Question

For the diprotic weak acid H2A, Ka1 = 3.6 × 10-6 and Ka2 = 8.0 × 10-9. What is the pH of a 0.0450 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?

Answer 1

H2A -------------------> HA- + H+

0.0450                           0         0

0.045 - x                         x         x

Ka1 = x^2 / 0.045 - x

3.6 × 10^-6 = x^2 / 0.045 - x

x = 4 x 10^-4

[H+] = 4 x 10^-4 M

pH = -log[H+] = -log(4 x 10^-4)

      = 3.4

[HA-] = 4 x 10^-4 M

[H2A] = 0.045 - x = 0.0446 M

[A^-2] = Ka2 = 8.0 × 10^-9 M