Answer
Complete ionic equation:
$2H^+(aq) + S{O_4}^{2-}(aq) + Ba^{2+}(aq) + 2OH^-(aq) -- \gt Ba_2SO_4(s) + 2H_2O(l)$
Net ionic equation:
$2H^+(aq) + S{O_4}^{2-}(aq) + Ba^{2+}(aq) + 2OH^-(aq) -- \gt Ba_2SO_4(s) + 2H_2O(l)$
Work Step by Step
1. Identify the salt produced by the mixture of this acid and this base.
$H_2SO_4$ gives $SO{_4}^{2-}$ after reaction.
$Ba(OH)_2$ gives $Ba^{2+}$ after reaction.
Therefore, the salt formed is : $BaSO_4$
2. Write the overall equation:
$Acid + Base --\gt Salt + Water$
$H_2SO_4(aq) + Ba(OH)_2(aq) -- \gt BaSO_4(s) + H_2O(l)$
* The equation will be balanced if we add a "2" to the water molecules:
$H_2SO_4(aq) + Ba(OH)_2(aq) -- \gt BaSO_4(s) + 2H_2O(l)$
** According to table 3.1, $BaSO_4$ is not soluble in water.
3. On the totally dissociated/ionized compounds, write the separate ions:
$2H^+(aq) + S{O_4}^{2-}(aq) + Ba^{2+}(aq) + 2OH^-(aq) -- \gt Ba_2SO_4(s) + 2H_2O(l)$
This is the complete ionic equation.
4. Remove the spectators ions:
$2H^+(aq) + S{O_4}^{2-}(aq) + Ba^{2+}(aq) + 2OH^-(aq) -- \gt Ba_2SO_4(s) + 2H_2O(l)$
** Notice, there are no repeated ions, so there are no spectators.
This is the net ionic equation.
