A solution of `K_(2)cr_(2)O_(7)` contanining `4.9 gL^(-1)` is used to tirate `H_(2)O_(2)` solution contaning `3.4 gL^(-1)` in acidic medium. What volume of `K_(2)Cr_(2)O_(7)` will be required to react with `20 mL` of `H_(2)O_(2)` solution ? Also calculate the strength of `H_(2)O_(2)` in terms of available oxygen.

Explanation

Strength =Norality`xx`Equivalent weight =`NxxEW`
`EW (K_(2)Cr_(2)O_(7))=(39xx2+52+2+16xx7)/6=294/6=49.0 g`
`:'` In acidic medium.
`Cr_(2)O_(7)^(2-)+14H^(o+)+6e^(ɵ)to2Cr^(3+)+7H_(2)O, n-fact or =6`
`N(K_(2)Cr_(2)O_(7))=("Strength")/(EW)=4.9/49=0.1 N`
`N(H_(2)O_(2))=("Strength")/(EW)=3.4/17=0.2 N`
`{{:(H_(2)O_(2)toO_(2)+2H^(o+)+2e^(ɵ)(n=2)),(EW=34/2=17 g):}}`
`Cr_(2)O_(7)^(2-)-=H_(2)O_(2)`
`mEq of Cr_(2)O_(7)^(2-)-=mEq of H_(2)O_(2)`
`N_(1)V_(1)=N_(2)V_(2)`
`0.1xxV_(1)=0.2xx20`
`V_(1)=(0.2xx20)/0.1=40 mL`
Volume strength of `H_(2)O_(2)`
`1N=5.6 L of O_(2)`(Volume strength of `H_(2)O_(2)`)
`0.2 N=5.6xx0.2=1.12 of H_(2)O_(2)`
it is writtens as `1.12` volume of `H_(2)O_(2)`.