Question

22.5 ml of 0.102N KMnO4 is required to oxidized 10ml. mohr's salt solution,determination the amount of Fe++ of that solution??

Answer 1

Approx...`1.0*mol*L^-1`....of `"Mohr's salt"`...

Explanation:

This is a redox titration...permanganate is reduced to `Mn^(2+)`...this is the probable reduction product...

`underbrace(MnO_4^(-))_"strongly purple" + 8H^+ +5e^(-) rarr`

`underbrace(Mn^(2+))_"almost colourless" +4H_2O(l)`

The colour change allows us to vizualize the stoichiometric endpoint of the reaction...

And `"ferrous ion"` is OXIDIZED to `"ferric ion..."`

`Fe^(2+) rarrFe^(3+) +e^(-)`

We add FIVE of the latter to ONE of the former....

`MnO_4^(-) +5Fe^(2+) + 8H^+ +5e^(-) rarrMn^(2+) +5Fe^(3+)+4H_2O(l)`

`MnO_4^(-) +5Fe^(2+) + 8H^+ rarrMn^(2+) +5Fe^(3+)+4H_2O(l)`

And so ONE EQUIV of permanganate oxidized FIVE EQUIV of ferrous ion....

`"Moles of permanganate"=22.5xx10^-3*Lxx0.102*mol*L^-1=0.002295*mol`

And so there are FIVE equiv of ferrous ion present....in a solution whose volume is `10*mL`

`[Fe^(2+)]=(5xx0.002295*mol)/(10*mLxx10^-3*L*mL^-1)=1.15*mol*L^-1`