Question

Fill in the left side of this equilibrium constant equation for the reaction of acetic acid HCH3CO2 with water.

Answer 1

Answer:

The left side equation is:

$Ka = \frac{[CH_{3}COO^{-}] [H_{3}O^{+}]}{[HCH_{3}COO] [H_{2}O]}$

Explanation:

For the reaction of acetic acid HCH3CO2 with the water, The equilibrium constant equation is

$HCH_{3}COO + H_{2}O \rightleftharpoons CH_{3}COO^{-} + H_{3}O$

The left side of this equilibrium constant equation will be written as shown below:

$Ka = \frac{[CH_{3}COO^{-}] [H_{3}O^{+}]}{[HCH_{3}COO] [H_{2}O]}$