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Help with a redox titration question! Watch

1. 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

Cheers!
2. (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

Cheers!
Do you have no ideas on how to start?
3. (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+
4. (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)
5. (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!
6. 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
7. (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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