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

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The ionization constant (IC) of hydrofluoric acid (HF) is 6.8 x 10-4. The ionization constant of its conjugate base, fluoride ion (F-), is calculated using the relationship Kw/Ka, where Kw is the ionic product of water (1.0 x 10-14) and Ka is the ionization constant of HF. This results in the basicity constant (Kb) for F- being equal to 1.4 x 10-11. The discussion clarifies that the conjugate base is indeed ionized in the context of its basicity equilibrium.

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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
 
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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 Kw/Ka?
 
Last edited:
yep, that's what is meant the ionization constant, in this problem. For the base it can also be called the hydrolysis constant, Kb
 

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