In the equilibrium CH_3COOH + HF rightlefthar | vedclass

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Question

(B) In the reaction $CH_3COOH + HF \rightleftharpoons CH_3COOH_2^+ + F^{-}$,$HF$ acts as an acid because it donates a proton $(H^+)$ to $CH_3COOH$.Aft

Vedclass · Accepted Answer

$F^{-}$ is the conjugate base of $HF$

Vedclass · Answer

(B) In the reaction $CH_3COOH + HF \rightleftharpoons CH_3COOH_2^+ + F^{-}$,$HF$ acts as an acid because it donates a proton $(H^+)$ to $CH_3COOH$.After losing a proton,$HF$ becomes $F^{-}$,which is its conjugate base.Similarly,$CH_3COOH$ acts as a base by accepting a proton to form $CH_3COOH_2^+$,which is its conjugate acid.Therefore,$F^{-}$ is the conjugate base of $HF$.

In the equilibrium $CH_3COOH + HF \rightleftharpoons CH_3COOH_2^+ + F^{-}$,which of the following statements is correct?

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