Question

A diprotic acid, H2A, has acid dissociation constants of Ka1 = 2.40× 10–4 and Ka2 = 2.70× 10–12. Calculate the pH and molar concentrations of H2A, HA–, and A2– at equilibrium for each of the solutions below.

.185M solution of NaHA

.185M solution of Na2A

Answer 1

Ka1 = 2.40 x 10^-4

pKa1 = -log Ka1 = -log (2.40 x 10^-4 )

pKa1 = 3.62

pKa2 = 11.57

a ) 0.185 M NaHA :

pH = pKa1 + pKa2 / 2

     = 3.62 + 11.57 / 2

pH = 7.60

[H+] = 10^-7.60 = 2.51 x 10^-8

[HA-] = 0.185 M

H2A   ------------> HA- +   H+

Ka1 = [HA-][H+] / [H2A]

[H2A] = [HA-][H+] / Ka1 = 0.185 x 2.51 x 10^-8 / 2.40 x 10^-4

[H2A] = 1.94 x 10^-5 M

HA-   -------------> H+   + A2-

Ka2 = [H+][A2-] / [HA-]

[A2-] = Ka2 x [HA-] / [H+] = 2.70 x 10^-12 x 0.185 / 2.51 x 10^-8

[A2-] = 1.99 x 10^-5 M

b) 0.185 M Na2A :

A2-    +   H2O   --------------> HA-   + OH-

0.185                                       0           0

0.185-x                                    x            x

Kb1 = x^2 / 0.185 - x

Kw / Ka2 = x^2 / 0.185 - x

3.70 x 10^-3 = x^2 / 0.185 - x

x = 0.0244

[HA-] = 0.0244 M

[OH-] = 0.0244 M

pOH = -log 0.0244 = 1.61

pH = 12.39

[A2-] = 0.161 M

[H2A] = [HA-][H+] / Ka1 = 0.0244 x 4.07 x 10^-13 / 2.40 x 10^-4

[H2A] = 4.14 x 10^-11 M