Question

A solution is 0.1 M in each KCl, KBr and $K{­}_{2}Cr{O}_4$ to this solution solid $AgN{O}_3$ is gradually added. Assume there is no change in the volume of the solution given:
Ksp (AgCl) = $1.7\times {10}^{-10}$
Ksp (AgBr) = $5.0\times {10}^{-13}$ and Ksp $\left(K_{2}Cr{O}_4\right)=1.9\times {10}^{-12}$
The concentration of first ion when the second ion starts precipitating is about:

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

Answer 1

The correct option is D 0.0003M

A solution is 0.1 M in each KCl, KBr and $K{­}_{2}Cr{O}_4$ to this solution solid $AgN{O}_3$ is gradually added. Assume there is no change in the volume of the solution given:
Ksp (AgCl) = $1.7\times {10}^{-10}$
Ksp (AgBr) = $5.0\times {10}^{-13}$ and Ksp $\left(K_{2}Cr{O}_4\right)=1.9\times {10}^{-12}$
$[Ag{]}^{+}$ to start the ppt. Of $c{l}^{-}$
$=\frac{Ksp\left(AgCl\right)}{\left[C{l}^{-}\right]}=\frac{1.7\times {10}^{-10}}{0.1}=1.7\times {10}^{-9}M$
When $C{l}^{-}$ starts precipitating
$\left[B{r}^{-}\right]=\frac{Ksp\left(AgBr\right)}{\left[A{g}^{+}\right]}=\frac{5\times {10}^{-18}}{1.7\times {10}^{-9}}=3\times {10}^{-4}M$