To find the solubility product $(K_{sp})$ of $$Ag$_2$C$_2$O$_4,$ we start with the dissolution equilibrium:$\\$Ag$_2$C$_2$O$_4 (s) \rightleftharpoons 2$Ag$^+ (aq) +$C$_2$O$_4^{2-} (aq) \\$Let $s$ be the solubility of $$Ag$_2$C$_2$O$_4$ in mol/L.$\\$From the stoichiometry of the reaction, we have:$\\ [$Ag$^+] = 2s \\ [$C$_2$O$_4^{2-}] = s \\$Given that the concentration of $$Ag$^+$ ions is $2.2 \times 10^{-4}$ mol/L, we have:$\\ 2s = 2.2 \times 10^{-4} \\$Solving for $s: \\ s = \frac{2.2 \times 10^{-4}}{2} = 1.1 \times 10^{-4}$ mol/L$\\$Now, we can calculate the solubility product $K_{sp}: \\ K_{sp} = [$Ag$^+]^2 [$C$_2$O$_4^{2-}] \\$Substitute the concentrations:$\\ K_{sp} = (2.2 \times 10^{-4})^2 (1.1 \times 10^{-4}) \\$Calculate:$\\ K_{sp} = (4.84 \times 10^{-8}) (1.1 \times 10^{-4}) \\ K_{sp} = 5.324 \times 10^{-12} \\$Therefore, the solubility product of $$Ag$_2$C$_2$O$_4$ is approximately $5.3 \times 10^{-12}. \\$This corresponds to Option 3.