Question

The diprotic acid,H2A has two ionizable protons with pKa1=4.175 and pKa2=10.235. Determine the pH, [H2A], [HA- ], [A-2 ], [OH-1 ] for the following three solutions.

a) A solution which is 0.1500M in H₂A.

b) A solution which is 0.1500M in NaA

c) A solution which is 0.1500M in Na₂A

Answer 1

For H2A

Ka1 = 6.7 x 10-5 ; Ka2 = 5.8 x 10^-11

a) [H2A] = 0.15 M

First dissociation,

H2A + H2O <==> H3O+ + HA-

let x amount dissociated

Ka1 = [H3O+][HA-]/[H2A]

6.7 x 10-5 = x^2/0.15

x = [H3O+] = 3.17 x 10^-3 M

pH = -log[H3O+] = 2.50

[OH-] = 1 x 10^-14/3.17 x 10^-3 = 3.15 x 10^-12 M

[H2A] = 0.15 - 3.17 x 10^-3 = 0.14683 M

second dissociation,

HA- + H2O <==> H3O+ + A^2-

extent of dissociation very small

Ka2 = 5.8 x 10^-11 = [H3O+][A^2-]/[HA-]

[A^2-] = 5.8 x 10^-11 M

[HA-] = 3.17 x 10^-3 M

b) [NaHA] = 0.15 M

HA- + H2O <==> H3O+ + A^2-

let x amount dissociated

Ka2 = [H3O+][A^2-]/[HA-]

5.8 x 10^-11 = x^2/0.15

x = [H3O+] = 2.95 x 10^-6 M

pH = -log[H3O+] = 5.53

[OH-] = 1 x 10^-14/2.95 x 10^-6 = 3.30 x 10^-9 M

[HA-] = 0.15 - 2.95 x 10^-6 = 0.15 M

[A^2-] = 2.95 x 10^-6 M

HA- hydrolyze

HA- + H2O <==> H2A + OH-

let x amount hydrolyzed

Kb2 = Kw/Ka1 = [H2A][OH-]/[HA-]

1 x 10^-14/6.7 x 10-5 = x^2/0.15

x = [H2A] = 4.73 x 10^-6 M

c) [Na2A] = 0.15 M

A^2- + H2O <==> HA- + OH-

let x amount reacted

Kb1 = Kw/Ka2 = [HA-][OH-]/[A^2-]

1 x 10^-14/5.8 x 10^-11 = x^2/0.15

x = [OH-] = 5.08 x 10^-3 M

pOH = -log[OH-] = 2.294

pH = 14 - pOH = 11.704

[A^2-] = 0.15 - 5.08 x 10^-3 = 0.14492 M

second hydrolysis

HA- + H2O <==> H2A + OH-

Kb2 = Kw/Ka1 = [H2A][OH-]/[HA-]

1 x 10^-14/6.7 x 10-5 = x^2/5.08 x 10^-3

x = [H2A] = 8.71 x 10^-7 M

[HA-] = 5.08 x 10^-3 M