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Titration calc watch

1. Question:
Hydrogen peroxide, H2O2, is sold as an aqueous solution. The concentration of H2O2(aq) can be determined by its reaction with acidified manganate(VII)
ions.

• Stage 1 – A 25.0 cm3 sample of H2O2(aq) is added to a 250 cm3 graduated flask.

• Stage 2 – Sufficient distilled water is added to the graduated flask to make 250 cm3 of diluted H2O2(aq).

• Stage 3 – A 10.0 cm3 sample of diluted H2O2(aq) is added to a conical flask.

• Stage 4 – The diluted sample has 25.0 cm3 of 1 mol dm–3 sulphuric acid added to it.

• Stage 5 – The contents of the flask are titrated against 0.0200 mol dm–3 MnO4
–.

In stage 5, the equation for the reaction between H2O2(aq) and acidified MnO4
– is shown below.

5H2O2(aq) + 6H+(aq) + 2MnO4 –> (aq) 2Mn2+(aq) + 5O2(g) + 8H2O(l)

so I worked out n(2Mn2+) correctly to be 5.71x10-4 and then the ratio of n(5h2o2) to be 1.4275x10-3 correctly. I need to work out the concentration of the undiluted H2O2 and can't do the ratio-ing out correctly :/
Any help would be appreciated
2. (Original post by tripodd)
Question:
Hydrogen peroxide, H2O2, is sold as an aqueous solution. The concentration of H2O2(aq) can be determined by its reaction with acidified manganate(VII)
ions.

• Stage 1 – A 25.0 cm3 sample of H2O2(aq) is added to a 250 cm3 graduated flask.

• Stage 2 – Sufficient distilled water is added to the graduated flask to make 250 cm3 of diluted H2O2(aq).

• Stage 3 – A 10.0 cm3 sample of diluted H2O2(aq) is added to a conical flask.

• Stage 4 – The diluted sample has 25.0 cm3 of 1 mol dm–3 sulphuric acid added to it.

• Stage 5 – The contents of the flask are titrated against 0.0200 mol dm–3 MnO4
–.

In stage 5, the equation for the reaction between H2O2(aq) and acidified MnO4
– is shown below.

5H2O2(aq) + 6H+(aq) + 2MnO4 –> (aq) 2Mn2+(aq) + 5O2(g) + 8H2O(l)

so I worked out n(2Mn2+) correctly to be 5.71x10-4 and then the ratio of n(5h2o2) to be 1.4275x10-3 correctly. I need to work out the concentration of the undiluted H2O2 and can't do the ratio-ing out correctly :/
Any help would be appreciated
you take 10 cm cube of diluted H2O2.
Let mol of 10 cm cube of H2O2 = x
then in the 250 cm cube of H2O2, you would have 25 times as much moles as there was in 10 cm cube,

so mol of 250 cm cube of H2O2 = 25x

then mol of 250 cm cube diluted H2O2 = mol of 25 cm cube of conc. H2O2

then mol of 25 cm cube of conc H2O2 = 25x = [conc H2O2] * 25/1000

then find out [conc H2O2]

[] = concentration
conc = concentrated
3. (Original post by shengoc)
you take 10 cm cube of diluted H2O2.
Let mol of 10 cm cube of H2O2 = x
then in the 250 cm cube of H2O2, you would have 25 times as much moles as there was in 10 cm cube,

so mol of 250 cm cube of H2O2 = 25x

then mol of 250 cm cube diluted H2O2 = mol of 25 cm cube of conc. H2O2

then mol of 25 cm cube of conc H2O2 = 25x = [conc H2O2] * 25/1000

then find out [conc H2O2]

[] = concentration
conc = concentrated
Don't really understand.. :/ I get the x25 to get the mol of 250cm(H2O2)

But then wouldn't I need to divide by 10 to get mol of 25cm? Ans. is 48.5 which I don't seem to get
4. (Original post by tripodd)
Don't really understand.. :/ I get the x25 to get the mol of 250cm(H2O2)

But then wouldn't I need to divide by 10 to get mol of 25cm? Ans. is 48.5 which I don't seem to get
nope, because when you have the mol of 10 cm cube, the unit is mol(it is the amount)!

then in 250 cm cube, you have 25 times more. the concentration would be this much mol divided by 10, but you don't want the concentration, you simply want to know the amount in that 250 cm cube!

see the difference between mole and mole per volume(ie amount divided by volume)

NB. there is no point of getting the answer if you couldn't follow through the logic of "back-titration"

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