Answer to Question #342584 in Chemistry for Larry

Question

4 Titration of 0.2121 g of pure Na2C2O4 required 43.31 mL of KMnO4. What is the molar concentration of the KMnO4 solution? The chemical reaction is MnO4- + C2O42- + H+ ----------> Mn2+ + CO2 + H2O (unbalanced equation) Atomic masses: Na = 23 C = 12 O = 1

Answer 1

Solution:

Balanced chemical equation:

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

According to stoichiometry:

Moles of MnO₄⁻ / 2 = Moles of C₂O₄²⁻ / 5

5 × Moles of KMnO₄ = 2 × Moles of Na₂C₂O₄

5 × Molarity of KMnO₄ × Volume of KMnO₄ = 2 × (Mass of Na₂C₂O₄ / Molar mass of Na₂C₂O₄)

Molarity of KMnO₄ = unknown

Volume of KMnO₄ = 43.31 mL × (1 L / 1000 mL) = 0.04331 L

Mass of Na₂C₂O₄ = 0.2121 g

Molar mass of Na₂C₂O₄ = 2×Ar(Na) + 2×Ar(C) + 4×Ar(O) = 2×23 + 2×12 + 4×16 = 134 (g mol⁻¹)

Therefore,

Molarity of KMnO₄ = (2 × Mass of Na₂C₂O₄) / (5 × Volume of KMnO₄ × Molar mass of Na₂C₂O₄)

Molarity of KMnO₄ = (2 × 0.2121 g) / (5 × 0.04331 L × 134 g mol⁻¹) = 0.01462 mol L⁻¹

Molarity of KMnO₄ = 0.01462 M

Answer: The molar concentration of the KMnO₄ solution is 0.01462 M