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Equivalent Mass

500 mL of a s...

Question

$500 m L$ of a solution of $C a C O_{3}$ and $N a O H$ was first titrated with $\frac{N}{10} H C l$ using phenolphthalein indicator. $18.5 m L$ of $H C l$ was required for the end point. After that methyl orange indicator was used and $2.5 m L$ of same $H C l$ was required for the next end point. Find out the mass of $C a C O_{3}$ in the mixture.

A

0.25 g

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B

0.025 g

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C

0.057 g

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D

0.875 g

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Solution

The correct option is B $0.025 g$
Phenolphthalein will only neutralise $50 %$ of $C a C O_{3}$ .
$∴ \frac{1}{2} Milliequivalents of C a C O_{3} + Milliequivalents of N a O H = Milliequivalents of H C l = \frac{1}{10} \times 18.5 = 1.85$

The methyl orange indicator will neutralise the rest $50 %$ of $C a C O_{3}$ .
$∴ \frac{1}{2} Milliequivalents of C a C O_{3} = Milliequivalents of H C l = \frac{1}{10} \times 2.5 = 0.25$
$∴ Milliequivalents of C a C O_{3} = 0.25 \times 2 = 0.50$
$\frac{W_{C a C O_{3}}}{M_{C a C O_{3}}} \times 2 \times 1000 = 0.5$
$∴ W_{C a C O_{3}} = \frac{0.5 \times 100}{2 \times 1000} = 0.025 g$

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500 m L

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