$25 m L$ of a mixture of $N a O H + N a_{2} C O_{3}$ , when titrated with $\frac{N}{10} H C l$ using phenolphthalein indicator required $25 m L H C l$ to decolourise phenolphthalein. At this stage, methyl orange was added and the addition of acid was continued. The second endpoint was reached after further addition of $5 m L$ of the acid. Calculate the amount of $N a_{2} C O_{3}$ and $N a O H$ in one litre of the solution.

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Solution

Between first and second end points,
$N a H C O_{3} + H C l \to N a C l + H_{2} C O_{3}$
$5 m L \frac{N}{10} H C l \equiv \frac{1}{2} N a_{2} C O_{3}$ present in $25 m L$ of a mixture
or $10 m L \frac{N}{10} H C l \equiv N a_{2} C O_{3}$ present in $25 m L$ of a mixture
$\equiv 10 m L \frac{N}{10} N a_{2} C O_{3} = 0.053 g N a_{2} C O_{3}$
Amount of $N a_{2} C O_{3}$ in one litre of mixture $= \frac{0.053}{25} \times 1000$
$= 2.12 g$
$(25 - 5) m L \frac{N}{10} H C l \equiv N a O H$ present in $25 m L$ of mixture
$\equiv 25 m L \frac{N}{10} N a O H$
$\equiv 0.08 g N a O H$
Amount of $N a O H$ in one litre of mixture $= \frac{0.08}{25} \times 1000$
$= 3.2 g$ .

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