Solve the following

$\mathrm{H}_{2} \mathrm{SO}_{4(\text { (at })}+\mathrm{H}_{2} \mathrm{O}_{(\eta)} \longrightarrow \mathrm{H}_{3} \mathrm{O}_{(\text {(a) })}^{+}+\mathrm{HSO}_{4(\text { aq })}^{-} ; \quad K_{a_{1}}>10$

$\mathrm{HSO}_{4(a q)}^{-}+\mathrm{H}_{2} \mathrm{O}_{(\ell)} \longrightarrow \mathrm{H}_{3} \mathrm{O}_{(a q)}^{+}+\mathrm{SO}_{4(a q)}^{-} ; \quad K_{a_{2}}=1.2 \times 10^{-2}$

It can be noticed that $K_{a_{1}} \gg X_{a_{2}}$

This is because a neutral H2SO4 has a much higher tendency to lose a proton than the negatively charged . Thus, the former is a much stronger acid than the latter.