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Help with a redox titration question!

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!

Reply 1

charco

Study Forum Helper

Original post by thesmallman

Do you have no ideas on how to start?

Reply 2

thesmallman

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+

Reply 3

charco

Study Forum Helper

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)

Reply 4

thesmallman

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!

Reply 5

charco

Study Forum Helper

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

Reply 6

username2097061

Original post by charco

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!