Find the volume of a 0.2 M solution of MnO₄⁻ that will react with 50 mL of 0.1 M solution of C₂O₄²⁻ in acidic medium____

V₁ = 10 mL

Detailed Explanation:

The balanced redox reaction between MnO₄⁻ and C₂O₄²⁻ in acidic medium is as follows:

2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 10CO₂ + 2Mn²⁺ + 8H₂O

From the stoichiometry of the reaction: - 2 moles of MnO₄⁻ react with 5 moles of C₂O₄²⁻. Using the molarity equation for titration:

M₁V₁ / n₁ = M₂V₂ / n₂

Where: - M₁ is the molarity of MnO₄⁻ (0.2 M), - V₁ is the volume of MnO₄⁻ solution (unknown), - n₁ is the number of moles of MnO₄⁻ (2), - M₂ is the molarity of C₂O₄²⁻ (0.1 M), - V₂ is the volume of C₂O₄²⁻ solution (50 mL), - n₂ is the number of moles of C₂O₄²⁻ (5). Using this, we calculate the volume of MnO₄⁻ solution (V₁):

0.2 × V₁ / 2 = 0.1 × 50 / 5

Solving for V₁:

V₁ = (0.1 × 50 × 2) / (0.2 × 5) = 10 mL

Therefore, the volume of MnO₄⁻ required is 10 mL.

Concept Behind the Solution:

This question involves the concept of stoichiometry in redox reactions and how to calculate the volume of a solution required for a reaction. The relationship between the volume and molarity of reactants in a redox reaction is governed by the balanced chemical equation.

The equation used here is based on the law of equivalence for titrations, where the number of equivalents of the oxidizing agent is equal to the number of equivalents of the reducing agent, allowing us to find the volume of MnO₄⁻ solution needed to react with the given amount of C₂O₄²⁻.