Question

$H_2S$ is passed into one $dm^3$ of a solution containing $0.1\, mole$ of $Zn^{2+}$ and $0.01\, mole$ of $Cu^{2+}$ till the sulphide ion concentration reaches $8.1 ×10^{-19}$ moles. Which one of the following statements is true?
[$K_{sp}$ of $ZnS$ and $CuS$ are $3 × 10^{-22}$ and $8 × 10^{-36}$ respectively]

• A. Only $ZnS$ precipitates
• B. Both $CuS$ and $ZnS$ precipitate
• C. Only $CuS$ precipitates
• D. No precipitation occurs

Answer 1

Answer: (D)

Precipitation occurs only when ionic product exceeds the value of solubility product.
$1 \,dm^3$ of a solution containing $0.1\, mole$ of $Zn^{2+}, 0.01\, mole$ of $Cu^{2+}$ and $8.1 × 10^{-19}\, mole$ of $S^{2+}$. Let us calculate the ionic product in each case.
Ionic product of $ZnS = [Zn^{2+}][S^{2-}]$
$= 0.1 × 8.1 × 10^{-19} = 8.1 × 10^{-20}$
$Ksp$ of $ZnS = 3 × 10^{-22}$
$∴$ Ionic product $> K_{sp}$
Ionic product of $CuS = [Cu^{2+}][S^{2-}]$
$= 0.01 × 8.1 × 10^{-19} = 8.1 × 10^{-21}$
But it has $K_{sp} = 8 × 10^{-36}$
$∴$ Ionic product $> K_{sp}$
Both $ZnS$ and $CuS$ have less $K_{sp}$ values than their respective ionic products so $ZnS$ and $CuS$ both precipitates.