Why do we add dilute sulphuric acid to dissolve mohr's salt?

Answer 1

For preparation of standard solution of Mohr salt {FeSO4.(NH4)2SO4.6H2O}, it's necessary to add dilute H2SO4 to prevent the Fe2+ ions of Mohr salt solution from undergoing oxidation (to Fe3+).

Just try to prepare the standard solution without adding this acid. You'll observe that your "Mohr solution" will become yellow in colour, i.e., it won't remain "Mohr salt solution" anymore, hence can't be used in titration.

And while doing the titration of the Mohr salt solution with KMnO4 (or even with K2Cr2O7), you need to add dilute H2SO4. The reason is well explained by the chemical equation involved:

reduction: (MnO4)- + 8H+ + 5e ----> Mn2+ + 4H2O

oxidation: Fe2+ ----> Fe3+ + e

net equation:

(MnO4)- + 8H+ + 5Fe2+ ----> Mn2+ + 4H2O + 5Fe3+

There is involvement of H+ ions in this reaction. This redox titration requires acidic medium. Moreover, KMnO4 acts as a very good oxidizing agent in the presence of acidic medium, than in the basic and neutral medium.

NOTE: Now a query may arise that for providing acidic medium, can HCl or HNO3 be used? (Let me tell you it's one of the favourite questions of external examiners)

The answer is NO, because HNO3 is a very strong oxidizing agent. So, the Fe2+ present will be oxidized both by KMnO4 and HNO3, which will result in an error in determination of volume of KMnO4 reacting with Mohr salt solution.

Also, HCl can't be used because Cl- from HCl will be oxidized by KMnO4:

HCl (aq) ---> H+ (aq) +Cl- (aq)

oxidation by KMnO4=> 2Cl- ----> Cl2 + 2e

So, some extra KMnO4 will be used up because it will oxidize both- Cl- as well as Fe2+, => error in titration.