Question

If a 0.10 M aqueous solution of CH3COOH exhibits a pH of 2.87, predict the pH of a 0.10 M aqueous solution of NaCH3COO at the same temperature.

Answer 1

Given that,

pH = 2.87

=> [H+] = 10^-2.87 = 1.35 x 10^-3 M

CH3COOH --------> CH3COO- + H+

0.1 - X.......................X................X

where X = 1.35 x 10^-3 M

Ka = [CH3COO-] [H+] / [CH3COOH]

=> Ka = X^2 / (0.1 - X) = (1.35 x 10^-3)^2 / (0.1 - 1.35 x 10^-3)

=> Ka = 1.845 x 10^-5

Kb for CH3COO- = 10^-14 / Ka = 10^-14 / 1.845 x 10^-5 = 5.42 x 10^-10

CH3COO- + H2O ------> CH3COOH + OH-

0.1 - X...................................X..............X

Kb = 5.42 x 10^-10 = X^2 / (0.1 - X)

=> X = 7.36 x 10^-6 M = [OH-]

pOH = - log [OH-] = 5.133

pH = 14 - pOH = 14 - 5.133 = 8.867