[Olympiad] The Ksp of Lead(II) iodide (PbI₂) is 7.1 × 10⁻⁹. What is the molar solubility of PbI₂ in pure water?
- 8.4 × 10⁻⁵ M
- 1.2 × 10⁻³ M
- 2.6 × 10⁻³ M
- 1.9 × 10⁻³ M
📝 Batafsil tushuntirish
PbI₂(s) ⇌ Pb²⁺ + 2I⁻. Ksp = s × (2s)² = 4s³.s = ∛(Ksp / 4) = ∛(7.1 × 10⁻⁹ / 4) = ∛(1.775 × 10⁻⁹) ≈ 1.21 × 10⁻³ M.