500 mL of 1.2 M KI solution is mixed with 500 mL of 0.2 M KMnO₄ solution in basic medium. The liberated iodine was titrated with standard 0.1 M Na₂S₂O₃ solution in the presence of starch indicator till the blue color disappeared. The volume (in L) of Na₂S₂O₃ consumed is ___

500 mL of 1.2 M KI solution is mixed with 500 mL of 0.2 M KMnO4 solution in basic medium. The liberated iodine was titrated with standard Na2S2O3 solution.

Q. 500 mL of 1.2 M KI solution is mixed with 500 mL of 0.2 M KMnO₄ solution in basic medium. The liberated iodine was titrated with standard 0.1 M Na₂S₂O₃ solution in the presence of starch indicator till the blue color disappeared. The volume (in L) of Na₂S₂O₃ consumed is ____ . (Nearest integer)

Correct Answer: 3

Explanation

Step 1: Calculate moles of KI and KMnO₄

Moles of KI = 1.2 × 0.5 = 0.6 mol

Moles of KMnO₄ = 0.2 × 0.5 = 0.1 mol

Step 2: Write the redox reaction in basic medium

2 KMnO₄ + 6 KI + H₂O → 2 MnO₂ + 6 KOH + 3 I₂

From the balanced equation:

2 mol KMnO₄ → 3 mol I₂

Step 3: Find moles of iodine liberated

0.1 mol KMnO₄ → (3/2) × 0.1 = 0.15 mol I₂

KI is in excess, so KMnO₄ is the limiting reagent.

Step 4: Reaction of iodine with sodium thiosulphate

I₂ + 2 Na₂S₂O₃ → 2 NaI + Na₂S₄O₆

1 mol I₂ requires 2 mol Na₂S₂O₃

Moles of Na₂S₂O₃ required = 2 × 0.15 = 0.30 mol

Step 5: Calculate volume of Na₂S₂O₃

Volume = moles / molarity = 0.30 / 0.1 = 3 L

Therefore, the required volume of Na₂S₂O₃ is 3 L.

Explanation

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