Question

Find out the solubility of $AgCl$ in $0.1MHCl$. Given that the ionic product of $AgCl$ is $1.8\times {10}^{-10}$.

• A. $1.8\times {10}^{-10}M$
• B. $1.8\times {10}^{-9}M$
• C. $1.34\times {10}^{-5}M$
• D. $1.25\times {10}^{-8}M$

Answer 1

The correct option is B $1.8\times {10}^{-9}M$

$HCl\to \underset{0.1M}{H^{+}}+\underset{0.1M}{C{l}^{-}}$
$AgCl\left(s\right)\rightleftharpoons \underset{\left(s\right)M}{A{g}^{+}(aq.)}+\underset{(s+0.1)M}{C{l}^{-}(aq.)}$
$K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]=\left(s\right)(s+0.1)\simeq s\times 0.1$
$s=\frac{K_{sp}}{0.1}=\frac{1.8\times {10}^{-10}}{0.1}=1.8\times {10}^{-9}M$