In the equilibrium CH_3COOH + HF rightlefthar | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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?

Similar Questions

Identify the conjugate acid and conjugate base for the $HCO_3{ }^{-}$ ion,respectively.

Which one of the following compounds is not a protonic acid?

Write the conjugate base of the following: $HF, CH_{3}NH_{3}^{+}, H_{3}PO_{4}, HPO_{4}^{2-}$

For any diprotic acid of the type $H_2A$,the relationship between the first $(Ka_1)$ and second $(Ka_2)$ ionization constants is:

The strongest Bronsted base among the following anions is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module