Silver sulfate, \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) , has a solubility product constant of \(1.0 \times 10^{-5} .\) The below diagram shows the products of a precipitation reaction in which some silver sulfate was formed. (Diagram Can't Copy) Which ion concentrations below would have led the precipitate to form? (A) \(\left[\mathrm{Ag}^{+}\right]=0.01 M\left[\mathrm{SO}_{4}^{2-}\right]=0.01 M\) (B) \(\left[\mathrm{Ag}^{+}\right]=0.10 M\left[\mathrm{SO}_{4}^{2-}\right]=0.01 M\) (C) \(\left[\mathrm{Ag}^{+}\right]=0.01 M\left[\mathrm{SO}_{4}^{2-}\right]=0.10 M\) (D) This is impossible to determine without knowing the total volume of the solution.

To determine the ion concentrations that lead to precipitation, we first write the solubility product expression for silver sulfate, which is \(K_{sp} = [\mathrm{Ag}^{+}]^{2} [\mathrm{SO}_{4}^{2-}]\). This expression is the product of the concentrations of the ions, each raised to the power of their stoichiometric coefficients in the balanced chemical equation.

Let's substitute the given values into the \(K_{sp}\) expression and see if they would lead to precipitation by comparing with the given \(K_{sp}\) value of \(1.0 \times 10^{-5}\).\n\n(A) For \([\mathrm{Ag}^{+}]=0.01 M\) and \([\mathrm{SO}_{4}^{2-}]=0.01 M\), \(K' = (0.01)^{2} \times 0.01 = 1 \times 10^{-6}\) which is less than \(K_{sp}\), so no precipitation will occur.\n\n(B) For \([\mathrm{Ag}^{+}]=0.10 M\) and \([\mathrm{SO}_{4}^{2-}]=0.01 M\), \(K' = (0.10)^{2} \times 0.01 = 1 \times 10^{-4}\) which is higher than \(K_{sp}\), so precipitation will occur. \n\n(C) For \([\mathrm{Ag}^{+}]=0.01 M\) and \([\mathrm{SO}_{4}^{2-}]=0.10 M\), \(K' = (0.01)^{2} \times 0.10 = 1 \times 10^{-5}\), equals to \(K_{sp}\), so no precipitation, or at equilibrium.

By comparing \(K'\) (calculated solubility product) with \(K_{sp}\), we can infer if a precipitate will form or not. If \(K'> K_{sp}\), then the solution is supersaturated and precipitation would occur; if \(K'\leq K_{sp}\), then the solution is unsaturated or at equilibrium and no precipitation would occur. The total volume of the solution is not necessary to determine unless we were asked to calculate the amount of precipitate formed.