Question

Calculate the volume of 0.05 KMnO₄ solution required to oxidise completely 2.70 g of oxalic acid in acidic solution.

Answer 1

The oxidation of oxalic acid by KMnO₄ is done by the chemical equation

2KMnO₄ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 3H₂O + 5[O]

H₂C₂O₄ + [O] → 2CO₂ + H₂O × 5

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

Step I. To oxidise completely 2.70 g of oxalic acid the number of moles of KMnO₄ required are:

Molar mass of H₂C₂O₄ = 2 × 1 + 2 × 12 + 4 × 16 = 90.0 g mol^–1

\(\therefore\)  No. of moles of H₂C₂O₄ contained in 2.70 g of oxalic acid are

= \(\frac{2.70}{90.0}\) = 0.03

5 moles of H₂C₂O₄ are oxidised by 2 moles of KMnO₄

\(\therefore\) 0.03 mole of it is oxidized by = \(\frac{2}{5}\) x 0.03

= 0.012 mole of KMnO₄

Step II. The volume of 0.05 KMnO₄ solution required is:

Now 0.05 mole of KMnO₄ are contained in 1000 cm³ of solution.

\(\therefore\) 0.012 mole of KMnO₄ will be contained in = \(\frac{1000}{0.05}\) x 0.012

= 240 cm³ of solution

Thus, the required volume of 0.05 M KMnO₄

Solution = 240 cm³