If the solubility product of PbS is 8 × 10 , then the solubility of PbS in pure water at 298 K is x — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

**Step 1: Write the dissolution equation and Ksp expression** PbS(s) ⇌ Pb²⁺(aq) + S²⁻(aq) Ksp = [Pb²⁺][S²⁻] = 8 × 10⁻²⁸ **Step 2: Set up solubility relationship** Let the solubility of PbS = s mol/L At equilibrium: - [Pb²⁺] = s - [S²⁻] = s **Step 3: Substitute into Ksp expression** Ksp = s × s = s² 8 × 10⁻²⁸ = s² **Step 4: Solve for solubility** s = √(8 × 10⁻²⁸) s = √8 × 10⁻¹⁴ **Step 5: Calculate √8** √8 = √(4 × 2) = 2√2 = 2 × 1.41 = 2.82 **Step 6: Find final solubility** s = 2.82 × 10⁻¹⁴ mol/L Since the answer format is x × 10⁻¹⁶ mol/L: s = 2.82 × 10⁻¹⁴ = 282 × 10⁻¹⁶ mol/L Therefore, x = 282, and the answer is 282.