Question

The solubility product of $AgI$ at $25^{\circ} C$ is $1.0 \times 10^{-16}\, mol ^{2} \, L ^{-2}$. The solubility of $AgI$ in $10^{-4} \, N$ solution of $KI$ at $25^{\circ} C$ is approximately (in mol L ${ }^{-1}$ )

• A. $ 1.0 \times 10^{-10} $
• B. $ 1.0 \times 10^{-8} $
• C. $ 1.0 \times 10^{-16} $
• D. $ 1.0 \times 10^{-12} $

Answer 1

Answer: (D)

$A g I \rightarrow A g^{+}+I^{-}$
For binary electrolyte
$K_{s p}=S^{2}$
where, $S=$ solubility in mol $/ L$
$1.0 \times 10^{-16}=S^{2}$
or $S=1 \times 10^{-8} \,mol / L$
Solubility of $Agl$ in $KI$ solution
$=1 \times 10^{-8} \times 10^{-4}$
$=1 \times 10^{-12} \,mol / L$