A sample of AgCl was treated with 5.00 ml of 2.0 M NaCO solution to give AgCO. The remaining solutio — Ionic Equilibrium Chemistry Question
Question
A sample of AgCl was treated with 5.00 ml of 2.0 M Na$_2$CO$_3$ solution to give Ag$_2$CO$_3$. The remaining solution contained 0.00355 g of Cl$^-$ ions per litre. The solubility product of AgCl is ($K_{sp} \text{ of Ag}_2\text{CO}_3 \text{ is } 8.0 \times 10^{-12}$).
💡 Solution & Explanation
The equilibrium is $2\text{AgCl}(s) + \text{CO}_3^{2-} \rightleftharpoons \text{Ag}_2\text{CO}_3(s) + 2\text{Cl}^-$, with $K_{eq} = \frac{[\text{Cl}^-]^2}{[\text{CO}_3^{2-}]} = \frac{K_{sp}(\text{AgCl})^2}{K_{sp}(\text{Ag}_2\text{CO}_3)}$. Given the remaining solution has $0.00355 \text{ g/L Cl}^-$, $[\text{Cl}^-] = \frac{0.00355}{35.5} = 10^{-4} \text{ M}$. Since very little $\text{CO}_3^{2-}$ reacted, $[\text{CO}_3^{2-}] \approx 2.0 \text{ M}$. Thus $K_{eq} = \frac{(10^{-4})^2}{2.0} = 5 \times 10^{-9}$. We have $K_{sp}(\text{AgCl})^2 = (5 \times 10^{-9})(8.0 \times 10^{-12}) = 40 \times 10^{-21} = 4 \times 10^{-20}$. Therefore, $K_{sp}(\text{AgCl}) = 2 \times 10^{-10}$.