Redox titration - calculating the molarity of the unknown

Question

If 28.00 mL of a KMnO₄ solution was used to titrate 25.00 mL of a Na₂C₂O₄ solution made by dissolving 1.02 g of sodium oxalate in 250.0 mL of water, what is the molarity of the potassium permanganate?

Answer 1

Write the balanced equation for the reaction in question which happens to be a redox. The easiest way to do this is by using the appropriate half reactions for oxidation and reduction.

MnO4- ===> Mn2+ ... reduction half reaction

H2C2O42- ===> CO2 ... oxidation half reaction

This redox will take place in the presence of acid (H+), so to balance we proceed as follows:

MnO₄^- + 8H⁺ + 5e- ===> Mn²⁺ + 4H₂O [x2 to balance electrons]

H₂C₂O₄^2- ===> 2CO2 + +2H+ + 2e- [x5 to balance electrons]

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2MnO₄^- + 16H⁺ + 10e- + 5H₂C₂O₄^2- ===> 2Mn²⁺ + 8H₂O + 10CO2 + 10H⁺ + 10e-

2MnO₄^- + 6H⁺ + 5H₂C₂O₄^2- ==> 2Mn²⁺ + 8H₂O + 10CO₂ ... balanced equation

moles Na2C2O4 present = 1.02 g x 1 mole/134 g = 0.00761 moles

moles KMnO4 needed to react with 0.00761 moles Na2C2O4 = 0.00761 moles x 2 mol MnO4-/5 mol C2O42- = 0.003045 moles KMnO4 needed

Molarity of KMnO4 = 0.003045 moles/0.02800 L = 0.109 M (to 3 significant figures)