Question

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

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

Answer 1

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

For $AgCl$ precipitation,
$\left[A{g}^{+}\right]=\frac{K_{sp}AgCl}{\left[C{l}^{-}\right]}=\frac{{10}^{-10}}{0.05}=2\times {10}^{-9}$
For $AgI$ precipation,
$\left[A{g}^{+}\right]=\frac{K_{sp}AgI}{\left[I^{-}\right]}=\frac{4\times {10}^{-16}}{0.05}=8\times {10}^{-15}$
Thus, $AgI$ will precipitate first. $AgCl$ will precipitate only when,
$\left[A{g}^{+}\right]=2\times {10}^{-9}$
Thus, $[I^{-}{]}_{Left}=\frac{4\times {10}^{-16}}{2\times {10}^{-9}}=2\times {10}^{-7}M$