# Write the complete molecular reaction, the complete (full) ionic equation, and lastly, the net...

## Question:

Write the complete molecular reaction, the complete (full) ionic equation, and lastly, the net ionic equation for the following:

{eq}\displaystyle \rm BaCl_2+NaOH \to {/eq}

## Double Displacement Reaction:

As the name suggests, the reaction between two reactants displaces each other as the reactivity series from new products is called a double displacement reaction.

The given reaction:

The reaction of barium chloride with sodium hydroxide where barium replaces sodium to form barium hydroxide and sodium replaces barium to form sodium chloride. Thus, showing a double displacement reaction.

In the reaction, precipitate formation occurred, also called precipitation reaction, as barium hydroxide is formed as a precipitate.

So, a balanced molecule reaction occurred as follows:

{eq}\rm BaCl_2 (aq) + 2NaOH(aq) \rightarrow Ba(OH)_2 (s) + 2NaCl(aq) {/eq}

Hence, reactions and products are dissociated from ions to form an ionic equation.

Thus, ionic equation is:

{eq}\rm Ba^{+2}(aq) + 2Cl^- (aq) + 2Na^+(aq) + 2OH^-(aq) \rightarrow Ba(OH)_2 (s) + 2Na^+(aq) + 2Cl^- (aq) {/eq}

The precipitate formation cannot dissociate to form ions.

To form the net ionic equation, cancel out that ion on both sides, which are called spectator ions.

Here, sodium and chloride ions are spectator ions.

Net ionic equation:

{eq}\rm Ba^{+2}(aq) + 2OH^-(aq) \rightarrow Ba(OH)_2 (s) {/eq}

Thus, the molecular reaction, ionic reaction, and net ionic reaction occurred, as shown above.