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Aisha_18 · Mar 10, 2009, 07:02 PM

A 25.0 cm3 portion of a solution containing Fe2+ and Fe3+ ions was acidified and titrated against potassium permanganate (VII) solution. 15.0 cm3 of a 0.0200 mol dm-3 solution of potassium manganate (VII) were required. A second 25.0cm3 portion was reduced with zinc and titrated against the same manganate solution. Calculate the concentration of Fe 2+ and Fe 3+ in the solution.

THis is what I have done

MOlarity of KMNO4 = 0.0200 moles dm-3
Therefore in 1000c3 = 0.0200 moles
The volume of KMNO4 in the first titration was 15.0cm3
So 1000cm3=0.0200 moles
15 cm3 = 0.0200/1000x15 = 0.0003 moles

Here's the part I am not too sure about.. dont really know but is Fe 5 times the amount of KMNO4?

0.0003 moles x 5 = 0.015 moles of Fe2+
molarity = 0.015 mole/25x10-3 = 0.06 M

THe second part o basically did the same time and for my Concentration I got out 0.0765 M
So for my FE3t I just subtracted 0.0765 from the 0.06 to give 0.016 M of Fe3+

Unknown008 · Mar 11, 2009, 03:52 AM

That sounds good to me.

To see how many moles of Fe2+ a mole of KMnO4 oxidises, write an equation!

$5Fe_^{2+} - 5e^-$ -> $5Fe_^{3+}$

$MnO_4^- + 8H^+ + 5e^-$ -> $Mn^{2+} + 4H_2O$

$5Fe^{2+} + MnO_4^- + 8H^+$ -> $5Fe^{3+} + 4 H_2O$

I guess you didn't type the volume of KMnO4 required in the second titration. If you did it like that, i would say that your work is good.