Find the pH and H2A of a solution

Question

For the diprotic weak acid H2A,  Ka1= 2.9 x 10^-6 and Ka2=6.7 x 10^-9.

What is the pH of a 0.0550 M solution of H2A?

Find pH and H2A

Answer 1

H₂A ---> H⁺ + HA^- ... Ka1 = 2.9x10^-6

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

2.9x10-6 = [H+][HA-]/[H₂A] = (x)(x)/0.0550 - x

x² = 1.6x10^-7 - 2.9x10^-6x

x² + 2.9x10^-6x - 1.6x10^-7 = 0

x = 3.99x10^-4 M = H⁺ = HA^- after 1st ionization

HA^- ---> H⁺ + A^2-

4x10^-4......4x10^-4...0....Initial

-x...........+x...........+x.....Change

4x10^-4-x...4x10^-4+x..x..Equilibrium

6.7x10^-9 = (4x10^-4 + x)(x)/4x10^-4 - x)

x = 6.7x10^-9 M (approximately) = [H⁺] = [A^2-] after 2nd ionization

The pH will be mostly determined by the 1st ionization where [H+] = 3.99x10-4 M

pH = -log 3.99x10^-4 = 3.4