Question

Consider a saturated solution of \(\mathrm{PbBr}_{2}(\mathrm{~s})\). If \(\left[\mathrm{Pb}^{2+}\right]\) is \(1.33 \times 10^{-5} \mathrm{M}\), find each of the following. a) \(\left[\mathrm{Br}^{-}\right]\) b) the \(K_{s p}\) of \(\mathrm{PbBr}_{2}(\mathrm{~s})\)

Short answer

[Br^-] = 2.66 x 10^{-5} M; K_{sp} = 9.41 x 10^{-15}.

Step-by-step solution

Understanding the Dissolution Reaction

First, let's write the equilibrium equation for the dissolution of lead(II) bromide, \(\mathrm{PbBr}_2 (s) \rightleftharpoons \mathrm{Pb}^{2+} (aq) + 2 \mathrm{Br}^- (aq).\)This tells us that for every mole of \(\mathrm{PbBr}_2\) that dissolves, one mole of \(\mathrm{Pb}^{2+}\) and two moles of \(\mathrm{Br}^-\) ions are produced.

Determine the Bromide Ion Concentration

Since the stoichiometry of the dissolution reaction shows that for every mole of \(\mathrm{Pb}^{2+}\) ions produced, two moles of \(\mathrm{Br}^-\) ions are produced, we can calculate the concentration of \(\mathrm{Br}^-\): \[[\mathrm{Br}^-] = 2 \times [\mathrm{Pb}^{2+}] = 2 \times 1.33 \times 10^{-5} = 2.66 \times 10^{-5} \text{ M}\].

Write the Expression for the Solubility Product

The solubility product, \(K_{sp}\), for \(\mathrm{PbBr}_2\) can be expressed as \[K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Br}^-]^2\].

Calculate the Solubility Product

Now, substitute the concentrations we have found into the \(K_{sp}\) expression:\[K_{sp} = (1.33 \times 10^{-5})(2.66 \times 10^{-5})^2\].Calculating this gives:\[K_{sp} = 1.33 \times 10^{-5} \times 7.08 \times 10^{-10} = 9.41 \times 10^{-15}\].

Key concepts

Saturation

Saturation occurs when a solution can no longer dissolve additional amounts of a substance at a given temperature and pressure. Once a solution is saturated, it reaches a dynamic equilibrium where the rate of dissolution equals the rate of precipitation. This means that, while some dissolving and precipitation continuously occur, the overall concentration of dissolved ions remains constant.
In the case of lead(II) bromide, \( \text{PbBr}_2 \), a solution is saturated when it holds the maximum amount of lead and bromide ions it can hold at equilibrium. Any additional \( \text{PbBr}_2 \) will form a solid precipitate that falls out of the solution.
It's important to know the saturation point because this tells us the maximum concentration of ions that can exist in the solution without forming a precipitate.

• The concept of saturation helps predict if a solution can dissolve more of a given solute.
• Understand that saturation is vital in cases like \( \text{PbBr}_2 \) to determine how much the solution dissolves.

Stoichiometry

Stoichiometry refers to the quantitative relationship between reactants and products in a chemical reaction. We can use stoichiometry to determine the concentrations of ions in a solution by looking at the chemical equation. For the dissolution of \( \text{PbBr}_2 \), the balanced equation is:\[ \text{PbBr}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{Br}^-(aq) \].
This equation tells us for every mole of \( \text{Pb}^{2+} \) produced, two moles of \( \text{Br}^- \) ions are formed.
Applying this stoichiometric relationship, if we have a \( [\text{Pb}^{2+}] = 1.33 \times 10^{-5} \text{ M} \), the concentration of bromide ions will be twice that amount: \( [\text{Br}^-] = 2 \times 1.33 \times 10^{-5} = 2.66 \times 10^{-5} \text{ M} \).

• Stoichiometry uses the coefficients in a balanced equation to predict concentrations.
• It's an essential tool for solving solubility problems like in this exercise.

Equilibrium

Equilibrium in chemistry is the state where the rates of the forward and reverse reactions are equal. In the context of a saturated solution, equilibrium is reached when the rate at which \( \text{PbBr}_2 \) dissolves in the solution is equal to the rate at which it precipitates. This balance means the concentration of dissolved ions remains constant.
The equation for the equilibrium of the dissolution of lead(II) bromide is:\[ \text{PbBr}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{Br}^-(aq) \].
At equilibrium, the concentrations of ions can be used to calculate the solubility product constant, \( K_{sp} \), which tells us about the solubility of the compound.
Understanding equilibrium helps predict how changes in conditions affect the solubility of a compound and guides in how to control precipitation and dissolution processes.

• Equilibrium involves maintaining a balance between dissolution and precipitation.
• The solubility product constant, \( K_{sp} \), is derived from concentrations at equilibrium.

Ion Concentration

Ion concentration measures how many ions are present in a given volume of solution, typically expressed in molarity (moles per liter). For example, in the dissolution of \( \text{PbBr}_2 \), we look at the respective ion concentrations for \( \text{Pb}^{2+} \) and \( \text{Br}^- \).
Given in the exercise, \( [\text{Pb}^{2+}] = 1.33 \times 10^{-5} \text{ M} \), we find \( [\text{Br}^-] = 2.66 \times 10^{-5} \text{ M} \).
These concentrations are crucial for calculating the solubility product,\( K_{sp} \).To find \( K_{sp} \) for \( \text{PbBr}_2 \), we use the formula:\[ K_{sp} = [\text{Pb}^{2+}][\text{Br}^-]^2 \].Substituting the concentrations gives us the numerical value for \( K_{sp} = 9.41 \times 10^{-15} \), reflecting the solubility level of \( \text{PbBr}_2 \) in water.

• Ion concentrations are used to calculate reaction attributes like \( K_{sp} \).
• Determining these concentrations allows insight into solution chemistry dynamics.