Question

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? Ksp(CuCl) = 1.0 × 10-6, Ksp(CuI) = 5.1 × 10-12.

Answer 1

Answer: Copper (I) iodide will precipitate first.

Explanation:

We are given:

$K_{sp}$ of CuCl = $1.0\times 10^{-6}$

$K_{sp}$ of CuI = $5.1\times 10^{-12}$

Concentration of $Cl^-\text{ ion}=0.021M$

Concentration of $I^-\text{ ion}=0.017M$

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

• For CuCl:

$K_{sp}=[Cu^+][Cl^-]$

Putting values in above equation, we get:

$1.0\times 10^{-6}=[Cu^+]\times 0.021$

$[Cu^+]=\frac{1.0\times 10^{-6}}{0.021}=4.76\times 10^{-5}M$

Concentration of copper (I) ion = $4.76\times 10^{-5}M$

• For CuI:

$K_{sp}=[Cu^+][I^-]$

Putting values in above equation, we get:

$5.1\times 10^{-12}=[Cu^+]\times 0.017$

$[Cu^+]=\frac{5.1\times 10^{-12}}{0.017}=3.00\times 10^{-10}M$

Concentration of copper (I) ion = $3.00\times 10^{-10}M$

For the precipitation of copper (I) ions, we need less concentration of copper (I) ions. So, copper (I) iodide will precipitate first.