Help with a metal sulfide molar solubility question! :)

Question

Calculate the molar solubility of a hypothetical metal sulfide, MS, in a solution saturated with H₂S at 25C. The pH of the solution is 3.44. Kspa for MS is 1.7 x 10-3.

 

I know that I need to use the pH to find the H₃O⁺ concentration... and I am pretty sure the rxn is:

MS + 2H₃O <-> H₂S + M + 2H₂O

 

But I am not sure what to do once I have the [H3O+]... Thanks!

Answer 1

H₂S is a weak diprotic acid and dissociates as follows in two steps.

I) H₂S(g) ⇔ HS^1-_(aq) + H⁺_(aq)

_{from this 1:1 dissociation} [H⁺] = [HS^1-]

Ka1 = [HS1-] [H⁺] / [H₂S]. =. 9.5 x 10^-8. (Ka1 and Ka2 values are taken from literature)

II) HS^1-(aq). ⇔ S^2-_(aq) +. H⁺_(aq)

Ka2. = [S^2-] [H⁺] / [HS^1-]. = 1.0 x 10^-14

Ka2 is very small, therefore, contribution to H⁺ concentration from the second dissociation is negligible.

[H⁺] = 10^-3.44 = 3.63 x 10^-4 (given)

If you substitute the known concentrations in Ka2 expression:

1.0 x10^-14 = 3.63 x 10^-4 [S^2-] / 3.63 x 10^-4

therefore, [S^2-] = 1.0 x 10^-14 = Ka2

Now we can look at the dissolution equilibrium of MS (in the presence of a common ion, S^2-, coming from H₂S).

MS_(s) ⇔ M^2-_(aq) + S^2-_(aq)

Ksp = [M²⁺] [S^2-] = 1.7 x 10^-3 (Given)

developing an ICE table at this point would be helpful.

I.conc. 0, 1.0x10^-14

C.conc. +X, +X

E.conc. +X, (1.0x10^-14 + X)

substitute the concentrations of M²⁺ and S^2- in the Ksp expression.

1.7 x10^-3 = X (1.0x10^-14 + X)

solving this equation using the quadratic solution should give a value for X as 0.041,

Answer: the solubility of MS in a saturated solution of H₂S is 0.041 mol/L