Answer:
The left side equation is:
[tex]Ka = \frac{[CH_{3}COO^{-}] [H_{3}O^{+}]}{[HCH_{3}COO] [H_{2}O]}[/tex]
Explanation:
For the reaction of acetic acid HCH3CO2 with the water, The equilibrium constant equation is
[tex]HCH_{3}COO + H_{2}O \rightleftharpoons CH_{3}COO^{-} + H_{3}O[/tex]
The left side of this equilibrium constant equation will be written as shown below:
[tex]Ka = \frac{[CH_{3}COO^{-}] [H_{3}O^{+}]}{[HCH_{3}COO] [H_{2}O]}[/tex]
