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Percentage Composition

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Question

A solution of $N a_{2} S_{2} O_{3}$ is standardized iodometrically against $0.167 g$ of $K B r O_{3}$ . This process requires $50 m L$ of the $N a_{2} S_{2} O_{3}$ solution. What is the normality of the $N a_{2} S_{2} O_{3}$ ?

0.2 N

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0.12 N

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0.72 N

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0.02 N

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Solution

The correct option is B $0.12 N$
$S_{2} O_{3}^{2 -} + B r O_{3}^{-} \to S O_{2} + B r^{-}$
Balance above reaction in ardic medium
$3 S_{2} O_{3}^{2 -} + 2 B r O_{3}^{-} + 6 H^{+} \to 6 S O_{2} + 2 B r^{-} + 3 H_{2} O$
Molar mass of $K B r O_{3} = 39 + 79.9 + 16 \times 3$
$= 166.9 \approx 167$ gm
No. of mole of $K B r O_{3} = \frac{0.167}{167} = 0.001$
$2$ mol of $B r O_{3}^{-}$ required 3 mol of $S_{2} O_{3}^{2 -}$
$0.001$ mol of $B r O_{3}^{-}$ required $\frac{3}{2} \times 0.001$
$0.0015 m o l$
Normally = $\frac{0.0015 \times 4}{50 \times 10^{- 3}} = 0.03 \times 4 = 0.12 N$

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Similar questions

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A solution of $N a_{2} S_{2} O_{3}$ is standardized iodometrically against $0.167 g$ of $K B r O_{3}$ . This process requires $50 m L$ of the $N a_{2} S_{2} O_{3}$ solution. What is the normality of the $N a_{2} S_{2} O_{3}$ ?