Question

Consider a diprotic acid, H2A, with the following Ka values. Ka1 = 0.01 Ka2 = 0.008 If you have a 0.01 M solution of H2A, what is [H3O+] and the pH ? Hint: The Ka values are too close together to ignore the second equilibrium.

Answer 1

The hydrolysis reaction can be written
H2A + H2O = AH^- + H3O+

The Ka values are too close together to ignore the second equilibrium.
Ka1 = [H3O+][AH-] / [H2A] = x² /(0.01 - x) = 0.01
→ x ≈ 0.01 M = [H3O+]

pH = -log [H3o+] = 2
→ y ≈ 1.68742*10^(-7) M

[H+] = x + y = 0.0448221 + 1.68742*10^(-7) = 0.0448222687 M
pH = -log[H+] = -log(0.0448222687) = 1.35