A saturated solution of H₂S in water has concentration of approximately 0.10 M. What is the pH of this solution and equilibrium concentrations of H₂S, HS^– and S^2-? Hydrogen sulphide is a diprotic acid and its dissociation constants are K_a1 = 9.1 x 10^-8, K_a2 = 1.3 x 10^-13 mol L^-1 respectively.

The second dissociation constant is very small than the first dissociation constant and therefore, the concentration of H₃O⁺ is obtained only from the first dissociation constant:
H₂S + H₂O < -------------> H₃O⁺ + HS^–
Initial concentration:
[H₂S] = 0.10
[H₃O⁺] = 0
[HS^–] = 0
Let the amount of H₂S dissociated be x. Therefore,
At equilibrium:
[H₂S] = 0.10 – x
[H₃O⁺] = x
[HS^–] = x
Hence,
K_a1 = [H₃O⁺] [HS^–] / [H₂S]
= 9.1 x 10^-8
x² / (0.10 – x) = 9.1 x 10^-8
assuming x << 0.10 x² / 0.10 = 9.1 x 10^-8
x² = 9.1 x 10^-8 x 0.1 = 9.1 x 10^-9
x = 9.5 x 10^-5
therefore,
[H₃O⁺] = [HS^–] = 9.5 x 10^-5 mol L^-1
pH = – log [H₃O⁺] = – log (9.5 x 10^-5)
= – log 9.5 + 5 = -0.98 + 5 = 4.02
[H₂S] = 0.10 – 9.5 x 10^-5 = 0.10 M
To calculate the concentration of S^2- ion, we have to consider the second dissociation:
HS^– + H₂O < -------> H₃O⁺ + S^2-
Initial concentration:
[HS^–] = 9.5 x 10^-5
[H₃O⁺] = 9.5 x 10^-5
[S^2-] = 0
Let the amount of H₂S dissociated be x. Therefore,
At equilibrium:
[HS^–] = 9.5 x 10^-5 – x
[H₃O⁺] = 9.5 x 10^-5 + x
[S^2-] = x
Hence,
K_a2 = [H₃O⁺] [S^2-] / [HS^–]
= (9.5 x 10^-5 + x) (x) / (9.5 x 10^-5 – x)
Assuming x to be very small:
9.5 x 10^-5 – x ≈ 9.5 x 10^-5 and
9.5 x 10^-5 + x = 9.5 x 10^-5
1.3 x 10^-13 = 9.5 x 10^-5 x (x) / 9.5 x 10^-5
X = 1.3 x 10^-13
[S^2-] = 1.3 x 10^-13thus,
[H₃O⁺] = 9.5 x 10^-5 M
[HS^–] = 9.5 x 10^-5 M
[S^2-] = 1.3 x 10^-13 M
pH = 4.02