Question

a) Calculate the pH of a 0.015M solution of formic acid, HCO2H?

Ka, HCO2H=1.8x10^-4    HCO2H(aq)-->HCO2^-(aq)+H^+(aq)

b) What is the concentration of HCO2H at equilibrium?

Answer 1

HCOOH + H₂O ---> HCOO^- + H₃O⁺

                   HCOOH                 HCOO^-                 H₃O⁺

initial                     0.015                      0                         0

change                    -x                         +x                      +x

equilibrium           0.015 - x                   x                         x

K_a = [HCOO^-][H₃O⁺] / [HCOOH]

1.8 x 10^-4 = x² / 0.015              (assume 0.015>>x)

x² = 2.7 x 10^-6

x = 1.643 x 10^-3 M

[H⁺] = 1.63 x 10^-3 M

pH = -log[H⁺]

pH = -log(1.63 x 10^-3)

pH = 2.78

Concentration of HCOOH at equilibrium:

[HCOOH] = 0.015 - 1.63 x 10^-3 = 0.013 M