Question

Redox titrations are used to determine the amounts of oxidizing and reducing agents in solution. For example, a solution of hydrogen peroxide, H2O2, can be titrated against a solution of potassium permanganate, KMnO4. The following equation represents the reaction: 2KMnO4(aq)+H2O2(aq)+3H2SO4(aq)→3O2(g)+2MnSO4(aq)+K2SO4(aq)+4H2O(l) A certain amount of hydrogen peroxide was dissolved in 100. mL of water and then titrated with 1.68 M KMnO4. What mass of H2O2 was dissolved if the titration required 21.3 mL of the KMnO4 solution?

Answer 1

2KMnO4(aq)+H2O2(aq)+3H2SO4(aq) ---- >

                         3O2(g)+2MnSO4(aq)+K2SO4(aq)+4H2O(l)

According to the balance reaction one mole of H2O2 will recats with 2 mole of KMnO4.

Molarity = moles / volume (in Liters) then

Number of moles = molarity * volume in L

Given that;

Molarity = 1.68 M

Volume = 21.3 ml = 0.0213 L

1000 ml = 1.00 L

Now calculate the moles of KMnO4 as follows:

Number of moles = molarity * volume in L

= 1.68 moles / L * 0.0213 L

= 0.035784 moles KMnO4

here 1 mole of H2O2, 2 moles of KMnO4 are required;

therefore:
moles H2O2

0.035784 moles KMnO4 * 1 mole H2O2 / 2 mole KMnO4

= 0.017892 Moles H2O2

Amount of H2O2 in g = molar mass * number of moles

= 0.017892 Moles H2O2 * 34.016 g / mole

= 0.609 g H2O2

molar mass H2O2 = 34.016 g / mole

Hence 0.609 grams of H2O2 was dissolved in 100.0 mL of water