Question

For the diprotic weak acid H2A, Ka1 = 3.3 × 10-6 and Ka2 = 8.0 × 10-9.

What is the pH of a 0.0500 M solution of H2A?

What are the equilibrium concentrations of H2A and A2– in this solution?

Answer 1

H2A -----------------> HA-   +    H+

0.05                           0            0

0.05 - x                      x             x

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

3.3 x 10^-6 = x^2 / 0.05 - x

x = 4.05 x 10^-4

[H+] = 4.05 x 10^-4 M

pH = -log[H+] = -log (4.05 x 10^-4)

pH = 3.39

[H2A] = 0.05 -4.05 x 10^-4 = 0.0496

[H2A] = 0.0496 M

[A2-] = Ka2 = 8.00 × 10-9. M