Hydrogen sulfide, H2S, is a weak acid that can be used to precipitate metal ions from solution selec — Equilibrium / Solubility Chemistry Question
Problem Context
Hydrogen sulfide, H2S, is a weak acid that can be used to precipitate metal ions from solution selectively by controlling the pH.
Acid Ionization Constants, H2S: K1 5.7×10–8, K2 1.3×10–13
Ksp: Bi2S3 1.6×10–72, MnS 3.0×10–11
Write equations to represent each of the ionization steps of H2S.
Model Answer
H2S ⇌ H+ + HS– K1 = 5.7×10–8
HS– ⇌ H+ + S2– K2 = 1.3×10–13
Write an equation to represent the overall ionization of H2S to form S2– and 2H+ and calculate the equilibrium constant for this process.
Model Answer
H2S ⇌ 2H+ + S2– K = 7.4×10–21
For a solution with [H2S] = 0.10 M, with [Bi3+] = [Mn2+] = 1.5 mM and [H+] = 10 mM, give the formula for the metal sulfide which precipitates first and calculate the percentage of it that will remain in solution at equilibrium.
Model Answer
Calculate sulfide ion concentration:
K = [H+]2[S2–] / [H2S] = (0.010)2[S2–] / (0.1) = 7.4×10–21 so [S2–] = 7.4×10–18
Now calculate Q and compare to K for each cation (with sulfide):
Bismuth: Ksp = [Bi3+]2[S2–]3 = 1.6×10–72
Q = (1.5×10–3)2(7.4×10–18)3 = 9.1×10–58
Q > Ksp so there will be a precipitate formed.
Manganese: Ksp = [Mn2+][S2–] = 3.0×10–11
Q = (1.5×10–3)(7.4×10–18) = 1.1×10–20
Q < Ksp so there will not be a precipitate formed.
Thus – the bismuth is the first metal sulfide to precipitate.
Now to calculate what percentage will remain in solution:
[Bi3+]2 = Ksp / [S2–]3 = 1.6×10–72 / (7.4×10–18)3 = 3.95×10–21 so [Bi3+] = 6.3×10–11
The percentage can be calculated using the ratio of the amount remaining in solution divided by the original amount:
% = (6.9×10–11 / 1.5×10–3) × 100 = 4.2×10–6 %
The pH of the solution is raised until the other metal sulfide begins to precipitate. Determine the pH of the solution at which the second metal sulfide begins to precipitate.
Model Answer
first determine the concentration of sulfide that will result in precipitation:
[S2–] = Ksp / [Mn2+] = (3.0×10–11) / (1.5×10–3) = 2.0×10–8
Now plug this value into the equation for K from Part (c):
[H+]2 = K[H2S] / [S2–] = (7.4×10–21)(0.10) / (2.0×10–8) = 3.7×10–14 and [H+] = 1.92×10–7 so pH = 6.7