We know that
HF < -----------> H+ + F–
Ka (acid) x Kb (conjugate base) = Kw
= 1.0 x 10-14
Ka = 6.8 x 10-4 (given)
Kb (F–) = 1.0 x 10-14 / 6.8 x 10-14
= 1.47 x 10-11
Hence, the dissociation constant of conjugate base of HF is:
Kb (F–) = 1.47 x 10-11
Calculate the dissociation constant of conjugate base of HF, Ka = 6.8 x 10-4.
Category: Ionic Equilibria
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