Why is `K_(a_(2)) ltlt K_(a_(1))` for `H_(2) SO_(4)` in water ?

Explanation

`H_(2)SO_(4)` is a dibasic acid, it ionizes in two stages and hence has two dissociation constants.
(i) `H_(2)SO_(4)(aq)+H_(2)O(l) rarr H_(3)O^(+) (aq) + HSO_(4)^(-)(aq) , K_(a_(1)) gt 10`
(ii) `HSO_(4)^(-) (aq) + H_(2)O(l) rarr H_(3)O^(+) (aq) + SO_(4)^(2-) (aq) , K_(a_(2)) = 1.2 xx 10^(-2)`
`K_(a_(2))` is less than `K_(a_(1))` because the negatively charged `HSO_(4)^(-)` ion has much less tendency to donate a proton to `H_(2)O` as compared to neutral `H_(2)SO_(4)`.