Question

For the diprotic weak acid H2A, Ka1 = 3.1 × 10-6 and Ka2 = 5.4 × 10-9. What is the pH of a 0.0650 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?

Answer 1

Let's do this by parts:

H2A <------> HA- + H+;..................Ka1 = 3.1 x 10^-6
HA- <------> A(^2-) + H+;...............Ka2 = 5.4 x 10^-9

For the concentrations of H+ and HA- you consider only the first reaction since the HA- product will only slightly dissociate (Ka2 is very small).

The equilibrium quotient for Ka1 looks like;

[H+][HA-] / [H2A] = 3.1 x 10^-6
[H+] = [HA-] So lets do some math to figure out [H+]

If x moles of H2A dissociate, then x moles of HA- and H+ are produced, and the concentration of H2A is reduced by x:

[H2A] = 0.0650 - x [considering x to be very small, we can ignore x here]
[H+] = x
[HA-] = x

x^2 / 0.0650 M = 3.1 x 10^-6
x = 0.00045 M
So the concentration or H+ and HA- after the first ionization is 0.00045 M

The new concentration of H2A will be:
0.0650 M - 0.00045 M
= 0.0646 M

The equilibrium quotient of Ka2 looks like this:
[H+][A2-] / [HA-] = 5.4 x 10^-9

[H+] = [A2-] so lets do some more algebra (NOT TAKING INTO ACCOUNT THE LAST IONIZATION).

x^2 / 0.00045 M = 5.4 x 10^-9
x = 1.56 x 10^-6 M
So the concentration of A2- and H+(for this ionization) is 1.56 x 10^-6 M

To solve for the pH we must first add the two concentration of H+:
0.00045 M + 1.56 x 10^-6 M = 4.52 x 10^-4 M

pH = -log(4.52 x 10^-4)
= 3.35

So the pH = 3.35
[H2A] = 0.0646 M
[A2-] = 1.56 x 10^-6 M