Question

The solubility of $AgCl$ in 0.1 $KCl$ solution $(K_{sp}=1.2\times {10}^{-10})$ is ___ mol/L.

Answer 1

Assume concentration o $A{g}^{+}=x$

$AgCl⟶A{g}^{+}+C{l}^{-}$

At equilibrium, $\left[A{g}^{+}\right]=x,\left[C{l}^{-}\right]=(x+0.1)$

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

Actually; $x<<<0.1$

So, term $(0.1+x)$ becomes $0.1$ only

$\therefore {10}^{-10}=0.1x$

$\Rightarrow x={10}^{-9}M$

So, solubility of $AgCl$ is ${10}^{-9}mol/L$