Question

A solution, containing $0.01MZ{n}^{+2}$ and $0.01MC{u}^{2+}$ is saturated by passing $H_{2}S$ gas. The $S^{-2}$ concentration is $8.1\times {10}^{-21}M$, $Ksp$ for $ZnS$ and $Cus$ are $3.0\times {10}^{-22}$ and $8.0\times {10}^{-36}$ respectively. Which of the following will occur in the solution :-

• A. $Zns$ will precipitate
• B. $Cus$ will precipitate
• C. Both $Zns$ and $Cus$ will precipitate
• D. Both $Z{n}^{2+}$ and $C{u}^{2+}$ will remain in the solution

Answer 1

The correct option is B $Cus$ will precipitate

Given

Molarity of $Z{n}^{2+}$= 0.01M

Molarity of $C{u}^{2+}$ = 0.01M

Conc. of $S^{2-}=8.1\ast {10}^{-21}M$

$K_{sp}ofZnS=3\ast {10}^{-22}$

$K_{sp}ofCuS=8\ast {10}^{-36}$

Solution

Ionic product of $ZnS=\left[Z{n}^{2+}\right]\ast \left[S^{2-}\right]$

Ionic product of $ZnS=0.01\ast (8.1\ast {10}^{-21})=8.1\ast {10}^{-23}$

Ionic product of $CuS=\left[C{u}^{2+}\right]\ast \left[S^{2-}\right]$

Ionic product of $CuS=0.01\ast (8.1\ast {10}^{-21})=8.1\ast {10}^{-23}$

Difference of solubility product and ionic product of CuS is very large as compare to difference of solubility product and ionic product of ZnS.

Therefore CuS will precipitate.

The correct option is B