Question

A diprotic acid, H2A, has acid dissociation constants of Ka1 = 3.66× 10–4 and Ka2 = 4.99× 10–12. Calculate the pH and molar concentrations of H2A, HA–, and A2– at equilibrium for each of the solutions below.

(a) a 0.121 M solution of H2A

pH = ? [H2A]=? [HA-]=? [A2-]=?

(b) a 0.121 M solution of NaHA

pH = ? [H2A]=? [HA-]=? [A2-]=?

(c) a 0.121 M solution of Na2A

pH = ? [H2A]=? [HA-]=? [A2-]=?

Answer 1

H2A -----------------> HA-   +    H+

0.121                          0            0

0.121 - x                      x             x

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

3.66 x 10^-4 = x^2 / 0.121 - x

x = 6.47 x 10^-3

[H+] = 6.47 x 10^-3 M

pH = -log[H+] = -log ( 6.47 x 10^-3)

pH = 2.19

[H2A] = 0.121 - 6.47 x 10^-3 = 0.1145

pH = 2.19

[H2A] = 0.114 M

[HA-] = 6.47 x 10^-3 M

[A2-] = Ka2 = 4.99 x 10^-12 M

b)

pH = pKa1 + pKa2 / 2

     = 3.44 + 11.30 / 2

pH = 7.37

[H+] = 10^-7.37 = 4.27 x 10^-8

[HA-] = 0.121 M

H2A   ------------> HA- +   H+

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

[H2A] = [HA-][H+] / Ka1 = 0.121 x 4.27 x 10^-8 / 3.66 x 10^-4

[H2A] = 1.41 x 10^-5 M

HA-   -------------> H+   + A2-

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

[A2-] = Ka2 x [HA-] / [H+] = 4.99 x 10^-12 x 0.121 / 4.27 x 10^-8

[A2-] = 1.42 x 10^-5 M

c) 0.121 M Na2A :

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

0.121                                       0           0

0.121-x                                    x            x

Kb1 = x^2 / 0.185 - x

Kw / Ka2 = x^2 / 0.185 - x

2.00 x 10^-3 = x^2 / 0.121 - x

x = 0.0146

[HA-] = 0.0146 M

[OH-] = 0.0146 M

pOH = -log 0.0146 = 1.84

pH = 12.16

[A2-] = 0.106 M

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

[H2A] = 2.76 x 10^-11 M