Chemistry - titation with old and new H2O2

[bjoyful]

I have been working on this assignment for several days, but I am not sure I am doing it 100% right. Any help will be appreciated!


1. Homework Statement :

1. For the NEW bottle of hydrogen peroxide, record and calculate the following:

(a) Volume of potassium permanganate used in each titration (mL):

(b) The volume of potassium permanganate required to titrate 10 mL of the new hydrogen peroxide (mL):

(c) Calculate the concentration of H2O2 from the volume of KMnO4 used and the stoichiometry of the reaction (MW of H2O2 is 34.01).

2. Repeat the calculations for the OLD bottle of H2O2:

(a) Volume of potassium permanganate used in each titration (mL):

(b) The volume of potassium permanganate required to titrate 10 mL of the old hydrogen peroxide (mL):

(c) Calculate the concentration of H2O2 from the volume of KMnO₄ used and the stoichiometry of the reaction (MW of H2O2 is 34.01).

3. Commercial hydrogen peroxide contains small amounts of organic compounds that are added to stabilize it. If these compounds react with permanganate, how would this affect your results?

2. Homework Equations :

I think I am suppose to be using the formula 5(C1xV1) H2O2 = 2(C2xV2) KMnO4. I know that the ratio is 2 to 5.

3. The Attempt at a Solution :

For the new bottle of H2O2: I started out with 50mL of potassium permanganate in my burette. I then mixed 10mL new hydrogen peroxide and 2mL sulfuric acid in the flask. When I titrate, the color turns purple at 31.95.

For the old bottle of H2O2: I started with 50mL of potassium permanganate in my burette. I then mixed 10mL old hydrogen peroxide and 2mL sulfuric acid in the flask. When I titrate, the color turns purple at 37.35.

I am uncertain for question "a" if the answer should be 50ml because I didn't really use 50mL. Once again, I am not sure if "b" is looking for 50.00-31.95=18.05 or something else. I am not sure if I am reading the question correctly. For "c", I am really confused! However, I have been trying to work through it! I started with the new bottle of H2O2 and used the formula 5(C1xV1) H2O2 = 2(C2xV2) KMnO4, I put 34.01 for C2 and 18.05 for V2 and multiplied both of them by 2, getting 104.12 as a total answer. I was putting 12ml for V1, but that is as far as I got.

I really need some help on this...I hope you can see that I have spent hours on this homework, and really need help - I am not looking for handouts!

Thanks!

 

[weatherhead]

... you have done titration calculations before, right?

 

[bjoyful]

Yes, I have, but it was earlier this semester...

 

[Borek]

What really matters is amount of titrant ADDED to titrated sample.

 

[GCT]

a)Simply record the volume amount required to obtain the titration equivalence point

b)Deduce the amount of volume of Potassium Permanganate required to obtain the equivalence point with a 10 mL solution of Hydrogen Peroxide using simple proportions

c)Write the reaction equation and use the molar equivalence numbers - the numbers in front of each molecule in the equation - as well as the molar masses to find your answer. Use the density of each solution to find the mass from the volumes and then find the values of the moles from the masses ; use the stoichiometry to find the respective number of moles within the experimental solution and then refer to how much of the volume was used for the titration - divide the moles by the volume and you've got your answer.

 

[Kushal]

the hypothesis behind this experiment is that the concentration of H2O2 is different in each of the bottles. H2O2 has a tendency to decompose into oxygen and water. In order t verify this you are asked to determine the concentration of H2O2 in each of the bottles using KMnO4.

i find proportion a much more efficient way to solve titration problems instead of formulae. you see which suits you best.

to calculate the concentration of H2O2, you have to know the concentration of the KMnO4, say x mol/L

but you used y ml (the volume of KMnO4 you read from the burette).

1000 ml contains x mol KMnO4
y ml contains (x/1000)*y mol KMnO4

this is the amount in mol of KMnO4 which reacted.

according to balanced equation, 2 mol of KMnO4 is supposed to react with 5 mol H2O2
but you have (x/1000)*y mol KMnO4 which reacted. how much H2O2 reacted with this amount (use proportion).

now the amount of H2O2 which reacted with that KMnO4 is present in a volume of 10 mL.
you can find its concentration from there.

once you get used to it, it becomes simple.:)