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Stoichiometric Calculations

A 200 mL so...

Question

A $200$ mL solution of $I_{2}$ is divided into two unequal parts. Part I reacts with hypo solution in acidic medium and requires $8$ mL of $2 M$ hypo solution for complete neutralization.
Part II was added with $300$ mL of $0.1 M$ $N a O H$ solution. Residual base required $30$ mL of $0.1 M$ $H_{2} S O_{4}$ solution for complete neutralization. Calculate the value of $20$ times the initial concentration of $I_{2}$ ?

A

1

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B

2

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C

3

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D

4

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Solution

The correct option is D $2$
First part:
Millimoles of $N a_{2} S_{2} O_{3}$ used $= 8 \times 2 \times 1$ ( $n$ -factor) $= 16$
$I_{2} + 2 N a_{2} S_{2} O_{3} \to 2 N a I + N a_{2} S_{4} O_{6}$
$1$ mol $2$ mol
Millimoles of $I_{2}$ used $= \frac{1}{2}$ Millimoles of $N a_{2} S_{2} O_{3} = \frac{16}{2} = 18$ ....(i)

Second part:
$3 I_{2} + 6 N a O H \to 5 N a I + N a I O_{3} + 3 H_{2} O$
Millimoles of $H_{2} S O_{4} =$ excess $N a O H = 30 \times 0.1 \times 2 = 6$
Millimoles of total $N a O H = 300 \times 0.1 \times 1 = 30$
Millimoles of $N a O H$ used $= 30 - 6 = 24$
Millimoles of $I_{2}$ used $= \frac{1}{2}$ millimoles of $N a O H$ used $= \frac{24}{2} = 12$ mmol of $I_{2}$ used
Total mmol of $I_{2}$ used $=$ Part I $+$ Part II $= 8 + 12 = 20$ mmol

Molarity of $I_{2} = \frac{mmol}{V_{m} L} = \frac{20}{200} = 0.1 M$

$20$ times the initial $M_{I_{2}} = 0.1 \times 20 = 2$

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