# A 21.8*mL volume of "potassium permanganate" whose concentration was 1.68*mol*L^-1 achieved a stoichiometric endpoint when a volume of 100*mL H_2O_2(aq) solution was added. What are [H_2O_2], and the mass of hydrogen peroxide used?

Oct 3, 2016

Hydrogen peroxide will reduce permanganate ion in aqueous solution. $\left[{H}_{2} {O}_{2}\right]$ $=$ $0.915 \cdot m o l \cdot {L}^{-} 1$; mass of hydrogen peroxide $=$ $3.11 \cdot g$.

#### Explanation:

$2 M n {O}_{4}^{-} \left(a q\right) + 5 {H}_{2} {O}_{2} \left(a q\right) + 6 {H}^{+} \left(a q\right) \rightarrow 2 M {n}^{2 +} \left(a q\right) + 5 {O}_{2} \left(g\right) + 8 {H}_{2} O \left(l\right)$

I am not going to go thru the derivation of this redox equation. Suffice to say it is balanced, and represents the reduction of permanganate by peroxide in acidic solution. The important thing to know in this rxn is that permanganate ion is VERY strongly coloured; it is a deep red/purple colour. $M {n}^{2 +}$, the reduction product, is almost colourless. And thus the reduction is self-indicating. These are very pretty reactions to do, and the endpoint can be easily visualized.

$\text{Moles of potassium permanganate}$ $=$ $21.8 \times {10}^{-} 3 L \times 1.68 \cdot m o l \cdot {L}^{-} 1$ $=$ $3.66 \times {10}^{-} 2 \cdot m o l .$

And given the stoichiometry, there were thus $\frac{5}{2} \times 3.66 \times {10}^{-} 2 \cdot m o l$ with respect to the hydrogen peroxide reagent.

And thus $\text{concentration of hydrogen peroxide}$ $=$ $\text{moles"/"volume}$ $=$ $\frac{\frac{5}{2} \times 3.66 \times {10}^{-} 2 \cdot m o l}{100 \times {10}^{-} 3 \cdot L}$ $=$ $0.915 \cdot m o l \cdot {L}^{-} 1$

In $100 \cdot m L$ there were thus $0.1 \cdot L \times 0.915 \cdot m o l \cdot {L}^{-} 1 \times 34.01 \cdot g \cdot m o {l}^{-} 1$ $=$ $3.11 \cdot g$ of peroxide reagent.

Thx to dk who pointed out an error.