Question

What is the pH of a 0.25 M solution of KHCOO? Ka (HCOOH) = 1.8 * 10-4

Please show all the steps because I'm doing something wrong and keep getting 2.17. Thank You

Answer 1

HCOOK(aq) HCOO^- (aq) + K⁺ (aq)

The ICE table can be constructed as,

                HCOO^- (aq) + H₂O(l) HCOOH(aq) + OH^- (aq)

I               0.25                                                       0                      0

C              -x                                                          +x                     +x

E            (0.25-x)                                                     x                      x

We have,

Kb = Kw/Ka = (1 x10^-14)/(1.8 x10^-4) = 5.6 x 10^-11

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

5.6 x 10^-11 = x²/(0.25-x)

(5.6 x 10^-11)(0.25-x) = x²

x   = 3.74 x 10^-6

[OH^-] = 3.74 x 10^-6 M

pOH = -log[OH^-] = -log(3.74 x10^-6) = 5.43

pH = 14-pOH = 14 - 5.43 = 8.57

pH = 8.57