Answer
Balanced equation:
$2HCH_3COO(aq) + Ba(OH)_2(aq) -> 2H_2O(l) + {Ba(CH_3COO)_2}(aq)$
Net ionic equation:
$2HCH_3COO(aq) + 2OH^-(aq) -> 2H_2O(l) + 2{CH_3COO}^-(aq)$
Work Step by Step
1. Write the equation:
$HCH_3COO(aq) + Ba(OH)_2(aq) -> H_2O(l) + {Ba(CH_3COO)_2}(aq)$
2. Balance the equation:
$2HCH_3COO(aq) + Ba(OH)_2(aq) -> 2H_2O(l) + {Ba(CH_3COO)_2}(aq)$
3. Write the net ionic equation:
$2HCH_3COO(aq) + Ba^{2+}(aq) + 2OH^-(aq) -> 2H_2O(l) + Ba^{2+}(aq) + 2{CH_3COO}^-(aq)$
4. Remove the spectator ions:
$2HCH_3COO(aq) + 2OH^-(aq) -> 2H_2O(l) + 2{CH_3COO}^-(aq)$
