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 (ag) →
3O2(g) + 2MnSO4(aq) + K2SO4(aq) + 4H2O(I)
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 14.3 mL of the KMnO4 solution?
To determine the mass of H2O2 dissolved, we first need to calculate the number of moles of KMnO4 that reacted with the H2O2.
From the balanced equation, we can see that the molar ratio between KMnO4 and H2O2 is 2:1. Therefore, the number of moles of KMnO4 that reacted is:
Moles of KMnO4 = Molarity of KMnO4 x Volume of KMnO4 (L)
Moles of KMnO4 = 1.68 mol/L x 0.0143 L
Moles of KMnO4 = 0.024024 mol
Since the molar ratio between KMnO4 and H2O2 is 2:1, the number of moles of H2O2 that reacted is half of the number of moles of KMnO4:
Moles of H2O2 = 0.024024 mol / 2
Moles of H2O2 = 0.012012 mol
Now, we can calculate the mass of H2O2 that was dissolved:
Molar mass of H2O2 = 2(1.008) + 2(15.999) = 34.014 g/mol
Mass of H2O2 = Moles of H2O2 x Molar mass of H2O2
Mass of H2O2 = 0.012012 mol x 34.014 g/mol
Mass of H2O2 = 0.408 g
Therefore, the mass of H2O2 that was dissolved in 100. mL of water is 0.408 g.