Question

A solution contains $0.1M$ ${Cl}^{-}$ and $0.001M$ $Cr{O}_4^{2-}$. If solid $Ag{NO}_3$ is gradually added to this solution, which will precipitate first, $AgCl$ or ${Ag}_{2}Cr{O}_4$? Assume that the addition causes no change in volume. Given
$K_{sp}AgCl=1.7\times {10}^{-10}{M}^2$
$K_{sp}{Ag}_{2}Cr{O}_4=1.79\times {10}^{-12}{M}^3$
What percent of ${Cl}^{-}$ remains in the solution when $Cr{O}_4^{2-}$ starts precipitating?

Answer 1

${Ag}^{+}$ ion concentration required for precipitation

For $AgCl$:

$\left[{Ag}^{+}\right]=\frac{K_{sp}AgCl}{\left[{Cl}^{-}\right]}=\frac{1.7\times {10}^{-10}}{0.1}=1.7\times {10}^{-9}M$

For ${Ag}_{2}Cr{O}_4:$

${\left[{Ag}^{+}\right]}^2=\frac{K_{sp}{Ag}_{2}Cr{O}_4}{\left[Cr{O}_4^{2-}\right]}=\frac{1.79\times {10}^{-12}}{\left[0.001\right]}$

$\left[{Ag}^{+}\right]={[1.79\times {10}^{-9}{M}^2]}^{1/2}=4.23\times {10}^{-5}M$

$AgCl$ will precipitate first because it required low conc. of ${Ag}^{+}$

Remaining conc. of $\left[{Cl}^{-}\right]=\frac{K_{sp}AgCl}{\left[{Ag}^{+}\right]}=\frac{1.7\times {10}^{-10}}{4.23\times {10}^{-5}}=4.01\times {10}^{-6}M$

% Remaining conc. of $\left[{Cl}^{-}\right]=\frac{4.01\times {10}^{-6}}{0.1}\times 100=4\times {10}^{-3}$%