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

Acid and base used to form this ion, respectively, : $CH_3COOH$ and $NaOH$, Complete ionic equation: $CH_3COOH(aq) + Na^+(aq) + OH^-(aq) -- \gt Na^+(aq) + CH_3COO^-(aq) + H_2O(l)$ Net ionic equation: $CH_3COOH(aq) + OH^-(aq) -- \gt + CH_3COO^-(aq) + H_2O(l)$
1. Identify the ions of the salt: $(NaCH_3COO)$: $Na^+$ and $CH_3COO^-$: To the cation, add a hydroxide ion: $NaOH$; this is the base. To the anion, add a hydrogen ion: $CH_3COOH$; this is the acid. 2. Now, write the balanced overall equation between them, which is: $Acid + Base -- \gt Salt + Water$ We already have the salt, so: $CH_3COOH(aq) + NaOH(aq) -- \gt NaCH_3COO(aq) + H_2O(l)$ 3. Write the complete ionic equation. - For the completely dissociated/ionized compounds, separate them by their ions: $CH_3COOH(aq) + Na^+(aq) + OH^-(aq) -- \gt Na^+(aq) + CH_3COO^-(aq) + H_2O(l)$ * NaOH is a strong base, and $NaCH_3COO$ is soluble according to the table 3.1, because it is a compound with the acetate ion. ** $CH_3COOH$ is a weak acid, so it is not completely dissociated. 4. Remove the repeated ions: $CH_3COOH(aq) + OH^-(aq) -- \gt + CH_3COO^-(aq) + H_2O(l)$ This is the net ionic equation.