JEE Main 2026 — Jan 28, Shift 1 — Chemistry solution

JEE Main2026Jan 28, Shift 1Chemistry

500 mL of 1.2 M KI solution is mixed with 500 mL of 0.2 M KMnO $_{4}$ solution in basic medium. The liberated iodine was titrated with standard 0.1 M Na $_{2}$ S $_{2}$ O $_{3}$ solution in the presence of starch indicator till the blue color disappeared. The volume (in L) of Na $_{2}$ S $_{2}$ O $_{3}$ consumed is ______. (Nearest integer)

Your answer

Solution

In basic medium, MnO $_{4}^{-}$ is reduced to MnO $_{2}$ (n-factor 3), and I $^{-}$ is oxidised to I $_{2}$ .

$MnO_{4}^{-} + I^{-} \to MnO_{2} + I_{2}$

Liberated I $_{2}$ is then titrated with thiosulphate: $I_{2} + 2 S_{2} O_{3}^{2 -} \to 2 I^{-} + S_{4} O_{6}^{2 -}$ .

Equivalents of KMnO $_{4}$ = equivalents of Na $_{2}$ S $_{2}$ O $_{3}$ (the same number of electrons is transferred through the redox chain):

$(0.2) (\frac{500}{1000}) (3) = (0.1) (V) (1)$

$0.3 = 0.1 V \Rightarrow V = 3$ L.

$∴$ Answer: $3$ L.

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