Chemical and Ionic Equilibrium - Result Question 125
68. The solubility product of $Ag _2 C _2 O _4$ at $25^{\circ} C$ is $1.29 \times 10^{-11} mol^{3} L^{-3}$. A solution of $K _2 C _2 O _4$ containing $0.1520$ mole in $500 $ $mL$ water is shaken at $25^{\circ} C$ 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 $0.0358$ mole 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 $Ag _2 CO _3$.
(1991,4 M)
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Answer:
Correct Answer: 68. $\left(9.67 \times 10^{-11}\right)$
Solution:
- Initial concentration of $K _2 C _2 O _4=\frac{0.152}{0.50}=0.304 $ $M$,
Also for the following equilibrium:
$\mathrm{Ag}_2 \mathrm{CO}_3(s)+ \underset{0.304-x}{\mathrm{K}_2 \mathrm{C}_2 \mathrm{O}_4(a q)} \rightleftharpoons \mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_4(s)+\mathrm{K}_2 \mathrm{CO}_3 $
$K=\frac{\left[\mathrm{CO}_3^{2-}\right]}{\left[\mathrm{C}_2 \mathrm{O}4^{2-}\right]} \times \frac{\left[\mathrm{Ag}^{+}\right]^2}{\left[\mathrm{Ag}^{+}\right]^2}=\frac{K{\text {sp }}\left(\mathrm{Ag}_2 \mathrm{CO}3\right)}{K{\text {sp }}\left(\mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_4\right)}$
$\begin{aligned} & \text { Given, } 0.304-x=0.0358 \\ & \Rightarrow x=0.2682 \\ & \Rightarrow K=\frac{0.2682}{0.0358}=7.5 \\ & K_{\text {sp }}\left(\mathrm{Ag}_2 \mathrm{CO}3\right)=K \times K{\text {sp }}\left(\mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_4\right) \\ & =7.5 \times 1.29 \times 10^{-11} \\ & =9.675 \times 10^{-11} \\ \end{aligned}$
BETA
