Question

In $1L$ saturated solution of $AgCl\left(K_{sp}\right(AgCl)=1.6\times {10}^{-10}]$, $0.1$ mol of $CuCl\left[K_{sp}\right(CuCl)=1.0\times {10}^{-6}]$ is added. The resultant concentration of $A{g}^{+}$ in the solution is $1.6\times {10}^{-x}$. The value of ${}^{\prime }{x}^{\prime }$ is:

Answer 1

$AgC{l}_5\leftrightharpoons A{g}^{+}+C{l}^{-}$

$Ksp=\frac{\left[A{g}^{+}\right]\left[{Cl}^{-}\right]}{1}=1.6\times {10}^{-10}$

$⟹\left[{Ag}^{+}\right]=\left[{Cl}^{-}\right]=\sqrt{1.6\times {10}^{-10}}=1.26\times {10}^{-5}mol/L$

We add $0.1$ mol of $CuCl$

$CuC{l}_5\leftrightharpoons C{u}^{+}+C{l}^{-}$

$Ksp=\frac{\left[{Cu}^{+}\right]\left[{Cl}^{-}\right]}{1}=1.0\times {10}^{-6}$

The $AgCl$ equilibrium shifts towards left due to commonj ion effect $\left[a^{-}\right]$

$⟹\left[{Cu}^{+}\right]=\left[a^{-}\right]=\sqrt{1.0\times {10}^{-6}}={10}^{-3}M$

$M\left(CuCl\right)=63.5+37.5\cong 99$

$S=99\times {10}^{-3}=0.099$ $g/L$

We add $0.1$ $mol⟶9.9g$ to $1$ $Liter$

Then $Ksp$ exceeds by $1000⟶$ Saturated solution.

$⟹\left[A{g}^{+}\right]\left[a^{-}\right]=1.6\times {10}^{-10}$

$\left[A{g}^{+}\right]=\frac{1.6\times {10}^{-10}}{{10}^{-3}}=1.6\times {10}^{-7}$ $\frac{mol}{liter}$

The concentration of $\left[A{g}^{+}\right]$ has reduced the value of $x=7$