Question

by titration with permanganate solution. The 13) Hydrogen peroxide, H2O2, can be determined by titration with permans reaction is: 2 MnO4 (aq) + 5 H2O2(aq) + 6 H+ + 502(g) + 2 Mn2(aq) + 8 H20 ate 5.0 mL of a H2Oz(aq) If it required 80.0 mL of 0.0220 molar MnO4 (aq) solution to titrate 5.0 m solution, what is the molarity of the H2O2(aq) solution?

Answer 1

Let us first calculate the number of moles of MnO4^- present in 80 mL of 0.022 M solution.

As the solution of MnO4^- is 0.022 M, it means that 0.022 moles of MnO4^- are present i the 1000 mL of solution. Therefore, we can consider

1000 mL = 0.022 moles

For 80 mL = (80 x 0.022) /1000 = 0.00176 moles

Therefore, 80 mL of 0.022 M solution of MnO4^- contains 0.00176 moles

It is clear from the balanced chemical equation that, 2 moles of MnO4^- require 5 moles of H2O2. Therefore, we can consider,

2 moles of MnO4^- = 5 moles of H2O2

therefore, 0.00176 moles of MnO4^- = (0.00176 x 5) / 2

= 0.0044 moles of H2O2

Thus, 0.0044 moles of H2O2 will be present in the 5 mL of solution.

To get the molarity of H2O2 solution we can consider,

5 mL = 0.0044 moles

For 1000 mL = (1000 x 0.0044) / 5

= 0.88 M

Therefore the molarity of H2O2 solution will be 0.88 M