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Primary, Secondary, Tertiary and Quaternary Carbon

0.4 g mixture...

Question

$0.4 g$ mixture of $N a O H$ and $N a_{2} C O_{3}$ and some inert impurities were first titrated with $\frac{N}{10} H C l$ using phenolphthalein as an indicator. $17.5 m L$ of $H C l$ was required at the end point. After this methyl orange was added and titrated. $1.5 m L$ of the same $H C l$ was required for the next end point. The weight percentage of $N a_{2} C O_{3}$ in the mixture is ______. (Rounded off to the nearest integer)

A

3.98

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B

4

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C

3.97

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D

3.9

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Solution

$1^{s t}$ end point reaction,
$N a O H + H C l \to N a C l + H_{2} O N a_{2} C O_{3} + H C l \to N a H C O_{3} + N a C l$
$∴ Milliequivalents of H C l = Milliequivalents of N a O H + Milliequivalents of N a_{2} C O_{3} \frac{1}{10} \times 17.5 \times 10^{- 3} = n_{N a O H} \times 1 + n_{N a_{2} C O_{3}} \times 1 n_{N a O H} + n_{N a_{2} C O_{3}} = 1.75 \times 10^{- 3} . . . (1)$

$2^{n d}$ end point reaction,
$N a H C O_{3} + H C l \to N a C l + H_{2} C O_{3}$
$∴ Milliequivalents of H C l = Milliequivalents of N a H C O_{3} \frac{1}{10} \times 1.5 \times 10^{- 3} = n_{N a H C O_{3}} \times 1 n_{N a H C O_{3}} = 0.15 \times 10^{- 3} \equiv n_{N a_{2} C O_{3}} n_{N a_{2} C O_{3}} = 0.15 \times 10^{- 3} w_{N a_{2} C O_{3}} = 0.15 \times 10^{- 3} \times 106 = 0.0159$

$Weight % of N a_{2} C O_{3} in the mixture = \frac{0.0159}{0.4} \times 100 \approx 4 %$

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Similar questions

0.4 g

N a O H

N a_{2} C O_{3}

and some inert impurities were first titrated with

\frac{N}{10} H C l

using phenolphthalein as an indicator.

17.5 m L

H C l

was required at the end point. After this methyl orange was added and titrated.

1.5 m L

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