Problem 3
Question
Lead chloride, \(\mathrm{PbCl}_{2}\), is moderately soluble with \(K_{\mathrm{ep}}\) equal to about \(1.7 \times 10^{-5}\). a. What is the solubility of lead chloride in pure water? (How many moles of \(\mathrm{PbCl}_{2}\) could be completely dissolved in one liter of solution?) __________ moles/L.. b. What would its solubility be in 0.10 M NaCl? (How many moles of PbCl_ could be completely dissolved in 1 L of solution?) __________ moles/L. c. The difference in these two results is caused by what is known as the common ion effect. State in words what the common ion effect predicts.
Step-by-Step Solution
Verified Answer
a. 0.0165 moles/L; b. 0.0017 moles/L; c. The common ion effect reduces solubility when one ion is already present.
1Step 1: Write the Solubility Product Expression
For lead chloride, \( \text{PbCl}_2 \), which dissolves in water as \( \text{PbCl}_2 (s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{Cl}^- (aq) \), the solubility product expression is \( K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 \). Given that \( K_{sp} = 1.7 \times 10^{-5} \).
2Step 2: Establish the Solubility in Pure Water
Assume the solubility of \( \text{PbCl}_2 \) in water is \( s \) moles/L. In pure water, \([\text{Pb}^{2+}] = s \) and \([\text{Cl}^-] = 2s \). Substitute into the \(K_{sp}\) expression to get: \( 1.7 \times 10^{-5} = s(2s)^2 = 4s^3 \).
3Step 3: Solve for Solubility in Pure Water
Solve for \( s \) in the equation \( 4s^3 = 1.7 \times 10^{-5} \). First, divide by 4: \( s^3 = 4.25 \times 10^{-6} \). Then take the cubic root: \( s = (4.25 \times 10^{-6})^{1/3} \approx 0.0165 \text{ moles/L} \).
4Step 4: Consider Solubility in 0.10 M NaCl
In a solution with 0.10 M \( \text{NaCl} \), \( \text{Cl}^- \) concentration from \( \text{NaCl} \) is already 0.10 M. Thus, \([\text{Cl}^-] = 0.10 + 2s \). Substitute into the \(K_{sp}\) expression: \( 1.7 \times 10^{-5} = s(0.10 + 2s)^2 \).
5Step 5: Solve the Complex Solubility Equation
Due to the complexity caused by \(s\) being significantly smaller compared to 0.10 in \(0.10 + 2s\), approximate \(0.10 + 2s \) as 0.10: \( 1.7 \times 10^{-5} \approx s(0.10)^2 = 0.01s \). Now solve for \( s \): \( s = \frac{1.7 \times 10^{-5}}{0.01} = 1.7 \times 10^{-3} \text{ moles/L} \).
6Step 6: Explain the Common Ion Effect
The common ion effect states that the solubility of a salt is reduced when one of its constituent ions is already present in the solution. For \( \text{PbCl}_2 \), the presence of \( \text{Cl}^- \) ions from \( \text{NaCl} \) reduces its solubility compared to pure water.
Key Concepts
Lead ChlorideSolubility Product ExpressionCommon Ion EffectKsp Calculation
Lead Chloride
Lead chloride, denoted as \( \text{PbCl}_2 \), is a salt compound composed of lead and chlorine. It is known to have moderate solubility in water, meaning it does not fully dissolve and more forms a saturated solution at equilibrium. Given its moderate solubility, lead chloride is often found as a solid when in contact with water. The chemical reaction that occurs as lead chloride dissolves in water can be represented as follows: \( \text{PbCl}_2 (s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{Cl}^- (aq) \). This equation shows that in an aqueous solution, lead chloride breaks down into lead ions \( \text{Pb}^{2+} \) and chloride ions \( \text{Cl}^- \).
Understanding the solubility behavior of lead chloride is essential when calculating the concentration of its ions in a solution, especially when external ions influence its solubility further.
Understanding the solubility behavior of lead chloride is essential when calculating the concentration of its ions in a solution, especially when external ions influence its solubility further.
Solubility Product Expression
The solubility of ionic compounds such as lead chloride is governed by the solubility product constant, commonly abbreviated as \( K_{sp} \). For lead chloride, the solubility product expression is derived from its dissociation reaction: \( K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 \). This equation signifies how the concentrations of the ions in solution relate to the solubility product. Each ion's concentration is raised to the power of its stoichiometric coefficient in the dissolution equation.
For example, since two chloride ions are formed for each formula unit of lead chloride that dissolves, the chloride concentration appears squared in the expression. The solubility product is temperature-dependent, but in this problem, it is set at \( 1.7 \times 10^{-5} \) at room temperature. Knowing the value of \( K_{sp} \), you can calculate the maximum concentration of ions that can exist in solution before a precipitate of lead chloride begins to form.
For example, since two chloride ions are formed for each formula unit of lead chloride that dissolves, the chloride concentration appears squared in the expression. The solubility product is temperature-dependent, but in this problem, it is set at \( 1.7 \times 10^{-5} \) at room temperature. Knowing the value of \( K_{sp} \), you can calculate the maximum concentration of ions that can exist in solution before a precipitate of lead chloride begins to form.
Common Ion Effect
The common ion effect is a shift in equilibrium caused by the addition of a substance containing an ion that is already a part of the equilibrium. This effect plays a critical role in solubility calculations, especially when a solution contains ions common to the solute.
In the example of lead chloride dissolving in water, adding sodium chloride \( \text{NaCl} \) introduces more chloride ions already present from lead chloride. This additional chloride shifts the equilibrium to the left according to Le Chatelier's principle, thus reducing the solubility of lead chloride. In practical terms, the presence of a common ion results in the precipitation of the solute. This phenomenon is exploited in analytical chemistry and industrial processes where altering the solubility of compounds can be beneficial for separation and purification techniques.
In the example of lead chloride dissolving in water, adding sodium chloride \( \text{NaCl} \) introduces more chloride ions already present from lead chloride. This additional chloride shifts the equilibrium to the left according to Le Chatelier's principle, thus reducing the solubility of lead chloride. In practical terms, the presence of a common ion results in the precipitation of the solute. This phenomenon is exploited in analytical chemistry and industrial processes where altering the solubility of compounds can be beneficial for separation and purification techniques.
Ksp Calculation
Calculating the solubility of a compound using its \( K_{sp} \) involves determining the concentrations of ions that can exist in a saturated solution. For lead chloride, when dissolved in pure water, the concentration of lead ions \( \text{Pb}^{2+} \) is thought to be equal to \( s \), and that of chloride ions \( \text{Cl}^- \) is \( 2s \). Substituting these into the equation \( K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 = 4s^3 \), you can solve for \( s \), the solubility in moles per liter.
In the presence of a common ion, such as added chloride from \( \text{NaCl} \), the equation becomes \( K_{sp} = s(0.10 + 2s)^2 \). Given that \( s \) is much smaller than the added chloride concentration, \( 2s \) can often be ignored in the background. Solving the simplified equation results in a much lower \( s \), indicating how the solubility of lead chloride decreases in the presence of \( \text{NaCl} \). This calculation is fundamental in predicting whether a precipitate will form in various chemical and environmental conditions.
In the presence of a common ion, such as added chloride from \( \text{NaCl} \), the equation becomes \( K_{sp} = s(0.10 + 2s)^2 \). Given that \( s \) is much smaller than the added chloride concentration, \( 2s \) can often be ignored in the background. Solving the simplified equation results in a much lower \( s \), indicating how the solubility of lead chloride decreases in the presence of \( \text{NaCl} \). This calculation is fundamental in predicting whether a precipitate will form in various chemical and environmental conditions.