Question

Consider a different titration for this exercise. Potassium permanganate (KMnO4) is the titrant and hydrogen peroxide (H2O2) the analyte according to the following balanced chemical equation. 2 MnO4− + 5 H2O2 + 6 H + → 2 Mn2+ + 5 O2 + 8 H2O

(a) What is the stoichiometry of MnO4− to H2O2? (ratio)

(b) Complete the following table for this titration. Data Table P3: Titration of hydrogen peroxide with potassium permanganate.

concentration of MnO4− 0.556 M

volume H2O2 solution 18.34 mL

mass H2O2 solution 18.42 g

volume of MnO4− solution 16.23 mL

Questions:

mmol of MnO4− : ?

mmol of H2O2: ?

mass of H2O2: ?

mass % of H2O2  in the original sample: ?

molarity of H2O2 in the original sample: ?

Answer 1

a)

The balanced chemical equaiton is -

2 MnO4− + 5 H2O2 + 6 H + → 2 Mn2+ + 5 O2 + 8 H2O

So, the ratio of MnO4- to H2O2 is 2/5

b)

Given that

concentration of MnO4− = 0.556 M

volume H2O2 solution = 18.34 mL

mass H2O2 solution = 18.42 g

volume of MnO4− solution = 16.23 mL

so,

1) mmol of MnO4- = molarity*volume in ml

= 0.556 M * 16.23 ml = 9.024 mmol

2)

From the balanced equation we know that the mole ratio of MnO4- and H2O2 = 2/5 so,

if we have 9.024 mmol of MnO4- we must have lets say x mmol of H2O2 so,

9.024 mmol / x mmol = 2/5

so, x = 9.024*5/2 mmol

= 22.56 mmol = 22.56/1000 mol

= 0.02256 mol

3) weight of H2O2 = molar mass*moles

and molar mass of H2O2 = molar mass of H*2 + molar mass of O*2

= 2g/mole*x + 16g/mole*2 = 34g/mole

moles of H2O2 = = 0.02256 mol

so, weight = 34 g/mole * 0.02256 mol = 0.767 g

4) mass % of H2O2 = (mass of H2O2/(total mass of sample))*100

= (0.767g/(18.42g))*100 = 4.164 %

5)

Molarity = moles/volume in L

volume = 16.23 ml = 16.23/1000 L = 0.01623 L

so, molarity = 0.02256 moles/0.01623 L = 1.39 M