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Electrolytes

0.848 g aqueo...

Question

$0.848$ g aqueous solution of a mixture containing $N a_{2} C O_{3}$ , $N a O H$ and an inert matter is titrated with $\frac{M}{2}$ $H C l$ . The colour of phenolphthalein disappears when $20$ mL of the acid has been added. Methyl orange is then added and $8.0$ mL more of the acid is required to give a red colour to the solution. The percentage of $N a_{2} C O_{3}$ is :

A

425

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B

12.5

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C

75

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D

50

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Solution

The correct option is D $50$
Let $x$ and $y$ are the mEq of $N a O H$ and $N a_{2} C O_{3}$ , respectively.
Phenolphthalein as indicator:
mEq of $N a O H = \frac{1}{2}$ mEq of $N a_{2} C O_{3} =$ mEq of $H C l$
$x + \frac{y}{2} = \frac{1}{2} \times 1 \times 20 = 10$ ....(i)
Methyl orange as indicator:
$\frac{m E q o f N a_{2} C O_{3} (l e f t)}{2} =$ mEq of $H C l$
$\frac{y}{2} = \frac{1}{2} \times 8 = 4$ ....(ii)
$y = 8, x = 6$
Weight of $N a_{2} C O_{3} = 8 \times 10^{- 3} \times \frac{106}{2} = 0.424 g$
$%$ of $N a_{2} C O_{3} = \frac{0.424}{0.848} \times 100 = 50$

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