Calculate the dissociation constant of conjugate base of HF, Ka = 6.8 x 10-4.

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

Category: Ionic Equilibria

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