First, check the standard equation:
CH3COOH+NaOH→H2O+CH3COONa
The ionic equation for the above:
CH3COOH(aq)+Na+(aq)+OHâ’(aq)→
CH3COOâ’(aq)+Na+(aq)+H2O(l)
Therefore, the net ionic equation is (remove the Na+ as it is present as both a reactant and a product):
CH3COOH(aq)+OHâ’(aq)→
CH3COOâ’(aq)+H2O(l)
(note the charge on both sides has to be the same, in this case 1.
