Question

Correct relation b/w dissociation constant's of a di-basic acid:

• A. $Ka_1=Ka_2$
• B. $Ka_1>Ka_2$
• C. $Ka_1<Ka_2$
• D. $Ka_1=\frac {1}{Ka_2}$

Answer 1

Answer: (B)

$Ka_1>Ka_2$

Answer 2

$H_2A ⇋ HA^- + HA^+$

$HA ⇋ A^- + H^+$

In the case of a dibasic acid, the initial step involves the removal of an H+ ion from H2A, followed by the removal of another H+ ion from HA−. Removing H+ from H2A is favored because it exhibits a significantly greater tendency to donate a proton compared to the negatively charged HA−, which is less likely to lose its proton.

So, the correct option is (B): $Ka_1>Ka_2$