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Uses of Nitric Acid

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A sample of $N a H C O_{3} + N a_{2} C O_{3}$ required $20 m l$ of $H C l$ using phenolphthalein as indicator and $35 m l$ more required if methyl orange is used as indicator. Then molar ratio of $N a H C O_{3}$ to $N a_{2} C O_{3}$ is?

A

\frac{1}{2}

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B

\frac{2}{3}

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C

\frac{3}{4}

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D

\frac{1}{3}

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Solution

The correct option is B $\frac{1}{2}$
$N a_{2} C O_{3} + H C l ⟶ N a H C O_{3} + N a c l$ (Phenolpthalein)
Let us assume, $[H C l] = 0.1 M$
Since, $1$ mole of $N a_{2} C O_{3}$ react with one mole of $H C l$ i.e., no. of moles of both will be same.
No. of moles of $H C l = m o l a r i t y \times v o l u m e (L)$
$= 0.1 \times \frac{20}{1000}$
$= 0.1 \times 0.02 = 0.002$ moles
no. of moles of $H C l$ =no. of moles of $N a_{2} C O_{3}$
$N a H C O_{3} + H C l ⟶ N a C l + H_{2} O + C O_{2}$
(methyl orange)
In this case, volume of $H C l$ is used= $20 + 35 = 55 m L$
no. of moles of $H C l = 0.1 \times \frac{55}{1000}$
$= 0.0055 =$ no. of moles of $N a H C O_{3}$
$\frac{m o l e s o f N a_{2} C O_{3}}{m o l e s o f N a H C O_{3}} = \frac{0.0020}{0.0055} = \frac{20}{55}$
$∴ \frac{m o l e s o f N a H C O_{3}}{m o l e s o f N a_{2} C O_{3}} = \frac{55}{20}$

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