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P16B.9* (a) A dilute solution of a weak electrolyte AB, which dissociates to A* +B, is prepared with an initial concentration c. Suppose that a fraction a of AB dissociates. Assuming that activities can be approximated by concentrations, show that the equilibrium constant K for dissociation may be written a'cA K=- AB (1-a)c (b) The conductivity of the solution described in (a) is measured as K, and the molar conductivity is then calculated as A = x/CAg. However, because the degree of dissociation, and hence the concentration of the ions, varies strongly with the initial concentration cAB» values of A calculated in this way also vary strongly with cA. Given that K can be expected to be proportional to the concentration of the ions, explain why a = AIAm where Am, is the molar conductivity in the limit of complete dissociation of AB. (c) Substitute this expression for a into the above expression for K. You now have two expressions for K: one in terms of a and one in terms of A/Am1- Equate these two expressions and hence show (by rearranging your expression) that *m,l 1 A_(1-a) 1 !! m 'm,l

P16B.9* (a) A dilute solution of a weak electrolyte AB, which dissociates to A* +B, is prepared with an initial concentration c. Suppose that a fraction a of AB dissociates. Assuming that activities can be approximated by concentrations, show that the equilibrium constant K for dissociation may be written a'cA K=- AB (1-a)c (b) The conductivity of the solution described in (a) is measured as K, and the molar conductivity is then calculated as A = x/CAg. However, because the degree of dissociation, and hence the concentration of the ions, varies strongly with the initial concentration cAB» values of A calculated in this way also vary strongly with cA. Given that K can be expected to be proportional to the concentration of the ions, explain why a = AIAm where Am, is the molar conductivity in the limit of complete dissociation of AB. (c) Substitute this expression for a into the above expression for K. You now have two expressions for K: one in terms of a and one in terms of A/Am1- Equate these two expressions and hence show (by rearranging your expression) that *m,l 1 A_(1-a) 1 !! m 'm,l

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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P16B.9* (a) A dilute solution of a weak electrolyte AB, which dissociates to A* +B, is prepared with an initial concentration c. Suppose that a fraction a of AB dissociates. Assuming that activities can be approximated by concentrations, show that the equilibrium constant K for dissociation may be written a'cA K=- AB (1-a)c (b) The conductivity of the solution described in (a) is measured as K, and the molar conductivity is then calculated as A = x/CAg. However, because the degree of dissociation, and hence the concentration of the ions, varies strongly with the initial concentration cAB» values of A calculated in this way also vary strongly with cA. Given that K can be expected to be proportional to the concentration of the ions, explain why a = AIAm where Am, is the molar conductivity in the limit of complete dissociation of AB. (c) Substitute this expression for a into the above expression for K. You now have two expressions for K: one in terms of a and one in terms of A/Am1- Equate these two expressions and hence show (by rearranging your expression) that *m,l 1 A_(1-a) 1 !! m 'm,l
Transcribed Image Text:P16B.9* (a) A dilute solution of a weak electrolyte AB, which dissociates to A* +B, is prepared with an initial concentration c. Suppose that a fraction a of AB dissociates. Assuming that activities can be approximated by concentrations, show that the equilibrium constant K for dissociation may be written a'cA K=- AB (1-a)c (b) The conductivity of the solution described in (a) is measured as K, and the molar conductivity is then calculated as A = x/CAg. However, because the degree of dissociation, and hence the concentration of the ions, varies strongly with the initial concentration cAB» values of A calculated in this way also vary strongly with cA. Given that K can be expected to be proportional to the concentration of the ions, explain why a = AIAm where Am, is the molar conductivity in the limit of complete dissociation of AB. (c) Substitute this expression for a into the above expression for K. You now have two expressions for K: one in terms of a and one in terms of A/Am1- Equate these two expressions and hence show (by rearranging your expression) that *m,l 1 A_(1-a) 1 !! m 'm,l
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