Problem 68 - Chemistry and Chemical Reactivity

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: \(\mathrm{PbCl}_{2}(\mathrm{s}) \rightleftarrows \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq}) \quad K=1.7 \times 10^{-5}\) \(\mathrm{PbF}_{2}(\mathrm{s}) \rightleftarrows \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{F}^{-}(\mathrm{aq}) \quad K=3.7 \times 10^{-8}\)

Step-by-Step Solution

Answer

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

Step 1: Write the solubility product expressions

First, we need to write down the solubility product expressions for both salts. For \( \mathrm{PbCl}_{2} \), the dissolution can be represented as: \[ \mathrm{PbCl}_{2} \rightleftarrows \mathrm{Pb}^{2+} + 2 \mathrm{Cl}^{-} \] and the equilibrium constant (K) is \( 1.7 \times 10^{-5} \).For \( \mathrm{PbF}_{2} \), the dissolution can be represented as: \[ \mathrm{PbF}_{2} \rightleftarrows \mathrm{Pb}^{2+} + 2 \mathrm{F}^{-} \] and the K is \( 3.7 \times 10^{-8} \).

Step 2: Set up the solubility expression equations

For both salts, set up the symmetry of the solubility expression, known as the solubility product (Ksp).For \( \mathrm{PbCl}_{2} \), \( K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^{-}]^{2} \).For \( \mathrm{PbF}_{2} \), \( K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{F}^{-}]^{2} \).

Step 3: Relate the ion concentration to the solubility

For \( \mathrm{PbCl}_{2} \), let \( s \) be the molarity of \( \mathrm{Pb}^{2+} \) at equilibrium. Then, \([ \mathrm{Cl}^{-} ] = 2s \). Substitute into the Ksp equation: \ \( K_{sp} = s \times (2s)^{2} = 4s^{3} \).For \( \mathrm{PbF}_{2} \), let \( s \) be the molarity of \( \mathrm{Pb}^{2+} \) at equilibrium. Then, \([ \mathrm{F}^{-} ] = 2s \). Substitute into the Ksp equation: \ \( K_{sp} = s \times (2s)^{2} = 4s^{3} \).

Step 4: Solve for solubility \( s \)

Solving for \( s \) gives us:For \( \mathrm{PbCl}_{2} \): \ \( 4s^{3} = 1.7 \times 10^{-5} \Rightarrow s^{3} = \frac{1.7 \times 10^{-5}}{4} \Rightarrow s = \sqrt[3]{\frac{1.7 \times 10^{-5}}{4}} \approx 1.65 \times 10^{-2} \).For \( \mathrm{PbF}_{2} \): \ \( 4s^{3} = 3.7 \times 10^{-8} \Rightarrow s^{3} = \frac{3.7 \times 10^{-8}}{4} \Rightarrow s = \sqrt[3]{\frac{3.7 \times 10^{-8}}{4}} \approx 2.11 \times 10^{-3} \).

Step 5: Compare the \( \mathrm{Pb}^{2+} \) concentrations

The solubility \( s \) represents the concentration of \( \mathrm{Pb}^{2+} \) in each solution. Since \( s = 1.65 \times 10^{-2} \) for \( \mathrm{PbCl}_{2} \) and \( s = 2.11 \times 10^{-3} \) for \( \mathrm{PbF}_{2} \), the concentration of \( \mathrm{Pb}^{2+} \) is greater in the beaker with \( \mathrm{PbCl}_{2} \).

Key Concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K \), is a key concept in understanding chemical reactions, especially at equilibrium. It describes the ratio of the concentrations of products to reactants, each raised to the power of their respective coefficients from the balanced chemical equation.
For slightly soluble substances like \( \mathrm{PbCl}_2 \) and \( \mathrm{PbF}_2 \), equilibrium constants measure how much the salt dissolves in water. When comparing these inorganic salts, the value of \( K \) helps determine their solubilities.
This constant provides insight into how far a reaction will proceed. A larger \( K \) value indicates more products are formed, suggesting that a greater amount of the salt dissolves.

• \( K = 1.7 \times 10^{-5} \) for \( \mathrm{PbCl}_2 \)
• \( K = 3.7 \times 10^{-8} \) for \( \mathrm{PbF}_2 \)

Comparatively, \( \mathrm{PbCl}_2 \) is more soluble than \( \mathrm{PbF}_2 \) because of its higher equilibrium constant.

Dissolution Process

The dissolution process describes how a solid substance dissolves in a solvent, forming a solution. For salts like \( \mathrm{PbCl}_2 \) and \( \mathrm{PbF}_2 \), this involves the breaking of ionic bonds in the solid and the interaction with water molecules.
In solution, the solid separates into its constituent ions:

• For \( \mathrm{PbCl}_2 \), it produces \( \mathrm{Pb}^{2+} \) and \( 2 \mathrm{Cl}^{-} \).
• For \( \mathrm{PbF}_2 \), it produces \( \mathrm{Pb}^{2+} \) and \( 2 \mathrm{F}^{-} \).

The extent of dissolution is represented by the solubility product (\( K_{sp} \). This is an expression that defines the product of the concentrations of the ions each raised to the power of their coefficients.
Understanding the dissolution process is crucial for predicting how much of a salt will dissolve, and it directly relates to the solubility equilibrium.

Chemical Equilibrium

Chemical equilibrium occurs when a reaction and its reverse proceed at the same rate, meaning no net change in the concentrations of reactants and products. For inorganic salts like \( \mathrm{PbCl}_2 \) and \( \mathrm{PbF}_2 \), equilibrium involves the solid salt dissolving into ions and the ions reforming the solid.
When \( K_{sp} \) is set up for these processes:

• \( \mathrm{PbCl}_2 \): \( K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^{-}]^2 \)
• \( \mathrm{PbF}_2 \): \( K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{F}^{-}]^2 \)

These expressions help in finding the concentration of ions at equilibrium.
Chemical equilibrium shows us that even seemingly insoluble salts dissolve to an extent. Determining when equilibrium is reached helps quantify solubility and predict changes with conditions.

Inorganic Salts

Inorganic salts such as \( \mathrm{PbCl}_2 \) and \( \mathrm{PbF}_2 \) consist of metal ions and nonmetals such as chlorides and fluorides. These salts typically form crystalline structures and dissolve in water to some extent, though many are only slightly soluble.
They dissolve to release ions which can participate in various chemical reactions. The solubility of inorganic salts is often limited by their equilibrium constants and is highly dependent on the specific ions involved.

• Chlorides tend to be more soluble than fluorides, as seen in the example of \( \mathrm{PbCl}_2 \) vs. \( \mathrm{PbF}_2 \).
• Knowing the solubility of different salts is essential in fields like chemistry and environmental science.

Understanding the behavior of inorganic salts helps in predicting the outcomes of their interactions and in industrial applications where solubility is crucial.