How many milliliters of 0.125 M `KMnO_4` are required to react completely with 25.0 mL of 0.250 M `FeSO_4` solution in the acidic medium ?

Explanation

The balanced chemical equation is :
`MnO_4^(-) +5Fe^(2+) +8H^(+) rarr Mn^(2+) +5Fe^(3+) +4H_2O`
From the balanced equation it is clear
`{:(1 mol KMnO_4 =, 5 mol FeSO_4),(158,5xx152):}`
Moles of `FeSO_4` present in 25 mL of 0.250 M solution :
`=(0.250)/1000xx25=0.00625` mol
Let us determine the number of moles of `KMnO_4` that must react
5 mol of `FeSO_4` react with `KMnO_4`
= 1mol
0.00625 mol of `FeSO_4` will react with `KMnO_4`
`=1/5xx0.00625=0.00125 mol`
Now, we are the calculate the volume of 1.25 M
`KMnO_4` solution which contain 0.00125 mol .
According to definition of morality , 0.125 M solution means that
`0.125 mol KMnO_4` is present in
= 1000 mL
0.00125 mol of `KMnO_4` is present in
`=1000/(0.125)xx0.00125=10 mL`
This can also be calculated as :
Morality `=("Moles"xx1000)/(Volume)`
`0.125=(0.00125xx1000)/(volume)`
`:. Volume = (0.00125 xx1000)/(0.125) = 10 ` mL