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

Explanation

`H_(2) SO_(4(aq)) + H_(2) O_((l)) to H_(3)O_((aq))^(+) + HSO_(4(aq))^(-), K_(a_(1)) gt 10`
`H_(2) SO_(4(aq))^(-) + H_(2) O_((l)) to H_(3)O_((aq))^(+) + SO_(4(aq))^(-)," " K_(a_(2)) = 1.2 xx 10^(-2)`
It can be noticed that `K_(a_(1)) gt gt K_(a_(2))`
This is because a neutral `H_(2)SO_(4)` has a much higher tendency to lose a proton than the negatively charged `HSO_(4)^(-)`. Thus, the former is a much stronger acid than the latter.