$20 g$ of sample of $B a {(O H)}_{2}$ is dissolved in $10 m L$ . Of 0.5 $N H C l$ solution. The excess of $H C l$ was titrated with $0.2 N N a O H$ . The volume of $N a O H$ used was 10 cc. Calculate the percentage of $B a {(O H)}_{2}$ in the sample.

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$1.28 %$

$0.64$

$3.74 %$

$2.52 %$

answer is B.

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Detailed Solution

Milli eq. of $H C l$ initially $= 10 \times 0.5 = 5$
Milli eq. of $N a O H$ consumed $=$ Milli eq. of $H C l$ in excess
$= 10 \times 0.2 = 2$
$∴$ Milli eq $H C l$ . consumed = Milli eq. of $B a {(O H)}_{2}$
$= 5 - 2 = 3$
$∴ e q . o f B a {(O H)}_{2} = \frac{3}{1000} = 3 \times 10^{- 3}$
Mass of $B a {(O H)}_{2} = 3 \times 10^{- 3} \times (171 / 2) = 0.02565 g$
$% B a {(O H)}_{2} = (0.2565 / 20) \times 100 = 1.28 %$