Question

The following experiment was carried out:

1.Prepare 250 cm3 of a solution of hydrogen peroxide by adding 7.5 cm3 of the stock solution supplied to the volumetric flask and diluting to the required volume.

2. Pipette 25.0 cm3 of this solution into a conical flask.

3. Add 10 cm3 of 1 mol dm-3 sulphuric acid in a measuring cylinder.

4. Titrate the mixture against the potassium manganate (VII) until a permanent pale pink colour appears in the conical flask.

5. Record the titre volume and repeat until concordant values are obtained.

**NB- stock solution of hydrogen peroxide is 0.02mol/dm^3

Results :

Trial 1- 18.4 cm3 (volume of KMnO4) used

Trial 2- 18.3cm3 (volume of KMnO4 used)

HENCE:

a. If the stock solution of hydrogen peroxide had a molar concentration of 0.882 mol dm-3, determine the molar concentration of the hydrogen peroxide solution prepared by you.

b. Determine the molar concentration of the hydrogen peroxide solution prepared by you using your titration results.

c.Determine the error in your results.

d. Hydrogen peroxide is sold using terms such as “20-volume”. This means that 1cm3 of hydrogen peroxide would produce 20 cm3 of oxygen upon decomposition. Using the molar concentration of the stock solution of 0.882mol dm-3, determine what designation you would assign this hydrogen peroxide.

Answer 1

H2O2 conc in stock sol = 0.02 mol/l

H2O2 moles in 7.5 ml = (0.02*7.5*10^-3)

= 0.00015 moles

Final volume of H2O2= 250 ml

conc of H2O2 in sol = 0.00015/(250*10^-3)

= 0.0006 mol/l

2.

Following reaction takes place upon addition of Acid and KMnO4

5H₂O₂ + 2KMnO₄ + 3H₂SO₄ → 5O₂ + 2MnSO₄ + K₂SO₄ + 8H₂O

25 ml of H2O2 sol was pippeted out

moles of H2O2 pippeted out = 25*10^-3 * 0.0006

= 0.000015

mol of acid = 10*10^-3 * 1

= .01

Volume of KMnO4 added = 18.4 ml

1 mol KMnO4 can react with 2.5 mol H2O2 + 1.5 mol H2SO4

moles of KMnO4 required =(0.000015/2.5)+(.01/1.5)

=0.0066 moles

conc of KMnO4= (0.0066/(18.3*10^-3)

=0.36 M

conc of KMnO4= 0.36 M

a.

If the stock solution of hydrogen peroxide had a molar concentration of 0.882 mol dm-3

H2O2 conc in stock sol = 0.822 mol/l

H2O2 moles in 7.5 ml = (0.822*7.5*10^-3)

= 0.00616 moles

Final volume of H2O2= 250 ml

conc of H2O2 in sol = 0.00616/(250*10^-3)

= 0.0246 mol/l

conc of H2O2= 0.822 mol/l

For 1 litre sol

moles of H2O2= 0.822

2 H2O2-> 2 H2O + O2

2 moles of H2O2 produce 1 mole of O2

0.822 moles -> 0.411 moles of O2

1 mol gas = 22.4 l at STP

0.411 mol of gas = 22.4*.411 =9.2 l

So 1 l H2O2 give 9.2 l O2

strength = 9.2 - Volume