Chemical and Ionic Equilibrium 2 Question 69
68. The solubility product of $\mathrm{Ag}{2} \mathrm{C}{2} \mathrm{O}{4}$ at $25^{\circ} \mathrm{C}$ is $1.29 \times 10^{-11} \mathrm{~mol}^{3} \mathrm{~L}^{-3}$. A solution of $\mathrm{K}{2} \mathrm{C}{2} \mathrm{O}{4}$ containing 0.1520 mole in $500 \mathrm{~mL}$ water is shaken at $25^{\circ} \mathrm{C}$ with excess of $\mathrm{Ag}{2} \mathrm{CO}{3}$ till the following equilibrium is reached
$$ \mathrm{Ag}{2} \mathrm{CO}{3}+\mathrm{K}{2} \mathrm{C}{2} \mathrm{O}{4} \rightleftharpoons \mathrm{Ag}{2} \mathrm{C}{2} \mathrm{O}{4}+\mathrm{K}{2} \mathrm{CO}{3} $$
At equilibrium, the solution contains 0.0358 mole of $\mathrm{K}{2} \mathrm{CO}{3}$. Assuming the degree of dissociation of $\mathrm{K}{2} \mathrm{C}{2} \mathrm{O}{4}$ and $\mathrm{K}{2} \mathrm{CO}{3}$ to be equal, calculate the solubility product of $\mathrm{Ag}{2} \mathrm{CO}_{3}$.
$(1991,4 \mathrm{M})$
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Solution:
- Initial concentration of $\mathrm{K}{2} \mathrm{C}{2} \mathrm{O}_{4}=\frac{0.152}{0.50}=0.304 \mathrm{M}$,
Also for the following equilibrium:
$$ \begin{gathered} \mathrm{Ag}{2} \mathrm{CO}{3}(s)+\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} \
