A diprotic acid, H2A, has acid dissociation constants of Ka1 = 1.56× 10–4 and Ka

Question

A diprotic acid, H2A, has acid dissociation constants of Ka1 = 1.56× 10–4 and Ka2 = 3.21× 10–11. Calculate the pH and molar concentrations of H2A, HA–, and A2– at equilibrium for each of the solutions below.

A diprotic acid, H2A, has acid dissociation constants of Kal = 1.56x 10-4 and Ka2-3.21x10-11. Calculate the pH and molar concentrations of H2A, HA-, and A2 at equilibrium for each of the solutions below (a) a 0.113 M solution of H2A [H,A] pH HA Number Number Number Number (b) a 0.113 M solution of NaHA pH HA Number Number Number Number (c) a 0.113 M solution of Na2A [H,A] HA Number Number Number Number

Explanation / Answer

  a)

H2A -----------------> HA-   +    H+

0.113                          0            0

0.113 - x                      x             x

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

1.56 x 10^-4 = x^2 / 0.113 - x

x =4 .12 x 10^-3

[H+] = 4.12 x 10^-3 M

pH = -log[H+] = -log ( 4.12 x 10^-3)

pH = 2.38

[H2A] = 0.186 - 4.12 x 10^-3 = 0.109

pH = 2.25

[H2A] = 0.109 M

[HA-] = 4.12 x 10^-3 M

[A2-] = Ka2 = 3.21 x 10^-11 M

b)

pH = 7.15

[H2A] = 5.13 x 10^-5 M

[HA-] = 0.113 M

[A2-] = 5.12 x 10^-5 M

c)

A2-   +    H2O    -------------> HA-   +    OH-

0.113                                        0            0

0.113-x                                     x             x

Kb1 = x^2 / 0.113 - x

Kw / Ka2 = x^2 / 0.113 - x

3.12 x 10^-4 = x^2 / 0.113 - x

x = 5.78 x 10^-3

[A2-] = 0.113 - 5.78 x 10^-3 = 0.107 M

[A2-] = 0.107 M

[HA-] = 5.78 x 10^-3 M

pOH = -log (5.78 x 10^-3 ) = 2.24

pH = 11.76

H2A   -------------> HA-   +   H+

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

          = 1.74 x 10^-12 x 5.78 x 10^-3 / 1.56 x 10^-4

[H2A] = 6.44 x 10^-11

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