To a 25 mL `H_(2)O_(2)` solution excess of an acidified solution of
potassium iodide was added. The iodine liberated required 20 mL of 0.3 N sodium thiosulphate solution Calculate the volume strength of `H_(2)O_(2)` solution.

Explanation

`2KI + H_(2)SO_(4) + H_(2)O_(2) rarr K_(2)SO_(4) + 2H_(2)O + I_(2)`
`2Na_(2)S_(2)O_(3) + I_(2) rarr Na_(2)S_(4)O_(6) + 2NaI`
Meq. of `Na_(2)S_(2)O_(3) = 20 xx 0.3 = 6 ("Nomality" xx "volume")`
Meq. of `Na_(2)S_(2)O_(3)` = Meq. of `I_(2)` = 6
Meq. of `I_(2)` = Meq. of `H_(2)O_(2) = 6`
Weight of `H_(2)O_(2) = Meq. xx E xx 10^(-3) = 6 xx 17 xx 10^(-3) = 0.102 g` (Eq. wt. of `H_(2)O_(2) = 34/2 = 17`)Strength of `H_(2)O_(2) = 0.102 xx 1000 /25 = 0.408 g//L`
Molarity of `H_(2)O_(2) = 0.408 g//L/molwt.(34) = 0.012 M`
`underset("underset("0.012 moles")2 mole")2H_(2)O_(2) rarr 2H_(2)O + underset("underset("0.06 moles")1 mole")O_(2)`
Volume of `O_(2)` at STP = 0.06 xx 22.4 L = 1.344 L
Hence the volume strength of `H_(2)O_(2) = 1.344 L`