Question

A saturated solution of $H_{2}S$ in $0.1$ $M$ $HCl$ at ${25}^{o}C$ contains $S^{2-}$ ion concentration of ${10}^{-23}$ $mol$ $L^{-1}$. The solubility product of some sulphides are $CuS={10}^{-44}$, $FeS={10}^{-14}$, $MnS={10}^{-15}$, $CdS={10}^{-25}$. If $0.01M$ solution of these salts in $1M$ $HCl$ are saturated with $H_{2}S$, which of these will be precipitated?

• A. All
• B. All except $MnS$
• C. All except $MnS$ and $FeS$
• D. Only $CuS$

Answer 1

The correct option is D Only $CuS$

$\left[S^{2-}\right]={10}^{-23}$ $mol.L^{-1}$

$\left[M^{2+}\right]={10}^{-2}M$

Ionic product, $K_{IP}=\left[M^{2+}\right]\left[S^{2-}\right]={10}^{-25}$,

$\therefore$ Ionic product is greater than $K_{sp}$ of $CuS$ and equal for $CdS$ and less for others.

Therefore, only CuS will precipitate out.