Question 5 of 6: Selective sulfide precipitation of Mn²⁺ and Cu²⁺
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2018 — 12-Mtl-B2, Hydrometallurgy and Electrometallurgy. Three hours, closed book, approved Sharp/Casio calculator only. Six numbered Problems, each worth 20 marks: Problems 1 and 2 are compulsory; the rubric asks for any 3 of the remaining 4 (Problems 3-6). All six Problems are answered here, since this set is a study resource rather than an exam script. Given constants: R = 8.314 J/(mol K); F = 96,485 C/g-eq; for all aqueous species, activities are taken equal to concentrations.
Note on the exam title
Nothing on the paper is a pyrometallurgy (roasting, smelting) question — the syllabus actually examined is aqueous flow-sheeting terminology, cyanide-complex electrochemistry (Eh-pH diagram reading), metal-hydroxide speciation/solubility, sulfide precipitation, and electrowinning energetics.
Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:
F. Habashi, A Textbook of Hydrometallurgy, 2nd ed. — leaching, solubility/speciation diagrams, solvent extraction, ion exchange, cementation and precipitation, electrowinning and electrorefining.
M. L. Free, Hydrometallurgy: Fundamentals and Applications, 1st ed. — Eh-pH (Pourbaix) diagram construction and reading, aqueous speciation, electrochemical cell design.
W. G. Davenport, M. King, M. Schlesinger & A. K. Biswas, Extractive Metallurgy of Copper, 4th ed. — electrowinning/electrorefining energy balances and current efficiency.
D. M. Marsden & C. I. House, The Chemistry of Gold Extraction, 2nd ed. — cyanide-complex electrochemistry, gold cementation and auxiliary oxidants (Problem 3).
T. Rosenqvist, Principles of Extractive Metallurgy, 2nd ed. — general extractive-metallurgy context.
Problem 5 — Selective sulfide precipitation of Mn²⁺ and Cu²⁺ (20 marks)
Find. (a) Whether Mn²⁺ and Cu²⁺ each precipitate as sulfide; (b) equilibrium [Cu²⁺] remaining; (c) whether MnS precipitates once $[\text{H}^+]$ is lowered to $10^{-7}$ M.
Approach. Combine the two H₂S dissociation steps into $[\text{S}^{2-}]=K_{a1}K_{a2}[\text{H}_2\text{S}]/[\text{H}^+]^2$, then compare the reaction quotient $Q=[\text{M}^{2+}][\text{S}^{2-}]$ against each metal's own $K_{sp}$: $Q \lt K_{sp}$ means undersaturated (no precipitate), $Q>K_{sp}$ means supersaturated (precipitates) — the three-way state distinguished in Problem 2(g).
(a) Sulfide ion concentration. $$[\text{S}^{2-}]=\frac{K_{a1}K_{a2}[\text{H}_2\text{S}]}{[\text{H}^+]^2}=\frac{(1\times10^{-7})(1.2\times10^{-13})(0.1)}{(0.003)^2}=\boxed{1.333\times10^{-16}\ \text{M}}$$
Reaction quotients. $$Q=[\text{M}^{2+}][\text{S}^{2-}]=(2\times10^{-4})(1.333\times10^{-16})=\boxed{2.667\times10^{-20}}\ \text{(same for both ions, equal starting moles)}$$ For Mn²⁺: $Q=2.667\times10^{-20} \lt K_{sp}(\text{MnS})=3\times10^{-14}$ → undersaturated, MnS does NOT precipitate. For Cu²⁺: $Q=2.667\times10^{-20} \gg K_{sp}(\text{CuS})=8\times10^{-37}$ → hugely supersaturated, CuS precipitates.
(b) Equilibrium [Cu2+] remaining. Once CuS(s) has precipitated to equilibrium, $[\text{Cu}^{2+}]=K_{sp}(\text{CuS})/[\text{S}^{2-}]$ (the same $[\text{S}^{2-}]$ from step 1, buffered by the large excess H₂S/H⁺ reservoir): $$[\text{Cu}^{2+}]=\frac{8\times10^{-37}}{1.333\times10^{-16}}=\boxed{6.0\times10^{-21}\ \text{M}}$$ — sixteen orders of magnitude below the initial $2\times10^{-4}$ M, i.e. essentially 100% of the copper is removed as CuS.
(c) MnS at [H+] = 1e-7 M (pH 7). Recompute $[\text{S}^{2-}]$ at the new, much lower $[\text{H}^+]$: $$[\text{S}^{2-}]=\frac{(1\times10^{-7})(1.2\times10^{-13})(0.1)}{(1\times10^{-7})^2}=\boxed{1.2\times10^{-7}\ \text{M}}$$ Since no MnS precipitated in part (a), $[\text{Mn}^{2+}]$ is still $2\times10^{-4}$ M, so $$Q_{\text{new}}=(2\times10^{-4})(1.2\times10^{-7})=\boxed{2.4\times10^{-11}} \;\gg\; K_{sp}(\text{MnS})=3\times10^{-14}$$ — now supersaturated: MnS WILL precipitate once the pH is raised to 7.
