A solution of 0.2 g of a compound containing Cu²⁺ and C₂O₄²⁻ ions on titration with 0.02 M - KMnO₄ in the presence of H₂SO₄ consumes 22.6 ml of the oxidant. The resultant solution is neutralized with

Question

A solution of 0.2 g of a compound containing Cu²⁺ and C₂O₄²⁻ ions on titration with 0.02 M - KMnO₄ in the presence of H₂SO₄ consumes 22.6 ml of the oxidant. The resultant solution is neutralized with Na₂CO₃, acidified with dilute acetic acid and treated with excess KI. The liberated iodine requires 11.3 ml of 0.05 M - Na₂S₂O₃ for complete reduction. If the molar ratio of Cu²⁺ to C₂O₄²⁻ in the compound is 1:x, the value of x is

TestHub · Accepted Answer

KMnO₄ = 0.02 × 0.0226 = 0.452 mmol \equiv 5 × 0.452 = 2.26 meq, so C₂O₄²⁻ = 2.26/2 = 1.13 mmol. Thiosulphate = 0.05 × 0.0113 = 0.565 mmol = Cu²⁺ (2Cu²⁺ → I₂ → 2S₂O₃²⁻). Ratio Cu : C₂O₄ = 0.565 : 1.13 = 1 : 2, so x = 2. Ans: 2.

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