A solution of 0.2 g of a compound containing Cu ⁺² and C ₂ O ₄⁻² ions on titration with 0.02 M KMnO ₄ in presence of H ₂ SO ₄ consumes 22.6 mL of the oxidant. The resultant solution is neutralized with Na ₂ CO ₃ , acidified with dilute acetic acid and treated with excess KI. The liberated iodine requires 11.3 mL of 0.05 ~N Na ₂ ~S ₂ O ₃ solution for complete reduction. Find out the molar ratio of Cu ⁺² to C ₂ O ₄⁻² i

Step-by-step solution

The mixture of Cu ²⁺ and C ₂ O ₄²⁻ are reacting separately first with KMnO ₄ solution and then with solid KI to liberate iodine. It can be seen that Cu ⁺² cannot be oxidised and C ₂ O ₄⁻² cannot be reduced. This is because Cu is already in its highest oxidation state +2 . Equivalents of KMnO ₄ solution = 0.02 5 22.6 1000 =2.26 10⁻³ moles of C ₂ O ₄²⁻= 2.26 10⁻³ 2 =1.13 10⁻³ This is because only C ₂ O ₄⁻² is oxidised by KMnO ₄ to CO ₂ (' n ' factor 2) Equivalents of Na ₂ ~S ₂ O ₃ solution = 0.05 11.3 1000 =5.65 10⁻⁴