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Salt of Strong Acid and Weak Base

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Question

A solution contains $N a_{2} C O_{3}$ and $N a H C O_{3}$ . $10 mL$ of this solution required $2.5 mL$ of $0.1 M H_{2} S O_{4}$ for neutralization using phenolphthalein as an indicator. Methyl orange is then added when a further $2.5 mL$ of $0.2 M H_{2} S O_{4}$ was required. The amount of $N a_{2} C O_{3}$ and $N a H C O_{3}$ respectively, in 1 litre of solution is:

A

5.3 g and 4.2 g

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B

3.3 g and 6.2 g

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C

4.2 g and 5.3 g

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D

6.2 g and 3.3 g

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Solution

The correct option is A $5.3 g and 4.2 g$
Lets say moles of $N a_{2} C O_{3}$ and $N a H C O_{3}$ in 10 ml mixture is x and y respectively
Phenolphthalein will reach end point after all the $N a_{2} C O_{3}$ is converted to $N a H C O_{3}$ so
$x \times 1 = 2.5 \times 10^{- 3} \times 0.1 \times 2$
$x = 5 \times 10^{- 4}$
Methyl orange will reach to the end point after all the $N a H C O_{3}$ is reacted that is $x + y$ mole,
Hence $x + y = 2.5 \times 10^{- 3} \times 0.2 \times 2$
$x + y = 10^{- 3} s o y = 5 \times 10^{- 4}$
Mass of $N a_{2} C O_{3}$ in 1000 ml solution $= \frac{2.5 \times 10^{- 3} \times 0.1 \times 2 \times 106 \times 1000}{10} = 5.3 g$
mass of $N a H C O_{3}$ in 1000 ml $= \frac{5 \times 10^{- 4} \times 84 \times 1000}{10} = 4.2 g$

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