Use K = 39, I = 127, O = 16. KIO3 reacts with excess KI in acid: IO3- + 5I- + 6H+ -> 3I2 + 3H2O. The iodine liberated is titrated with 0.5 M Na2S2O3 (I2 + 2S2O3^2- -> 2I- + S4O6^2-). The volume of thiosulfate needed for 2.14 g of KIO3 is:

JEE Main levelSingle correct optionDifficulty 2 of 3rx132

Use K = 39, I = 127, O = 16. $KIO X_{3}$ reacts with excess $KI$ in acid: $IO X_{3} X^{-} + 5 I X^{-} + 6 H X^{+} 3 I X_{2} + 3 H X_{2} O$ . The iodine liberated is titrated with $0.5 M$ $Na X_{2} S X_{2} O X_{3}$ ( $I X_{2} + 2 S X_{2} O X_{3} X^{2 -} 2 I X^{-} + S X_{4} O X_{6} X^{2 -}$ ). The volume of thiosulfate needed for $2.14 g$ of $KIO X_{3}$ is:

n (KIO X_{3}) = 2.14/214 = 0.01

, giving

0.03

mol

I X_{2}

, which needs

0.06

mol

S X_{2} O X_{3} X^{2 -}

V = 0.06/0.5 = 0.12 L = 120 mL