Only looking for the answer to (c)! Already have the answers for (a) and (b)! Wh
ID: 508618 • Letter: O
Question
Only looking for the answer to (c)! Already have the answers for (a) and (b)!
What is the fraction of association (a) for the following potassium propionate solutions? Ignore activities. The K_a of propanoic acid is 1.34 times 10^-5. (a)2.00 times 10^-1MK(C_2H_5CO_2) alpha = (b) 2.00 times 10^-2 M K(C_2H_5CO_2) (c) 2.00 times 10^-12 M K(C_2H_5CO_2) Incorrect. The concentration of OH^- resulting from the reaction of C_2H_5CO_2^- with water is negligible compared to the amount of OH-ions present from the autoionization of water (1.0 times 10^-7 M). Starting with the equilibrium constant expression, find an expression for [C_2H_2CO_2^-] in terms of [C_2H_5CO_2H], K_b = [C_2H_5CO_2H] [OH^-]/[C_2H_5CO^-]Explanation / Answer
C2H5CO2- (aq) + H2O -----------------> C2H4CO2H (aq) + OH- (aq)
2 x 10^-12 0 0
2 x 10^-12 - x x x
Kb = [C2H4CO2H][OH-] / [C2H5CO2-]
1.0 x 10^-14 / 1.34 x 10^-5 = x^2 / (2x 10^-12) - x
7.46 x 10^-10 = x^2 / (2x 10^-12) - x
x = 1.995 x 10^-12
[OH-] = 1.995 x 10^-12 M
fraction of dissociation = 1.995 x 10^-12 / 2.0 x 10^-12
= 0.9975
fraction of dissociation = 1.00
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.