Why is $K_{a_{2}} << K_{a_{1}}$ for $H_{2}SO_{4}$ in water?

(N/A) $H_{2}SO_{4(aq)} + H_{2}O_{(l)} \to H_{3}O_{(aq)}^{+} + HSO_{4(aq)}^{-}; \quad K_{a_{1}} > 10$
$HSO_{4(aq)}^{-} + H_{2}O_{(l)} \to H_{3}O_{(aq)}^{+} + SO_{4(aq)}^{2-}; \quad K_{a_{2}} = 1.2 \times 10^{-2}$
It can be observed that $K_{a_{1}} >> K_{a_{2}}$.
This is because a neutral $H_{2}SO_{4}$ molecule has a much higher tendency to lose a proton compared to the negatively charged $HSO_{4}^{-}$ ion. The electrostatic attraction between the positively charged proton and the negatively charged $HSO_{4}^{-}$ ion makes the removal of the second proton significantly more difficult.