Problem 83 - General Chemistry: Principles and Modern Applications

Question

A mixture of \(\mathrm{PbSO}_{4}(\mathrm{s})\) and \(\mathrm{PbS}_{2} \mathrm{O}_{3}(\mathrm{s})\) is shaken with pure water until a saturated solution is formed. Both solids remain in excess. What is \(\left[\mathrm{Pb}^{2+}\right]\) in the saturated solution? For \(\mathrm{PbSO}_{4}, K_{\mathrm{sp}}=1.6 \times 10^{-8} ;\) for \(\mathrm{PbS}_{2} \mathrm{O}_{3}, K_{\mathrm{sp}}=4.0 \times 10^{-7}\).

Step-by-Step Solution

Answer

The concentration of \(\mathrm{Pb}^{2+}\) in the saturated solution is \(\sqrt[3]{4.0 \times 10^{-7}}\).

Step 1: Identify solubility from \(K_{sp}\)

First, it is noted that the solubility of \(\mathrm{PbSO}_{4}\) and \(\mathrm{PbS}_{2}\mathrm{O}_{3}\) is driven by their \(K_{sp}\) values. The higher the \(K_{sp}\), the greater the solubility. Here, \(K_{\mathrm{sp}}\) for \(\mathrm{PbS}_{2}\mathrm{O}_{3}\) is \(4.0 \times 10^{-7}\), which is larger than \(K_{sp}\) for \(\mathrm{PbSO}_{4}\) (\(1.6 \times 10^{-8}\)), so \(\mathrm{PbS}_{2}\mathrm{O}_{3}\) will dissolve more. Hence, the \(\mathrm{Pb}^{2+}\) ion concentration in the solution will be predominantly from \(\mathrm{PbS}_{2}\mathrm{O}_{3}\). The solubility of \(\mathrm{PbSO}_{4}\) can be essentially ignored due to the larger solubility of \(\mathrm{PbS}_{2}\mathrm{O}_{3}\).

Step 2: Calculate \(\mathrm{Pb}^{2+}\) concentration

Secondly, the dissolution reaction for \(\mathrm{PbS}_{2}\mathrm{O}_{3}\) in water can be represented as: \(\mathrm{PbS}_{2}\mathrm{O}_{3}(\mathrm{s}) \rightarrow \mathrm{Pb}^{2+} (\mathrm{aq}) + 2\mathrm{SO}_{3}^{2-} (\mathrm{aq})\). The \(K_{\mathrm{sp}}\) expression for this reaction is \(K_{\mathrm{sp}} = \left[\mathrm{Pb}^{2+}\right][\mathrm{SO}_{3}^{2-}]^2\). As both solids are in excess, it can be assumed that the \([\mathrm{SO}_{3}^{2-}]\) would also reach its equilibrium concentration predominantly from \(\mathrm{PbS}_{2}\mathrm{O}_{3}\). Hence, we can simplify and solve the expression for \(\left[\mathrm{Pb}^{2+}\right]\): \(\left[\mathrm{Pb}^{2+}\right] = \sqrt[3]{K_{sp}} = \sqrt[3]{4.0 \times 10^{-7}}\) which gives \(\left[\mathrm{Pb}^{2+}\right]\).

Key Concepts

Chemical Equilibrium

Understanding the concept of chemical equilibrium is vital for grasping why certain compounds dissolve to the extent they do in water. At chemical equilibrium, the rate at which a substance dissolves in water is equal to the rate at which it precipitates out of the solution. This balance does not mean that the reactants and products are at the same concentration, but rather that their concentrations have stabilized and are no longer changing.

When a solid like \(\mathrm{PbSO}_4\) or \(\mathrm{PbS}_2\mathrm{O}_3\) is introduced to water, it starts to dissolve, and its ions begin to disperse in the solution. If there is an excess of the solid, as in the provided exercise, the solution can reach a saturation point where it cannot dissolve more of the substance. At this moment, equilibrium is established between the undissolved solid and the dissolved ions.

Importantly, the equilibrium does not just depend on the amount of solid added, but on the solubility product constant (\(K_{sp}\)). This constant is specific to each compound and temperature, and it dictates the maximum concentration of ions that can coexist in equilibrium with the undissolved solid.

Concentration of Ions

The concentration of ions in a solution reflects how much of a compound has dissolved. In the field of chemistry, concentration is often expressed in moles per liter (M). The solubility product constant (\(K_{sp}\)) directly relates to the concentration of the ions of a sparingly soluble compound at equilibrium.

The exercise provided presents the dissolution of lead compounds where lead ions (\(\mathrm{Pb}^{2+}\)) are one of the products. In such a chemical scenario, knowing the \(K_{sp}\) allows us to calculate the equilibrium concentration of these ions. A higher \(K_{sp}\) suggests a greater solubility and thus a higher concentration of ions at equilibrium.

The relationship of ion concentration in these lead compounds is also governed by the stoichiometry of the dissolution reaction. In the case of \(\mathrm{PbS}_2\mathrm{O}_3\), the reaction produces two sulfate ions for every lead ion, which is crucial in calculating the final concentrations of these ions using the known \(K_{sp}\) value.

Solubility and Saturation

Solubility is a measure of how much a substance can dissolve in a solvent at a given temperature and pressure. Saturation, in contrast, occurs when a solution contains the maximum amount of solute that can be dissolved. Beyond this point, any additional substance will remain undissolved.

In the context of the problem we are discussing, we are looking at a saturated solution where any additional substance cannot dissolve because we have reached equilibrium between the dissolved ions and the undissolved solid. As it's a dynamic equilibrium, individual ions continually enter and exit the solution phase, but overall, the concentrations remain constant.

Improving understanding of solubility requires recognizing factors like temperature, the common ion effect, and the presence of competing equilibria. However, as long as the equilibrium constant remains the same, the maximum solubility under those conditions does not change, providing us with a clear method to calculate the concentration of ions present in a saturated solution.