Calculate the pH of a 0.025 M CH3COO- solution? (Ka for CH3COOH = 1.8x10-5)

Answer 1

CH3COO- will act as a base when in H2O:

CH3COO- + H2O ==> CH3COOH + OH-

Then, Kb = [CH3COOH][OH-] / [CH3COO-]

So we need the Kb for CH3COO-

Kb x Ka = 1x10^-14 and Kb = 1x10^-14 / Ka = 1x10^-14 / 18x10^-5 = 5.56x10^-10

Kb = 5.56x10^-10 = (x)(x) / 0.025 - x (assume x is small relative to 0.025 and ignore it)

5.56x10^-10 = x² / 0.025

x² = 1.39x10^-11

x = 3.73x10^-6 (this is insignificant relative to 0.025 so above assumption was valid)

[OH-] = 3.73x10^-6

pOH = -log 3.73x10^-6 = 5.43

pH = 14 - pOH = 14 - 5.43

pH = 8.57