Redox titration, determination of iron in iron tablets

Hey guys, basically tommorow I'm starting my practical investigation and looks like my first experiment is using titration to find the mass of iron in an iron tablet.

So I have Potassium Permanganate, Sulphuric Acid, Iron tablets.

Permanganate solution is of known concentration which is 0.001 mol/l, so starting off i dissolve the tablet in sulphuric acid right instead of water, but what volume though?

Hey guys, basically tommorow I'm starting my practical investigation and looks like my first experiment is using titration to find the mass of iron in an iron tablet.

So I have Potassium Permanganate, Sulphuric Acid, Iron tablets.

Permanganate solution is of known concentration which is 0.001 mol/l, so starting off i dissolve the tablet in sulphuric acid right instead of water, but what volume though?

I found using permanganate gave me crap experimental results so hopefully you will do better, I used 20cm^3 but it's worth expermenting. I followed a colorimetric anaylsis technique in the end, I would look up dichromate titrations also.

So I can use like 20 cm3 of sulphuric acid and then 80cm3 of deionised water just to beef up the volume?

Also now the irons dissolved is the 100cm3 taken to be the v value in the calculation.

I'm pretty crap with the calculations lol. I have the C value for Permanganate, and using titration I'll get the V value. So using N=c x v i've got number of moles of permanganate. Then from here how do i get the number of moles of Fe??

Thanks.

Well you work out the balanced equation (or just remember it as it is a pretty standard equation) then use the relationship 5Fe:KMnO4 so the concentration of Fe will be 5 times the concentration of MnO4 ions. Then this concentration multiplied by 0.1 is the conc Fe in moll-1 in your 100cm^3 soloution or multiply by Fe GFM (55.8g) to find the Nass of Fe in the tablet.