Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    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
    Offline

    11
    ReputationRep:
    (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
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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
    Offline

    11
    ReputationRep:
    (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"
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: March 20, 2011
The home of Results and Clearing

1,284

people online now

1,567,000

students helped last year

University open days

  1. London Metropolitan University
    Undergraduate Open Day Undergraduate
    Sat, 18 Aug '18
  2. Edge Hill University
    All Faculties Undergraduate
    Sat, 18 Aug '18
  3. Bournemouth University
    Clearing Open Day Undergraduate
    Sat, 18 Aug '18
Poll
A-level students - how do you feel about your results?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.