Question 10

**Setting up the mixture** - Ba²⁺: 200 mL × 0.010 M = 2 mmol - IO₃⁻: 100 mL × 0.10 M = 10 mmol - Total volume = 300 mL Initial concentrations: [Ba²⁺] = 2/300 = 6.67 × 10⁻³ M, [IO₃⁻] = 10/300 = 3.33 × 10⁻² M. Ionic product = (6.67 × 10⁻³)(3.33 × 10⁻²)² ≈ 7.4 × 10⁻⁶ ≫ Ksp = 1.58 × 10⁻⁹, so Ba(IO₃)₂ precipitates. **Precipitation (Ba²⁺ is limiting):** Ba²⁺ + 2 IO₃⁻ → Ba(IO₃)₂(s) 2 mmol Ba²⁺ consumes 4 mmol IO₃⁻, leaving 6 mmol IO₃⁻ in 300 mL: [IO₃⁻] = 6/300 = 0.020 M (excess, common ion) **Solubility of Ba(IO₃)₂ in this solution:** Let solubility = s. Then [Ba²⁺] = s, [IO₃⁻] = 0.020 + 2s ≈ 0.020. Ksp = s (0.020)² s = 1.58 × 10⁻⁹ / 4.0 × 10⁻⁴ = 3.95 × 10⁻⁶ mol dm⁻³ So X = 3.95.

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