# Question #ff6ba

Mar 8, 2015

The mass of hydrogen peroxide dissolved is $\text{2.04 g}$.

So, what you're essentially doing is performing a redox titration - the potentiometric titration of hydrogen peroxide to be precise.

It is of the utmost importance - and I can't stress this enough - to always check to see if your chemical equation is balanced. In your case, the form written in the question is clearly not balanced, The balanced chemical equation looks like this

$2 K M n {O}_{4 \left(a q\right)} + 5 {H}_{2} {O}_{2 \left(a q\right)} + 3 {H}_{2} S {O}_{4 \left(a q\right)} \to 2 M n S {O}_{4 \left(a q\right)} + {K}_{2} S {O}_{4 \left(a q\right)} + 8 {H}_{2} {O}_{\left(l\right)} + 5 {O}_{2 \left(g\right)}$

The net ionic equation will give you a better idea of what's going on in this reaction

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

Basically, you're using potassium permanganate to oxidize hydrogen peroxide to oxygen gas in an acidified solution (that's what the sulfuric acid is for).

You should treat this problem like you would any other titration calculation- figure out how many moles of added potassium permanganate are needed to completely oxidize the hydrogen peroxide.

Of course, go about doing this by using the mole ratio that exists between potassium permanganate and hydrogen peroxide - 2 moles of the former will react with 5 moles of the latter.

So, the number of moles of $K M n {O}_{4}$ added is

$C = \frac{n}{V} \implies {n}_{K M n {O}_{4}} = C \cdot V = \text{1.68 M" * 14.3 * 10^(-3)"L}$

${n}_{K M n {O}_{4}} = \text{0.02402 moles}$

This means that the number of moles you had in your hydrogen peroxide solution was

$\text{0.02402 moles " KMnO_4 * ("5 moles " H_2O_2)/("2 moles " KMnO_4) = "0.06005 moles } {H}_{2} {O}_{2}$

Now use hydrogen peroxide's molar mass to determine how many grams of hydrogen peroxide you would have needed to prepare your solution

$\text{0.06005 moles " H_2O_2 * "34.0 g"/("1 mole " H_2O_2) = "2.04 g } {H}_{2} {O}_{2}$