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)-»O2(g)+2MnSO4(aq)+K2SO4(aq)+4H2O(l)

A certain amount of hydrogen peroxide was dissolved in 100. \rm mL of water and then titrated with 1.68 M \rm KMnO_4. How much \rm H_2O_2 was dissolved if the titration required 12.8 mL of the \rm KMnO_4 solution?

To determine the amount of H2O2 dissolved in 100 mL of water, we need to use the stoichiometry of the balanced chemical equation provided.

The balanced equation is:

2KMnO4(aq) + H2O2(aq) + 3H2SO4(aq) -> O2(g) + 2MnSO4(aq) + K2SO4(aq) + 4H2O(l)

From the equation, we can see that the stoichiometric ratio between KMnO4 and H2O2 is 2:1.

Given that 1.68 M KMnO4 was titrated and 12.8 mL of KMnO4 solution was used, we can calculate the number of moles of KMnO4 used as follows:

Number of moles KMnO4 = concentration (M) x volume (L)
= 1.68 M x (12.8 mL / 1000 mL)
= 0.021504 mol

Since the stoichiometric ratio between KMnO4 and H2O2 is 2:1, the number of moles of H2O2 used is half the number of moles of KMnO4 used.

Number of moles H2O2 = 0.021504 mol / 2
= 0.010752 mol

To find the amount of H2O2 dissolved in 100 mL of water, we divide the number of moles of H2O2 by the volume in liters:

Amount of H2O2 = Number of moles / Volume (L)
= 0.010752 mol / 0.100 L
= 0.10752 mol/L

Therefore, the amount of H2O2 dissolved in 100 mL of water is 0.10752 mol/L.

0.532