Question

A 25.00 mL H2O2 solution required 22.50 mL of 0.01881 M MnO4 - for titration to the endpoint. Given that the original H2O2 was diluted 1 in 20 before titration with the MnO4 - , (i) what is the molarity of the diluted H2O2? (ii) What is the molarity of the original H2O2 solution? (iii) What is the % w/w of H2O2 in the original solution, assuming a solution density of 0.9998 g/mL?

Answer 1

Solution:- The balanced equation is.....

5H₂O₂ + 6H⁺ + 2MnO₄^-  ----->2Mn²⁺ + 5O_{2 +} 8H₂O

moles of MnO₄^- used in titration = 0.02250 L x 0.01881 mol/L = 0.0004232 mol MnO₄^-

there is 5:2 mol ratio between H₂O₂ and MnO₄^-, so the moles of hydrogen peroxide would be...

0.0004232 mol MnO₄^- x 5 mol H₂O₂/2 mol MnO₄^- = 0.001058 mol H₂O₂

concentration of diluted H₂O₂ solution = 0.001058 mol/0.025 L = 0.04232 M

For hydrogen peroxide it says. original H₂O₂ was diluted 1 in 20 means water was added to it's 1 mL to make the volume 20 mL. We could calculate the original concentration using dilution equation, M₁V₁ = M₂V₂

1.00 mL(M₁) = 20.0 mL(0.04232 M

M₁ = 0.8464 M

So, the original concentration(molarity) of H₂O₂ is 0.8464 M.

we get 20 mL diluted solution by 1 mL of hydrogen peroxide. How many mL of it would give 25.00 mL of diluted solution?

25.00 mL x 1/20 = 1.25 mL

mass of original solution = 1.25 mL x 0.9998 g/mL = 1.25 g

mass of H₂O₂ present in original solution would be....

1.25 mL x (1L/1000 mL) x (0.8464 mol/1L) x (34.02 g/1mol) = 0.0360 g

%w/w of H₂O₂ = (0.0360/1.25) x 100 = 2.88%