500 mL of 1.2 M KI solution is mixed with 500 mL of 0.2 M solution in basic medium. The liberated iodine was titrated with standard 0.1 M solution in the presence of starch indicator till the blue color disappeared. The vol

Question

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) [2026]

Smart Achievers · Accepted Answer

(3)

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

$I_{2} + S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + I^{-}$

${gram eq of KMnO}_{4} = {gram eq of Na}_{2} S_{2} O_{3}$

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

$V = 3 L$

Solution

Ans.
(3)

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

$I_{2} + S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + I^{-}$

${gram eq of KMnO}_{4} = {gram eq of Na}_{2} S_{2} O_{3}$

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

$V = 3 L$

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Solution

Ans. (3) MnO4-+I-→MnO2+I2 I2+S2O32-→S4O62-+I- gram eq of KMnO4=gram eq of Na2S2O3 0.2×5001000×3=0.1×V×1 V=3 L

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Ans.
(3)

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

$I_{2} + S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + I^{-}$

${gram eq of KMnO}_{4} = {gram eq of Na}_{2} S_{2} O_{3}$

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

$V = 3 L$