Question

Neither \(\mathrm{PbCl}_{2}\) nor \(\mathrm{PbF}_{2}\) is appreciably soluble in water. If solid \(\mathrm{PbCl}_{2}\) and solid \(\mathrm{PbF}_{2}\) are placed in equal amounts of water in separate beakers, in which beaker is the concentration of \(\mathrm{Pb}^{2+}\) greater? Equilibrium constants for these solids dissolving in water are as follows: \(\mathrm{PbCl}_{2}(\mathrm{s}) \rightleftarrows \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq}) \quad K=1.7 \times 10^{-5}\) \(\mathrm{PbF}_{2}(\mathrm{s}) \rightleftarrows \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{F}^{-}(\mathrm{aq}) \quad K=3.7 \times 10^{-8}\)

Short answer

The concentration of \(\mathrm{Pb}^{2+}\) is greater in the \(\mathrm{PbCl}_{2}\) beaker.

Step-by-step solution

Understand the problem statement

We need to determine in which beaker the concentration of \(\mathrm{Pb}^{2+}\) ions is higher. We are given equilibrium constants \(K\) for the dissolution of \(\mathrm{PbCl}_{2}\) and \(\mathrm{PbF}_{2}\) in water. These values will help us compare the solubility of the two compounds.

Write the dissolution reactions and expressions for solubility product

For \(\mathrm{PbCl}_{2}\), the dissolution reaction is: \(\mathrm{PbCl}_{2}(\mathrm{s}) \rightleftarrows \mathrm{Pb}^{2+}(\mathrm{aq}) + 2\mathrm{Cl}^{-}(\mathrm{aq})\). The expression for the solubility product constant is \(K = [\mathrm{Pb}^{2+}][\mathrm{Cl}^{-}]^2\). For \(\mathrm{PbF}_{2}\), the dissolution reaction is: \(\mathrm{PbF}_{2}(\mathrm{s}) \rightleftarrows \mathrm{Pb}^{2+}(\mathrm{aq}) + 2\mathrm{F}^{-}(\mathrm{aq})\). The expression for the solubility product constant is \(K = [\mathrm{Pb}^{2+}][\mathrm{F}^{-}]^2\).

Relate concentration to solubility product

Since both solids dissociate to give one part of \(\mathrm{Pb}^{2+}\) and two parts of halide, if \(s\) is the solubility of \(\mathrm{PbCl}_{2}\) and \(\mathrm{PbF}_{2}\) respectively, then: \([\mathrm{Pb}^{2+}] = s\). Substitute these into their equilibrium expressions: For \(\mathrm{PbCl}_{2}\), \(K = s \cdot (2s)^2 = 4s^3\); For \(\mathrm{PbF}_{2}\), \(K = s \cdot (2s)^2 = 4s^3\).

Calculate solubility for both salts

Solving for \(s\) from the equation \(4s^3 = K\), the solubility \(s\) for both salts can be calculated as: \(s = \sqrt[3]{\frac{K}{4}}\). Substitute the given values: For \(\mathrm{PbCl}_{2}\), \(s = \sqrt[3]{\frac{1.7 \times 10^{-5}}{4}}\). For \(\mathrm{PbF}_{2}\), \(s = \sqrt[3]{\frac{3.7 \times 10^{-8}}{4}}\).

Compare the solubilities

Calculate each using the equation from the previous step. Since \(1.7 \times 10^{-5}\) is greater than \(3.7 \times 10^{-8}\), the solubility of \(\mathrm{PbCl}_{2}\) is greater than that of \(\mathrm{PbF}_{2}\). Thus, the concentration of \(\mathrm{Pb}^{2+}\) ions in the \(\mathrm{PbCl}_{2}\) beaker is greater than in the \(\mathrm{PbF}_{2}\) beaker.

Key concepts

Solubility Product

The solubility product constant, denoted as \( K_{sp} \), is an essential concept in understanding chemical equilibrium in solutions. It is a type of equilibrium constant that applies to the dissolution of sparingly soluble salts in water. Specifically, \( K_{sp} \) gives us an idea of how much a solid can dissolve in water to form a saturated solution, where the solid is in equilibrium with its ions in the solution.

When a solid dissolves in water, it disassociates into its constituent ions. For a generic salt \( AB \), which dissolves according to the reaction \( AB(s) \rightleftarrows A^+(aq) + B^-(aq) \), the solubility product is expressed as:
\[ K_{sp} = [A^+][B^-]. \]

For compounds that dissolve into multiple ions, like \( \mathrm{PbCl}_2 \) and \( \mathrm{PbF}_2 \), the expressions are slightly different, such as for \( \mathrm{PbCl}_2 \):
\[ K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl^-}]^2. \]

Knowing the \( K_{sp} \) helps predict the extent of dissolution and the concentration of ions in a saturated solution.

Dissolution Reactions

Dissolution reactions are the processes where solid substances dissolve into a solvent, typically water, resulting in ions. These reactions are fundamental to a broader understanding of solubility and equilibrium in chemistry.

For example, the dissolution of lead chloride \( \mathrm{PbCl}_2(s) \) in water can be represented by the equation:
\( \mathrm{PbCl}_2(s) \rightleftarrows \mathrm{Pb}^{2+}(aq) + 2\mathrm{Cl}^-(aq) \).

Similarly, dissolving lead fluoride \( \mathrm{PbF}_2(s) \) follows:
\( \mathrm{PbF}_2(s) \rightleftarrows \mathrm{Pb}^{2+}(aq) + 2\mathrm{F}^-(aq) \).

In these reactions:

• The solid compound separates into its constituent ions in the aqueous solution.
• The reverse of this reaction can also occur, where ions come together to form the solid again, establishing an equilibrium.

Understanding these equations helps in calculating the ion concentrations when the solution reaches equilibrium, guided by the solubility product constants.

Lead Chloride

Lead chloride, \( \mathrm{PbCl}_2 \), is an example of a slightly soluble ionic compound. Though it doesn't dissolve well in water, it's important in understanding chemical equilibria.

- **Dissolution**: When \( \mathrm{PbCl}_2 \) dissolves, it splits into one \( \mathrm{Pb}^{2+} \) ion and two \( \mathrm{Cl}^- \) ions. The equilibrium expression for its dissolution is \( K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl^-}]^2 \).

- **Solubility Expression**: Given its solubility product constant \( (K = 1.7 \times 10^{-5}) \), we can calculate its solubility (\( s \)) using the equation:
\[ 4s^3 = K. \]

In this specific exercise, you would find \( s \) by solving \( s = \sqrt[3]{\frac{1.7 \times 10^{-5}}{4}} \), indicating the concentration of \( \mathrm{Pb}^{2+} \) in a saturated solution. Thus, \( \mathrm{PbCl}_2 \) has a higher solubility than other salts with lower \( K_{sp} \) like \( \mathrm{PbF}_2 \).

Lead Fluoride

Lead fluoride, \( \mathrm{PbF}_2 \), is another sparingly soluble compound, similar in context but not in dissolution rate, as compared to lead chloride.

Let’s break down its characteristics:

• **Dissolution**: When \( \mathrm{PbF}_2 \) dissolves, it produces \( \mathrm{Pb}^{2+} \) ions and \( \mathrm{F}^- \) ions, following the equation \( K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{F}^-]^2 \).
• **Solubility**: The solubility expression for \( \mathrm{PbF}_2 \) is derived from its equilibrium constant \( (K = 3.7 \times 10^{-8}) \), using \( 4s^3 = K \).
• **Calculation**: This involves finding \( s \) with \( s = \sqrt[3]{\frac{3.7 \times 10^{-8}}{4}} \).

With its lower \( K_{sp} \) value compared to \( \mathrm{PbCl}_2 \), \( \mathrm{PbF}_2 \) has a lesser tendency to dissolve, leading to a lower concentration of \( \mathrm{Pb}^{2+} \) ions in solution. This is why in a comparative scenario, as detailed in the exercise, the beaker with \( \mathrm{PbCl}_2 \) holds a higher concentration of lead ions than the one with \( \mathrm{PbF}_2 \).