## Chemistry: The Molecular Science (5th Edition)

$CH_3COOH(aq) + H_2O(l) \lt -- \gt CH_3COO^-(aq) + H_3O^+(aq)$ $K_a = \frac{[H_3O^+][CH_3COO^-]}{[CH_3COOH]}$
1. Write the ionization chemical equation: - Since $CH_3COOH$ is an acid, write the reaction where it donates a proton to a water molecule: $CH_3COOH(aq) + H_2O(l) \lt -- \gt CH_3COO^-(aq) + H_3O^+(aq)$ 2. Now, write the $K_a$ expression: - The $K_a$ expression is the concentrations of the products divided by the concentration of the reactants: $K_a = \frac{[Products]}{[Reactants]}$ $K_a = \frac{[H_3O^+][CH_3COO^-]}{[CH_3COOH]}$ *** We don't consider the $[H_2O]$, because it is the solvent of the solution.