To derive the net ionic equation for the reaction, we start by analyzing the given total ionic equation. The total ionic equation represents all of the ions present in a reaction:
[tex]\[
6 \text{Na}^+ + 2 \text{PO}_4^{3-} + 3 \text{Ca}^{2+} + 6 \text{Cl}^- \rightarrow 6 \text{Na}^+ + 6 \text{Cl}^- + \text{Ca}_3(\text{PO}_4)_2
\][/tex]
A net ionic equation excludes the spectator ions, which are ions that do not participate directly in the reaction. From the total ionic equation above, we can identify the spectator ions:
- [tex]\(\text{Na}^+\)[/tex] (Sodium ion): appears on both the reactant and product sides.
- [tex]\(\text{Cl}^-\)[/tex] (Chloride ion): appears on both the reactant and product sides.
These spectator ions can be canceled out from both sides of the equation. Removing the spectator ions, the simplified (net ionic) equation involves only the ions that participate in forming the precipitate:
[tex]\[ 2 \text{PO}_4^{3-} + 3 \text{Ca}^{2+} \rightarrow \text{Ca}_3(\text{PO}_4)_2 \][/tex]
Thus, the net ionic equation for the reaction is:
[tex]\[
2 \text{PO}_4^{3-} + 3 \text{Ca}^{2+} \rightarrow \text{Ca}_3(\text{PO}_4)_2
\][/tex]
This equation shows the formation of the solid calcium phosphate ([tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex]) from the combination of phosphate ions ([tex]\(\text{PO}_4^{3-}\)[/tex]) and calcium ions ([tex]\(\text{Ca}^{2+}\)[/tex]).
