A 150 mL of solution of $I_{2}$ is divided into two unequal parts. $1^{s t}$ part reacts with hypo solution in acidic medium. 15 mL of 0.4 M hypo was consumed. $2^{n d}$ part was added to 100 mL of 0.3M hot NaOH solution. Residual base required 10 mL of 0.3M $H_{2} S O_{4}$ solution for complete neutralization. What was the initial concentration of $I_{2}$ ?
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Detailed Solution

$I_{2} + 2 N a_{2} S_{2} O_{3} \to 2 N a I + N a_{2} S_{4} O_{6}$
Millimoles of $N a_{2} S_{2} O_{3}$ consumed = 6
Millimoles of $I_{2}$ consumed = 3
$3 I_{2} + 6 N a O H \to 5 N a I + N a I O_{3} + 3 H_{2} O$
Millimole of $I_{2}$ reacted with $N a O H = \frac{30 - 6}{2} = 12$
Total millimole of $I_{2}$ consumed in this reaction = 15
Initial concentration of $I_{2} = \frac{15}{150} = 0.1 M$