Question

The solubility product of $AgCl$ in water is $1.5\times {10}^{-10}$. Calculate its solubility in $0.01M$ $NaCl$ aqueous solution.

Answer 1

$AgCl\to A{g}^{+}+C{l}^{-}$

$S$ $S$

After addition of 0.01M $NaCl$, $\left[C{l}^{-}\right]=0.01+S$

$K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]=\left(S\right)(S+0.01)=S^2+0.01S$

We can neglect $S^2$ term as $S<<1$

So $K_{sp}=1.5\times {10}^{-10}=0.01S$

$S=1.5\times {10}^{-8}$