Question

On passing $H_2S$ into a solution containing both $Zn^{2+}$ and $Cu^{2+}$ in acidic medium, only $CuS$ gets precipitated. This is because

• A. $K_{sp}$ of $CuS < K_{sp}$ of $ZnS$
• B. $K_{sp}$ of $CuS = K_{sp}$ of $ZnS$
• C. $K_{sp}$ of $CuS > K_{sp}$ of $ZnS$
• D. $CuS$ is more stable than $ZnS$.

Answer 1

Answer: (A)

Since the solubility product of $CuS$ is less than that of $ZnS$, CuS gets precipitated.
$K _{ sp }( CuS )< K _{ sp }( ZnS )$
$K _{ sp }( CuS )=8 \times 10^{-37}$
$K _{ sp }( Zn S )=1.2 \times 10^{-23}$