Explain ionization and ionization constant in di- and polyprotic acids.

As an example,the ionization of a dibasic acid $H_2X$ in aqueous solution is represented in two steps:
$(i) \ H_2X_{(aq)} + H_2O_{(l)} \rightleftharpoons H^{+}_{(aq)} + HX^{-}_{(aq)}$
$(ii) \ HX^{-}_{(aq)} + H_2O_{(l)} \rightleftharpoons H^{+}_{(aq)} + X^{2-}_{(aq)}$
If the equilibrium constants for these steps are $K_{a1}$ and $K_{a2}$ respectively,then:
$K_{a1} = \frac{[H^{+}][HX^{-}]}{[H_2X]}$ and $K_{a2} = \frac{[H^{+}][X^{2-}]}{[HX^{-}]}$
The overall reaction is the sum of $(i)$ and $(ii)$:
$H_2X_{(aq)} + 2H_2O_{(l)} \rightleftharpoons 2H^{+}_{(aq)} + X^{2-}_{(aq)}$
The overall equilibrium constant $K_a$ is given by:
$K_a = \frac{[H^{+}]^2 [X^{2-}]}{[H_2X]} = K_{a1} \times K_{a2}$
For any polyprotic acid,the overall dissociation constant is the product of the individual ionization constants:
$K_a = K_{a1} \times K_{a2} \times K_{a3} \dots$
Generally,$K_{a1} > K_{a2} > K_{a3} \dots$ because it is progressively more difficult to remove a positively charged proton from a negatively charged ion.