Answer
Complete ion reaction:
$CH_3COO^-(aq) + H^+(aq) + Na^+(aq) + OH^-(aq) -- \gt Na^+(aq) + CH_3COO^-(aq) + H_2O(l)$
Net ionic reaction.
$H^+(aq) + OH^-(aq) -- \gt H_2O(l)$
Work Step by Step
1. Identify the salt produced by the mixture of this acid and this base.
$CH_3COOH$ gives $CH_3COO^-$ after reaction.
$NaOH$ gives $Na^+$ after reaction.
Therefore, the salt formed is: $NaCH_3COO$
2. Write the overall equation:
$Acid + Base --\gt Salt + Water$
$CH_3COOH(aq) + NaOH(aq) -- \gt NaCH_3COO(aq) + H_2O(l)$
* The equation is already balanced.
** According to table 3.1, $NaCH_3COOH$ is soluble in water.
3. Since we have a strong base in NaOH, for the aqueous compounds, write the separate ions:
$CH_3COO^-(aq) + H^+(aq) + Na^+(aq) + OH^-(aq) -- \gt Na^+(aq) + CH_3COO^-(aq) + H_2O(l)$
This is the complete ionic equation.
4. Remove the spectators ions:
$H^+(aq) + OH^-(aq) -- \gt H_2O(l)$
This is the net ionic equation.
