To find the net ionic equation for the reaction represented by the given 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 need to follow these steps:
1. Identify the spectator ions: Spectator ions are those ions that appear unchanged on both sides of the equation. In this case, the spectator ions are [tex]\( \text{Na}^+ \)[/tex] and [tex]\( \text{Cl}^- \)[/tex] because they appear on both the reactant and product sides of the equation in identical forms.
2. Remove the spectator ions from the total ionic equation:
[tex]\[ 6 \, \text{Na}^+ \text{(spectator ion)} + 2 \, \text{PO}_4^{3-} + 3 \, \text{Ca}^{2+} + 6 \, \text{Cl}^- \text{(spectator ion)} \longrightarrow 6 \, \text{Na}^+ \text{(spectator ion)} + 6 \, \text{Cl}^- \text{(spectator ion)} + \text{Ca}_3(\text{PO}_4)_2 \][/tex]
3. Simplify the equation by removing the spectator ions:
[tex]\[ 2 \, \text{PO}_4^{3-} + 3 \, \text{Ca}^{2+} \longrightarrow \text{Ca}_3(\text{PO}_4)_2 \][/tex]
Therefore, the net ionic equation for the reaction is:
[tex]\[ 2 \, \text{PO}_4^{3-} + 3 \, \text{Ca}^{2+} \longrightarrow \text{Ca}_3(\text{PO}_4)_2 \][/tex]
This equation shows the formation of calcium phosphate from phosphate ions ([tex]\( \text{PO}_4^{3-} \)[/tex]) and calcium ions ([tex]\( \text{Ca}^{2+} \)[/tex]).
