Question

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

Short answer

The solubility product constant, \(K_\text{sp}\), for \(\mathrm{PbBr}_{2}\) is approximately \(3.91 \times 10^{-5}\).

Step-by-step solution

Write the balanced dissolution equation

The balanced chemical equation for the dissolution of \(\mathrm{PbBr}_{2}\) in water is: \[\mathrm{PbBr}_{2}(s) \rightleftharpoons \mathrm{Pb^{2+}}(aq) + 2 \mathrm{Br^{-}}(aq)\]

Determine the equilibrium concentrations of the ions

We are given the concentration of \(\mathrm{Pb^{2+}}\) ions, which is \(2.14 \times 10^{-2} M\) in the saturated solution. The dissolution equation shows that for every one \(\mathrm{Pb^{2+}}\) ion, there are two \(\mathrm{Br^{-}}\) ions. Thus, the concentration of \(\mathrm{Br^{-}}\) ions in the saturated solution will be twice the concentration of \(\mathrm{Pb^{2+}}\) ions: \[ [\mathrm{Br^{-}}] = 2 \times [\mathrm{Pb^{2+}}] = 2 \times 2.14 \times 10^{-2} M = 4.28 \times 10^{-2} M \]

Calculate \(K_\text{sp}\)

The solubility product constant, \(K_\text{sp}\), is calculated using the equilibrium concentrations of the ions: \[ K_\text{sp} = [\mathrm{Pb^{2+}}] [\mathrm{Br^{-}}]^2 \] Plug in the equilibrium concentrations of \(\mathrm{Pb^{2+}}\) and \(\mathrm{Br^{-}}\): \[ K_\text{sp} = (2.14 \times 10^{-2})(4.28 \times 10^{-2})^2 \] Now, calculate the value of \(K_\text{sp}\): \[ K_\text{sp} \approx 3.91 \times 10^{-5} \] So, the solubility product constant, \(K_\text{sp}\), for \(\mathrm{PbBr}_{2}\) is approximately \(3.91 \times 10^{-5}\).

Key concepts

Equilibrium Concentration

When a sparingly soluble ionic compound like lead(II) bromide (\textbf{PbBr}\(_2\)) dissolves in water, it dissociates into its constituent ions to some extent. The point at which no more of the solid can dissolve at a given temperature is known as the saturation point. At this stage, an equilibrium is established between the undissolved substance and the dissolved ions.

This leads us to the concept of equilibrium concentration, which is the concentration of a solute in a saturated solution when equilibrium is reached. In the context of our exercise, the equilibrium concentration of \textbf{Pb}\(^{2+}\) ions is given as \(2.14 \times 10^{-2}\) \textbf{M}. To find the equilibrium concentration of bromide, \textbf{Br}\(^{-}\), we apply stoichiometry from the dissolution equation, which indicates that each \textbf{Pb}\(^{2+}\) ion is accompanied by two \textbf{Br}\(^{-}\) ions.

Dissolution of Ionic Compounds

Dissolution refers to the process in which ionic compounds dissolve in a solvent like water. In this process, the ionic lattice breaks down and the individual ions are surrounded by solvent molecules. This occurs due to the attractive forces between the ions and the solvent molecules, which are strong enough to overcome the ionic bonds within the compound.

The equation representing the dissolution of \textbf{PbBr}\(_2\) into its ions is an essential piece in understanding how equilibrium is reached and how to calculate the solubility product constant (\textbf{K}\(_{\text{sp}}\)). The balanced dissolution equation is \textbf{PbBr}\(_2\)(s) \(\rightleftharpoons\) \textbf{Pb}\(^{2+}\)(aq) + 2 \textbf{Br}\(^{-}\)(aq). This tells us that the solid ionic compound dissociates into its ions in a specific molar ratio.

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward reaction (dissolution) and the reverse reaction (precipitation) of a reversible reaction become equal, and the concentrations of reactants and products remain constant over time. It's important to distinguish that this doesn't mean the amounts of reactants and products are equal, but rather that their rates of transformation are. For the dissolution of \textbf{PbBr}\(_2\), equilibrium is described by the solid being in balance with its dissolved ions.

At equilibrium, the reaction has a specific constant value known as the equilibrium constant. For dissolution reactions that result in a saturation solution, the equilibrium constant is called the solubility product constant, represented by \textbf{K}\(_{\text{sp}}\), which is calculated from the equilibrium concentrations of the dissolved ions. It provides a measure of how much the ionic compound can dissolve in a solvent.

Stoichiometry

Stoichiometry, which comes from the Greek words for 'element' and 'measure', is a quantitative relationship between the amounts of reactants and products in a chemical reaction. It helps us predict the amounts of substances consumed and produced in a given reaction.

In our exercise, stoichiometry is used to determine the concentration of the bromide ions (\textbf{Br}\(^{-}\)) based on the given concentration of lead ions (\textbf{Pb}\(^{2+}\)). The stoichiometric coefficients from the balanced dissolution equation provide the molar ratio needed for these calculations. Since two bromide ions form for each lead ion, the concentration of \textbf{Br}\(^{-}\) is twice that of \textbf{Pb}\(^{2+}\). Understanding stoichiometry is crucial when dealing with solubility equilibriums, as it connects the dissolution process with the calculation of \textbf{K}\(_{\text{sp}}\). By mastering stoichiometry, students can adeptly handle a variety of chemical reactions and equilibriums in their studies.