Find the equilibrium concentration of H2A in this solution.

Question

For the diprotic weak acid H2A, ka1=2.9 x 10^-6 and ka2= 6.7 x 10^-9. There is a pH of a 0.0550 M solution of H2A. What is the equilibrium concentration of H2A in this solution?

Answer 1

This question was posted earlier, and I forgot to answer the second part, i.e. asking for the equilibrium concentration of H2A. Here is the completion of that question.

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

0.0550...0......0.....Initial

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

0.0550-x..x...x.....Equilibrium

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

[H₂A] @ equilibrium = 0.0550 - 3.99x10^-4 = 0.0550 - 4x10^-4 = 0.0550 - 0.0004 = 0.0546 M