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Neutralisation Reaction

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Question

A solution contains $N a_{2} C O_{3}$ and $N a H C O_{3}$ . $10 m L$ of the solution required $2.5 m L$ of $0.1 M H_{2} S O_{4}$ for neutralisation using phenolphthalein as indicator. Methyl orange is then added when a further $2.5 m L$ of $0.2 M H_{2} S O_{4}$ was required. Calculate the amount of $N a_{2} C O_{3}$ and $N a H C O_{3}$ in one litre of the solution.

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Solution

$2.5 m L$ of $0.1 M H_{2} S O_{4} = 2.5 m L$ of $0.2 N H_{2} S O_{4}$
$= \frac{1}{2} N a_{2} C O_{3}$ present in $10 m L$ of mixture
So,
$5 m L$ of $0.2 N H_{2} S O_{4} = N a_{2} C O_{3}$ present in $10 m L$ of mixture
$\equiv 5 m L$ of $0.2 N N a_{2} C O_{3}$ $
$\equiv \frac{0.2 \times 53}{1000} \times 5 = 0.053 g$
Amount of $N a_{2} C O_{3} = \frac{0.053}{10} \times 1000 = 5.3 g / L$ of mixture
Between first and second end points,
$= 2.5 m L$ of $0.2 M H_{2} S O_{4}$ used
$= 2.5 m L$ of $0.4 N H_{2} S O_{4}$ used
$= 5 m L$ of $0.2 N H_{2} S O_{4}$ used
$\equiv \frac{1}{2} N a_{2} C O_{3} + N a H C O_{3}$ present in $10 m L$ of mixture
$(5 - 2.5) m L 0.2 N H_{2} S O_{4}$
$\equiv N a H C O_{3}$ present in $10 m L$ of mixture
$\equiv 2.5 m L 0.2 N N a H C O_{3}$
$\equiv \frac{0.2 \times 84}{1000} \times 2.5 = 0.042 g$
Amount of $N a H C O_{3} = \frac{0.042}{10} \times 1000 = 4.20 g / L$ of mixture.

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