A solution contains 0.036 mol L^-1 Cu²⁺ and 0.044 mol L^-1 Fe²⁺. A solution containing sulfide ions is added to selectively precipitate one of the metal ions from solution. At what concentration of sulfide ion will a precipitate begin to form? What is the identity of the precipitate? K_sp(CuS) = 1.3 × 10^-36, K_sp(FeS) = 6.3 × 10^-18. A) 1.4 × 10^-16 mol L^-1, FeS B) 3.6 × 10^-35 mol L^-1, CuS C) 3.6 × 10^-35 mol L^-1, FeS D) 1.4 × 10^-16 mol L^-1, CuS E) 1.4 × 10^-16 mol L^-1, both CuS and FeS will precipitate

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Quizplus · Accepted Answer

To determine the concentration of sulfide ions needed to precipitate each metal ion, use the Ksp expression: [M^2+][S^2-] = Ksp. For CuS, [S^2-] = Ksp(CuS) / [Cu^2+] = 1.3 × 10^-36 / 0.036 = 3.6 × 10^-35 mol L^-1. For FeS, [S^2-] = Ksp(FeS) / [Fe^2+] = 6.3 × 10^-18 / 0.044 = 1.4 × 10^-16 mol L^-1. CuS will precipitate first at 3.6 × 10^-35 mol L^-1.

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