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Common Ion Effect

A beaker cont...

Question

A beaker contains a saturated solution of copper(I) chloride. Copper(I) chloride is slightly soluble salt with a solubility product of $1.2 \times 10^{- 6}$ .
Which of the following salts when added to the solution would precipitate copper(I) chloride?

A

Sodium chloride

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B

Potassium bromide

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C

Silver(I) nitrate

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D

Lead(II) acetate

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E

Magnesium iodide

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Solution

The correct option is B Sodium chloride
A beaker contains a saturated solution of copper(I) chloride. Copper(I) chloride is slightly soluble salt with a solubility product of $1.2 \times 10^{- 6}$ .
The addition of sodium chloride to the solution would precipitate copper(I) chloride.
Sodium chloride is a strong electrolyte. It completely dissociates to provide chloride ions.
Chloride ion is a common ion. It suppresses the solubility of copper(I) chloride.
With increase in chloride ion concentration, the the ionic product of copper(I) chloride exceeds its solubility product. This results in precipitation of copper(I) chloride.

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