To find the net ionic equation for the given reaction, we first need to understand what a net ionic equation represents. The net ionic equation includes only the ions that directly participate in the chemical reaction, leaving out the spectator ions that do not change during the reaction.
Given the total ionic equation:
[tex]\[ 6 \text{Na}^{+} + 2 \text{PO}_4^{3-} + 3 \text{Ca}^{2+} + 6 \text{Cl}^{-} \longrightarrow 6 \text{Na}^{+} + 6 \text{Cl}^{-} + \text{Ca}_3(\text{PO}_4)_2 \][/tex]
We can observe that both [tex]\( \text{Na}^+ \)[/tex] and [tex]\( \text{Cl}^- \)[/tex] ions appear on both the reactant and the product sides of the equation. These ions do not participate in the formation of the precipitate, [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex], and thus can be classified as spectator ions. Removing these spectator ions gives us the net ionic equation.
So, let's remove the spectator ions:
1. Identify the ions involved in the formation of the precipitate:
[tex]\[ 2 \text{PO}_4^{3-} + 3 \text{Ca}^{2+} \longrightarrow \text{Ca}_3(\text{PO}_4)_2 \][/tex]
2. Write the net ionic equation:
[tex]\[ 2 \text{PO}_4^{3-} + 3 \text{Ca}^{2+} \longrightarrow \text{Ca}_3(\text{PO}_4)_2 \][/tex]
Therefore, the net ionic equation is:
[tex]\[ 2 \text{PO}_4^{3-} + 3 \text{Ca}^{2+} \longrightarrow \text{Ca}_3(\text{PO}_4)_2 \][/tex]
This is the balanced net ionic equation for the reaction, and it shows the ions that directly form the precipitate, eliminating the spectator ions from the total ionic equation.
