Question

The concentration of \(\mathrm{Pb}^{2+}\) in a solution saturated with \(\mathrm{PbBr}_{2}(s)\) is \(2.14 \times 10^{-2} M\). Calculate \(K_{\text {sn }}\) for \(\mathrm{PbBr}\).

Short answer

The solubility product constant, \(K_{\text{sn}}\), for \(\mathrm{PbBr}_{2}\) is \(3.90 \times 10^{-5}\).

Step-by-step solution

Write the dissociation equation of \(\mathrm{PbBr}_{2}\)

When \(\mathrm{PbBr}_{2}(s)\) dissolves in water, it dissociate into its constituent ions. Write down the balanced chemical equation for the dissolution of \(\mathrm{PbBr}_{2}(s)\): \[\mathrm{PbBr}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + 2\mathrm{Br}^-(aq)\]

Set up the \(K_{\text{sn}}\) expression for \(\mathrm{PbBr}_{2}\)

The solubility product constant, \(K_{\text{sn}}\), expression for the dissolution of \(\mathrm{PbBr}_{2}(s)\) can be written as: \[K_{\text{sn}} = [\mathrm{Pb}^{2+}][\mathrm{Br}^-]^2\]

Find the concentration of \(\mathrm{Br}^-\) ions

From the balanced chemical equation for the dissolution of \(\mathrm{PbBr}_{2}(s)\), we can see that for every \(\mathrm{Pb}^{2+}\) ion produced, there are 2 \(\mathrm{Br}^-\) ions produced. Since the concentration of \(\mathrm{Pb}^{2+}\) is given as \(2.14 \times 10^{-2} \mathrm{M}\), the concentration of \(\mathrm{Br}^-\) ions will be: \[[\mathrm{Br}^-] = 2 \times [\mathrm{Pb}^{2+}] = 2 \times 2.14 \times 10^{-2} \mathrm{M} = 4.28 \times 10^{-2} \mathrm{M}\]

Calculate \(K_{\text{sn}}\) for \(\mathrm{PbBr}_{2}\)

Substitute the given concentration of \(\mathrm{Pb}^{2+}\) and the calculated concentration of \(\mathrm{Br}^-\) into the \(K_{\text{sn}}\) expression: \[K_{\text{sn}} = (2.14 \times 10^{-2})(4.28 \times 10^{-2})^2 = 3.90 \times 10^{-5}\] So, the solubility product constant, \(K_{\text{sn}}\), for \(\mathrm{PbBr}_{2}\) is \(3.90 \times 10^{-5}\).

Key concepts

Ion Concentration

Understanding ion concentration is essential when discussing the solubility of ionic compounds in water. Ion concentration is measured in moles per liter (M), expressing the amount of ions in a given volume of solution. In the context of solubility product constants, knowing the concentration of each ion that results from the dissolution of an ionic compound allows you to calculate the solubility product constant, often abbreviated as Ksp.

For instance, in a solution saturated with lead(II) bromide (PbBr2), the concentration of lead ions, Pb2+, can be experimentally determined through various analytical methods. Once the concentration of one type of ion is known, the stoichiometry of the compound's dissociation can be used to calculate the concentration of the other ions present. In the given exercise, the concentration of bromide ions, Br-, was found by doubling the concentration of lead ions, reflecting the mole ratio from the compound's dissociation equation in the water.

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction and its reverse reaction proceed at the same rate. As a result, the concentration of the reactants and products remains constant over time. Equilibrium doesn’t mean that the reactants and products are equal in concentration, but that their concentrations have stabilized at a particular ratio.

For the solubility of ionic compounds, equilibrium is reached when the rate at which the substance dissolves equals the rate at which it precipitates from the solution. The solubility equilibrium for an ionic compound is represented by the dissolution reaction, with the solid substance on one side and the ions it breaks into on the other. When students solve problems involving the dissolution of ionic compounds, they are often working with systems at equilibrium, meaning the concentrations of the ions are constant and can be used to calculate the solubility product constant, Ksp.

Dissolution of Ionic Compounds

The dissolution of ionic compounds in water is a process where the compounds break apart into their constituent ions. This process is governed by the characteristics of the solvent and the solute as well as temperature and pressure. In water, the polar molecules interact with the positive and negative ions of the compound, stabilizing them in solution.

Each ionic compound has a specific solubility product constant (Ksp) at a given temperature, which quantitatively expresses the extent to which the compound can dissolve in water. A higher Ksp value indicates a more soluble compound. When calculating the Ksp from the concentrations of the resulting ions, students commonly use the stoichiometry of the compound's dissociation to determine the correct exponents in the Ksp expression, as each ion's contribution to the Ksp depends on its coefficient in the balanced dissolution equation.