500 mL of 1.2 M KI is completely reacted with 500 mL of 0.2 M KMnO₄ in basic medium.

I^- is oxidised to I₂. The liberated I₂ reacts with 0.1 M Na₂S₂O₃.

Find the volume of Na₂S₂O₃ solution required.

Moles of KMnO₄:

= 0.2 × 0.5

= 0.1 mol

In basic medium:

MnO₄^- + 2H₂O + 3e^- → MnO₂ + 4OH^-

One mole KMnO₄ accepts 3 electrons.

Therefore:

Total electrons accepted:

= 0.1 × 3

= 0.3 mol e^-

Oxidation:

2I^- → I₂ + 2e^-

Moles of I₂ formed:

= 0.3 / 2

= 0.15 mol

Reaction with thiosulphate:

I₂ + 2S₂O₃^2- → 2I^- + S₄O₆^2-

Moles of Na₂S₂O₃ required:

= 2 × 0.15

= 0.30 mol

Volume:

V = n/M

= 0.30 / 0.10

= 3 L

Answer:

3 L