Question

For the diprotic weak acid H2A, Ka1 = 2.6

Answer 1

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

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

First equilibrium

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

I              0.0700                                   0                      0

C               -x                                        +x                   +x

E             0.0700-x                                +x                    +x

Ka1 = x^2 / (0.0700-x) = 2.6 x 10^-6

x^2 = 2.6 x 10^-6

x = 1.6 x 10^-3 M

Second equilibrium

                HA-             <----------->     A2- +         H+

I             1.6 x 10^-3                               0                .3 x 10^-3

C               -y                                        +y                   +y

E             1.3 x 10^-3 - y                        +y         1.3 x 10^-3 + y

Ka2 = (y) ( 1.3 x 10^-3 + y) / (1.3 x 10^-3 - y ) = 5.2 x 10^-9

y = 5.2 x 10^-9

Thus, [H2A] = 0.0700 - x = 0.0700 - (1.3 x 10^-3) = 0.0687 M

[A2-] = +y = 5.2 x 10^-9 M

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pH = -log [H+]

[H+] = 1.3 x 10^-3 + y

As, y is very small value, y = 5.2 x 10^-9, it can be ignored.

pH = -log (1.3 x 10^-3)

pH = 2.89