Answer
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) $
Work Step by Step
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.
