/32 methyl acetate by a solution of A1C1 3 containing 1/32 gram-molecule per litre was 0*00216 at 100. Assuming proportionality between the velocity-constant and the concentration of HC1, calculate the degree of hydrolysis of the A1C1 3 solution. Ans. 8-7 per cent. PROBLEM 320. At 100 the following figures were obtained in the catalysis of the inversion of a 7*5 per cent, cane sugar 148 HYDEOLYSIS- PROBLEMS solution- by N/1000 HC1, A being the angle of, rotation at the time t, tQ 22 40 oo minutes A + 10-66 4- 2-43 - 0-49 D - 3-63. At 100 the figures for the catalysis of the same sugar solution by an A1C1 3 solution containing 1/32 molecule per litre were 15 26 oo minutes A + 10-78 - 0-16 - 2-36 - 3'53. Assuming proportionality between the velocity-constant and the HC1 concentration, calculate the degree of hydrolysis of the A1C1 3 solution. Ans. 8 '03 per cent. PROBLEM 321. The partition-coefficient of the weak acid hydroxyazobenzene, C 6 H 5 N : N . C 6 H 4 OH, between benzene and water is 539. 1000 c.c. of an aqueous solution contain- ing O'Ol gram-equivalent of the acid and 0-012 gram-equiva- lent of Ba(OH) 9 was shaken with 60 c.c. of benzene. The concentration of the acid in the benzene layer was found to be 0-0537 gram in 50 c.c. Calculate the hydrolysis-constant of the barium salt and the degree of hydrolysis in a O'Ol equivalent N-solution of the pure salt. Ans. K H = 2-34 x 10 ~, *x = 1-53 per cent. PROBLEM 322 At 25 the partition-coefficient of ^?-nitrani- line between benzene and water is 9*0. 1000 c.c. of an aqueous solution containing "05035 equivalent of HC1 and 0-00693 equivalent of the base _p-nitraniline was shaken with 59 c.c. of benzene till equilibrium was established. 50 c.c. of the benzene solution then contained 0-2165 gram of -nitraniline. Calculate the hydrolysis-constant of ^-nitrani- line hydrochloride and the degree of hydrolysis in a 0*05035 N pure aqueous solution of the salt. Ans. K a = 10-67 x 10 - 2 , x = 74-2 per cent. PROBLEM 323. At 25 the solubility of cinnamic acid in water is 0'00331 gram-molecule per litre, and its dissociation- constant is 3-55 x 10 ~ 5 . The ionic product of water is 1-2 x 10 ~ 14 . The solubility of cinnamic acid in a solution containing O'Ol gram-molecule of aniline per litre is 0-00804 gram-molecule per litre. Assuming the unhydrolysed aniline cinnamate to be 93 per cent, dissociated, calculate the dis- sociation-constant of aniline. A , K.AO v, in - 10 GHAPTEE X ELECTROMOTIVE FORCE ELECTEODE POTENTIAL. NOEMAL POTENTIAL. CONCEN- TEATION CELLS. ELECTEOMOTIVE FOECE OF GALVANIC ELEMENTS. DIFFUSION POTENTIAL. OXIDATION-EE- DUCTION POTENTIAL. AFFINITY OE MAXIMUM WOEK OF A EEACTION IN A GALVANIC ELEMENT. GIBBS- HELMHOLTZ EQUATION Electrode Potential HPHE electromotive force of a metal electrode (i.e. one J- which furnishes only positive ions) against a solution of a salt of the metal is BT, P ET, C (i) ~^ l g^= -SF lo & c- Here e denotes the potential difference between the metal and the solution, and must be taken with its proper sign. It is positive if the metal is charged positively with respect to the solution and negative if the metal is negatively charged. P is the electrolytic solution pressure (or tension) of the metal, p the osmotic pressure of the metal ions in the solu- tio^ and C and c the concentrations of the metal ions in granafcions per litre, corresponding to the osmotic pressures P andbp. n is the valency of the metal ion or the number of gram-equivalents in a gram-ion. F is one faraday or 96540 coulombs, the charge carried by one gram-equivalent of any ion. T is the absolute temperature and E the gas- constant. When e is expressed in volts (i.e. when the unit of energy is the volt-coulomb or joule) the numerical value of E is 8-32. On changing from natural to common logarithms the above expression, therefore, becomes 149 150 NORMAL POTENTIAL 8-32 x 2-302 x T, P n x 96540 g ? 1-984 x lO- 4 x T, P 1-984 x 10-* x T, G - log n & c 2-032 x ET For a temperature of 18 C. = 291 absolute, is, therefore, 0-0577, and for 25 C. = 298 absolute, it is 0-0591. For ordinary temperatures the value 0'058 may be employed, so that we obtain 0-058, P 0-058, G (2) = - log = log - . n & p n & c For an electrode which furnishes only negative ions (e.g. C1 2 , Br 2 , etc.) the corresponding expression is 0-058, P 0-058, G (3) - + ^r log ^= + -ir log c- Normal Potential The electromotive force, e , of an electrode against a solu- tion which is normal with respect to the corresponding ions (i.e. contains 1 gram-ion per litre) is called the normal potential (or electrolytic potential) of the electrode substance. Thus for a metal electrode we obtain from (i), since c =^ 1, ET . log < 0-058 and, for any other concentration c, from (2) 0-058 (5) nF logo. E.M.F. of Concentration Cell The E.M.F. of a concentration cell of the form Metal is metal salt solution in which the concentra- tion of the metal ions is c metal salt solution in which the concentra- tion of the metal ions is c l Metal E ~ e - if the diffusion potential at the junction of the two solutions is neglected, e is the electrode potential at the junction ELECTEOMOTIVE FOECE 151 Metal/solution of ionic cone. c. is the electrode potential at the junction Metal/solution of ionic cono. Cj. From (i), therefore, or, at ordinary temceiature, from (2) ,, 0-058 , c E = log E.M.F. of Galvanic Element The electromotive force of a galvanic element of the form Metal M of j solution of salt valency n of metal M in which the cone, of the metal ions is c is, from (i) and (5), (7) E = e - ei RT, C and solution of salt of metal M. 1 in which the cone, of the metal ions is c l RT Metal valency of . log. C - e 01 - are the potential differences between the metals e and e 01 the an *! are e poen ncs ew M and M x and their respective solutions, and normal potentials of the metals M and M r Diffusion-potential The diffusion-potential at the junction of two solutions of different concentrations of the same binary electrolyte, com- posed- of two univalent ions, is 1 C -1 A RT < 8 > 6 = *7T7 c is the concentration of the ions in the concentrated solution and Cj that in the dilute solution. l c and 1 A are the ionic conductivities of the positive and negative ions respectively 152 OXIDATION-KEDUCTION POTENTIAL (p. 104). The sign of e is that of the potential of the dilute solution with respect to the concentrated solution. It should be noted that the potential of the dilute solution is of the same sign as the faster moving ion. At the junction of two solutions of different binary electro- lytes of the same ionic concentration, and when each electro- lyte gives only univalent ions, the diffusion-potential is (9) .- where l c and 1 A are the ionic conductivities of the ions of the one electrolyte, and l' c and l' A those of the other. Oxidation-reduction Potential Every oxidation process in which ions take part can be re- garded as a taking-up of positive charges or a giving-up of negative charges, and, conversely, every reduction as a giving-up of positive or a taking-up of negative charges. The oxidation of ferrous to ferric ions, for example, may be represented by the equation Fe + = Fe . If such a process takes place at an indifferent (unattack- able) electrode, a potential-difference between the electrode and the solution is developed. The magnitude of this potential- difference depends on the concentrations of the substances taking part in the reaction. Thus for the reaction mA + nB + nF = pC + qD, in which m molecules of A and n molecules of B are converted into p molecules of C and q molecules of D by taking up n faradays of positive electricity, the electrode-potential e is given by the equation The square brackets denote the concentrations of the en- closed substances. e is the normal potential of the electrode reaction. It is the potential of the electrode against the solution when all the substances A, B, C and D, which de- termine the potential, are present in unit concentration. In the numerator of the logarithmic expression are the concen- trations of those substances which are formed by the taking- up of positive charges (i.e. the higher state of oxidation). GIBBS- HELMHOLTZ EQUATION 153 Affinity of Reaction The affinity of a chemical reaction (p. 67) which takes place reversibly in a galvanic element is (n) A = nFE volt-coulombs or joules, where E is the electromotive force of the element in volts, F = 96540 coulombs, and n the number of faradays which flow through the cell during the transformation of the quanti- ties of the reacting substances given by the chemical equation representing the cell-reaction. Thus for the reaction 2H 2 + O 2 = 2H 2 n = 4, since 4 equivalents of hydrogen or of oxygen are transformed in the formation of 2 molecules of water and 1 faraday flows through the cell during the transformation of each equivalent. Since 1 joule = 0*2387 calorie, we obtain from (u) (12) A = nFE x 0-2387 = nE x 96540 x 0-2387 = nE x 23040 calories. Gibbs-Helmholtz Equation If A is the maximum work obtainable from a chemical reaction at absolute temperature T and Q is the heat evolved during the reaction (as measured in a calorimeter), then ac- cording to the Gibbs-Helmholtz equation, (13) A = Q + T^. -Tm is the temperature-coefficient of the maximum work, or the amount by which A is altered for a temperature-change of 1, and may be positive, negative or zero. A and Q should, of course, be expressed in the same unit of energy. From (n) and (13) we obtain from which the electromotive force E of a galvanic element at T absolute may be calculated from Q, the heat of the re- action taking place in the element, and the temperature- 154 ELECTKOMOTIVE FOEGE EXAMPLES J 777 coefficient of the electromotive force -^jp. To express E in volts, Q must be expressed in volt-coulombs. If Q is given in calories, then, since 0*2387 calorie = 1 volt-coulomb, Q + T^ Q 4-T^ nFx 0-2387 dT " n x 96540 x 02387 dT E.M.F. Examples PROBLEM 324. The normal potential of Cd is = - 0-420 volt and of Ag e 01 =- 0*798 volt, both referred to the N-JJ- electrode as zero. Neglecting the diffusion-potential at the junction of the two electrolytes, calculate the electromotive force E of the cell 2%--- Cd | c = 0-5 mol. N-Cd(N0 3 ) 2 | c, = 0-1 N-AgNO 3 | Ag at 18, given that the degree of dissociation of the Cd(NO 3 ) 2 solution is a = 0'48 and of the AgN0 3 solution a l = 0-81. SOLUTION 324. The concentration of the Cd --ions in the c = 0*5 mol. N-solution is ac = 0-48 x 0'5 gram-ion per litre. The concentration of the Ag --ions in the c l = 0-1 N-AgN0 3 solution is a l c 1 = 0-81 x O'l gram-ion per litre. From (7), therefore, RT. E = + lo ' ac ~ e and, since n for Cd = 2 and n l for Ag = 1, we obtain at 18 E = - 0-420 + 0-029 log (0-48 x 0-5) - 0*798 - 0-058 log (0-81 x 0-1) = - 0-420 + (0-029 x -0-6198) -0-798- (0-058 x - 1-091) = - 0-420 - 0-018 - 0-798 + 0-063 = - 1*173 volt. The Cd is, therefore, the negative pole, or, in the cell the current flows from the Cd to the Ag. PROBLEM 325. The electromotive force of the cell Hg Hg 2 Cl 2 in N-KC1 HgOin N-NaOH Hg is 0-168 volt at 18. The degree of dissociation of N-KC1 and of N-NaOH is 0-72. The ionic conductivity of K- is Z c = 64-9, ELECTKOMOTIVB FORCE EXAMPLES 155 of OF 1 A = 65-4, of Na- l' a = 43-6 and of OH' l\ = 174 at 18. Taking the concentration of the Hg '-ions in a saturated solution of Hg 2 Cl 2 in N-KG1 (normal calomel electrode) as c 3 x 10 ~ 20 gram-ion per litre, calculate the solubility- product of Hg(OH) 2 at 18. SOLUTION 325. According to (9) the diffusion-potential at the junction of the two electrolytes KC1 and NaOH is ;...- 0,58 , og = 0,58 lo g = 0-058 log gjjj = 0-0198 volt. Since OH' is by far the fastest moving ion the KC1 solution will be negative with respect to the NaOH solution, and the diffusion-potential must, therefore, be subtracted from the total electromotive force of the cell to obtain the electromotive force of the Hg concentration-cell alone. The latter is, therefore, E = 0-168 - 0-0198 = 0-1482 volt. . If c l is the concentration of the Hg - --ions in the NaOH solution saturated with HgO, we obtain from (6), since n = 2, ^OJ^c 3x10-^ C l C l .-. logCi = 2eT-684 and ^ = 4-83 x 10 ~ 2(5 gram-ion per litre. The solubility-product of Hg(OH) 2 is (p. 121) L = [Hg--][OH'f. In N-NaOH saturated with HgO we have just found [Hg ] = Cj = 4-83 x 10 ~ 26 , and, since the degree of dis- sociation of the N-NaOH is 0*72, the concentration of the OH'-ions is 1 x 0'72, or [OH'J = 0'72. Therefore L = (4-83 x 10 - 26 ) (0-72) 2 = 2-5 x 10 - 26 . PROBLEM 326. The E.M.F. of the cell Ag | c = 0-01 N-AgN0 3 c i = ' 001 N-AgNO 3 | Ag is E = 0*0579 volt at 25. Assuming that the interposition of saturated NH 4 N0 3 solution completely eliminates the 156 ELECTEOMOTIVE FOKCE EXAMPLES diffusion-potential, and that the 0*001 N-AgN0 3 solution is completely dissociated, calculate the degree of dissociation and the concentration of the Ag --ions in the 0*01 N-AgN0 3 solution. SOLUTION 326. If a is the degree of dissociation of the c = 0-01 N-solution, the concentration of the Ag --ions is ac. Since- the c l = 0*001 N-solution is completely dissociated, the concentration of the Ag --ions in it is c r For the E.M.F. of the Ag concentration-cell we therefore obtain from (6) ET, and, since n = 1, at 25 C 0-0579 = 0-059 log - x 0-01 0-001 /. a = 0-958. The concentration of the Ag --ions in the c = O'Ol N-solu- tion is [Ag '] = ac 0*958 x O'Ol = 0*00958 gram-ion per litre. PROBLEM 327. The normal potential of silver is e Ag = + 0*771 volt (hydrogen standard), that of chlorine at atmo- spheric pressure is C12 = + 1-366 volt, and that of bromine at atmospheric pressure is e Br2 = + 0'99 volt. The solubility- product of AgCl at 25 is L^ = 2 x 10 ~ 10 , and that of AgBr L 2 = 2 x 10 ~ 13 . What is the affinity of silver, in joules and calories, (a) to chlorine under atmospheric pressure, (b) to bromine-vapour under atmospheric pressure ? SOLUTION 327. (a) The affinity of silver to chlorine at atmospheric pressure is denned as the maximum work which can be gained by the reversible combination of 1 molecule of chlorine at atmospheric pressure with 2 equivalents of metallic silver to form two molecules of solid silver chloride, according to the equation C1 2 + 2Ag = 2AgCl. This reaction proceeds reversibly, and therefore yields the maximum work, in a galvanic element of the type Ag saturated solution of AgCl Pt charged with C1 2 at atmospheric pressure. When the element is in action, Ag goes into solution as Ag'-ions at one electrode and C1 8 as Cl'-ions at the other, and ELECTROMOTIVE FORCE EXAMPLES 157 solid AgCl is formed. If E denotes the E.M.F. of fcho ele- ment, then by the solution of 1 molecule of C1 2 and 2 equi- valents of Ag the electrical energy A = 2# x 96540 joules is produced. E can be calculated from the two separate electrode potentials, e x between the silver electrode and the solution and e 2 between the chlorine electrode and the solu- tion. From (i) and (3) RT, [Ag-] - + * and -' C Ag and C C12 are the concentrations of the Ag - and Cl'-ions corresponding to the electrolytic solution pressures of silver and of chlorine at atmospheric pressure. The square brackets, as usual, denote the concentrations of the ions at the electrodes. Since in both cases n = 1, we obtain from (7) ET [Ag-] ET C C12 ET ET ET = -jr log, [Ag ] [Cl'J - -p log. M - -jr log, C C12 e Ag - C12 . The product [Ag ] [01'] is equal to the solubility-product L 1 of silver chloride, since the solution is saturated with AgCl. Therefore ET E = -p log e L 1 + e^ - C12 and, at 25, E = 0-059 log 2 x 10- 10 + 0-771 - 1-366 = 0-059 x - 9-699 - 0-595 = - 1-167 volt. The negative sign means that the silver electrode, which in the calculation was taken as positive ( + ej, is the negative pole. In the calculation of A the sign of E is of no conse- quence ; it is only the absolute value that is required. The affinity of silver to chlorine at atmospheric pressure is, therefore, 158 ELECTKOMOTIVE FORCE EXAMPLES A = 2 x 1-167 x 96540 = 225300 joules or, in heat units, A = 2 x 1-167 x 23040 = 53760 calories. (b) For the affinity of silver to bromine-vapour at atmo- spheric pressure we have the corresponding equation A = ZE x 96540 joules, where E is the E.M.F. of the galvanic element Ag solution saturated with AgBr Pt charged with Br- vapour at atmospheric pressure. As above 7? *~.1 T ^ = ~ 10 g* L 2 + Ag ~ *Br 2 = 0-059 log 2 x 10 - 13 + 0-771 - 0-99 = 0-969 volt, and A = 2 x 0-969 x 96540 = 187100 joules. = 2 x 0-969 x 23040 = 44640 calories. PBOBLEM 328. The solubility of silver chloride in water at 25 0. is S = 1-4 x 10 ~ 5 gram-molecules per litre. The solubility in ammonia solutions is greater, and is proportional to the total concentration of the ammonia. The factor of proportionality is k = 0*05. The increased solubility in ammonia depends on the formation of a complex ion accord- ing to the scheme Ag -f nNH 3 = Ag(NH 3 ) n -. How many molecules of ammonia take part in the formation of this com- plex cation, i.e. what is the value of n, and what is the t dis- sociation-constant K = nf AJJJ \ 3 .-| f tne complex ion ? SOLUTION 328. In any solution which is saturated with silver chloride the product of the concentrations of the ions, [Ag*] [OF], must, for a given temperature, have a definite constant value L, which is called the solubility-product of silver chloride. In a saturated aqueous solution which con- tains only silver chloride, [Ag ] = [01'] = S, and, therefore, L = S 2 , it we assume that the dissociation of the dissolved salt is complete. The greater solubility of silver chloride in solutions of ammonia compared with pure water is caused by ELECTROMOTIVE FORCE EXAMPLES 159 the formation of a complex cation by the addition of ammonia - to the Ag '-ions according to the scheme Ag- + wNH 3 = Ag(NH 3 ) n -; this diminishes the concentration of the silver ions, and, therefore, requires an increase in the concentration of the chlorine ions, so that the value of the solubility-product may be maintained. The application of the law of mass-action to the formation of the complex cation leads to the equilibrium- equation [Ag-][NH 3 ]- _ w [Ag(NH 3 ) n -] = n must, of course, be a whole number. The total concentration of silver in the solution, a, is equal to the sum of the concentrations of the Ag'-ions and the Ag(NH 3 ) n -ions, if we assume that the salts AgCl and the Ag(NH 3 ) n Cl are practically completely dissociated. Accord- ing to the conditions of the problem, the solubility of AgCl in the ammonia solution is (2) a = [Ag ] + [Ag(NH 8 ) n -] = fc(NH 3 ). Here (NH 3 ) denotes the total concentration of ammonia in the solution, and, therefore, (3) (NH 3 ) _ [NHJ + [NH 4 -] + [A g (NH 3 ) n -], since the total ammonia in the solution is equal to the sum of the undissociated ammonia, the ammonium ions and the ammonia in the complex ion. From (2) we therefore obtain (4) [Ag-] + [Ag(NH 3 ) n -] = When the concentration of ammonia is sufficiently great, the concentration of the silver ions, [Ag*], which is always less than , that is, less than 1'25 x 10 ~ 5 , may be neglected in comparison with the total concentration of silver a. Under these circumstances we obtain from (2) a= [Ag(NH 3 ) M -], and hence from (4) a = &[NH 3 ] + &[NH 4 ] + Jena, a(l - kn) =. M[NH 8 ] + [NH 4 ]} and 160 ELECTROMOTIVE FORCE EXAMPLES By dividing the equation [A g -][cr] = i by equation (1) written in the form [Ag.][NH 3 ]* _ a ' we obtain L "K' Every molecule of AgCl which dissolves produces one Ag(NH 3 ) n --ion and one Cl'-ion. The concentrations of these ions are, therefore, equal and [01'] = a, and (6) becomes w - and, therefore, (7) a- When equations (5) and (7) are compared, it is evident, since -, 7 and A-^ are constants, that for different con- 1 Kn \ K centrations of ammonia the two equations can be simultane- ously satisfied only (1) if the concentration of the ammonium ions, [NH 4 ], is negligible compared with the concentration of the undissociated ammonia molecules, [NHJ, a condition which is satisfied in the case of weak bases like ammonia, and (2) if n = 2. From (5) and (7) we then obtain and 2fr 2 1-4 2 x 10 - 10 x k* (O05) 2 = 5-4 x 10 ~ 8 . The silver-ammonia complex ion has, therefore, the com- position Ag(NH 3 ) 2 , and its dissociation -constant at 25 C. is 54 x 10 ~ 8 . ELECTROMOTIVE FORCE EXAMPLES 161 PKOBLEM 329 (cf. preceding problem). What are the E.M.F.'s, E l and E 2 of the cells Ag/c N-AgNO 3 /c' N-KC1, AgCl/Ag and c N-AgNO C N-NH c' N-KC1, AgCl Ag if o = 0-01, c' = 1 and C = 1 ? SOLUTION 329. According to Nernst's formula (equation (6)) the E.M.F. of a concentration cell is nF =' [Me-] 2 ' if the diffusion-potential at the point of contact of the two electrolytes, in which the electrodes are placed, is neglected. n is the valency of the cation, [Me*^ and [Me ] 2 the concentra- tions of the cation at the two electrodes. The cells described in the problem are to be regarded as concentration cells with respect to the Ag cation, and, since n = 1 in this case, the formula becomes The concentrations of the silver ions, [Ag ], in the various solutions may be calculated as follows. In the c N-AgNO 3 solution the silver ion concentration, [Ag 'Jj, is approximately equal to the total concentration, since the salt may be regarded as practically completely dissociated, .-.[Ag ], = <;. In c' N-KC1 solution, which is saturated with AgCl, the product of the silver ion concentration, [Ag -]. 2 , and the chlorine ion concentration, [OF], must be equal to L, the solubility-product of AgCl (cf. problem 328). If the KC1 is regarded as completely dissociated, [Cl'J = c'. We obtain, therefore, [Ag -] 2 rCl'] = L, and, for the E.M.F. of the first cell at 25 C., BT, cc' n 162 ELECTROMOTIVE FORCE EXAMPLES P - 0-069 log ax g. 10 - 0-454 volt. The concentration of the silver ions, [Ag ], in the C N -ammonia solution follows from the equation (cf. problem 328) a) [Ag-J[NHJ_ g [Ag(NH 3 ) 2 -J Since K has a relatively small value (5*4 x 10 ~ 8 ), the complex-formation between the silver ions and the excess of ammonia may be regarded as practically complete, i.e. practically the whole of the silver in the solution is present as the complex ion Ag(NH 3 ) 2 ', .-. [Ag(NH 3 ) 2 -] = c. The concentration of the free ammonia is, therefore, [NHJ = C-2o, and, from equation (1), the concentration of the silver ions is For .Eg, the E.M.F. of the second cell, we therefore obtain = 0*0264 volt. The addition of C = 1 N-ammonia to the 0-01 N-AgNO 3 solution has, therefore, diminished the E.M.F. by 0'454 - 0-0264 - 0-428 volt. The sign of the E.M.F., however, re- mains unchanged. PROBLEM 330. The E.M.F. of the hydrogen- oxygen cell at 25 is E l = 1-23 volt, and the E.M.F. of the cell Ag/saturated solution of Ag 2 O/H 2 is E< 2 = 1-18 volt. What is the affinity of silver to oxygen under atmospheric pressure (a) at 25 and (b) at 35, if the molecular heat of formation of Ag 2 O is Q = 6400 cals. ? SOLUTION 330.- -(a) The E.M.F. E of any galvanic element ELECTROMOTIVE FORCE EXAMPLES 163 which works reversibly is a measure of the affinity A of the reaction which produces the current ; for A = nE x 96540 volt-coulombs, if n denotes the number of equivalents of ions which enter, or are deposited from, the solution during the transformation of the quantities of the reacting substances given by the equation representing the cell-reaction. The E.M.R of the hydrogen- oxygen cell H 2 /any aqueous solution/O 2 is, therefore, a measure of the affinity of the reaction (1) 2H 2 + 2 = 2H 2 0, and the E.M.F. of the cell H 2 /saturated solution of Ag 2 O/Ag is a measure of the affinity of the reaction (2) H 2 + Ag 2 O = H 2 O + 2Ag. By multiplying (2) by 2 and subtracting from (1) we obtain (3) 2 + 4Ag = 2Ag 2 0, or, for the affinities, A 3 = AI *A or, collecting the logarithmic terms on one side and dividing throughout by RTjF, JL(l + x) inc e c'(2 + 3x)xjL BT/F' 166 ELECTEOMOTIVB FOECE EXAMPLES e = 2-718 is the base of the natural logarithms. The right-hand side of equation (4) contains only known quanti- ties. If, for brevity, we call its value A t we obtain from (4) 1 + x = #(2 + 3x)A = 2Ax + 3Ax 2 , 6A ' \V~ 64 ; ' 34' The value of A is obtained from the equation eO-ei />' 0-748-02^3 10-3 A = e -WF x - = (2-718) 0.059 x J _ c 1 = (2-718) 18 x ID- 3 = 6-56 x 10 4 . Since A is very large, x must, according to equation (4) be very small. If in this equation we neglect the term in x 2 as small compared with x, we obtain approximately The reduction of the ferric chloride to ferrous chloride is, therefore, practically complete. PROBLEM 332. With potassium iodide mercuric ions form the complex ion HgI 4 ". The dissociation-constant of the complex ion is K = fe ^ j L = 5 x 10~ 81 . In presence of metallic mercury equilibrium between mercurous and mercuric ions is reached when I & 2 j = JT, = 120. (1) How [ H g ' J many grams of mercurous iodide dissolve on shaking solid mercurous iodide with a litre of c = 1 N-potassium iodide solution, if the solubility of Hg 2 I 2 is a = 3'1 x 10~ 10 gram- molecule per litre? (2) What is the concentration of a KI ELECTROMOTIVE FORCE EXAMPLES 167 solution which dissolves equal weights (in grams) of Hg in the form of mercurous and mercuric salts ? SOLUTION 332. (1) In any solution which is saturated with mercurous iodide the equation [Hg 2 ] [17 = L must be satisfied, where L is the solubility-product of Hg 2 I 2 . In a pure saturated aqueous solution of Hg 2 I. ; , in which the salt may be regarded as completely dissociated, [Hg 2 - ] = a, and [I'] = 2a, (1) .-. [Hg 2 .-][IT = 4a3 = . Mercurous ions react in presence of metallic mercury accord- ing to the equation Hg 2 * = Hg + Hg , until equilibrium is reached when -_( 2 )gtq = *, If the solution contains potassium iodide, the potassium iodide reacts with the mercuric ions to form the complex ion HgI 4 ", according to the equation The equilibrium equation for this reaction is, therefore, In the case of the potassium iodide solution which is satu- rated with mercurous iodid , equations (1), (2) and (3) must be simultaneously satisfied. In these equations [Hg 2 ], [Hg ], [HgI 4 "J and [I'] are unknown. An additional equa- tion is, therefore, required for their solution. This is obtained as follows from the initial concentration of KI, which is known. If the concentration of the undissociated molecules is neglected, i.e. if all the salts are regarded as completely dissociated, the total iodine in the solution is equal to the sum of the iodine present as I'- and HgI 4 "-ions. Of the four iodine atoms contained in HgI 4 ", only two are derived from the dissolved Hg 2 I 2 , and the other two from the KI originally present. The concentration of the free iodine ions is, there- fore, (4) [I'] = c - 2[HgI 4 "]. By dividing (1) by (2) we obtain 168 ELECTROMOTIVE FORCE EXAMPLES (5) [Hg ] [IT = |, and from (3) and (5) (6) ^[rp- From (6) and (4) we obtain (7) (7) is a quadratic equation containing only one unknown quantity [I'J, which may, therefore, be calculated : , 8) rri _ KK, + According to (7) If x grams of mercurous iodide dissolve, and if M is the 2; molecular weight of Hg 2 I 2 , there are formed equivalents of HgI 4 ", if the concentrations of the Hg 2 - - and Hg -ions in the concentrated potassium iodide solution are neglected. From (7) we therefore obtain (9) x = ^ ^ l ^ grams. Substituting the numerical values in (8) we obtain rri = - 5 x io~ ai x 120 4 x 4 x (3-1 x 10 " 10 ) 3 (5 x 10 ~ 31 x 12Q) 2 5 x 10- 31 x 120 x 1 16 x 16 x (3-1 x 10 - 10 ) 6 2 x 4 x (3'1 x 10 ~ lo f = - 0-126 + V(0-126) 2 + U = 0-392, and from (9) = 554(1 - 0-392) = 4 (2) The ratio of mercurous to mercuric ions is always K-i= 120. [Hg- ], may, therefore, be neglected in compari- ELECTEOMOTIVE FORCE EXAMPLES 169 son with [Hg 2 * ], and the whole of the mercuric mercury may be regarded as contained in the complex ion. If, therefore, equal weights of mercury dissolve in the mercuric and mer- curous forms, the weight of mercury in the mercuric complex HgI 4 " must be equal to that in the mercurous ions, or (10) [HgI 4 "] = From (3) and (10) [Hg ][!']* = and from this equation and (2) [I']' = From (10) and (1) we obtain = 2-18 x 10 ~ 14 , and from (4) c = [I'] + 2[HgI 4 "] = 740 x 10 ~ s + 2-18 x 10 ~ 14 = 7*40 x io~ 8 N. In all potassium iodide solutions of measurable concentration, therefore, the mercurous iodide dissolves chiefly as complex mercuric salt. PROBLEM 333. The normal potential of a copper electrode against cupric ions is e = + 0*329 volt, the constant for the equilibrium between cupric and cuprous ions in presence of metallic copper is [Cu -]' 2 /[Cu ] = K = 0-6 x 10 ~ 3 and the solubilities of cuprous chloride and bromide are a = 1-1 x 10 " 3 and b = 2*04 x 10 ~ 4 gram-molecules per litre respec- tively. What is the potential of a copper electrode (1) in a c = 0-1 N - HC1 solution which is saturated with cuprous chloride, and (2) in a c = 0-1 N - HBr solution which is saturated with cuprous bromide ? (3) What is the normal potential e' of a copper electrode against cuprous ions ? SOLUTION 333. (1) and (2). In the solution which is sat- urated with CuCl, and which at the same time is in equi- librium with metallic copper, let the concentrations of the cupric (Cu ) and cuprous (Cu ) ions be x and x respectively, and in the corresponding CuBr solution let the concentrations 170 ELECTKOMOTIVE FOECE EXAMPLES of the cupric and cuprous ions be y and y respectively. Then the potential of a copper electrode, regarded as a cupric electrode, against the CuCl solution is (1) l = c + ~log e x, and against the CuBr solution pm (2)c 2 = Pb , localised at the negative electrode, and the other, Pb-----Pb-- f localised at the positive electrode. The potential difference ELECTEOMOTIVE FORCE EXAMPLES 173 between the two electrodes must be equal to the difference of the single potential differences between the separate elec- trodes and the solution, therefore, from (5) and (10), (3) = c + log, [Pb ' '] - o The lead peroxide electrode is here regarded as an oxidation electrode. E, e and e' are given ; the required concentra- tion of plumbic ions, [Pb ''], may, therefore, be determined, if [Pb ] can be calculated. This can be done from the given solubility, a, of lead sulphate, with which the solution is saturated. In a saturated pure aqueous solution of lead sulphate, [Pb ] [SO 4 "] = L = a 2 , if we assume that the dissociation is complete.* In molecular N-H 2 S0 4 the concentration of the SO 4 "- ions would be [S0 4 "] = 1, if the dissociation of the acid were complete ; if the degree of dissociation is assumed to be approximately 50 per cent., [SO/'] = 0-5 and [Pb ] = _L Therefore, from (3), and, changing to common logarithms, The sign of E in the above equation must be taken as negative, because, according to (3), E is the potential differ- ence between the lead electrode and the lead peroxide electrode, and as the lead electrode is the negative pole, E = - 1-88. - 2 x 7-5 = - 16-35 .'. [Pb ---- ] = 4*5 x 10 ~ 17 gram-ion per litre. (2) During the discharge the concentration of the sulphuric acid diminishes, and, according to reaction (1), to such an * This assumption is, however, only approximately correct. 174 ELECTEOMOTIVE FOECE EXAMPLES extent that for every equivalent of lead which goes into solution at the negative electrode one molecule of H 2 S0 4 is deposited as PbSO 4 . By a discharge of one ampere-hour, therefore, c = 0-037 molecules of H 2 S0 4 are so deposited yoo Ttvj and removed from the solution. The 100 c.c. of molecular N-H 2 SO 4 , which originally contained Ol molecule H 2 SO 4 , contain after the discharge only Ol - 0*037 = 0-063 mole- cules. The concentration of the H 2 S0 4 is, therefore, only 0-63 molecular normal after the discharge. If we assume that the degree of dissociation of this acid is, approximately, 60 per cent., the concentration of the SO 4 "-ions is [SO/'] = 0-63 x 0-6 = 0-38, and the concentration of the Pb --ions [Pb ] = a . 0*38 The concentration of the Pb ---- -ions also changes during the discharge with the change in concentration of the acid. In the solution, which is saturated with lead peroxide, the equation must always be satisfied, where L 1 denotes the solubility- product of plumbic hydroxide, and, since [H'][OH'J = K lt we obtain In two solutions, which we may distinguish by the suffixes 1 and 2, we have, therefore, [Pb-- ],_[!!], ], [H-y Before the discharge the H'-ion concentration of the H 2 S0 4 is [H-] t = 2 x 0*5 = 1, if we assume the dissociation to take place only according to the equation H 2 SO 4 = 2H- + SO 4 " ; after the discharge it is [H-] 2 = 2 x O38 = 0'76. Therefore = 4-5 x 10 ~ 17 x 0-33 = 1-5 x 10 ~ 17 gram-ion per litre. (3) The E.M.F. of the accumulator after the discharge follows from the equation (cf. (3)) ELECTEOMOTIVE FOECE PROBLEMS 175 r 2P A ' 2F 1 [PFT ^ = ' " + W lo & [ pb ""I- ^ lo & [Pb ] = 1-80 + 0-12 + 0-0295 log 1-5 x 10 - 17 - 0-059 log 1>58 x 1Q ~ 8 0*38 = 1-92 - 0-496 + 0-436 = r86 volt. E.M.F. Problems for Solution PROBLEM 335. The equivalent conductivity at 18 of a - 02 equivalent N-ZnCl 2 solution is 94 r.o., and the equiva- lent conductivity of ZnCl 2 at infinite dilution is 112 r.o. The normal potential of Zn (referred to the N-H- electrode as zero) is - 0-770 volt. What is the potential of Zn against a 0-02 equivalent N-ZnCl 2 solution at 18? Ans. - 0-830 volt. PROBLEM 336. At 25 the E.M.F. of the cell Zn/0-5 mol. N-ZnSO 4 /0-05 mol. N-ZnSO 4 /Zn is 0-018 volt. Neglecting the diffusion-potential at the junc- tion of the electrolytes, and assuming the dilute solution to be 35 per cent, dissociated, calculate the degree of dissociation of the concentrated solution. Ans. 0-142. PROBLEM 337. The E.M.F. of the cell Ag/0-1 N-AgNO 3 /saturated NH 4 NO 3 /0'01 N-AgN0 3 /Ag is 0-0556 volt at 25. At 25 the specific conductivity of 0-1 N-AgNO 3 is 109-3 x 10 ~ 4 r.o., and of O'Ol N-AgNO 3 12-53 x 10 ~ 4 r.o. On the assumption that the conductivity is a true measure of the ionic concentration, calculate the E.M.F. of the Ag concentration -cell given above, and com- pare with the value found. Ans. Calculated value = 0-0555 volt. PROBLEM 338. The E.M.F. of a Daniell cell in which the Cu - and Zn --ion concentrations are equal is 1-1 volt at 18. What is the E.M.F. at 18 of a Daniell cell in which 176 ELECTROMOTIVE FORCE PEOBLEMS the Cu --ion concentration is 0-0005 and the Zn centration is 0-5 gram-ion per litre ? Ans. 1-013 volt. PROBLEM 339. At 25 the E.M.F. of the cell -ion con- Pb 0.01 mol. N-Pb(NO 3 ) 2 saturated NH 4 N0 3 is - 0-469 volt, and that of the cell Pb 001' mol. N-Pb(C10 3 ) 2 N-calomel electrode N-calomel electrode is - 0-463 volt. The NH 4 NO 3 solution eliminates the dif- fusion-potentials. If the Pb(C10 3 ) 2 is assumed to be com- pletely dissociated, what is the degree of dissociation of the Pb(N0 3 ) 2 ? Ans. 62*1 per cent. PROBLEM 340. The E.M.F. of the cell Pb 0-01 mol. N-Pb(N0 3 ) 2 NH 4 N0 3 is - 0-469 volt at 25. The Pb(NO 3 ) 2 is 62 per cent, dissoci- ated. What is the normal potential of Pb referred to the N-calomel electrode as zero ? Ans. - 0-405 volt. PROBLEM 341. At 25 the E.M.F. of the cell saturated N-calomel electrode * Ag f\ i XT A AT r\ 0-lN-AgN0 0-1 N-calomel electrode is 0-396 volt, and that of the cell saturated solution of Ag(C 2 H 3 2 ) saturated O'l N-calomel NH 4 NO 3 electrode is 0-383 volt. 0-1 N-AgNO 3 is 82 percent, dissociated. Calculate (a) the Ag - ion concentration in the saturated solution of silver acetate and (6) the degree of dissociation of the saturated solution, given that the solubility of the silver acetate at 25 is 0-0664 gram-molecule per litre. Ans. (a) 0-04938 gram-ion per litre, (b) 0'7435. PROBLEM 342. The E.M.F. of the cell Ag/AgCNS in 0-1 N-KCNS/AgBr in 0-1 N-KBr/Ag ELECTROMOTIVE FORCE PROBLEMS 177 is 0'015 volt at 25. The solubility of AgBr in pure aqueous solution is 7'2 x 10~ 7 gram- molecule per litre. Neglecting the diffusion-potential and assuming that all the salts are completely dissociated, calculate (a) the solubility- product and (b) the solubility of AgCNS at 25. Ans. (a) 0-93 x 10- 12 , (b) 0;96 x 10~ 6 gram-molecule per litre. PROBLEM 343. The E.M.F. of the cell N - AgNOi Ag is - 0-198 volt at 25. Neglecting the diffusion-potential, and taking the 01 N-AgNO 3 as 83 per cent, dissociated and the O'Ol N -K 2 C a 4 as completely dissociated, calculate (a) the solubility-product and (b) the solubility of silver oxalate. Ans. (a) 1-3 x 10~ n , (b) 1-48 x 10- 4 gram-molecule per litre. PROBLEM 344. What is the E.M.F. at 25 of the cell H^O-S N-HC1/01 N-NaOH/H 2 if the H 2 at each electrode is under atmospheric pressure and if the diffusion-potential is neglected? The degree of dissociation of 0'5 N-HC1 = 0-87, of 01 N-NaOH = 0'9 and the ionic-product of water is 1*2 x 10~ 14 . Ans. 0-738 volt. PROBLEM 345. What is the E.M.F. at 25 of the cell H 2 /0-5 N-fonnic acid/1 N-acetic acid/H 2 if the diffusion-potential is neglected? The dissociation- constant of formic acid is 127 x 10 - 5 and of acetic acid 1-8 x 10- 5 . Ans. 0-0451 volt. PROBLEM 346. The E.M.F. of the cell A I AgCl in Ag 01 N-KC1 saturated NH 4 NO 3 01 N-AgNO, Ag is - 0-450 volt at 25. 01 N-KC1 is 85 per cent, dissoci- ated and 0-1 N-AgNO 3 82 per cent. Calculate (a) the solubility-product and (b) the solubility of AgCl. 12 178 ELECTROMOTIVE FORCE PROBLEMS ^ Ans. (a) 1-65 x 10 - 10 , (b) 1-28 x 10 - 5 gram-molecule per litre. PROBLEM 347. The E.M.F. of the cell Cu/CuS0 4 + 100 H 2 O/ZnS0 4 + 100 H 2 O/Zn is 1-0960 volt at C. and 1-0961 volt at 3 C. What is the heat of the reaction taking place in the cell ? Ans. 63390 cals. PROBLEM 348. The E.M.F. of the combination Cl 2 /molten PbCl 2 /Pb is given by the formula E = 1-263 - [0-000679 (t - 498)] volts, where t is the temperature C. Calculate the heat of formation of lead chloride at 498 C. Ans. 82500 cals. PROBLEM 349. At 18 the potential of a Cu electrode against a 0'005 mol. N-Cu(NO 3 ) 2 solution is 0'266 volt, referred to the N-H- electrode as zero. Assuming that the Cu(NO 3 ) 2 solution is completely dissociated, calculate the normal potential of Cu against Cu --ions. Ans. 0-333 volt. PROBLEM 350. The normal potential of Zn referred to the N-H- electrode as zero is - 0*770 volt and of Cu 0-329 volt. When excess of Zn is added to a solution of CuSO 4 the Zn displaces the Cu till equilibrium is reached. What is the ratio of the concentration of the Zn - to the Cu *-ions at equilibrium ? (At equilibrium the potentials of the Zn and Cu against the solution are equal.) Ans. [Zn -]/[Cu ] = 7'94 x 10 37 . PROBLEM 351. The normal potential of Ag referred to the N-calomel electrode as zero is 0-488 volt at 18. Taking the absolute potential of the calomel electrode as 0-56 volt (Hg positive), calculate the electrolytic solution pressure of Ag in atmospheres. Ans. 2-04 x 10 ~ 17 . PROBLEM 352. When metallic copper is shaken with a solution of a cupric salt in absence of air at 20, the reaction BLECTEOMOTIVE FOKCB PKOBLEMS 179 Cu + Cu = 2 Cu proceeds till equilibrium is established. The concentrations of the Cu - and Cu --ions are then such that [Cu -]/[Cu -] 2 = 2-02 x 10 4 . If the normal potential of Cu against Cu --ions is 0*329 volt (hydrogen standard), what is the normal potential of Cu against Cu '-ions ? Ans. 0-454 volt. PEOBLEM 353. When CuCl is dissolved in KC1 solutions a complex salt of the formula (KCl) n *(CuCl) 1D is formed, which dissociates giving the ions K and Cu m Cl' m + u . |Two very dilute solutions of CuCl in 0*1 N-KC1 were prepared, the one containing 4 times as much CuCl as the other. The E.M.F. of the cell Cu dilute solution of CuCl in KC1 cone, solution of CuCl in KC1 Cu was 0-0351 volt at 18. On the assumption that practically the whole of the dissolved CuCl is present as complex salt, and that, on account of the very large excess of KC1 over CuCl, the Cl'-ion concentration in the two solutions is the same, and that the dissociation of the KC1 and of the com plex salt is complete, calculate the value of m. Ans. 0-995 (.-. 1, as it must be a whole number). PROBLEM 354 (cf. preceding problem). Two solutions of KC1, one 0-21 N and the other O'l N, and each containing 0-0002 gram-molecule CuCl per litre, were prepared. The E.M.F. of the cell Cu/dilute KC1 solution/cone. KC1 solution/Cu was 0*0374 volt at 18. Taking the value of m found above and on the assumptions made in the preceding problem, cal- culate the value of n. Ans. ^L- n - 2, .-. n - 1. m PROBLEM 355.- What is the total E.M.F at 18 of the eel Hj/0-1 N-HCl/0-001 N-HC1/H 2 if the H 2 at each electrode is at atmospheric pressure? The 0-1 N-HC1 is 92 per cent, dissociated and the O'OOl N-HC1 is completely dissociated. The ionic conductivity of H- is 318 and of Cl' 65-4. Ans. 0-039 volt, 180 ELECTKOMOTIVE FORCE PEOBLEMS PROBLEM 356. A saturated solution of AgN0 2 contains 0-0265 gram-molecule per litre at 25. The E.M.F. of the cell Ag AgN0 3 solution containing 0-0224 gram-ion Ag per litre of AgN0 2 saturated solution Ag is 0-011 volt at 25. Calculate (a) the solubility - product of AgNO 2 , (b) the degree of dissociation of the saturated solution. Ans. (a) 2-13 x 10 ~ 4 , (b) 0'551. PROBLEM 357. When AgNO 2 is dissolved in KNO 2 solu- tions the complex anion Ag m (NO' 2 ) u is formed. The E.M.F. at 25 of the cell 0-584 N-KNO 0-05 N-AgNO 0-584 N-KNO^ 025 N-AgN0 2 is 0-0170 volt, and of the cell Ag 0-379 N-KN0 2 0-025 N-AgNO 0-219 N-KN0 2 I A 0-025 N-AgN0 2 I Ag is - 0-0295 volt. Assuming that practically the whole of the dissolved AgN0 2 exists as complex ion, and that the NO' 2 -io Q concentrations are proportional to the KNO 2 con- centrations, calculate the values of m and n and hence the formula of the complex ion. Ans. m = 1-04(.'. 1), ~ = 2-1, .-. n = 2 and formula of com- plex ion is Ag(NO 2 ) 2 . PROBLEM 358. What is the diffusion-potential at 18 at the junctions (a) 0-001 N-HCl/0-001 N-KC1 and (b) 0-001 N-KCl/0-001 N-KOH, if all the electrolytes are assumed to be completely dissoci- ated ? The ionic conductivities of H-, K *, 01' and OH' are 318, 64-9, 65-4 and 174 respectively at 18. Ans. (a) 0-0272 volt (HC1 negative), (b) 0'0153 volt (KC1 negative). PROBLEM 359 (cf. preceding problem). What is the total E,M.F. at 16 of the cell ELECTROMOTIVE FORCE PROBLEMS 181 H 2 /0-001 N-HCl/0-001 N-KCl/0-001 N-KOH/H 2 if the H 2 at each electrode is under atmospheric pressure and all the electrolytes are completely dissociated ? The ionic product of water is 1'2 x 10 ~ 14 . Ans. 04199 volt. PROBLEM 360. The E.M.F. of the cell Cu/N-CuS0 4 /N-ZnS0 4 /Zn is ri volt at 18. What is the maximum work (a) in joules, (b) in calories obtainable at 18 by the reversible displacement of Cu by Zn according to the equation CuS0 4 + Zn = ZnS0 4 + Cu? Ans. (a) 212300, (b) 50670. PROBLEM 361. The E.M.F. at 25 of the cell Pt 0-0337 mol. N-T1(N0 3 ) 3 0-0216 mol. N-T1N0 3 0-42 N-HNO, N-H- H 2 at atmos. pr. is 1-2008 volt. Taking the potential of H 2 /N-H- as zero, calculate the normal potential of the thallic-thallous ion elec- trode. Assume that the concentrations of Tl - and Tl --ions are proportional to the concentrations of the corresponding nitrates and neglect the diffusion-potential. The HNO 3 is added to prevent hydrolysis of the T1(NO 3 ) 3 . Ans. 1-195 volt. PROBLEM 362 (cf. preceding problem). What is the ratio of [Tl ] to [Tl ] at which the potential of a thallic-thallous ion electrode is zero (referred to the normal hydrogen elec- trode as standard) ? Ans. [Tl -]/[Tl ] = 6-1 x 10 ~ 42 . CHAPTER XI DIFFUSION RADIOACTIVITY Diffusion WHEN a dissolved substance diffuses from a place of higher concentration c to a place of lower concen- tration c - dc, the amount which diffuses in unit time through unit cross-section is proportional to the concentration-gradient --, and is, therefore, equal to D-^> (Fick's law), where dx is the thickness of the diffusion-layer. The factor of propor- tionality D is called the diffusion-coefficient of the dissolved substance. Diffusion Examples PROBLEM 363. An electrolytic trough contains a small rotating platinum cathode and a large platinum anode. The electrolyte is sulphuric acid which contains. r 0-0005 gram of iodine per c.c. At all potentials between 0'2 and 1-0 volt a constant current (residual current) of C = O'OOl ampere passes through the cell. What is the thickness 8 of the layer which clings by adhesion to the rotating cathode, if the area of the cathode surface is A = 0'2 sq. cm. and the diffusion- coefficient of iodine is D = 0*6 cm. 2 per day ? What is the value of the residual current if the sulphuric acid contains 0-0008 gram of iodine per c.c. ? SOLUTION 363. Every electrolyte is decomposed by the passage of the electric current. For every electrolyte, how- ever, a definite minimum E.M.F. is necessary for the free separation of the decomposition products at the electrodes. This minimum E.M.F. is called the decomposition-potential of the electrolyte. If the E.M.F. applied is less than the decomposition-potential, a current can pass only if the de- 182 DIFFUSION EXAMPLES 183 composition - products at the electrodes are continuously removed by inter-action with some substance present in the solution. In the present case, for example, the hydrogen evolved at the cathode is " depolarised " by the iodine in the solution with formation of hydrogen iodide. The velocity with which this reaction takes place determines the strength of the current which flows through the electrolyte when an E.M.F. less than the decomposition - potential, is applied. According to Nernst's theory, the velocity of the chemical reaction between iodine and hydrogen at the electrode is very great, whilst the diffusion of the iodine in the solution, by which the supply of iodine at the electrode is maintained, takes place slowly. The effective velocity of the depolarising reaction is, therefore, determined by the velocity of diffusion of the iodine to the electrode. When the electrode of area A is rotated, a layer of solution is formed, which clings to the surface of the electrode in con- sequence of adhesion and rotates with it. The concentration of the iodine in this layer diminishes regularly from the value c (grams per c.c.), the concentration of the iodine in the rest of the solution, to the value at the electrode. The thick- ness of this adhesion layer depends on the velocity of rotation. If 8 is the thickness of the layer for a definite velocity of rotation, then, according to Fick's law, there diffuses to the electrode in the time dt the quantity of iodine DA\--~~ ' dt grams, where D denotes the diffusion-coefficient of iodine. If M is the 64iiivalent weight of iodine, the number of equivalents of hydrogen depolarised by the diffused iodine is DAc dt. In the time dt the current C liberates the quantity of hydrogen C ' c dt, where c denotes the quantity liberated by unit current in unit time, that is, by unit quantity of electricity. In the stationary condition, therefore, ~ . , f DAc, "~8M dt> or the required thickness = 3f6 " If C is given in amperes and if e denotes the number of equivalents of hydrogen liberated by one ampere in one 184 DIFFUSION EXAMPLES second, that is, by one coulomb of electricity, then D must be expressed in cm. 2 /second. In the present example )t6 =7-0 x 10- 6 cm. 2 /sec.,^ =0-2sq.cm.,c = 0-0005 gm. per c.c., C = O'OOl amp., M = 127, = ---, equiv., . x 7-0 x 10-" x 0-2 x 0-0005 x 96540 127 x 0-001 = 5*3 x 10 ~ 4 cm. If the concentration of iodine in the solution is c' instead of c, then, if C' is the residual current in this case, r,, 7 , DAc' 7jt -.'- -gy-dt, and C'= DAc> = 7-0 x 10 " 6 x 0-2 x 0-0008 x 96540 5-3 x 10-* x 127 = 0*0016 amp. PEOBLEM 364 (cf. preceding problem). If the residual current in the preceding problem is C = 0'05 ampere when, instead of iodine, the sulphuric acid contains 0-01 gram of bromine per c.c., what is the diffusion-coefficient of bromine ? SOLUTION 364. If, instead of iodine, the solution con- tains bromine of concentration c = O'Ol, and if the re- sidual current is C = 0'05 ampere, the equation C,= DAc applies in this case too, if D is the diffusion-coefficient and M the equivalent weight of bromine. We obtain, therefore, 0-05 x 5-3 x 10 ~ 4 x 80 Ac 0-2 x 0-01 x 96540 = 1-1 x 10 ~ 5 cm. 2 /second, = 0*95 cm. 2 /day. EADIOACTIVITY EXAMPLES 185 Radioactivity A radioactive substance gradually loses its activity C accord- ing to the law (Rutherford) C-Cf-u, or log, G = log, G - Xt, G is the initial activity, C the activity after time , e the base of the natural logarithms and A. a constant for the particular substance. Radioactivity Example PROBLEM 365. In a space filled with radium emanation the current strength under the influence of a strong potential difference, which is sufficient to produce a saturation current, is, at a definite point of time, G = 35-4 (in arbitrary units). After t l = 12 hours the current strength has diminished to G! = 32-4, and after t, = 200 hours to G a = 8'10. After what time has the current strength the value G 3 = 10, and after what time has it sunk to half its original value ? SOLUTION 365. The current strength at any time is pro- portional to the amount of radium emanation present at that time. The decomposition of the radium emanation follows the exponential law Taking logarithms of both sides we obtain The value of \ may be obtained from each of the equations and log C, = log C - ***. A o From the former we obtain 2-3 log S) 2-3 log ^ A = Ll = ** = 2-3 x_uuouu = . 00738 ^ 12 12 12 186 RADIOACTIVITY EXAMPLES and from the latter The mean value of X is. therefore, 0-00738. The time, t 3 , after which the current strength has the value C 2 = 10, is obtained from the equation 2-3 log > 2-3 log and the time, t 4 , after which the current strength has sunk to half its original value, from the equation = 2-3 log 2 2-3 x 0-301 -- = -0-00738- = 93 ' 8 INDEX Theoretical matter ig indicated by ordinary type, solved problems by italics, and problems for solution by thick type. Affinity of a reaction, 67, 153 ; 81, 156, 162 ; 96, 181. Birnolecular reaction, velo3ity of, 54 ; 56 ; 61. Boiling-point, elevation of, 29 ; 31 ; 39. Degree of association of liquids, 41 ; 41 ; 43. dissociation of electrolytes, 103 ; 27 1 29, 32, 105-117, 155 ; 12 32-40, 136-142, 175-181. Density, 14 ; 14 ; 17. - of mixtures, 14 ; 14 ; 17. Diffusion, 182 ; 182. Diffusion-potential, 151 ; 154 ; 179-181. Dilution law, 104; 105-117; 136-142. Dissociation-constant of electrolytes, 104; 105-117 ; 136-142. Dissociation of gases, 7 ; 7 ; 11. Electrode potential, 149 ; 154-175; 175-181. Electrolytic dissociation, 103-105; 105-117 ; 136-142. Electromotive force, 149 ; 154-175; 175-181. of concentration-cell, 150; 154 ; 175. galvaaic element, 151 ; 154 ; 175. Equilibrium-constant, 62 ; 70-88 ; 91-96. and temperature, 65 ; 70-88 ; 91-96. Equivalent conductivity. 103 ; 105 ; 136-142. Faraday's laws, 3*00 ; 102 ; 134. Fick's law of diffusion, 182 ; 182. Free energy, change of, 67, 153 ; 81, 166, 162; 96, 181. Freezing-point, lowering of, 28 ; 29 ; 35. Gas laws, 1 ; 3 ; 9. Gibbs-Helmholtz equation, 153 ; 162 ; 178. Heating effect of current, 100 ; 101. Heat of reaction, 46; 47 ; 49. Hess's law, 45 ; 47 ; 49. Hydrolysis of salts, 128 ; 130 ; 144-148. Ionic conductivity, 104; 105-117; 136-142. Mass-action, law of, b'2 ; 70-88 ; 91-96. Maximum work, 67, 153; 81, 156, 162; 96, 181. Mixture formula, 14, 20 Migration numbers, 117 ; 119 ; 135. Molecular conductivity, 104 ; 136-142. elevation of boiling-point, 29 ; 31 ; 39. 187 188 INDEX Molecular lowering of freezing-point, 28 ; 29 ; 35. " weight from boiling-point, 29 ; 31 ; 39. freezing-point, 28 ; 29; 35. ' vapour-pressure, 24 ; 25 ; 32. * of liquids, 41; 41; 43. Monomolecular reaction, velocity of, 53 ; 55 ; 60 Normal potential, 150 ; 154-175 ; 175-181 Ohm's law, 100 ; 101. Osmotic pressure, 2 ; 9 ; 12. Oxidation-reduction potential, 152 ; 164 ; 181 Partition law, 67 ; 89 ; 96. Quantity of electricity, 100. Radioactivity, 185 ; 185. Refractivity, 19 ; 20 : 21. of mixtures, 20 ; 21 ; 22. Solubility, 121 ; 122 ; 142-146. of gases, 68 ; 90 ; 98. Solubility-product, 121; 122,154-175; 142-146, 176-180 Specific conductivity, 103 ; 705, 122 ; 136-144 -- volume, 14; 14; 17. of mixtures, 14 ; 14 ; 17. Surface tension, 41 ; 41 ; 43. Thermochemistry, 45 ; 47 ; 49. Transport numbers, 117 ; 119 ; 135. Vapour-pressure, lowering of, 24; 25; 32. and temperature, 25 ; 25 ; 32. Velocity of migration, 10 1 ; 139. reaction, 53; 55, 108; 60, 147. and temperature, 58. a PRINTED IN GREAT BRITAIN AT THE UNIVERSITY PRESS, ABERDEEN 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. This book is due on the last date stamped below, or on the date to which renewed. Renewed J?ooks O subject to immediate recall. I69C7 3 2 , THE UNIVERSITY OF CALIFORNIA LIBRARY