Question

A diprotic acid, H2A, has acid dissociation constants of Ka1 = 3.61× 10–4 and Ka2 = 4.02× 10–12. Calculate the pH and molar concentrations of H2A, HA–, and A2– at equilibrium for each of the solutions below.

(a) a 0.124 M solution of H2A

pH [H2A] [HA-] [A2-]

(b) a 0.124 M solution of NaHA

pH [H2A] [HA-] [A2-]

(c) a 0.124 M solution of Na2A

pH [H2A] [HA-] [A2-]

Answer 1

The reactions are

H2A <------> HA- + H+;..................Ka1 = 3.61*10^-4
HA- <------> A(^2-) + H+;...............Ka2 = 4.02*10^-12

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).

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] = 0.124 - x
[H+] = x
[HA-] = x

Ka1 = [H+]*[HA-]/[H2A] = x²/(0.124 - x)

3.61*10^-4*(0.124- x) = x²
x² + 3.61*10^-4*x - 0.124*3.61*10^-4 = 0
x = 0.0065

[H+] = x = 0.0065
pH = -log(0.0065) = 2.19
[H2A] = 0.124 - x = 0.1175 M

For the second reaction, Ka2 = [A(^2-)]*[H+]/[HA-] = [A(^2-)]*x/x;

[A(^2-] = Ka2
[A(^2-) = 4.02*10^-12 M