Final answer:
To calculate the solubility product constant (Ksp) for PbI2, convert the given solubility from grams per liter to molarity, then use the dissolution equation to determine the concentration of ions in solution and apply these to the Ksp expression.
Step-by-step explanation:
The solubility of lead (II) iodide in water is 0.62 grams per 1.0 L of solution. To find the solubility product constant (Ksp) for PbI₂, we first need to convert the solubility in grams per liter to molarity (moles per liter). The molar mass of PbI₂ is approximately 461 g/mol.
So, 0.62 grams per liter is equivalent to 0.62 g / 461 g/mol = 1.34 x 10⁻³ mol/L (molar solubility of PbI₂). The dissolution equation is PbI₂(s) ⇌ Pb₂+(aq) + 2I-(aq), which means for every mole of PbI₂ that dissolves, 1 mole of Pb²⁺ and 2 moles of I- are produced.
Hence, the equilibrium concentrations will be 1.34 x 10⁻³ M for Pb2+ and 2 x 1.34 x 10⁻³ M for I-, which is 2.68 x 10⁻³ M. Using these concentrations, Ksp can be calculated using the expression Ksp = [Pb²⁺][I-]2 = (1.34 x 10⁻³) x (2.68 x 10⁻³)² = 9.6 x 10⁻⁹.