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Equivalent Mass

A definite am...

Question

A definite amount of $B a C l_{2}$ was dissolved in $H C l$ solution of unknown normality. $20$ mL of this solution was treated with $21.4$ mL of $0.1 N N a O H$ for complete neutralization. Further $20$ mL of the solution was added to $50$ mL $0.1 N$ $N a_{2} C O_{3}$ and the precipitate was filtered off. The filtrate reacted with $10.5$ mL of $0.08 N$ $H_{2} S O_{4}$ using phenolphthalein as indicator. The strength of $B a C l_{2}$ in mixture is :

A

6.136 g L^{- 1}

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B

6.196 g L^{- 1}

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C

6.146 g L^{- 1}

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D

6.138 g L^{- 1}

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Solution

The correct option is A $6.136 g L^{- 1}$
Milliequivalents of $N a O H =$ Milliequivalents of $H C l$

$21.4 \times 0.1 = 20 \times N$ $\Rightarrow N_{H C l} = \frac{21.4}{200}$

Milliequivalents of $H_{2} S O_{4} = 10.5 \times 0.08$ $= \frac{1}{2}$ milliequivalents of $N a_{2} C O_{3}$
(Phenolphthalein indicator)

Milliequivalents of unused $N a_{2} C O_{3} = 2 \times 10.5 \times 0.08$

Milliequivalents of used $N a_{2} C O_{3} =$ milliequivalents of $B a C l_{2} +$ milliequivalents of $H C l$

$(50 \times 0.1) - (2 \times 10.5 \times 0.08) =$ milliequivalents of $B a C l_{2} + N_{H C l} \times V_{H C l}$

$3.32 =$ milliequivalents of $B a C l_{2} + [\frac{21.4}{200} \times 20]$

Milliequivalents of $B a C l_{2} = 3.32 - 2.14$ $= \frac{1.18}{20}$ $= \frac{59}{1000}$

Weight of $B a C l_{2} = 59 \times 10^{- 3} \times \frac{208}{2} = 6.136 g L^{- 1}$

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