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Equivalent Mass

50 mL of a so...

Question

$50 m L$ of a solution of $N a_{2} C O_{3}$ and $N a H C O_{3}$ requires $17.5 m L$ of $0.1 M H_{2} S O_{4}$ for neutralisation using phenolphthalein as indicator. Methyl orange is then added when further $25 m L$ of $0.2 M H_{2} S O_{4}$ was required. Calculate the weight of $N a_{2} C O_{3}$ and $N a H C O_{3}$ respectively in the solution.

A

1.454 g, 0.221 g

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B

0.667 g, 0.118 g

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C

0.892 g, 1.651 g

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D

0.371 g, 0.546 g

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Solution

The correct option is D $0.371 g, 0.546 g$
Phenolphthalein will only neutralise $50 %$ of $N a_{2} C O_{3}$ .
$\frac{1}{2} Milliequivalents of N a_{2} C O_{3} = Milliequivalents of H_{2} S O_{4} = 0.1 \times 17.5 \times 2 = 3.5$
$∴ Milliequivalents of N a_{2} C O_{3} = 7.0$
$\frac{W_{N a_{2} C O_{3}}}{M_{N a_{2} C O_{3}}} \times 2 \times 1000 = 7.0$
$∴ W_{N a_{2} C O_{3}} = \frac{7.0 \times 106}{2 \times 1000} = 0.371 g$

The methyl orange indicator will neutralise the rest $50 %$ of $N a_{2} C O_{3}$ and original $N a H C O_{3}$ .
$∴ \frac{1}{2} Milliequivalents of N a_{2} C O_{3} + Milliequivalents of N a H C O_{3} = Milliequivalents of H_{2} S O_{4} = 0.2 \times 25 \times 2 = 10$
$∴ Milliequivalents of N a H C O_{3} = 10 - 3.5 = 6.5$
$\frac{W_{N a H C O_{3}}}{M_{N a H C O_{3}}} \times 1 \times 1000 = 6.5$
$∴ W_{N a H C O_{3}} = \frac{6.5 \times 84}{1 \times 1000} = 0.546 g$

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