Question

A student used an average of 11.28 mL of

0.008500 mol/L KMnO4 (aq) to titrate 10.00
mL of diluted acidified hydrogen peroxide.
Determine the concentration of the stock
hydrogen peroxide in mol/L if it was diluted by
a factor of 30. (Record your answer to four
decimal places)

Answer 1

0.7191 mol/L

Step-by-step explanation:

To solve it, we need to use the information given to determine the number of moles of potassium permanganate (KMnO4) that were used in the titration. The concentration of the KMnO4 solution is 0.008500 mol/L and the average volume used in the titration was 11.28 mL, so the number of moles of KMnO4 used is (0.008500 mol/L) * (11.28 mL) * (1 L / 1000 mL) = 0.00009588 mol.

The balanced chemical equation for the reaction between potassium permanganate and hydrogen peroxide in an acidic solution is:

2MnO4- + 5H2O2 + 6H+ -> 2Mn2+ + 5O2 + 8H2O

According to this equation, two moles of MnO4- react with five moles of H2O2. This means that for every two moles of MnO4- that react, five moles of H2O2 are consumed.

Since we have 0.00009588 moles of MnO4-, we can expect that (5 moles H2O2 / 2 moles MnO4-) * 0.00009588 moles MnO4- = 0.0002397 moles of H2O2 were consumed in the reaction.

The volume of the diluted hydrogen peroxide solution that was titrated was 10.00 mL, so its concentration is (0.0002397 mol) / (10.00 mL) * (1000 mL / L) = 0.02397 mol/L.

Since this solution was diluted by a factor of 30, the concentration of the stock hydrogen peroxide solution must be 30 times greater than the concentration of the diluted solution: 30 * 0.02397 mol/L = 0.7191 mol/L.

Answer 2

Answer: The concentration of the stock hydrogen peroxide solution is 0.0086 mol/L (rounded to four decimal places).

Step-by-step explanation:

5 H2O2 + 2 KMnO4 + 3 H2SO4 → K2SO4 + 2 MnSO4 + 8 H2O + 5 O2

moles H2O2 = (0.008500 mol/L) x (11.28 mL/1000 mL) x 30 = 0.009684 mol

Since 10.00 mL of the diluted solution was titrated, the number of moles of H2O2 in the undiluted (stock) solution is:

moles H2O2 = (0.009684 mol/11.28 mL) x 10.00 mL = 0.008577 mol

concentration = moles H2O2/volume of stock solution = 0.008577 mol/L