Answer to Question #338505 in Chemistry for dlo

Question

Titration of 0.2121 g of pure Na2C2O4 (134.00 g/mol) required 43.31 mL

of KMnO4.What is the molar concentration of the KMnO4 solution?

Answer 1

Solution:

Balanced chemical equation:

5Na₂C₂O₄ + 2KMnO₄ + 8H₂SO₄ → 2MnSO₄ + 5Na₂SO₄ + K₂SO₄ + 8H₂O + 10CO₂

According to stoichiometry:

Moles of Na₂C₂O₄ / 5 = Moles of KMnO₄ / 2

or:

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

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 × 134.00 g mol⁻¹) = 0.01462 mol/L

Molarity of KMnO₄ = 0.01462 M

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