A solution contains 0.021 M Cl⁻ and 0.017 M 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^-10M, CuI B) 3.0 × 10^-10 M, CuCl C) 4.8 × 10^-5 M, CuCl D) 4.8 × 10^-5 M, CuI E) No precipitate will form at any concentration of copper(I).

Question

Quizplus · Accepted Answer

To determine at what concentration of copper(I)ion a precipitate will begin to form, we need to compare the Ksp values of copper(I) chloride (CuCl) and copper(I) iodide (CuI) with their respective ion product (IP) values in the solution:
- For CuCl: Ksp = 1.0 × 10^-6, IP = [Cu+][Cl^-] = (unknown concentration)[0.021 M] = 0.021x
- For CuI: Ksp = 5.1 × 10^-12, IP = [Cu+][I^-] = (unknown concentration)[0.017 M] = 0.017x

The ion product (IP) value will exceed the Ksp value and precipitation will occur when x is equal to the solubility product constant (Ksp) for the compound. The compound with the smaller Ksp value will be the first to precipitate. Therefore, we need to find the concentration of copper(I)ion where either Ksp is exceeded.

For CuCl: 1.0 × 10^-6 = 0.021x^2, solving for x gives x ≈ 1.5 × 10^-3 M
For CuI: 5.1 × 10^-12 = 0.017x^2, solving for x gives x ≈ 3.0 × 10^-6 M

Therefore, when the concentration of copper(I)ion is about 3.0 × 10^-6 M, copper(I) iodide (CuI) will begin to precipitate first because its Ksp value is smaller. Hence, the identity of the precipitate is CuI. Thus, the best choice is A.

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