Question

For the diprotic weak acid H2A, Ka1 = 2.4 × 10-6 and Ka2 = 7.2 × 10-9. What is the pH of a 0.0650 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?

Answer 1

H2A -----------------> HA-   +    H+

0.065                           0            0

0.065 - x                      x             x

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

2.4 x 10^-6 = x^2 / 0.065 - x

x = 3.94 x 10^-4

[H+] = 3.94 x 10^-4 M

pH = -log[H+] = -log (3.94 x 10^-4)

pH = 3.40

[H2A] = 0.065 - 3.94 x 10^-4 = 0.0446

[H2A] = 0.0646 M

[A2-] = Ka2 = 7.2 × 10^-9 M