Question

$20g$ of a sample of $Ba(OH{)}_2$ is dissolved in $10mL$ of $0.5NHCl$ solution. The excess of $HCl$ was titrated with $0.2NNaOH.$ The volume of $NaOH$ required was $10mL$. Calculate the percentage of $Ba(OH{)}_2$ in the sample.
(Given molar mass of $Ba$ is $137gmo{l}^{-1}$)

• A. $12.8\%$
• B. $1.28\%$
• C. $50\%$
• D. $33.33\%$

Answer 1

The correct option is B $1.28\%$

$\text{Milliequivalents of}HCl\text{present initially}=10\times 0.5=5$
$\text{Milliequivalents of NaOH consumed = Milliequivalents of}HCl\text{in excess}=10\times 0.2=2$
$\therefore \text{Milliequivalents of}HCl\text{consumed}=\text{Milliequivalents of}Ba(OH{)}_2=5-2=3$
$\therefore \text{Equivalents of}Ba(OH{)}_2=3/1000=3\times {10}^{-3}$
$\text{Molecular Mass of}Ba(OH{)}_2=137+(16+1)\times 2=171gmo{l}^{-1}$
$\text{Mass of}Ba(OH{)}_2=\frac{3\times {10}^{-3}\times 171}{2}=0.2565g$
$\%Ba(OH{)}_2=\frac{0.2565}{20}\times 100=1.28\%$