25 ml of household bleach solution was mixed with 30 ml of 0.5 M KI solutions and 10 ml of 4 M acetic acid solution. In the titration of the liberated iodine 48 ml of 0.25 N ${Na}_{2} S_{2} O_{3}$ was used to reach the end point. Which of the following is(are) correct? The reactions involved are as follows:
$KI + {CH}_{3} {CO}_{2} H \to_{}^{} {CH}_{3} {CO}_{2} K + HI$ ; $2 HI + NaOCl \to_{}^{} I_{2} + NaCl + H_{2} O$

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Molarity of household bleach solution is 0.24 M

Milli equivalents of acetic acid left are 25

Milli equivalents of I₂ liberated are 12

${Na}_{2} S_{2} O_{3}$ reacts with $I_{2}$ and bleach solution both

answer is A, B, C.

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Detailed Solution

$2 H^{+} + {OCl}^{-} + 2 I^{-} \to_{}^{} I_{2} + {Cl}^{-} + 2 H_{2} O I_{2} + 2 {Na}_{2} S_{2} O_{3} \to_{}^{} 2 NaI + {Na}_{2} S_{4} O_{6} n_{hypo} = 2 n_{I_{2}} = 2 n_{NaOCl} 48 \times 0.25 = 2 \times 25 \times M \frac{M}{2} \times 100 = \frac{100}{4} \times 10^{- 2} \times 48 M = 0.24 M (1) Correct KI + {CH}_{3} COOH \to_{}^{} {CH}_{3} COOK + HI t = 0 mm 15 40 {(me)}_{HAc left} = 25 (2) correct {(me)}_{I_{2}} = {(me)}_{hypo} = 48 \times \frac{100}{4} \times 10^{- 2} = 12 (3) correct (4) correct$