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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.Determine the normality of the $N a_{2} S_{2} O_{3}$ solution.
(Given: Molar mass of $K B r O_{3} = 167 g / m o l; N a_{2} S_{2} O_{3} = 158 g / m o l)$

0.2 N

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0.72 N

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0.02 N

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0.12 N

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Solution

The correct option is D $0.12 N$
Here the chemical reaction looks like
$N a_{2} S_{2} O_{3} + K B r O_{3} + K I + H^{+} \to N a_{2} S_{4} O_{6} + K B r + N a I + H_{2} O$

From the equation we can say that
n-factor of $N a_{2} S_{2} O_{3}$ = 1
n-factor of $K B r O_{3}$ = 6
Applying law of equivalence here,
eq. of $K B r O_{3}$ = eq. of $N a_{2} S_{2} O_{3}$
$\frac{0.167}{(\frac{167}{6})} = N \times 50 \times 10^{- 3}$
$= 0.12 N$
Therefore, the correct answer is option (b)

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A solution of Na2S2O3 is standardized iodometrically against 0.167 g of KBrO3. This process requires 50 mL of the Na2S2O3 solution.Determine the normality of the Na2S2O3 solution. (Given: Molar mass of KBrO3 = 167 g/mol;Na2S2O3 = 158 g/mol)