Final answer:
The molar solubility of lead(II) sulfate (PbSO4) in a solution with Na2SO4 at a concentration of 1.0 × 10—3 M is determined by the Ksp value of 1.8 × 10—8. Due to the common ion effect, the molar solubility (s) of PbSO4 in this case is 1.8 × 10—5 M.
Step-by-step explanation:
To calculate the molar solubility of lead(II) sulfate (PbSO4) in a solution that already contains sodium sulfate (Na2SO4), it is important to recognize that the sulfate ions from Na2SO4 will affect the solubility of PbSO4. The solubility product constant (Ksp) for PbSO4 is given as 1.8 × 10⁻₈.
The dissolution of lead sulfate can be represented as:
PbSO4 (s) → Pb2+ (aq) + SO42– (aq)
At equilibrium, the concentrations of the ions can be shown by:
Ksp = [Pb2+][SO42–]
Since sodium sulfate is present in the solution, the concentration of sulfate ions is already high and remains constant at 1.0 × 10⁻₃ M.
So for PbSO4:
Ksp = (s)(1.0 × 10⁻₃) = 1.8 × 10⁻₈
Therefore, the molar solubility (s) of PbSO4 is:
s = ⁷/₈ⁱ⁸ × 10⁻₅ M