Question

A commercial sample containing hydrogen peroxide was titrated with a standard 0.010 M potassium permanganate in an acidic solution. If 2.00 mL of the sample took 20.00 mL of potassium permanganate, what is the concentration of hydrogen peroxide in the sample in ppm (mg/L)?

Answer 1

Balanced chemical equation is

5H₂O_2(aq) + 2MnO₄^-_(aq) + 6H⁺_(aq)   5O_2(g) + 2Mn²⁺_(aq) + 8H₂O_(l)

no. of mole = molarity X volume of solution in liter

20 ml = 0.02 liter

no.of mole of pottasium permagnate used for titration = 0.010 X 0.02 = 0.0002 mole

according to reaction 5 mole of hydrogen peroxide react with 2 mole of KMnO₄ molar retio between H₂O₂ and

KMnO₄ is 5:2 therefore to react with 0.0002 mole of  KMnO₄ required H₂O₂ = 0.0002 X 5/2 = 0.0005 mole

molar mass of H₂O₂ = 34.0147 g/mol

gm of compound = no. of mole X molar mass

gm of H₂O₂ = 0.0005 X 34.0147 = 0.017 gm

0.017 gm = 17 mg

2 ml sample contain 17 mg H₂O₂ then 1000 ml sample contain H₂O₂ = 17 x 1000 / 2 = 8500 mg

1L sample contain 8500 mg H₂O₂

therefore concentration of H₂O₂ sample = 8500 mg/L = 8500 ppm