Question

The solubility product of ${Ag}_2{C}_2{O}_4$ at ${25}^{o}C$ is $1.29\times {10}^{-11}{mol}^3{L}^{-3}$. A solution of $K_2{C}_2{O}_4$ containing $0.1520mol$ in $500mL$ of water is shaken with excess of ${Ag}_2{CO}_3$ till the following equilibrium is reached.
${Ag}_2{CO}_3+K_2{C}_2{O}_4\rightleftharpoons {Ag}_2{C}_2{O}_4+K_2{CO}_3$
At equilibrium, the solution contains $0.0358mole$ of $K_2{CO}_3$. Assuming the degree of dissociation of $K_2{C}_2{O}_4$ and $K_2{CO}_3$ to be equal, calculate the solubility product of ${Ag}_2{CO}_3$.

Answer 1

${Ag}_2{CO}_3+K_2{C}_2{O}_4\rightleftharpoons {Ag}_2{C}_2{O}_4+K_2{CO}_3$
Initial $0.1520$ $0$
At equi. $(0.1520-0.0358)$ $0.0358$
$=0.1162$
Conc. $2\times 0.1162$ $2\times 0.0358$
$=0.23324M$ $=0.0716M$
$K_{sp}{Ag}_2{C}_2{O}_4={\left[{Ag}^{+}\right]}^2\left[C_2{O}_4^{2-}\right]$
$\left[{Ag}^{+}\right]={\left[\frac{K_{sp}{Ag}_2{C}_2{O}_4}{\left[C_2{O}_4^{2-}\right]}\right]}^{1/2}$
$K_{sp}{Ag}_2{CO}_3={\left[{Ag}^{+}\right]}^2\left[C{O}_3^{2-}\right]$
$\left[{Ag}^{+}\right]={\left[\frac{K_{sp}{Ag}_2{CO}_3}{\left[C{O}_3^{2-}\right]}\right]}^{1/2}$
So, ${\left[\frac{K_{sp}{Ag}_2{C}_2{O}_4}{\left[C_2{O}_4^{2-}\right]}\right]}^{1/2}={\left[\frac{K_{sp}{Ag}_2{CO}_3}{\left[C{O}_3^{2-}\right]}\right]}^{1/2}$
or $K_{sp}{Ag}_2{CO}_3=\frac{K_{sp}{Ag}_2{C}_2{O}_4\times \left[C{O}_3^{2-}\right]}{\left[C_2{O}_4^{2-}\right]}$
$=K_{sp}{Ag}_2{C}_2{O}_4\frac{\left[K_2{CO}_3\right]}{\left[K_2{C}_2{O}_4\right]}=1.29\times {10}^{-11}\times \frac{0.0716}{0.2324}=3.97\times {10}^{-12}{mol}^3{L}^{-3}$