Question

A solution is a mixture of $0.01$ M $KI$ and $0.1$ M $KCl$. If solid $AgN{O}_3$ is added to the solution, what is the $\left[I^{-}\right]$ when $AgCl$ begins to precipitate?
$\left[K_{sp}\right(AgI)=1.5\times {10}^{-16};[K_{sp}\left(AgCl\right)=1.8\times {10}^{-10}]$

• A. $3.5\times {10}^{-7}$
• B. $6.1\times {10}^{-8}$
• C. $2.2\times {10}^{-7}$
• D. $8.3\times {10}^{-8}$

Answer 1

Answer: (D)

$8.3\times {10}^{-8}$

$K_{sp}\left(AgCl\right)=\left[{Ag}^{+}\right]\left[{Cl}^{-}\right]$

$KCl\to$ Strong electrolyte $\Rightarrow \left[KCl\right]=\left[{Cl}^{-}\right]=0.1M$

So, concentration of ${Ag}^{+}$ when $AgCl$ starts precipitating is

$\left[{Ag}^{+}\right]=\frac{K_{sp}\left(AgCl\right)}{\left[{Cl}^{-}\right]}=\frac{1.8\times {10}^{-10}}{0.1}=1.8\times {10}^{-9}$

now $K_{sp}\left[AgI\right]=\left[{Ag}^{+}\right]\left[I^{-}\right]$

$\left[I^{-}\right]=\frac{1.5\times {10}^{-16}}{1.8\times {10}^{-9}}=8.3\times {10}^{-8}M$