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Neutralisation Reaction

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Question

A solution contained $N a_{2} C O_{3}$ and $N a H C O_{3}$ . $15 m l$ of the solution required $5 m l$ of $\frac{N}{10} H C l$ for neutralization using phenolphthalein as an indicator. Addition of methyl orange required a further $15 m l$ of the acid for neutralization. The amount of $N a_{2} C O_{3}$ present in the solution is:

A

21.2 g

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B

2.12 g

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C

0.212 g

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D

4.24 g

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Solution

The correct option is B $21.2 g$
${N a}_{2} {C O}_{3}$ can be estimated directly from $h p h$
$∵$ in $h p h$ only ${N a}_{2} {C O}_{3} \to {N a H C O}_{3}$ conversion take place
moles of ${N a}_{2} {C O}_{3} = 5 \times \frac{1}{10} \times 10^{- 3}$ moles
$= 5 \times 10^{- 4}$ moles
weight of ${N a}_{2} {C O}_{3}$ present $= 0.053 g$

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