Question

For the diprotic weak acid H2A, Ka1 = 2.7 × 10-6 and Ka2 = 7.9 × 10-9. What is the pH of a 0.0750 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?

Answer 1

As given the ka2 is very low so the H+ concentration will be mainely due to first dissoication of diprotic acid

H2A --> HA- + H+

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

2.7 × 10-6 = x^2 / [0.075-x]

x <<1 so can be ignored in the denominator

x^2 = 2.7 X 10^-6 X 0.075 = 0.2025 X 10^-6

x = 4.5 X 10^-4 = [H+]

pH = -log[H+] = 3.35

[H2A] left = 0.075 - 0.00045 = 0.07455

b) HA- --> A-2 + H+

Ka2 = [A2-] [H+] / [HA-]

7.9 × 10-9 = y^2 / [0.00045-y]

we may ignore x in the denominator , y<<1

7.9 × 10-9 = y^2 / 0.00045

y^2 = 0.003555 X 10^-9

y = 0.188 X 10^-5

So [A-2] = 1.88 X 10^-6