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xeze [42]
3 years ago
8

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 t

he 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.
Chemistry
1 answer:
lidiya [134]3 years ago
7 0

<u>Answer:</u> Copper (I) iodide will precipitate first.

<u>Explanation:</u>

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.

  • <u>For CuCl:</u>

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

  • <u>For CuI:</u>

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.

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