Question

A diprotic acid, H2A, has acid dissociation constants of Ka1 = 1.56× 10–4 and Ka2 = 2.28× 10–12. Calculate the pH and molar concentrations of H2A, HA–, and A2– at equilibrium for each of the solutions below.

(a) a 0.203 M solution of H2A

(b) a 0.203 M solution of NaHA

(c) a 0.203 M solution of Na2A

Answer 1

a)

H2A -----------------> HA-   +    H+

0.203                           0            0

0.203 - x                      x             x

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

1.56x10^-4 = x^2 / 0.203 - x

x = 5.55 x 10^-3

[H+] = 5.55 x 10^-3 M

pH = -log[H+] = -log (5.55 x 10^-3)

pH = 2.26

[H2A] = 0.203 - 5.55 x 10^-3 = 0.197

[H2A] = 0.197 M

[A2-] = Ka2 = 2.28 × 10^-12 M