At 310 K, the solubility of CaF in water is 2.34 × 10 g /100 mL. The solubility product of CaF is __ — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

**Step 1: Convert solubility to mol/L** Solubility = 2.34 × 10⁻³ g/100 mL Convert to g/L: 2.34 × 10⁻³ × (1000/100) = 2.34 × 10⁻² g/L Convert to mol/L using molar mass (78 g/mol): Molarity = (2.34 × 10⁻²)/78 = 3.0 × 10⁻⁴ mol/L **Step 2: Write the dissolution equilibrium** CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq) **Step 3: Determine ion concentrations** If solubility = s = 3.0 × 10⁻⁴ mol/L, then: - [Ca²⁺] = s = 3.0 × 10⁻⁴ mol/L - [F⁻] = 2s = 6.0 × 10⁻⁴ mol/L **Step 4: Apply Ksp expression** Ksp = [Ca²⁺][F⁻]² Ksp = (3.0 × 10⁻⁴)(6.0 × 10⁻⁴)² Ksp = (3.0 × 10⁻⁴)(3.6 × 10⁻⁷) Ksp = 1.08 × 10⁻¹⁰ mol³/L³ **Step 5: Express in required form** 1.08 × 10⁻¹⁰ ≈ 1.1 × 10⁻¹⁰, which rounds to **1.0 × 10⁻¹⁰** Therefore, the answer is **1.0 × 10⁻¹⁰** (or **0** if expressed as