Hydrogen peroxide solution `(20 mL)` reacts quantitatively with a solution of `KMnO_(4) (20 mL)` acidified with dilute of `H_(2)SO_(4)`. The same volume of the `KMnO_(4)` solution is just decolourised by `10 mL` of `MnSO_(4)` in neutral medium simultaneously forming a dark brown precipitate of hydrated `MnO_2)`. The brown precipitate is dissolved in 10 mL of 0.2 M sodium oxalate under boiling condition in the presence of dilute `H_(2)SO_(4)`. Write the balanced equations involved in the reactions and calculate the molarity of `H_(2)O_(2)`.

Explanation

The given reactions are:
`underset((ppt.))(MnO_(2))+Na_(2)C_(2)O_(4)+2H_(2)SO_(4)rarrMnSO_(4)+CO_(2)+Na_(2)SO_(4)+2H_(2)O`
`therefore` Meq.of `MnO_(2)="Meq.of" Na_(2)C_(2)O_(4)`
`=10xx0.2xx2=4`
`because Mn^(4+)+2erarMn^(2+)`
`therefore` valenece factor of `MnO_(2)=2`
`therefore mM` of `MnO_(2)=(4)/(2)=2`
Now, `2KMnO_(4)+3MnSO_(4)+2H_(2)Orarrunderset((ppt.))(5MnO_(2))+K_(2)SO_(4)+2H_(2)O`
Since, eq.of wt. of `MnO_(2)` is derived from `KMnO_(4)` and `MnSO_(4)` both, thus it is better to proceed by mole concept.
`mM` of `MnO_(2)xx(2)/(5)=(4)/(5)`
Also
`5H_(2)O_(2)+2KMnO_(4)+3H_(2)SO_(4)rarr2MnSO_(4)+K_(2)SO_(4)+8H_(2)O+5O_(2)`
`therefore mM` of `H_(2)O_(2)=mM` of `KMnO_(2)xx(5)/(2)=(4)/(5)xx(5)/(2)=2`
`therefore Mxx20=2`
`M_(H_(2)O_(2))=0.1`