Answer:
The equilibrium chemical equation of dissociation of an acetic acid is given as:
[tex]CH_3COOH(aq)+H_2O(l)\rightleftharpoons CH_3COO^-(aq)+H_3O^+(aq)[/tex]
Explanation:
The value of dissociation constant of an acetic acid = [tex]K_a=1.80\times 10^{-5}[/tex]
The equilibrium chemical equation is given as:
[tex]CH_3COOH(aq)+H_2O(l)\rightleftharpoons CH_3COO^-(aq)+H_3O^+(aq)[/tex]
The expression of an equilibrium constant will be given as:
[tex]K_c=\frac{[CH_3COO^-][H_3O^+]}{[CH-3COOH][H_2O]}[/tex]
[tex]K_a=K_c\times [H_2O]=\frac{[CH_3COO^-][H_3O^+]}{[CH-3COOH]}[/tex]
As we know that, water is a pure substance ,so:
[tex][H_2O]=1[/tex]
[tex]K_a=\frac{[CH_3COO^-][H_3O^+]}{[CH-3COOH]}[/tex]
