Question

20 g a sample of $Ba(OH{)}_2$ is dissolved in 10 mL of 0.5 N HCL solution: The excess of HCL was titrated with 0.2 N NaOH. The volume of NaOH used was 20 cc. Calculate the percentage of $Ba(OH{)}_2$ in the sample.

Answer 1

Excess $HCl$ left after reaction with $Ba(OH{)}_2=$

$20mL0.2NNaOH=xmL0.5NHCl$

$V_1{N}_1=V_2{N}_2$

$20\times 0.2=x\times 0.5$

$x=8mL$

Only $10-8=2mLHCl$ reacted with the $Ba(OH{)}_2$ sample.

$2mL0.5NHCl=\frac{2\times 0.5}{1000}=0.001moles$

Now, $Ba(OH{)}_2+2HCl\to BaC{l}_2+H_{2}O$

Ratio is $1:2$

$x:0.001$

$x=0.0005moles$

Weight of $Ba(OH{)}_2$ in sample

$=molarweight\times 0.0005$

$=171\times 0.0005g=0.0855g$

%$Ba(OH{)}_2$ in sample = $\frac{weightofBa(OH{)}_{2}inthesample}{weightofthesample}\times 100$

$=\frac{0.0855}{20}\times 100$

$=0.4275$ %