A volume of 1.0 L of solution which was in equilibrium with solid mixture of AgCl and was found to c — Ionic Equilibrium Chemistry Question
Question
A volume of 1.0 L of solution which was in equilibrium with solid mixture of AgCl and $\text{Ag}_2\text{CrO}_4$ was found to contain $1 \times 10^{-4}$ moles of $\text{Ag}^+$ ions, $1.0 \times 10^{-6}$ moles of $\text{Cl}^-$ ions and $8.0 \times 10^{-4}$ moles of $\text{CrO}_4^{2-}$ ions. $\text{Ag}^+$ ions are added slowly to the above mixture (keeping the volume constant) till $8.0 \times 10^{-7}$ moles of AgCl got precipitated. How many moles of $\text{Ag}_2\text{CrO}_4$ were precipitated simultaneously?
💡 Solution & Explanation
Initially, $K_{sp}(\text{AgCl}) = (10^{-4})(10^{-6}) = 10^{-10}$ and $K_{sp}(\text{Ag}_2\text{CrO}_4) = (10^{-4})^2(8.0 \times 10^{-4}) = 8.0 \times 10^{-12}$. When $8.0 \times 10^{-7} \text{ moles}$ of AgCl precipitates, the remaining $[\text{Cl}^-] = 1.0 \times 10^{-6} - 0.8 \times 10^{-6} = 2.0 \times 10^{-7} \text{ M}$. The new equilibrium $[\text{Ag}^+] = \frac{10^{-10}}{2.0 \times 10^{-7}} = 5.0 \times 10^{-4} \text{ M}$. Using $K_{sp}(\text{Ag}_2\text{CrO}_4)$, the new $[\text{CrO}_4^{2-}] = \frac{8.0 \times 10^{-12}}{(5.0 \times 10^{-4})^2} = 3.2 \times 10^{-5} \text{ M}$. Moles of $\text{Ag}_2\text{CrO}_4$ precipitated = initial - final = $(8.0 \times 10^{-4}) - (0.32 \times 10^{-4}) = 7.68 \times 10^{-4} \text{ moles}$. Therefore, correct answer is A.