Question

A solution is made of $0.1M$ $C{l}^{-}$ and ${10}^{-4}MCr{O}_4^{2-}$. If solid $AgN{O}_3$ is gradually added to this solution, then what will be the concentration of $C{l}^{-}$ when $A{g}_{2}Cr{O}_4$ begins to precipitate?
$\left[K_{sp}\right(AgCl)={10}^{-10}{M}^2;K_{sp}(A{g}_{2}Cr{O}_4)={10}^{-12}{M}^3]$

• A. ${10}^{-6}M$
• B. ${10}^{-4}M$
• C. ${10}^{-5}M$
• D. ${10}^{-9}M$

Answer 1

The correct option is A ${10}^{-6}M$

When $A{g}_{2}Cr{O}_4$ begins to precipitate, $\left[A{g}^{+}{]}^2\right[Cr{O}_4^{2-}]={10}^{-12};[A{g}^{+}]={10}^{-4};as[Cr{O}_4^{2-}]={10}^{-4}M$.

So, at this time:

$\left[C{l}^{-}\right]=\frac{k_{sp}}{\left[A{g}^{+}\right]}=\frac{{10}^{-10}}{{10}^{-4}}={10}^{-6}M$.