Hydrogen peroxide solution (20 ml) reacts quantitavely with a solution of `KMnO_(4)` (20 ml) acidified with dilute `H_(2)SO_(4)`. The same volume of the `KMnO_(4)` solution is just decolourized 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 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 volume strength of `H_(2)O_(2)`.

Explanation

Meq. Of `MnO_(2) = " Meq. of " Na_(2)C_(2)O_(4)`
` = 10 xx 0.2 xx 2 = 4`
` (C_(2)O_(4)^(2-) to 2CO_(2), ` n- factor = 2)
Millimoles of `MnO_(2) = 4/2 = 2`

or `2MnO_(4)^(-) + 3Mn^(2+) to 5MnO_(2) ` (Mole ratio is reciprocal of n-factor ratio )
`("Millimoles of " MnO_(4)^(-))/("Millimoles of " MnO_(2)) = 2/5`
Millimoles of `MnO_(4)^(-) = 2/5 xx ` Millimoles of `MnO_(2)`
` = 2/5 xx 2 = 4/5`
`2KMnO_(4) + 5H_(2)O_(2) + 3H_(2)SO_(4) to 2 MnSO_(4) + K_(2)SO_(4) + 8H_(2)O + 5O_(2)`
` ("Millimoles of " H_(2)O_(2))/("Millimoles of " MnO_(4)^(-)) = 5/4 `
` :. ` Millimoles of `H_(2)O_(2) = 5/2 xx 4/5 = 2`
`N_(H_(2)O_(2)) xx 20 = 2 xx 2 ("n- factor for " H_(2)O_(2) = 2) `
`:. N_(H_(2)O_(2)) = 0.2`
Volume strength of `H_(2)O_(2) = 5.6 xx N_(H_(2)O_(2))`
` = 5.6 xx 0.2 = 1.12`