Problem 75 - Chemistry: The Central Science in SI Units, Global Edition

Question

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 \(\operatorname{AgCl}\left(K_{s p}=1.8 \times 10^{-10}\right) ?\)

Step-by-Step Solution

Answer

The first compound to precipitate when dilute \(\mathrm{AgNO}_{3}\) is added to the solution is \(\mathrm{Ag}_{2}\mathrm{CO}_{3}\), as its ion product (Q) exceeds its solubility product (\(K_{sp}\)) first.

Step 1: Write the precipitation reactions and \(K_{sp}\) expressions for each silver salt

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. \(\operatorname{AgCl}\rightleftharpoons\mathrm{Ag}^{+}+\mathrm{Cl}^{-}\), \(K_{sp} = [\mathrm{Ag}^{+}] [\mathrm{Cl}^{-]\)

Step 2: Calculate the ion product (Q) for each silver salt

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 \(\operatorname{AgCl}\)

Step 3: Compare Q values with the respective \(K_{sp}\) values

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.

Key Concepts

Solubility Product Constant (Ksp)

The solubility product constant, denoted as \(K_{sp}\), is a crucial concept in understanding precipitation reactions. It reflects how much of a compound can dissolve in water, effectively showing the solubility limits for sparingly soluble salts. When a salt dissolves in water, it dissociates into its ions. The \(K_{sp}\) value is obtained by taking the product of the concentrations of these ions, each raised to the power of its coefficient in the balanced equation of dissolution. For example, if the dissolved form of a salt is represented as \(aA + bB \rightleftharpoons cC + dD\), the expression for \(K_{sp}\) would be \([C]^c[D]^d / [A]^a[B]^b\).
- Higher \(K_{sp}\) indicates greater solubility.- If the ionic product exceeds \(K_{sp}\), a precipitate will form as the solution becomes supersaturated.For instance, in our scenario, the \(K_{sp}\) for \(\mathrm{Ag}_{2}\mathrm{CrO}_{4}, \mathrm{Ag}_{2}\mathrm{CO}_{3}, \operatorname{AgCl}\) defines the limits at which these compounds stay soluble or start precipitating.

Ion Product (Q)

The ion product, symbolized as \(Q\), is used to predict the formation of a precipitate in a solution. \(Q\) is calculated similarly to \(K_{sp}\), except it uses the current concentrations of ions in a solution at a given moment. This helps determine whether the solution is undersaturated, saturated, or supersaturated, directly affecting precipitation outcomes.
- If \(Q < K_{sp}\), the solution is undersaturated, and no precipitate forms.- If \(Q = K_{sp}\), the solution is at equilibrium.- If \(Q > K_{sp}\), the solution is supersaturated, leading to precipitation.In learning about \(Q\), you understand how dynamic conditions in a solution can push it to form a solid phase. Step 2 of the exercise showcased \(Q\) calculations for various salts to see which precipitates when \(AgNO_3\) is introduced.

Chemical Equilibrium

Chemical equilibrium is a state in which the concentrations of reactants and products remain constant over time, as seen in reversible reactions. This balance is a result of the forward and reverse reactions occurring at the same rate. Conceptually, it's akin to a Zen-like state where despite ongoing activity, there appears to be no visible change.
Understanding and applying chemical equilibrium principles is essential in predicting whether a reaction has reached a state where the reactants and products sustain each other without further changes. This is crucial for determining the conditions under which different phases - like solids and liquids - coexist. In precipitate reactions, equilibrium happens when the ion product \(Q\) equals the solubility product constant \(K_{sp}\). Step 3 of the exercise uses this equilibrium principle to predict which compound comes out of solution first, comparing \(Q\) to \(K_{sp}\) for the different silver salts.
- Equilibrium doesn't mean equal concentrations, but rather consistent concentrations over time.- Seasons like summer and winter symbolize equilibrium; they balance changes even with ongoing Earthly activities.