Question

A diprotic acid, H2A, has acid dissociation constants of Ka1 = 3.48× 10–4 and Ka2 = 3.21× 10–12. Calculate the pH and molar concentrations of H2A, HA–, and A2– at equilibrium for each of the solutions below.(a) a 0.153 M solution of H2A.(b) a 0.153 M solution of NaHA.(c) a 0.153 M solution of Na2A

Answer 1

part A ) .(a) a 0.153 M solution of H2A

H2A ----------------------------> HA- + H+

0.153 0 0

0.153-x x x

Ka1 = x^2 / 0.153-x

3.48 x 10^-4 = x^2 / 0.153-x

x^2 + 3.48 x 10^-4 x - 5.32 x 10^-5 = 0

x = 7.12 x 10^-3

[HA-] = 7.12 x 10^-3 M --------------------->

[H2A] = 0.153 - (7.12 x 10^-3 )

= 0.146 M

[H2A] = 0.146 M --------------------->

[A-2] = Ka2

[A-2] = 3.21 x 10^-12 M--------------------->

pH = -log [H+]

pH = -log (7.12 x 10^-3)

pH = 2.15 --------------------->