A sample of fuming `H_2SO_4` contains `H_2SO_4,SO_3` and `SO_2.2.0g` of the above sample was dissolved in water to make a 500 mL solution. 50 " mL of " the above solution on titration in presence of methyl orange requires 42.4 " mL of " 0.1 N NaOH. On the other hand , 100 " mL of " the same sample solution requires 1.85 " mL of " 0.1 N of `I_2` where `I_2` is reduced to `I^(ɵ)` ions. Determine the percentage composition of oleum sample. (In methyl orange,
`SO_2+H_2OtoH_2SO_3overset(NaOH)toHSO_3^(ɵ))`

Explanation

NaOH displaced only one proton from `H_2SO_3` obtained from `SO_2` in methyl orange as indicator.
" Eq of "`SO_2=Eq` of `H_2SO_3`
Ew of `(H_2SO_3)=(Mw)/(x)(x=1)`
Let x g of `H_2SO_4,y` g of `SO_3`, and z g of `SO_2` in `(2.0)/(500)` mL mixture.
(i). `x+y+z=2.0`
(ii). In 50 mL sample
m" Eq of "`NaOH=mEq` of `(SO_3+H_2SO_4+SO_2)`
`42.4xx0.1=(((x)/(2))/((98)/(2))+((y)/(2))/((80)/(2))+((z)/(10))/((64)/(1)))xx1000`
(iii). In 100 mL sample : only `SO_2` reacts with `I_2(I_2` is used as oxidising agent).
`I_2+underset(x=4)(SO_2)tounderset(x=6)(SO_4^(2-)+2I^(ɵ)`
`mEq` of `SO_2=mEq` of ` I_2`
`1000((z)/(5))/((64)/(2))=1.85xx0.101`
`z=0.0298g`
`%` of `z=(0.0298)/(2)xx100=1.495%`
From (ii)
`(x)/(49)+(y)/(40)+(z)/(64)=0.042`
Substituting `z=0.0298`
`x=160` `%x(H_2SO_4)=80`
`y=0.368" "% y (SO_3)=1.85`
`z=0.0298 %z(SO_2)=1.495`