Stoichiometry
9.0 Volume strength peroxide solution
9.0 Volume strength peroxide solution
The concentration of ${H_2}{O_2}$ is usually represented in terms of volume. If a sample of ${H_2}{O_2}$ is labeled as $‘x'$ volume, it means that $1$ volume of ${H_2}{O_2}$ solution gives $‘x'$ volumes of ${O_2}$ gas at STP on complete decomposition.
Consider the decomposition of ${H_2}{O_2}$ as,
\[\mathop {2{H_2}{O_2}}\limits_{2 \times 34g} \xrightarrow{\Delta }2{H_2}O + \mathop {{O_2}}\limits_{22.4{\text{ L at STP}}} \]
$\because $ $22400$ ml of ${O_2}$ gas is liberated by $68g$ of ${H_2}{O_2}$ solution
$\therefore $ $x$ ml of ${O_2}$ gas will be liberated by = $\frac{{68x}}{{22400}}$ = $\frac{{17x}}{{5600}}g{\text{ of }}{{\text{H}}_{\text{2}}}{O_2}$
It means that $\frac{{17x}}{{5600}}g$ of ${H_2}{O_2}$ will be present in $1\ ml$ of solution.
$\therefore $ $1000$ ml of solution contains ${H_2}{O_2}$ = $\frac{{17x}}{{5600}} \times 1000 = \frac{{17x}}{{5.6}}$
Strength $\left( {g{\text{ }}{L^{-1}}} \right) = $ Normality $ \times $ Equivalent weight
$\frac{{17x}}{{5.6}} = N \times \frac{{34}}{2}$ $(\because n - factor{\text{ }}of{\text{ }}{H_2}{O_2} = {\text{ }}2)$
$x{\text{ }} = {\text{ }}5.6 \times N$
i.e., $\boxed{{\text{Volume strength of }}{{\text{H}}_{\text{2}}}{O_2} = 5.6 \times Normality}$
