Why is $K{a}_2<<K{a}_1$ for $H_{2}S{O}_4$ in water?

$H_{2}S{O}_4$ is dibasic acid. It ionises in two stages and hence has two dissociation constants as given below:

(i) $H_{2}S{O}_{4\left(aq\right)}+H_2{O}_{\left(l\right)}\to H_3{O}_{\left(aq\right)}^{+}+HS{O}_{4\left(aq\right)}^{-};K{a}_1>10$

(ii) $HS{O}_{4\left(aq\right)}^{-}+H_2{O}_{\left(l\right)}\to H_3{O}_{\left(aq\right)}^{+}+S{O}_{4\left(aq\right)}^{2-};K{a}_2=1.2\times {10}^{-2}$

$K{a}_1$ is greater than $K{a}_2$, i.e., tendency to move towards the products is greater in (i) than in (ii).

This is because the negatively charged $HS{O}_4^{-}$ ion has much less tendancy to donate a proton to $H_{2}O$ as compared to neutral $H_{2}S{O}_4$.