Answer to Question #15627 in Physical Chemistry for Matthias

Question

This is a question I am presently doing right now which I found in my homework. I am quite sure that in (a) I have to use the Hasselbach equation though I am not sure how to work it out. If anyone knows any of the answers, and how you got to them, please feel free to share. Any help is greatly appreciated :)

CH3COO- (aq) + H2O (l) <==> CH3COOH (aq) + OH- (aq)

A 0.01mol dm^-3 solution has a pH of 8.87

a. Calculate [H+] and [OH-] in the solution

b. Find [CH3COOH] in the solution.

c. Calculate the acid dissociation constant of CH3COOH.

Thanks! :)

Answer 1

a.
pH = -log[H⁺]

[H⁺] = 10^-8.87 = 1.5*10^-9

pH + pOH = pK_w = 14

14 - 8.87 = 5.13

pOH = -log[OH^-]

[OH^-] = 4.4*10^-6

b.
[OH^-] = [CH₃COO^-]

c.
K_b for CH₃COO^- = K_w/K_a
K_w/K_a = (CH₃COOH)(OH^-)/(CH₃COO^-)
K_a = 4.4*10^-6