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    Hey guys- would really appreciate some help on this redox titration qs!!


    A 25.0 cm3 aliquot of a solution containing Fe2+ and Fe3+ ions was acidified and titrated against 0.0200 M potassium
    manganate (VII) solution, requiring 15.0 cm3. Zn reduces Fe3+ to Fe2+ and a second aliquot was reduced by zinc and
    after filtering off the excess zinc, was titrated with the same potassium manganate solution, requiring 19.0 cm3.
    Calculate the concentrations of the Fe2+ and Fe3+ in the solution


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    (Original post by thesmallman)
    Hey guys- would really appreciate some help on this redox titration qs!!


    A 25.0 cm3 aliquot of a solution containing Fe2+ and Fe3+ ions was acidified and titrated against 0.0200 M potassium
    manganate (VII) solution, requiring 15.0 cm3. Zn reduces Fe3+ to Fe2+ and a second aliquot was reduced by zinc and
    after filtering off the excess zinc, was titrated with the same potassium manganate solution, requiring 19.0 cm3.
    Calculate the concentrations of the Fe2+ and Fe3+ in the solution


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    Do you have no ideas on how to start?
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    (Original post by charco)
    Do you have no ideas on how to start?
    Yeahh, I think I've worked out [Fe2+]:

    MnO4- + 8H+ + 5Fe2+---> 5Fe3+ + Mn2+ + 4H2O

    no. of moles of MnO4- = 0.02 X (15/1000)= 0.0003
    no. of moles of Fe2+ = 0.0003 X 5= 0.0015
    [Fe2+]= 0.0015/(25/1000)= 0.06 mol dm^(-3)

    However, I have no idea how to approach the second part i.e. calculating Fe3+
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    (Original post by thesmallman)

    A 25.0 cm3 aliquot of a solution containing Fe2+ and Fe3+ ions was acidified and titrated against 0.0200 M potassium manganate(VII) solution, requiring 15.0 cm3. Zn reduces Fe3+ to Fe2+ and a second aliquot was reduced by zinc and after filtering off the excess zinc, was titrated with the same potassium manganate solution, requiring 19.0 cm3.

    Calculate the concentrations of the Fe2+ and Fe3+ in the solution

    Yeahh, I think I've worked out [Fe2+]:

    MnO4- + 8H+ + 5Fe2+---> 5Fe3+ + Mn2+ + 4H2O

    no. of moles of MnO4- = 0.02 X (15/1000)= 0.0003
    no. of moles of Fe2+ = 0.0003 X 5= 0.0015
    [Fe2+]= 0.0015/(25/1000)= 0.06 mol dm^(-3)

    However, I have no idea how to approach the second part i.e. calculating Fe3+
    The zinc added changes ALL of the iron(III) to iron(II).

    Hence the second titration finds total moles of iron(II), which equals initial moles of iron(III) + initial moles of iron(II)
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    (Original post by charco)
    The zinc added changes ALL of the iron(III) to iron(II).

    Hence the second titration finds total moles of iron(II), which equals initial moles of iron(III) + initial moles of iron(II)
    So the answer is 0.016 mol dm^(-3)??

    Thanks for ur help!
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    A 25.0 cm3 aliquot of a solution containing Fe2+ and Fe3+ ions was acidified and titrated against 0.0200 M potassium manganate(VII) solution, requiring 15.0 cm3. Zn reduces Fe3+ to Fe2+ and a second aliquot was reduced by zinc and after filtering off the excess zinc, was titrated with the same potassium manganate solution, requiring 19.0 cm3.

    Calculate the concentrations of the Fe2+ and Fe3+ in the solution

    (Original post by thesmallman)

    So the answer is 0.016 mol dm^(-3)??

    Thanks for ur help!
    The short-cup way of doing this is to say that an "extra" 4.00 cm3 were needed for the second titration...

    Therefore the iron(III) must have contained 5 x 0.004 x 0.0020 mol = 4 x 10-4 mol

    This is in 25ml...

    Therefore the concentration = 4 x 10-4/0.025 = 0.016 mol dm-3
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    (Original post by charco)
    The short-cup way of doing this is to say that an "extra" 4.00 cm3 were needed for the second titration...

    Therefore the iron(III) must have contained 5 x 0.004 x 0.0020 mol = 4 x 10-4 mol

    This is in 25ml...

    Therefore the concentration = 4 x 10-4/0.025 = 0.016 mol dm-3
    But surely the assumption of 4 cm^3 is only true if the original solution was only Fe 2+ ions, but it's not, it's Fe 2+ and 3+?

    Do you mind explaining further the method for this?

    EDIT: Don't worry, I thought that the MnO4 reacted with both Fe2+ and Fe3+ ions, and thus was confused with your method. I now understand!
 
 
 
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