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Correct relation between dissociation constants of a di-basic acid :
A
$K_{a1} = K_{a2}$
B
$K_{a1} \gt K_{a2}$
C
$K_{a1} \lt K_{a2}$
D
$K_{a1} = \dfrac{1}{K_{a2}}$
Detailed Solution
A dibasic acid $H_2A$ ionises in two steps: $H_2A \rightleftharpoons H^+ + HA^-$ ($K_{a1}$), then $HA^- \rightleftharpoons H^+ + A^{2-}$ ($K_{a2}$).
It is much harder to remove a proton from the negatively charged $HA^-$ ion than from the neutral $H_2A$, because the proton is attracted by the negative charge.
Also, the $H^+$ from the first step suppresses the second step (common ion effect).
So $K_{a1} \gt K_{a2}$ (e.g. for $H_2S$, $K_{a1} \approx 10^{-7}$ and $K_{a2} \approx 10^{-13}$).
It is much harder to remove a proton from the negatively charged $HA^-$ ion than from the neutral $H_2A$, because the proton is attracted by the negative charge.
Also, the $H^+$ from the first step suppresses the second step (common ion effect).
So $K_{a1} \gt K_{a2}$ (e.g. for $H_2S$, $K_{a1} \approx 10^{-7}$ and $K_{a2} \approx 10^{-13}$).