A solution contains 0.021 mol L^-1 Cl^- and 0.017 mol L^-1 I^-.A solution containing copper(I)ions is added to selectively precipitate one of the ions.At what concentration of copper(I)ion will a precipitate begin to form? What is the identity of the precipitate? K_sp(CuCl)= 1.0 × 10^-6,K_sp(CuI)= 5.1 × 10^-12. A)3.0 × 10^-10mol L^-1,CuI B)3.0 × 10^-10 mol L^-1,CuCl C)4.8 × 10^-5 mol L^-1,CuCl D)4.8 × 10^-5 mol L^-1,CuI E)3.0 × 10^-5 mol L^-1,both CuCl and CuI are in the precipitate

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Quizplus · Accepted Answer

The concentration of Cu+ needed to start precipitating CuI is calculated using Ksp(CuI) = [Cu+][I-]. Solving for [Cu+] gives 3.0 × 10^-10 mol L^-1, which is lower than the concentration needed to precipitate CuCl, so CuI will precipitate first.

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