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Limiting Reagent or Reactant

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Question

A solution containing $N a_{2} C O_{3}$ and $N a O H$ requires $300 m L$ of $0.1 N H C l$ using phenolphthalein as an indicator. Methyl orange is then added to the above-titrated solution when a further $25 m L$ of $0.2 N H C l$ is required. The amount of $N a O H$ present in the original solution is:

A

0.5 g

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B

1 g

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C

2 g

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D

4 g

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Solution

The correct option is C $1 g$
$300 m L H C l$ of $0.1 N$ neutralises entire amount of $N a O H$ and $1 / 2$ of $N a_{2} C O_{3}$ . Remaining $1 / 2$ of $N a_{2} C O_{3}$ is neutralised by $25 m L$ of $0.2 N H C l$ , i.e., $50 m L$ of $0.1 N H C l$ .
Thus, $250 m L$ of $0.1 N H C l$ is required to neutralise $N a O H$ completely.
$N_{1} V_{1} (N a O H) = N_{2} V_{2} (H C l)$
$= 0.1 \times 250$
$= 25$
$W_{N a O H} = \frac{E N V}{1000} = \frac{40 \times 25}{1000} = 1 g$ .

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