Question

Determine the pH of a 0.20 M KCHO2 (potassium formate) aqueous solution. [Ka for HCHO2 = 1.8 x 10-4]. Use ice table.

Answer 1

Kb = Kw / Ka = 1.00 * 10^-14 / 1.8 * 10^-4 = 0.555 * 10^-10

In aqueous solution, [KCHO2] = [CHO2^-] = 0.20 M

Consider a hydrolysis reaction of CHO2^-

CHO2^- + H2O. -----> OH^- + HCHO2

Initial concentration 0.20 0 0

Change in concentration. - x +x +x

Final concentration . 0.20 - x x x

Kb = [OH^-][HCHO2] / [CHO2^-] = (x)(x) / (0.20 - x)

Kb = x^2 / (0.20 - x) = x^2 / 0.20

Since Kb is very small hence 0.20 - x = ~ 0.20

x^2 = 0.20 * Kb = 0.20 * 0.555 * 10^-10 = 0.111 * 10^-10

x = √(0.111 * 10^-10) = 0.333 * 10^-5

Since [OH^-] = x M then [OH^-] = 0.333 * 10^-5 M

pOH = - log [OH^-] = - log(0.333 * 10^-5) = - log 0.333 + 5 log 10 = - (-0.477) + 5 * 1 = 5.477

pH = 14.00 - pOH = 14.00 - 5.477 = 8.52

pH = 8.52