A solution contains three anions with the following concentrations: $0.20 \mathrm{M} \mathrm{CrO}_{4}^{2-}, 0.10 \mathrm{M} \mathrm{CO}_{3}^{2-},$ and \(0.010 \mathrm{M} \mathrm{Cl}^{-}\). If a dilute \(\mathrm{AgNO}_{3}\) solution is slowly added to the solution, what is the first compound to precipitate: $\mathrm{Ag}_{2} \mathrm{CrO}_{4}\left(K_{s p}=1.2 \times 10^{-12}\right), \mathrm{Ag}_{2} \mathrm{CO}_{3}\left(K_{s p}=8.1 \times 10^{-12}\right)$ or \(\operatorname{AgCl}\left(K_{s p}=1.8 \times 10^{-10}\right) ?\)

The precipitation reactions and corresponding \(K_{sp}\) expressions for \(\mathrm{Ag}_{2}\mathrm{CrO}_{4}\), \(\mathrm{Ag}_{2}\mathrm{CO}_{3}\) and \(\operatorname{AgCl}\) are: 1. \(\mathrm{Ag}_{2}\mathrm{CrO}_{4}\rightleftharpoons2\mathrm{Ag}^{+}+\mathrm{CrO}_{4}^{2-}\), \(K_{sp} = [\mathrm{Ag}^{+}]^2 [\mathrm{CrO}_{4}^{2-}]\) 2. \(\mathrm{Ag}_{2}\mathrm{CO}_{3}\rightleftharpoons2\mathrm{Ag}^{+}+\mathrm{CO}_{3}^{2-}\), \(K_{sp} = [\mathrm{Ag}^{+}]^2 [\mathrm{CO}_{3}^{2-}]\) 3. \(\operatorname{AgCl}\rightleftharpoons\mathrm{Ag}^{+}+\mathrm{Cl}^{-}\), \(K_{sp} = [\mathrm{Ag}^{+}] [\mathrm{Cl}^{-]\)

To calculate the ion product (Q) for each silver salt, multiply the concentrations of the ions in each reaction. Since the \(\mathrm{Ag}^+\) concentration is the same for both \(\mathrm{CrO}_{4}^{2-}\) and \(\mathrm{CO}_{3}^{2-}\) reactions, we can set a variable for this concentration, let's call it x. With this, we can express their Q values as follows: 1. Q = \((x)^2 (0.20)\) for \(\mathrm{Ag}_{2}\mathrm{CrO}_{4}\) 2. Q = \((x)^2 (0.10)\) for \(\mathrm{Ag}_{2}\mathrm{CO}_{3}\) 3. Q = \((x)(0.010)\) for \(\operatorname{AgCl}\)

Compare the Q values of each reaction with its respective \(K_{sp}\) value. If Q > \(K_{sp}\), precipitation will occur. Otherwise, the solution will be unsaturated. The first compound to precipitate will be the one that exceeds its \(K_{sp}\) first. 1. For \(\mathrm{Ag}_{2}\mathrm{CrO}_{4}\), Q = \((x)^2 (0.20)\) and \(K_{sp} = 1.2 \times 10^{-12}\) 2. For \(\mathrm{Ag}_{2}\mathrm{CO}_{3}\), Q = \((x)^2 (0.10)\) and \(K_{sp} = 8.1 \times 10^{-12}\) 3. For \(\operatorname{AgCl}\), Q = \((x)(0.010)\) and \(K_{sp} = 1.8 \times 10^{-10}\) We can see that the ratio of Q over \(K_{sp}\) for each reaction would be: 1. \(\frac{(x)^2 (0.20)}{1.2 \times 10^{-12}}\) 2. \(\frac{(x)^2 (0.10)}{8.1 \times 10^{-12}}\) 3. \(\frac{(x)(0.010)}{1.8 \times 10^{-10}}\) Since \(x\) is the same value in all ratios, we can compare these ratios directly to see which one exceeds its \(K_{sp}\) first. The ratio for \(\mathrm{Ag}_{2}\mathrm{CO}_{3}\) is the closest to 1, with both \(\mathrm{Ag}_{2}\mathrm{CrO}_{4}\) and \(\operatorname{AgCl}\) having larger ratios. This means that \(\mathrm{Ag}_{2}\mathrm{CO}_{3}\) will precipitate first when dilute \(\mathrm{AgNO}_{3}\) is added to the solution.