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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Solution:
- Initial concentration of $K _2 C _2 O _4=\frac{0.152}{0.50}=0.304 M$,
Also for the following equilibrium:
$$ \begin{gathered} Ag _2 CO _3(s)+K _2 C _2 O _4(a q) \rightleftharpoons Ag _2 C _2 O _4(s)+K _2 CO _3 \
