If 6.8 * 10(adsbygoogle = window.adsbygoogle || []).push({}); ^{-4}is ionisation constant(IC) of HF, what is the IC of its conjugate base.

HF +H_{2}O -----> H_{3}O^{+}+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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# Homework Help: Ionisation constants

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