500 mL of 1.2 M KI solution is mixed with 500 mL of $0.2\,\mathrm{M}$ $\ce{KMnO4}$ solution in basic medium. The liberated iodine was titrated with standard $0.1\,\mathrm{M}$ $\ce{Na2S2O3}$ solution in the presence of starch indicator till the blue color disappeared. The volume (in L) of $\ce{Na2S2O3}$ consumed is ______. (Nearest integer)

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Redox Reactions and Electrochemistry

500 mL of 1.2 M KI solution is mixed with 500 mL of

0.2 M

KMnO X_{4}

solution in basic medium. The liberated iodine was titrated with standard

0.1 M

Na X_{2} S X_{2} O X_{3}

solution in the presence of starch indicator till the blue color disappeared. The volume (in L) of

Na X_{2} S X_{2} O X_{3}

consumed is ______. (Nearest integer)

Answer: 3

In basic medium permanganate is reduced to

MnO X_{2}

, a three-electron change, so each mole of

KMnO X_{4}

supplies 3 equivalents:

0.2 M \times 0.500 L \times 3 = 0.30

equivalents, which oxidise iodide to an equal number of equivalents of iodine. In the titration

I X_{2} + 2 S X_{2} O X_{3} X^{2 -} 2 I X^{-} + S X_{4} O X_{6} X^{2 -}

, so thiosulfate reacts equivalent-for-equivalent and 0.30 equivalents of

Na X_{2} S X_{2} O X_{3}

are required. At 0.1 M this is

V = \frac{0.30}{0.1} = 3 L

. The volume of thiosulfate consumed is 3 L.

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Integer answer