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Equivalent Mass

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Question

A solution contains $N a_{2} C O_{3} a n d N a H C O_{3}$ .10 ml of the solution requires 2.5 ml of 0.1M $H_{2} S O_{4}$ for neutralisation using phenolphthalein as an indicator. Methyl orange is added when further 2.5 ml of 0.2 M $H_{2} S O_{4}$ is required. The amount of $N a H C O_{3}$ in one litre of the solution is

A

2.1g

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B

8.4 g

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C

5.3 g

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D

4.2 g

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Solution

The correct option is D 4.2 g
$N o . m e q N a_{2} C O_{3} \equiv N o . m e q N a H C O_{3} \equiv 2.5 \times 0.1 \times 2 = 0.5 m e q N o . o f m e q N a H C O_{3} + N o . o f M e q N a H C O_{3} = 2.5 \times 0.2 \times 2 = 1.0 m e q (f o r m e d f r o m N a_{2} C O - 3) 0.5 + N o . o f m e q N a H C O_{3} = 1.0 N o . o f m e q N a H C O_{3} = 0.5 m e q \frac{W}{84} \times 1000 = 0.5, W = \frac{0.5 \times 84}{1000} 10 m l o f s o l u t i o n h a s \frac{0.5 \times 84}{1000} g 100 m l - \frac{0.5 \times 84}{1000} \times \frac{1000}{10} = 4.2 g / l i t$

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Similar questions

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Equivalent Mass

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