For any diprotic acid of the type H_2A,the re | vedclass

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(C) For a diprotic acid $H_2A$,the ionization occurs in two steps:$1$. $H_2A(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + HA^-(aq)$ with constant $Ka

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$Ka_2 < Ka_1$

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(C) For a diprotic acid $H_2A$,the ionization occurs in two steps:$1$. $H_2A(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + HA^-(aq)$ with constant $Ka_1$.$2$. $HA^-(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + A^{2-}(aq)$ with constant $Ka_2$.In the second step,the removal of a proton $(H^+)$ from a negatively charged species $(HA^-)$ is much more difficult due to electrostatic attraction between the positive proton and the negative ion. Therefore,$Ka_1$ is always greater than $Ka_2$ ($Ka_1 > Ka_2$ or $Ka_2 < Ka_1$).

For any diprotic acid of the type $H_2A$,the relationship between the first $(Ka_1)$ and second $(Ka_2)$ ionization constants is:

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