For the diprotic weak acid H2A, Ka1=2.8×10−6 and Ka2=7.7×10−9 . What is the pH of a 0.0450 M solution of H2A ?

Question

What are the equilibrium concentrations of H2A and A^2- in this solution?

Answer 1

This turns out to be an exercise in math more than in chemistry. I'm not going to go through the quadratic and the rest of the math, but will illustrate how to solve this. In fact, because the two Ka values differ by almost 1000, the contribution to the pH of the 2nd ionization is minimal and probably of little consequence in actual practice. But, here goes anyway.....

H₂A <==> H⁺ + HA^- Ka1 = 2.8x10^-6

HA- <==> H+ + A- Ka2 = 7.7x10^-9

(1) Ka1 = 2.8x10^-6 = [H+][HA-]/[H₂A] = x²/0.0450-x and assuming x is small relative to 0.045, we can ignore it

x² = 1.26x10^-7

x = 3.56x10^-4 M = [H+] from the first ionization

HA- ==> H+ + A- ............

.0445......3.56x10-4....3.56x10-4.....Initial

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

0.0445-x..3.56x10^-4+x...3.56x10^-4+x...Equilb

7.7x10^-9 = (3.56x10^-4)²/0.0445 - x

Solve for x which will be the [H+] from the 2nd ionization

Add to first and take -log [H+] to find pH