Question

Solubility of $AgCl$ in $0.1$ M $AgN{O}_3$ is ${10}^{-9}M.$ Its solubility in $0.1$ M $HCl$ is :

• A. ${10}^{-9}M$
• B. ${10}^{-10}M$
• C. ${10}^{-5}M$
• D. none of these

Answer 1

The correct option is A ${10}^{-9}M$

$AgCl\rightleftharpoons A{g}^{+}+C{l}^{-}$
$0.1$ $x$
Now,

$K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]=0.1\times {10}^{-9}={10}^{-10}$

$\therefore$ Now, $AgCl\rightleftharpoons A{g}^{+}+C{l}^{-}$
$y$ $0.1$

$K_{sp}=y\times 0.1={10}^{-10}$

$\therefore y={10}^{-9},$

$\therefore S={10}^{-9}M$.