A solution of `K_2Cr_2O_7` containing `4.9gL^(-1)` is used to titrate `H_2O_2` solution containing `3.4gL^(-1)` in acidic medium. What volume of `K_2Cr_2O_7` will be required to react with 20 " mL of " `H_2O_2` solution? Also calculate the strength of `H_2O_2` in terms of available oxygen.

Explanation

Strength `=NxxEw`
`Ew(K_2Cr_2O_7)=(39xx2+52xx2+16xx7)/(6)=(294)/(6)=49.0g`
`[6e^(-)+Cr_2O_7^(2-)to2Cr^(3+)](n=6)`
`N(K_2Cr_2O_7)=("strength")/(Ew)=(4.9)/(49)=0.1N`
`N(H_2O_2)=("strength")/(Ew)=(3.4)/(17)=0.2N`
`{:[(H_2O_2toO_2underset((n=2))(2H^(o+))+2e^(-)),(Ew=(34)/(2)=17g)]:}`
`Cr_2O_7^(2-)-=H_2O_2`
`mEq-=mEq`
`N_1V_1-=N_2V_2` ltBrgt `0.1xxV_1=0.2xx20`
`V_1=(0.2xx20)/(0.1)=40mL`
Volume strength of `H_2O_2`
`1N=5.6L of O_2` (volume strength of `H_2O_2`)
`0.2N=5.6xx0.2=1.12L of H_2O_2`
It is written as 1.12 volume `H_2O_2`.