. Determine the molar solubility and value of Ksp for each compound in water at 25°C. Include the chemical equation for each equilibrium, the expression for Ksp (in terms of molarities and x), and show all work in calculations.
Compound 2 PbCl2
Chemical equation:
Molar solubility (x) = Ksp = 2.
The Correct Answer and Explanation is:
Compound: Lead(II) chloride (PbCl₂)
Chemical Equation for the Dissolution Equilibrium:
PbCl2(s)⇌Pb2+(aq)+2Cl−(aq)\text{PbCl}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2\text{Cl}^- (aq)
This represents the dissociation of solid lead(II) chloride in water into its ions.
Let the molar solubility be x (mol/L).
From the dissociation equation:
- [Pb²⁺] = x
- [Cl⁻] = 2x (because 2 moles of Cl⁻ are produced per mole of PbCl₂)
Ksp Expression (Solubility Product):
Ksp=[Pb2+][Cl−]2K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 Ksp=(x)(2x)2=4x3K_{sp} = (x)(2x)^2 = 4x^3
Given: Ksp=1.7×10−5K_{sp} = 1.7 \times 10^{-5} at 25°C
Now solve for x (molar solubility): Ksp=4×3⇒x3=1.7×10−54=4.25×10−6K_{sp} = 4x^3 \Rightarrow x^3 = \frac{1.7 \times 10^{-5}}{4} = 4.25 \times 10^{-6} x=4.25×10−63≈0.0162 mol/Lx = \sqrt[3]{4.25 \times 10^{-6}} \approx 0.0162 \, \text{mol/L}
✅ Final Answers:
- Molar Solubility (x): ≈ 0.0162 mol/L
- Ksp: 1.7×10−51.7 \times 10^{-5}
Explanation:
The solubility product constant (Ksp) is an equilibrium constant that describes how much of an ionic compound dissolves in water to form a saturated solution. For sparingly soluble salts like lead(II) chloride (PbCl₂), only a small amount dissolves, establishing a dynamic equilibrium between the undissolved solid and its dissolved ions.
When PbCl₂ dissolves, it breaks into one Pb²⁺ ion and two Cl⁻ ions. We represent this as: PbCl2(s)⇌Pb2+(aq)+2Cl−(aq)\text{PbCl}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2\text{Cl}^- (aq)
The Ksp expression reflects the concentrations of the dissolved ions: Ksp=[Pb2+][Cl−]2K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2
Let x represent the molar solubility — the number of moles of PbCl₂ that dissolve per liter. Since each mole of PbCl₂ gives one mole of Pb²⁺ and two moles of Cl⁻, the ion concentrations are:
[Pb²⁺] = x and [Cl⁻] = 2x. Substituting into the Ksp equation gives: Ksp=x(2x)2=4x3K_{sp} = x(2x)^2 = 4x^3
Given the Ksp value is 1.7×10−51.7 \times 10^{-5}, solving this equation gives: x=4.25×10−63≈0.0162 mol/Lx = \sqrt[3]{4.25 \times 10^{-6}} \approx 0.0162 \, \text{mol/L}
This means only about 0.0162 moles of PbCl₂ dissolve in one liter of water at 25°C. The solubility is limited by the common ion effect and the low Ksp value. Knowing molar solubility is crucial in predicting precipitation, determining ion concentrations, and understanding chemical equilibria in aqueous systems.
