Question

•A 25.00 mL volume of commercial hydrogen peroxide (H2O2) solution was diluted to 250.0 mL in a volumetric flask. Then 25.00 mL of the diluted solution were mixed with 200 mL of water and 20 mL of H2SO4 and titrated with 0.02123 M KMnO4. The first pink color was observed with 27.66 mL of titrant. A blank prepared from water in place of H2O2 required 0.04 mL to give visible pink color. Find the molarity of the commercial H2O2.

Answer 1

The balanced chemical equation is

3H2O2 + 2KMnO4 → 3O2 + 2MnO2 + 2KOH + 2H2O

Volume of KMnO4 consumed by H2O2 = volume -blank = 27.66-0.04 = 27.62 mL

Number of moles of KMnO4 in (please note that this is from the volumetric flask)= (0.02123 x 27.62)/1000 = 0.000586 moles

(0.000586) x 3/2 =0.000879 moles is present in the 25 mL or 0.00879 moles in the 250 mL volumetric flask

Concentration of H2O2 in the volumetric flask = (0.00879/250)1000 = 0.03518 M

This was prepared by taking 25 mL from the commercial bottle.

M1V1 = M2V2

M1 =?

V1=25 mL

M2= 0.03518 MM

V2 = 250 mL

V1 = 250*0.03518 M/25 = 27.62 M = 0.35182 M

0.35182 M is the H2O2 commercial available ones concentration