HELP PROBLEM

[destain]

To determine the amount of MnO2 in an ore of manganese, 5.25g of the ore was allowed to react with 100ml of 0.250M oxalic acid ( H2C2O4 ).
The remaining amount of oxalic acid is determined by titration with tri-iodide ion.
20mL of the resulting solution is then titrated with 0.1M solution of the triiodide ion. The average titre found was 12.54ml. Determine the percentage by mass of MnO2 present in the ore. Molar mass of MnO2 = 87

SO i've got to the amount of reacted MnO2 to be n=0.023746mol but that's wrong.

[Somye]

Is it 0.015595?

[charmanderp]

I'm such an idiot! I've just had an epiphany which I'm really hoping is correct.

The reaction between oxalic acid and triiodide would be a redox reaction. C in oxalic acid is oxidised from +3 to +4  in CO2and I in triiodide is reduced from -1/3 to -1 in I-.

The half-reactions are as such.

Oxidation: H2C2O4(aq) --> 2CO2(g) + 2H+(aq) + 2e-

Reduction: I3-(aq) + 2e- --> 3I-(aq)

So the overall equation would be:

H2C2O4(aq) + I3-(aq) --> 2CO2(g) + 2H+(aq) + 3I-(aq)

So therefore the amount of H2C2O4 required is equal to the amount of I3- required. The amount of triiodide is 0.001254mol

n(H2C2O4) in excess=n(I3-)
= 0.1*0.01254
= 0.001254mol

But then we need to multiply that by 5, seeing as we've only taken a 20ml aliquot from 100ml of oxalic acid added.

n(H2C2O4) reacted = n(H2C2O4) added - n(H2C2O4) excess*5
=0.1*0.25-0.001254*5
=0.01873mol.

Please tell me that's correct :p

[destain]

yup thats right but i didnt do the x5 bit to go back to 100ml because it confused me....i thought about it but then....

so the theory of back titration is that...you add something to let it react and you titrate which bit?
the bit that you reacted so in this case you titrate the 100ml just that they took 20ml?


ALSO IF ANYONE CAN ANSWER...oxidation numbers
when there are three variables what do you do... do you split them up or just do it somehow

E.g KCr2O7
do you go 2K+ + Cr2O72-?

and if you do split them up, sometimes it doesn't work... like 2NaH2PO4?
and also VO(SO4) and VO2(NO3)

how would you do it and what method...

[charmanderp]

I can't believe I didn't recognise that it was a redox titration :/ You react the excess left over from the initial reaction.

You really need to explain your O.N question a little bit more clearly.