Question

At ${25}^o$C, the $K_{sp}$ value of AgCl is $1.8\times {10}^{-10}$. If ${10}^{-5}$ moles of $A{g}^{+}$ are added to solution then $K_{sp}$ will be ?

Answer 1

The equation for reaction is

$AgC{l}_{\left(s\right)}⟶A{g}_{\left(aq\right)}^{+}+C{l}_{\left(aq\right)}^{-}$

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

Given, $K_{sp}=1.8\times {10}^{-10}$

$\Rightarrow 1.8\times {10}^{-10}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$

$\Rightarrow S^2=1.8\times {10}^{-10}$

$\Rightarrow S=1.34\times {10}^{-5}$

When ${10}^{-5}$ moles of $A{g}^{+}$ are added

$K_{sp}=[S+{10}^{-5}]\left[S\right]$

$=(1.34\times {10}^{-5}+{10}^{-5})(1.34\times {10}^{-5})$

$=3.13\times {10}^{-25}$

The value of $K_{sp}$ is $3.13\times {10}^{-25}$ .