Question

When `100 mL` of an aqueous of `H_(2)O_(2)` is titrated with an excess of `KI` solution in dilute `H_(2)O_(2)`, the liberated `I_(2)` required `50 mL` of `0.1 M Na_(2) S_(2)O_(3)` solution for complete reaction. Calculate the percentage strength and volume strength of `H_(2)O_(2)` solution.

Answer 1

Rections involved are as follwos
`{:(2I^(ɵ)toI_(2)+cancel(2e^(ɵ))......(i)),(2H^(o+)+cancel(2e^(ɵ))+H_(2)O_(2)to2H_(2)O....(ii)),(ulbar(H_(2)O_(2)+2I^(ɵ)toI_(2)+H_(2)O....(iii))),(cancel(2e^(ɵ))+I_(2)to2I^(ɵ).....(iv)),(2S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+cancel(2e^(ɵ))....(v)),(ulbar(2S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+2I^(ɵ))):}`
For Eq. (v) `n`-factor =`2/2=1`
`H_(2)O_(2)-=I^(ɵ)-=I_(2)-=S_(2)O_(3)^(2-)`
`mEq of H_(2)O_(2)-=mEq of I_(2)-=mEq of S_(2)O_(3)^(2-)`
`N_(1)V_(1)(mL)-=N_(2)V_(2)(mL)`
`N_(1)xx100-=0.1xx1("n-factor")xx50 mL`
`N(H_(2)O_(2))-=1.7% of H_(2)O_(2)`
`0.05 NH_(2)O_(2)-=1.7xx0.05=0.085 % of H_(2)O_(2)`
`1NH_(2)O_(2)=5.6` volume of `O_(2)`
`0.05 NH_(2)O_(2)=5.6xx0.05=0.28` volume of `O_(2)`
Volume strength of `H_(2)O_(2)=0.28` volume