Q.Why is Ka2 > Ka1 for H2SO4 in water?

Sulphur forms a range of oxoacids of different oxidation states and structures, built mainly around tetrahedral $SO_4$-type units with terminal S=O bonds and ionisable S-OH bonds — sulphurous acid ($H_2SO_3$), sulphuric acid ($H_2SO_4$), and peroxo-acids such as peroxodisulphuric acid ($H_2S_2O_8$). Passing sulphur trioxide into water is the general route from an acidic oxide to its oxoacid, $SO_3 + H_2O \rightarrow H_2SO_4$, an exothermic hydration (industrially $SO_3$ is instead absorbed into concentrated $H_2SO_4$ to avoid forming a hard-to-condense mist). Because concentrated sulphuric acid is a strong, non-volatile acid, it can displace a more volatile acid from its salt, for example liberating hydrogen fluoride gas from calcium fluoride, $CaF_2 + H_2SO_4 \rightarrow CaSO_4 + 2HF$. For a diprotic oxoacid like $H_2SO_4$, the two ionisation steps, $H_2SO_4 \rightarrow H^+ + HSO_4^-$ and $HSO_4^- \rightarrow H^+ + SO_4^{2-}$, differ greatly in acid strength: the first proton leaves an electrically neutral molecule and is lost almost completely, giving a very large $K_{a1}$, while the second proton must leave an already negatively charged $HSO_4^-$ ion, which holds it ba …