Ionisation Constants of HF & Its CB: Why 10-14?

[chound]

If 6.8 * 10^-4 is ionisation constant(IC) of HF, what is the IC of its conjugate base.

HF +H₂O -----> H₃O⁺ +F^-
So F^- is the conjugate base(CB)
I figured the IC of CB would be the reciprocal of IC of HF. But the answer given is the same but multiplied by Ionic product of water i.e. 10^-14

Please explain why

 

[siddharth]

Ionisation of the conjugate base? Isn't it already ionised?

Perphaps you mean the basicity constant for the equilibirium
F^- + H₂O ----> HF + OH^-

In that case, won't it be K_w/K_a?

 

[GCT]

yep, that's what is meant the ionization constant, in this problem. For the base it can also be called the hydrolysis constant, Kb