How many millilitres of 0.02 M `KMnO_(4)` solution would be required to exactly titrate 25 mL of 0.2 M `Fe(NO_(3))_(2)` solution in acidic medium ?

Explanation

Method - 1 : Mole concept method
Starting with 25 mL of 0.2 M `Fe^(2+)`, we can write :
Millimoles of `Fe^(2+)=25xx0.2" " .......(1)`
and in volume V (in milliliters) of the `KMnO_(4)`,
Millimoles of `MnO_(4)^(-)=V (0.02) " " .....(2)`
the balanced reaction is :
`MnO_(4)^(-)+5Fe^(2+)+8H^(+)rarrMn^(2+)+5Fe^(3+)+4H_(2)O`
This requires that at the equivalent point,
`("m.moles of " MnO_(4)^(-))/(1)=("m.moles of " Fe^(2+))/(5)`
`:. (V(0.02))/(1)=((25)(0.2))/(5) " " `(from (1) & (2))
`:. ` V=50 mL.
Method - 2 : Equivalent Method :
At the equivalence point,
milliequivalents of `MnO_(4)^(-)`=milliequivalents of `Fe^(2+)`
`M_(1)xxvf_(1)xxV_(1)=M_(2)xxvf_(2)xxV_(2)`
`0.02xx5xxV_(1)=0.2xx1xx25 " " (.:' MnO_(4)^(-)rarrMn^(2+) , v.f. = 5, Fe^(2+)rarrFe^(3+), v.f.=1)`