A 0.1 M − KMnO 4 solution is used for the following titration. What volume of the solution will be required to react with 0.158 g of Na 2 S 2 O 3 ?→ S O 3 2 − + M n O 4 − + H 2 O → M n O

Question

A 0.1 M − KMnO 4 ​ solution is used for the following titration. What volume of the solution will be required to react with 0.158 g of Na 2 ​ S 2 ​ O 3 ​ ?→ S O 3 ​ 2 − + M n O 4 ​ − + H 2 ​ O → M n O 2 ​ ( s ) + SO 4 ​ 2 − + O H −

TestHub · Accepted Answer

Given: Mass of Na₂S₂O₃ = 0.158 g

Molarity of KMnO₄ = 0.1 M

In S₂O₃²⁻, the average oxidation state of S = +2. It is oxidized to SO₄²⁻ where S = +6, so the n-factor = 8 per mole of Na₂S₂O₃.

Moles of Na₂S₂O₃ = $0.158/158 = 0.001 m o l$

Equivalents of Na₂S₂O₃ = $0.001 \times 8 = 0.008 e q$

For KMnO₄ → MnO₂ (basic medium), n-factor = 3

Normality = $0.1 \times 3 = 0.3 N$

Volume of KMnO₄ required:

$V = \frac{Eq u i v a l e n t s}{N or ma l i t y} = \frac{0.008}{0.3} = 0.02667 L = 26.67 m L$

Answer: (B) 26.67 mL

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