A solution contains 0.036 M Cu²⁺ and 0.044 M 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 M, FeS B) 3.6 × 10^-35 M, CuS C) 3.6 × 10^-35 M, FeS D) 1.4 × 10^-16 M, CuS E) No precipitate will form at any concentration of sulfide ion.

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Quizplus · Accepted Answer

The precipitate will begin to form when the product of the metal ion concentration and the sulfide ion concentration equals the solubility product constant (Ksp) of the least soluble compound. For CuS, Ksp = 1.3 × 10^-36, and for FeS, Ksp = 6.3 × 10^-18. Since CuS has a much lower Ksp, it will precipitate first. To find the concentration of sulfide ions ([S^2-]) needed for CuS to begin precipitating, we use the formula Ksp = [Cu^2+][S^2-]. Substituting the given values, 1.3 × 10^-36 = 0.036 × [S^2-], solving for [S^2-] gives approximately 3.6 × 10^-35 M.

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