The solubility of CdSO in water is 8.0 × 10 mol L . Its solubility in 0.01 M H SO solution is ...... — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

**Step 1: Find Ksp of CdSO₄ in pure water** For CdSO₄ ⇌ Cd²⁺ + SO₄²⁻ Given solubility = 8.0 × 10⁻⁴ mol/L At equilibrium: [Cd²⁺] = [SO₄²⁻] = 8.0 × 10⁻⁴ M Ksp = [Cd²⁺][SO₄²⁻] = (8.0 × 10⁻⁴)² = 6.4 × 10⁻⁷ **Step 2: Analyze the H₂SO₄ solution** 0.01 M H₂SO₄ provides [SO₄²⁻] = 0.01 M (from complete dissociation) This is the common ion effect—added SO₄²⁻ shifts equilibrium left. **Step 3: Set up equilibrium in 0.01 M H₂SO₄** Let solubility of CdSO₄ = s mol/L [Cd²⁺] = s [SO₄²⁻] = 0.01 + s ≈ 0.01 M (since s << 0.01) **Step 4: Apply Ksp expression** Ksp = [Cd²⁺][SO₄²⁻] 6.4 × 10⁻⁷ = s × 0.01 s = 6.4 × 10⁻⁷ / 10⁻² = 6.4 × 10⁻⁵ mol/L **Step 5: Express in required form** s = 6.4 × 10⁻⁵ = 64 × 10⁻⁶ mol/L Rounding to nearest integer: **64** Therefore, the answer is 64.00.