A diprotic acid H2A has acid dissociation constants of Kai 4

A diprotic acid, H2A, has acid dissociation constants of Kai 4.46x 10-4 and Ka2 3.63x 10-12. Calculate the pH and molar concentrations of H2A, HA, and A* at equilibrium for each of the solutions below (a) a 0.203 M solution of H2A H,A] pH HA Number Number Number Number (b) a 0.203 M solution of NaHA H,A] [A2-] pH HA Number Number Number Number (c) a 0.203 M solution of Na2A 1H2 [A2-1 pH H,A HA Number Number Number Numher

Solution

a)

H2A -----------------> HA-   +    H+

0.203                          0            0

0.203 - x                      x             x

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

4.46 x 10^-4 = x^2 / 0.203 - x

x = 9.295 x 10^-3

[H+] = 9.295 x 10^-3 M

pH = -log[H+] = -log ( 9.295 x 10^-3)

pH = 2.03

[H2A] = 0.203 - 9.30 x 10^-3= 0.194

pH = 2.03

[H2A] = 0.194 M

[HA-] = 9.30 x 10^-3 M

[A2-] = Ka2 = 3.63 x 10^-12 M

b ) 0.203 M NaHA :

pH = pKa1 + pKa2 / 2

     = 3.35 + 11.44 / 2

pH = 7.40

[H+] = 10^-7.60 = 4.03 x 10^-8

[HA-] = 0.203 M

H2A   ------------> HA- +   H+

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

[H2A] = [HA-][H+] / Ka1 = 0.203x 4.03 x 10^-8 / 4.46 x 10^-4

[H2A] = 1.83 x 10^-5 M

HA-   -------------> H+   + A2-

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

[A2-] = Ka2 x [HA-] / [H+] = 3.63 x 10^-12 x 0.203 / 4.03 x 10^-8

[A2-] = 1.83 x 10^-5 M