Question

For the diprotic weak acid H2A, Ka1 = 4.0 × 10-6 and Ka2 = 6.1 × 10-9. What is the pH of a 0.0800 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?

Answer 1

pH of solution

H2A + H2O <==> H3O+ + HA-

Ka1 = [HA-][H3O+]/[H2A]

4 x 10^-6 = x^2/0.08

x = [H3O+] = 5.65 x 10^-4 M

pH = -log[H3O+] = 3.25

[H2A] equilibrium = 0.08 - 5.65 x 10^-4 = 0.0794 M

HA- + H2O <==> H3O+ + A^2-

Ka2 = [H3O+][A^2-]/[HA-]

6.1 x 10^-9 = x^2/5.65 x 10^-4

x = [A^2-] = 1.85 x 10^-6 M