Question

A solution is $0.1M$ in each $KCl,KBr$ and $K_{2}Cr{O}_4$. To this solution solid $AgN{O}_3$ is gradually added. Assume no change in the volume of the solution.
Given: $K_{sp}\left(AgCl\right)=1.7\times {10}^{-10}$
$K_{sp}\left(AgBr\right)=5\times {10}^{-13}$
and $K_{sp}\left(A{g}_{2}Cr{O}_4\right)=1.9\times {10}^{-12}$

The concentration of first ion when the second ion starts precipitating is:

• A. $0.1M$
• B. $0.03M$
• C. $0.003M$
• D. $0.0003M$

Answer 1

The correct option is D $0.0003M$

$\left[A{g}^{+}\right]\text{to start the precipitation of}C{l}^{-}=\frac{K_{sp}\left(AgCl\right)}{\left[C{l}^{-}\right]}=\frac{1.7\times {10}^{-10}}{0.1}=1.7\times {10}^{-9}M$

$\left[A{g}^{+}\right]\text{to start the precipitation of}B{r}^{-}=\frac{K_{sp}\left(AgBr\right)}{\left[B{r}^{-}\right]}=\frac{5\times {10}^{-13}}{0.1}=5\times {10}^{-12}M$

$\left[A{g}^{+}\right]\text{to start the precipitation of}Cr{O}_{{}_4}^{2-}={\left[\frac{K_{sp}\left(A{g}_{2}Cr{O}_4\right)}{\left[Cr{O}_{{}_4}^{2-}\right]}\right]}^{1/2}={\left[\frac{1.9\times {10}^{-12}}{0.1}\right]}^{1/2}=4.36\times {10}^{-6}M$

Order of precipitation is $B{r}^{-}$, $C{l}^{-}$, $Cr{O}_{{}_4}^{2-}$.

$\text{When}C{l}^{-}\text{starts precipitating then}\left[B{r}^{-}\right]=\frac{K_{sp}\left(AgBr\right)}{\left[A{g}^{+}\right]}=\frac{5\times {10}^{-13}}{1.7\times {10}^{-9}}=3\times {10}^{-4}M$