The solubility product of AgCO at 25°C is . A solution of KCO containing 0.15 moles in 500 ml water — Ionic Equilibrium Chemistry Question
Question
The solubility product of Ag$_2$C$_2$O$_4$ at 25°C is $2.3 \times 10^{-11} \text{ M}^3$. A solution of K$_2$C$_2$O$_4$ containing 0.15 moles in 500 ml water is shaken at 25°C with excess of Ag$_2$CO$_3$ till the following equilibrium is reached:<br><br>Ag$_2$CO$_3$ + K$_2$C$_2$O$_4 \rightleftharpoons$ Ag$_2$C$_2$O$_4$ + K$_2$CO$_3$<br><br>At equilibrium, the solution contains 0.035 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$_3$,
💡 Solution & Explanation
At equilibrium, moles of $\text{K}_2\text{CO}_3 = 0.035$ and moles of remaining $\text{K}_2\text{C}_2\text{O}_4 = 0.15 - 0.035 = 0.115$. The equilibrium constant $K$ for the reaction is $\frac{[\text{K}_2\text{CO}_3]}{[\text{K}_2\text{C}_2\text{O}_4]} = \frac{0.035}{0.115} = \frac{7}{23}$. Also, $K = \frac{[\text{CO}_3^{2-}]}{[\text{C}_2\text{O}_4^{2-}]} = \frac{[\text{Ag}^+]^2[\text{CO}_3^{2-}]}{[\text{Ag}^+]^2[\text{C}_2\text{O}_4^{2-}]} = \frac{K_{sp}(\text{Ag}_2\text{CO}_3)}{K_{sp}(\text{Ag}_2\text{C}_2\text{O}_4)}$. Thus, $K_{sp}(\text{Ag}_2\text{CO}_3) = K \times K_{sp}(\text{Ag}_2\text{C}_2\text{O}_4) = \frac{7}{23} \times (2.3 \times 10^{-11}) = 7 \times 10^{-12} \text{ M}^3$.