0.2640 g of sodium oxalate are dissolved in a flask and titrated with 30.75 ml of potassium permanganate till pink endpoint.

Question

How many moles of sodium oxalate are present in the flask?

How many moles of potassium permanganate have been titrated into flask to reach endpoint?

Molarity of potassium permanganate?

Answer 1

This is a redox reaction that takes place in an acidic environment, so H₂SO₄ should be present. Assuming that is the case, then the sodium oxalate is present as oxalic acid (H₂C₂O₄) and it is then oxidized by the potassium permanganate (KMnO₄). The KMnO₄ is purple, and at the end point it turns pink. The balanced equation for this chemical redox reaction is:

5H₂C₂O₄ + 3H₂SO₄ + 2KMnO₄ ==> 10CO₂ + 2MnSO₄ + 8H₂O + K₂SO₄

...moles oxalic acid = 0.2640 g sodium oxalate x 1 mole/134 g = 0.001970 moles sodium oxalate x 1 mole oxalic acid/mole sodium oxalate = 0.001970 moles oxalic acid

...moles KMnO₄ = 0.001970 mol H₂C₂O₄ x 2 mol KMnO₄/5 mol H₂C₂O₄) = 0.0007881 moles KMnO₄

...molarity of KMnO₄ = 0.0007881 moles/0.03075 L = 0.02563 M