Chemistry Pemanganate Titration?

Question

25.00cm^2 of a solution containing iron (II) and iron (III) sulphate required 20.50cm^2 of 0.02 mol dm^-3 pottasium pemanganate for oxidation. After complete reduction by zinc amalgum, 25.00cm^3 of the solution required 34.80cm^3 of the same pemanganate solution. Calculate the concentration of iron (II) and iron (III) ions present in the given solution. [RAM; H:1 ; O:16 ; Fe : 56].

I would appreciate an answer with a full explaination. I am totally lost with this. Thank you in advance.

Answer 1

You start with a mixture of Fe+2 and Fe+3
KMnO4 will oxidise all of Fe+2 to Fe+3.
So a1M1V1=a2M2V2 for the titration where a is the number of e per molecule of reactant, M the molarity and V the volume.

You don't specify the conditions of the titration. For acidic conditions the a of Mn is 5 because it is reduced from +7 to +2
For iron it is 1

So 5*0.02*20.5=1*[Fe+]*25 thus
[Fe+2]=0.082 M in your initial solution

You oxidize all so you end up with only Fe+3.
Then you reduce all so you end up with only Fe+2
You titrate again so you will get the total concentration of iron

using the same equation;
5*0.02*34.8=1*[Fe+2]total*25
Thus [Fe+2]total=0.139 M

But [Fe+2]total=[Fe+2]+[Fe+3] =>
0.139 =0.082+[Fe+3] =>
[Fe+3]=0.057 M