Question

Di chloroacetic acid (Cl2HCCOOH) has a pka of 1.3 .What is the pH of a 0.125 M solution of potassium dichloroacetate?

Answer 1

Ka = 10^-pKa = 10^-1.3 = 0.05

Cl2HCCOO- + H2O -------------------------->Cl2HCCOOH + OH-

0.125                                                             0                       0---------------> initial

0.125-x                                                           x                       x------------> equilibrium

Kb = [Cl2HCCOOH] [OH-][Cl2HCCOO-]

Kb = x^2 / 0.125-x                                           

Kw / Ka = x^2 / 0.125-x

1.0 x 10^-14 / 0.05 = x^2 / 0.125-x

2 x 10^-13 = x^2 / 0.125-x

x^2 + 2 x 10^-13 x - 2.5 x 10^-14 =0

x = 1.58 x 10^-7

x = [OH-] =1.58 x 10^-7 M

pOH = -log [OH-] = -log (1.58 x 10^-7 )

pOH = 6.80

pH + pOH =14

pH = 7.20