Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

20g of sample of $Ba(OH)_2$ is dissolved in 50c.c of 0.1N HCl solution.The excess of HCl has titrated with 0.1NaOH.The volume of NaOH used was 20c.c.Calculate percentage of $Ba(OH)_2$ in the sample

$\begin{array}{1 1}(a)\;1.28\%& (b)\;12.8\% \\(c)\;25.6\%&(d)\;0.128\%\end{array}$

Can you answer this question?

1 Answer

0 votes
Eq.of $Ba(OH)_2$(pure)=Eq.of HCl used
$\Rightarrow $Total Eq.of HCl taken-eq.of HCl left
$\Rightarrow$ Eq.of HCl taken-eq.of NaOH used for back titration.
$\large\frac{W}{171/2}$$=50\times 0.1\times 10^{-3}-0.1\times 20\times 10^{-3}$
Percentage purity of $Ba(OH)_2$ in the sample =$\large\frac{0.256}{20}$$\times 100=1.28\%$
$\therefore$ The option is (a) 1.28%
answered Mar 24, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App