# Question #2f795

Jan 9, 2017

Here's what I got.

#### Explanation:

Ther equilibrium reaction given to you describes the partial ionization of acetic acid, $\text{CH"_3"COOH}$, in aqueous solution

${\text{CH"_ 3"COOH"_ ((aq)) rightleftharpoons "CH"_ 3"COO"_ ((aq))^(-) + "H}}_{\left(a q\right)}^{+}$

We can talk about a partial ionization here because acetic acid is a weak acid; in other words, acetic acid does not ionize completely in aqueous solution to produce acetate anions, ${\text{CH"_ 3"COO}}^{-}$, and hydrogen cations, ${\text{H}}^{+}$, which you'll often see referred to as hydronium cations, ${\text{H"_3"O}}^{+}$.

Consequently, the equilibrium will lie to the left, meaning that most of the molecules of acetic acid present in solution will not donate their acidic proton, i.e. they will remain unionized.

By comparison, hydrochloric acid, $\text{HCl}$, is a strong acid because it ionizes almost completely in aqueous solution.

We don't write the ionization of hydrochloric acid as an equilibrium reaction because the equilibrium lies so far to the right that for all intended purposes, the reaction is considered to be irreversible.

So, the ionization of acetic acid is indeed reversible because at equilibrium, the rate at which acetic acid molecules give up their acidic proton to form acetate anions is equal to the rate at which acetate anions pick up protons to reform acetic acid molecules.