Final answer:
To find the Ksp for PbCl₂, we sum the ΔG° values of Pb²+ and Cl⁻ to find the reaction's ΔG°, then use the equation ΔG° = -RT ln(Ksp) to solve for Ksp. The calculation yields a Ksp value of 1.2 × 10⁻⁴, which corresponds to answer choice (a).
Step-by-step explanation:
To determine the solubility product, Ksp, for PbCl₂, we can use the provided standard free energy changes (ΔG°) for Pb²+(aq) and Cl⁻(aq). The dissolution of PbCl₂ in water is represented by the following chemical equation:
PbCl₂(s) ↔ Pb²+(aq) + 2Cl⁻(aq)
The relation between the standard free energy change for a process (ΔG°) and the equilibrium constant (K) at standard conditions (298 K and 1 atm) is given by the equation:
ΔG° = -RT ln(K)
Where R is the gas constant (8.314 J/(mol·K)) and T is the temperature in Kelvin. The equilibrium constant for the dissolution of PbCl₂ is the solubility product Ksp. We will calculate ΔG° for the overall reaction by adding the ΔG° values of Pb²+ and 2Cl⁻, and then use this value to find Ksp.
The ΔG° for the formation of Pb²+(aq) and 2Cl⁻(aq) from PbCl₂(s) is a sum of the individual ΔG° values:
ΔG°(reaction) = ΔG°(Pb²+) + 2 * ΔG°(Cl⁻)
ΔG°(reaction) = -24.3 kJ/mol + 2 * (-131.2 kJ/mol) = -286.7 kJ/mol
Convert kJ to J: -286.7 kJ/mol * 1000 J/kJ = -286700 J/mol
Now, solving for Ksp:
-RT ln(Ksp) = ΔG°(reaction)
ln(Ksp) = -ΔG°(reaction) / (RT)
ln(Ksp) = 286700 J/mol / (8.314 J/(mol·K) * 298 K)
ln(Ksp) ≈ 11.722
Ksp = e^{11.722} ≈ 1.2 × 10⁻⁴
Therefore, the correct answer is (a) 1.2 × 10⁻⁴.