A diprotic acid, H₂A, has Ka1 = 3.4 × 10⁻⁴ and Ka2 = 6.7 × 10⁻⁹. What is the pH of a 0.20 M solution of H₂A?

Answer 1

H₂A ==> H⁺ + HA^- ... Ka1 = 3.4x10^-4

HA^- ==> H⁺ + A^2- ... Ka2 = 6.7x10^-9

Because Ka2 is more than 1000 times less than Ka1, we can ignore it's contribution to the final [H⁺] and hence its contribution to the pH.

Ka1 = 3.4x10^-4 = [H⁺][HA^-] / [H2A] = (x)(x) / 0.2 - x (assuming x is small we can ignore it in the denominator)

3.4x10^-4 = x² / 0.2

x² = 6.8x10^-5

x = 8.2x10^-3 M = [H⁺] (note: this is about 4% of 0.2 M so ignoring it is justified according to the 5% rule)

pH = -log [H⁺] = -log 8.2x10^-3

pH = 2.1