Question

A solution is a mixture of $0.05MNaCl$ and $0.05MNaI$. The concentration of iodide ion in the solution when $AgCl$ just starts precipitating is equal to:
$K_{sp}\left(AgCl\right)=1\times {10}^{-10}{M}^2;K_{sp}\left(AgI\right)=4\times {10}^{-16}{M}^2$

• A. $2\times {10}^{-7}M$
• B. $2\times {10}^{-8}M$
• C. $4\times {10}^{-6}M$
• D. $8\times {10}^{-15}M$

Answer 1

The correct option is A $2\times {10}^{-7}M$

$K_{sp}\left(AgCl\right)>K_{sp}\left(AgI\right)$
Hence $AgI$ is precipitated before $AgCl$
$\left[A{g}^{+}\right]$ to precipitated, $=\frac{1\times {10}^{-10}}{0.05}=2\times {10}^{-9}M$
$\left[I^{-}\right]$ at this stage $=\frac{K_{sp}\left(AgI\right)}{\left[A{g}^{+}\right]}=\frac{4\times {10}^{-16}}{2\times {10}^{-9}}=2\times {10}^{-7}M$