[Olympiad] The solubility product of Calcium fluoride (CaF₂) is 3.9 × 10⁻¹¹. What is its molar solubility in a 0.01 M NaF solution?
- 6.2 × 10⁻⁶ M
- 3.9 × 10⁻⁷ M
- 2.1 × 10⁻⁴ M
- 3.9 × 10⁻⁹ M
📝 Batafsil tushuntirish
CaF₂(s) ⇌ Ca²⁺ + 2F⁻. Ksp = [Ca²⁺][F⁻]².Let [Ca²⁺] = s. [F⁻] = 2s + 0.01 ≈ 0.01.
3.9 × 10⁻¹¹ = s * (0.01)² => s = 3.9 × 10⁻¹¹ / 10⁻⁴ = 3.9 × 10⁻⁷ M.