Question

Answer this question instead please!

what is the percent hydrolysis of 0.100 M potassium formate solution

Sorry posted wrong question!

The answer for this one is 0.0024 percent!

Answer 1

We know Ka for formic acid (HCOOH) = 0.00018 = 1.8 * 10-4

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

Kb = 5.55 * 10^-11

Consider a hydrolysis reaction of formate ion

HCOO^- + H2O ----> OH^- + HCOOH

Initial concentration 0.100 0 0

Change in concentration. -x +x +x

Final concentration 0.100 -x x x

Kb = [OH^-][HCOOH] / [HCOO^-]

5.55 * 10^-11 = (x)(x) / (0.100 -x)

Since formic acid is a weak acid hence 0.100 - x ~ 0.100

x2 = 5.55 *10^-11 *0.1 = 5.55 * 10^-12

x = √(5.55*10^-12) = 2.356 * 10^-6

Since [OH^-] = x and x = 2.356 * 10^-6

[OH-] = 2.356 * 10^-6 M

% hydrolysis = (2.356 * 10^-6 M)*100 / 0.100 M = 0.002356% ~ 0.0024%