Being a dibasic acid, \(\mathrm{H}_2 \mathrm{SO}_4\) ionized in two stages as follows:
\(\begin{aligned}
&\mathrm{H}_2 \mathrm{SO}_4(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{HSO}_4^{-}(a q) \mathrm{K}_{\mathrm{a}_1}>10 \\
&\mathrm{HSO}_4^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{SO}_4^{2-}(a q) \\
&\mathrm{K}_{\mathrm{a}_2}=1.2 \times 10^{-2}
\end{aligned}\)
\(\mathrm{K}_{\mathrm{a}_2} \ll \mathrm{K}_{\mathrm{a}_1}\) because negatively charged \(\mathrm{HSO}_4^{-}\)has much less tendency to donate a proton to \(\mathrm{H}_2 \mathrm{O}\) as compared to neutral \(\mathrm{H}_2 \mathrm{SO}_4\).