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Introduction to Oxidation and Reduction

For the react...

Question

For the reaction given below:
$N a_{2} C O_{3} (a q) + 2 H C l (a q) \to 2 N a C l (a q) + C O_{2} (g) + H_{2} O (l)$
If $38.55$ ml of $H C l$ is required to titrate $2.150$ g of $N a_{2} C O_{3}$ as given, then concentration of $H C l$ solution is $y \times 10^{- 2} (m o l . L^{- 1})$ . Find the nearest integral value of $y$ .

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Solution

The number of equivalents of $N a_{2} C O_{3} = \frac{M a s s}{E q u i v a l e n t m a s s} = \frac{2.150}{53} =$ the number of equivalent of $H C l$
Therefore, $\frac{N V}{1000} = \frac{N \times 38.55}{1000}$
Hence, $\frac{N \times 38.55}{1000} = \frac{2.150}{53}$
Thus, the normality of $H C l$ is equal to $1.05 N$ $= 105 \times 10^{- 2} N$
Since,
Molarity = Normality in case of HCl

So, the value of $y =$ 105

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