Question

The solubility (in mol $L^{-1}$) of $AgCl$ ($K_{sp}=1.0\times {10}^{-10}$) in a $0.1$M $KCl$ solution will be:

Answer 1

Let us assume $\left[A{g}^{+}\right]=x$

then, at equilibrium concentration , ve see that, common ion is $c{l}^{-}$

so, the reaction is $Agcl\to A{g}^{+}+c{l}^{-}$

$K_{sp}=\left(A{g}^{+}\right)\left(c{l}^{-}\right)$

$=\left(x\right)(0.1+x)$

given, $k_{sp}=1.0\times {10}^{-10}$ and also $x<<0.1$

$\to x={10}^{-9}m$

$\therefore$ solubility of Agcl is ${10}^{-9}mol/litre.$