$25 m L$ of a mixture of $N a O H$ and $N a_{2} C O_{3}$ when titrated with $N / 10 H C l$ using phenolphthalein indicator required $25 m L H C l$ . The same volume of mixture when titrated with $N / 10 H C l$ using methyl orange indicator required $30 m L$ of $H C l$ . Calculate the amount of $N a_{2} C O_{3}$ and $N a O H$ in one litre of this mixture.

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Solution

When phenolphthalein is the indicator, whole of $N a O H$ has been neutralised and carbonate converted into bicarbonate, i.e.,
$N a O H + H C l \to N a C l + H_{2} O$

$N a_{2} C O_{3} + H C l \to N a H C O_{3} + N a C l$

So, $25 m L \frac{N}{10} H C l \equiv N a O H + 1 / 2 N a_{2} C O_{3}$ present in $25 m L$ of mixture.

In another titration, when methyl orange is the indicator, whole of $N a O H$ has been neutralised and carbonate converted into carbonic acid, i.e.,

$N a_{2} C O_{2} + 2 H C l \to 2 N a C l + H_{2} C O_{3}$

$30 m L \frac{N}{10} H C l \equiv N a O H + N a_{2} C O_{3}$ present in $25 m L$ of mixture

Hence,
$(30 - 25) m L \frac{N}{10} H C l \equiv \frac{1}{2} N a_{2} C O_{3}$ present in $25 m L$ of mixture

Hence,
$10 m L \frac{N}{10} H C l \equiv N a_{2} C O_{3}$ present in $25 m L$ of mixture

$\equiv 10 m L \frac{N}{10} N a_{2} C O_{3}$ solution

Amount of $N a_{2} C O_{3} = \frac{53 \times 10}{10 \times 1000} = 0.053 g$

This amount of $N a_{2} C O_{3}$ is present in $25 m L$ of mixture.

This amount present in one litre of mixture,

$= \frac{0.053}{25} \times 1000 = 2.12 g$

$(30 - 10) m L \frac{N}{10} H C l \equiv N a O H$ present in $25 m L$ of mixture

$\equiv 20 m L \frac{N}{10} N a O H$

Amount of $N a O H$ in $25 m L$ of mixture $= \frac{40 \times 20}{10 \times 1000} = 0.08 g$

The amount present in one litre of mixture $= \frac{0.08}{25} \times 1000 = 3.20 g$ .

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