Question

A solution contains a mixture of sulphuric acid and oxalic acid, 25 ml of the solution requires 35.54 ml of 0.1 M NaOH for neutralization and 23.45 ml of 0.02 M KMnO 4 for oxidation. Calculate the molarity of solution with respect to sulphuric acid and oxalic acid.

Answer 1

Important points to note:

- Sulphuric acid is a strong oxidizing agent and thus is not oxidised by the permanganate.

- Oxalic acid is an organic acid and thus it is oxidised by the permanganate.

- There are two things that happen in this reaction : Oxidation of oxalic acid and neutralisation of the acid mixture.

EQUATIONS FOR THE NEUTRALIZATION

1.) H₂SO₄ + 2 NaOH —> Na₂SO₄ + 2 H₂O

Mole ratio is 1:2

2.) H₂C₂O₄ + 2 NaOH — > Na₂C₂O₄ + 2 H₂O

Mole ratio is 1 : 2

In both, the mole ratio is 1 : 2

Moles of NaOH in 35.54 ml of 0.1 M is :

(35.54/1000) × 0.1 = 0.003554moles

Since mole ratio is 1 : 2

Moles of the acid solution is :

0.003554 / 2 = 0.001777 moles

Molarity of the acid solution :

0.001777 moles are in 25 ml

How many in 1000 ml

(100/25) × 0.001777 = 0.07108 M


OXIDATION OF OXALIC ACID

2 MnO₄⁻ + 5 CO₂O₄²⁻ + 16 H⁺ — > 2 Mn₂ + 10 CO₂ + 8 H₂O

The mole ratio is 2 : 5

Moles of MnO₄ in 23.45ml of 0.02 M KMnO₄

(23.45 / 1000) × 0.02 = 0.000469 moles.

Moles of oxalic acid = 5/2 × 0.000469 = 0.0011725 moles

Molarity of oxalic acid:

0.0011725 moles are in 25 ml

How many are in 1000 ml

(1000 / 25) × 0.0011725 = 0.0469 M

Molarity of sulphuric acid is thus :

Total molarity of the acid solution - molarity of oxalic acid.

0.07108 M - 0.0469 = 0.02418 M

ANSWER :

Molarity of Oxalic acid = 0.0469 M

Molarity of Sulphuric acid = 0.02418 M