Hydrogen
10.0 Volume strength of Hydrogen Peroxide
10.0 Volume strength of Hydrogen Peroxide
It is defined as volume of $O_2$ evolved at STP in $ml$ that is obtained per $ml$ of $H_2O_2$ solution i.e., if $1$ litre solution of $H_2O_2$ gives $10$ litres of oxygen at STP, Volume strength of $H_2O_2$ is $10$ volume.
Let us say a $H_2O_2$ sample is labeled ‘$V$’ volume. Then, $$2{H_2}{O_2} \to 2{H_2}O + {O_2}$$
$22400\ ml$ $O_2$ is obtained by $68\ g$ $H_2O_2$
$\therefore$ $V\ ml$ $O_2$ is obtained by $$\frac{{68}}{{22400}} \times {V_g}$$
If $$1\ ml\ H_2O_2 = \frac{{68}}{{22400}} \times {V}\ g\ H_2O_2$$
Then, $$1000\ ml\ H_2O_2 = \frac{{68}}{{22.4}} \times V\;g/litre$$
Note:
$${{\text{Strength of }}{H_2}{O_2}{\text{ = }}\left( {\frac{{68}}{{22.4}}} \right)V}$$$${{\text{Molarity }} = {\text{ }}\left( {\frac{V}{{11.2}}} \right)}$$$${{\text{Normality }} = {\text{ }}\left( {\frac{V}{{5.6}}} \right)}$$
Illustration 2. What is the percentage strength, normality and molarity of $10\ V$ $H_2O_2$ solution?
Solution: $$\begin{equation} \begin{aligned} 2{H_2}{O_2} \to 2{H_2}O + {O_2} \\ 2(2 + 32) = 68g \to 22.4litre \\\end{aligned} \end{equation} $$ Thus $22.4\ litres$ of oxygen at STP is obtained from $68\ g$.
$10\ litres$ of oxygen at STP is obtained from $$\frac{{68}}{{22.4}} \times 10g$$
The concentration of $10$ volume $H_2O_2$ solution is $$30.35{\text{ }}g/litre = 3.035{\text{ g/100ml }} = {\text{ }}3.035\% $$
Normality of $10$ volume $H_2O_2$ solution $$\begin{equation} \begin{aligned} 2{H_2}{O_2} \to 2{H_2}O + {O_2} \\ 1litre{\text{ }}10litre = \frac{{10}}{{22.4}}moles \\\end{aligned} \end{equation} $$
$$\begin{equation} \begin{aligned} 1litre{\text{ of the solution gives }}\frac{{10}}{{22.4}}moles{\text{ of }}{O_2} \\ \therefore 1litre{\text{ of solution contains }}\frac{{10 \times 2}}{{22.4}}moles{\text{ of }}{H_2}O \\ {\text{Thus, }}Molarity = \frac{{20}}{{22.4}} = 0.893 \\ and{\text{ }}Normality = \frac{{20 \times 2}}{{22.4}} = 1.786 \\ \therefore {\text{Volume strength of 1 normal }}{H_2}{O_2}{\text{ is }}5.6 \\\end{aligned} \end{equation} $$
