Two solutions A and B each of 100 L was made by dissolving 4 g of NaOH and 9.8 g of H SO in water, r — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

**Step 1: Calculate moles of NaOH in Solution A** - Molar mass of NaOH = 40 g/mol - Moles of NaOH = 4 g ÷ 40 g/mol = 0.1 mol in 100 L - Concentration of NaOH = 0.1 mol ÷ 100 L = 0.001 M **Step 2: Calculate moles of H₂SO₄ in Solution B** - Molar mass of H₂SO₄ = 98 g/mol - Moles of H₂SO₄ = 9.8 g ÷ 98 g/mol = 0.1 mol in 100 L - Concentration of H₂SO₄ = 0.1 mol ÷ 100 L = 0.001 M **Step 3: Calculate moles in the mixture** - Moles of NaOH from 40 L of Solution A = 0.001 M × 40 L = 0.04 mol - Moles of H₂SO₄ from 10 L of Solution B = 0.001 M × 10 L = 0.01 mol - H₂SO₄ provides 0.02 mol of H⁺ (diprotic acid) **Step 4: Determine excess base** - Neutralization: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O - NaOH needed = 0.02 mol; NaOH available = 0.04 mol - Excess NaOH = 0.04 - 0.02 = 0.02 mol **Step 5: Calculate [OH⁻] and pOH** - Total volume = 40 + 10 = 50 L - [OH⁻] = 0.02 mol ÷ 50 L = 0.0004 M = 4 × 10