Question

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

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

Answer 1

Answer: (D)

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

Given that a solution is $0.01MKI$ and $0.1MKCl$.

When $\text{AgCl}$ begins to precipitate,

$\left[A{g}^{+}\right]\left[C{l}^{-}\right]=1.8\times {10}^{-10}=K_{sp}$ of $AgCl$

$\left[A{g}^{+}\right]=1.8\times {10}^{-9}$ as $\left[C{l}^{-}\right]=0.1$ which comes from $KCl$.

$K_{sp}$ of $AgI$ is $1.5\times {10}^{-16}$.

So, at this time, $\left[I^{-}\right]=\frac{K_{sp}}{\left[A{g}^{+}\right]}=\frac{1.5\times {10}^{-16}}{1.8\times {10}^{-9}}=8.3\times {10}^{-8}M$

Hence, the concentration of $I^{-}$ when $AgCl$ begins to precipitate is $8.3\times {10}^{-8}M$.

Hence, the correct option is $\text{D}$