Homework Answers
b) 0.133 M NaHA :
pH = pKa1 + pKa2 / 2
= 3.3925 + 10.385 / 2
pH = 6.89
[H+] = 10^-6.89 = 1.288 x 10^-7
[HA-] = 0.133 M
H2A ------------> HA- + H+
Ka1 = [HA-][H+] / [H2A]
[H2A] = [HA-][H+] / Ka1 = 0.133 x 1.288 x 10^-7 / 4.05 x 10^-4
[H2A] = 4.23 x 10^-5 M
HA- -------------> H+ + A2-
Ka2 = [H+][A2-] / [HA-]
[A2-] = Ka2 x [HA-] / [H+] = 4.12 x 10^-11 x 0.133 / 1.288x 10^-7
[A2-] = 4.25 x 10^-5 M
c) 0.133 M Na2A :
A2- + H2O --------------> HA- + OH-
0.133 0 0
0.133-x x x
Kb1 = x^2 / 0.133 - x
Kw / Ka2 = x^2 / 0.133 - x
2.427 x 10^-4 = x^2 / 0.133 - x
x = 0.00568
[HA-] = 0.00568 M
[OH-] = 0.00568 M
pOH = -log 0.00568 = 2.25
pH = 11.75
[A2-] = 0.127 M
[H2A] = [HA-][H+] / Ka1 = 0.00568 x 1.778 x 10^-12 / 4.05 x 10^-4
[H2A] = 2.49 x 10^-11 M
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