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Equivalent Mass

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Question

A solution contains a mixture of $N a_{2} C O_{3}$ and $N a O H .$ Using phenolphthalein as indicator $35 m L$ of mixture required $20 m L$ of $0.80 N H C l$ for the end point. With methyl orange, $35 m L$ of solution required $37.5 m L$ of the same $H C l$ for the end point. What is the weight of $N a_{2} C O_{3}$ and $N a O H$ respectively in the mixture?

A

1.484 g, 0.08 g

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B

1.5 g, 0.5 g

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C

0.08 g, 1.484 g

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D

0.5 g, 1.5 g

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Solution

The correct option is A $1.484 g, 0.08 g$
Suppose milliequivalents of $N a_{2} C O_{3}$ and $N a O H$ are $x$ and $y$ respectively.

When phenolphthalein is used as an indicator, only $50 %$ of $N a_{2} C O_{3}$ will be neutralised.
$\frac{1}{2} Milliequivalents of N a_{2} C O_{3} + Milliequivalents of N a O H = Milliequivalents of H C l$
$∴ \frac{x}{2} + y = 20 \times 0.80$
$\frac{x}{2} + y = 16 . . . . . (1)$

When methyl orange is used as an indicator, $100 %$ of $N a_{2} C O_{3}$ will be neutralised.
$Milliequivalents of N a_{2} C O_{3} + Milliequivalents of N a O H = Milliequivalents of H C l$
$∴ x + y = 37.5 \times 0.80$
$x + y = 30 . . . . . (2)$

From equation $(1)$ and $(2)$ ,
$x = 28, y = 2$

$∴ Milliequivalents of N a_{2} C O_{3} = 28.0$
$\frac{W_{N a_{2} C O_{3}}}{M_{N a_{2} C O_{3}}} \times 2 \times 1000 = 28.0$
$∴ W_{N a_{2} C O_{3}} = \frac{28.0 \times 106}{2 \times 1000} = 1.484 g$

$∴ Milliequivalents of N a O H = 2.0$
$\frac{W_{N a O H}}{M_{N a O H}} \times 1 \times 1000 = 2.0$
$∴ W_{N a O H} = \frac{2.0 \times 40}{1 \times 1000} = 0.08 g$

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