Question

In a saturated solution of $AgCl(K_{sp}=1.6\times {10}^{-10}$ at ${25}^{\circ }C)$, the $\left[A{g}^{+}\right]=1.3\times {10}^{-5}mol/L$. Then enough potassium chloride is added to this solution so that $\left[C{l}^{-}\right]=0.020M$. The solubility of $AgCl$ in this solution of potassium chloride is:

• A. $3.2\times {10}^{-8}mol/L$
• B. $8\times {10}^{-11}mol/L$
• C. $8\times {10}^{-8}mol/L$
• D. $8\times {10}^{-9}mol/L$

Answer 1

The correct option is D $8\times {10}^{-9}mol/L$

$Ksp\left[AgCl\right]=1.6\times {10}^{-10}Ksp=\left[{Ag}^{+}\right]\left[{Cl}^{-}\right]\left[{Ag}^{+}\right]=1.3\times {10}^{-5}\frac{mol}{L}$

Then, $\left[{Cl}^{-}\right]=0.020M$

now, $\left(AgCl\right)Ksp=\left[{Ag}^{+}\right]\left[{Cl}^{-}\right]1.6\times {10}^{-10}=\left[{Ag}^{+}\right]\left(0.02\right)8\times {10}^{-9}=\left[{Ag}^{+}\right]$

$⟶$ Solubility of $AgCl$ in $KCl$ solution is $8\times {10}^{-9}M$