The solubility product of PbI is 8.0 × 10 . The solubility of lead iodide in 0.1 molar solution of l — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

**Step 1: Write the dissolution equation and Ksp expression** PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq) Ksp = [Pb²⁺][I⁻]² = 8.0 × 10⁻⁹ **Step 2: Set up initial and equilibrium concentrations** Initial [Pb²⁺] from Pb(NO₃)₂ = 0.1 M Let solubility of PbI₂ = s mol/L At equilibrium: - [Pb²⁺] = 0.1 + s ≈ 0.1 M (s is negligible) - [I⁻] = 2s **Step 3: Apply the Ksp expression** Ksp = [Pb²⁺][I⁻]² 8.0 × 10⁻⁹ = (0.1)(2s)² 8.0 × 10⁻⁹ = (0.1)(4s²) 8.0 × 10⁻⁹ = 0.4s² **Step 4: Solve for s** s² = (8.0 × 10⁻⁹)/0.4 s² = 2.0 × 10⁻⁸ s = √(2.0 × 10⁻⁸) = 1.41 × 10⁻⁴ mol/L **Step 5: Express in the required format** s = 1.41 × 10⁻⁴ = 141 × 10⁻⁶ mol/L Therefore, x = 141, and the answer is **141.00**.