Question

Calculate solubility of $AgCl$ in $0.1MNaCl$ at ${25}^{o}C$ if its solubility product at same temperature is $2.0\times {10}^{-10}mo{l}^2{L}^{-2}$.

• A. $2.0\times {10}^{-7}mol{L}^{-1}$
• B. $2.0\times {10}^{-9}mol{L}^{-1}$
• C. $2.5\times {10}^{-7}mol{L}^{-1}$
• D. $3.0\times {10}^{-9}mol{L}^{-1}$

Answer 1

The correct option is B $2.0\times {10}^{-9}mol{L}^{-1}$

$AgCl$ and $NaCl$ has univalent cation and anion.
We know,
In presence of common ion
$Solubility=\frac{K_{sp}}{\text{Concentration of common ion}}$

Putting the values,
$=\frac{2.0\times {10}^{-10}}{0.1}$
$=2.0\times {10}^{-9}mol{L}^{-1}$