Question

Solid $AgN{O}_3$ is added to a solution which is $0.1M$ in $C{l}^{-}$ and $0.1M$ in $Cr{O}_4^{2-}$. If $K_{sp}$ values for $AgCl$ and $A{g}_{2}Cr{O}_4$ solutions are $1.7\times {10}^{-10}$ and $1.9\times {10}^{-12}$ respectively. Find the minimum concentration of $C{l}^{-}$ when $A{g}_{2}Cr{O}_4$ starts precipitating.

• A. $1.1\times {10}^{-7}M$
• B. $2.1\times {10}^{-8}M$
• C. $3.9\times {10}^{-5}M$
• D. $5.1\times {10}^{-9}M$

Answer 1

The correct option is C $3.9\times {10}^{-5}M$

The concentration of $A{g}^{+}$ required to precipitate $AgCl$ and $A{g}_{2}Cr{O}_4$ can be calculated as follows:
$K_{sp}\left(AgCl\right)=\left[A{g}^{+}\right]\left[C{l}^{-}\right]\left[A{g}^{+}\right]=\frac{1.7\times {10}^{-10}}{0.1}=1.7\times {10}^{-9}M.$
To precipitate $A{g}^{+}$ in $A{g}_{2}Cr{O}_4$ :
$K_{sp}\left(A{g}_{2}Cr{O}_4\right)=\left[A{g}^{+}{]}^2\right[Cr{O}_4^{2-}\left]\right[A{g}^{+}]=\sqrt{\frac{1.9\times {10}^{-12}}{0.1}}=4.35\times {10}^{-6}M.$
Since $\left[A{g}^{+}\right]$ required to precipitate $AgCl$ is less than that required to precipitate $A{g}_{2}Cr{O}_4,AgCl$ will precipitate first.
Now since $\left[A{g}^{+}\right]=4.35\times {10}^{-6}M$ when $A{g}_{2}Cr{O}_4$ starts precipitating, $\left[C{l}^{-}\right]$ at that state can be calculated as follows:
$K_{sp}\left(AgCl\right)=\left[A{g}^{+}\right]\left[C{l}^{-}\right]1.7\times {10}^{-10}=(4.35\times {10}^{-6})\left[C{l}^{-}\right]\left[C{l}^{-}\right]=3.9\times {10}^{-5}M$