Question

what is the net ionic equation for the reaction that is represented by the following total ionic equation?
6na⁺ + 2po₄³⁻ + 3ca²⁺ + 6cl⁻ ⟶ 6na⁺ + 6cl⁻ + ca₃(po₄)₂
2na₃po₄ + 3cacl₂ ⟶ 6nacl + ca₃(po₄)₂
2po₄³⁻ + 3ca²⁺ ⟶ ca₃(po₄)₂
2po₄³⁻ + 3ca²⁺ + 6cl⁻ ⟶ 6na⁺ + ca₃(po₄)₂
2p⁵⁺ + 8o²⁻ + 3ca²⁺ ⟶ ca₃(po₄)₂

Explanation:

Step1: Identify spectator ions

Spectator ions are ions that appear on both sides of the total ionic equation. Here, \( \mathrm{Na}^+ \) and \( \mathrm{Cl}^- \) are spectator ions as they are present on both reactant and product sides.

Step2: Remove spectator ions

To get the net ionic equation, we remove the spectator ions (\( 6\mathrm{Na}^+ \) and \( 6\mathrm{Cl}^- \)) from the total ionic equation \( 6\mathrm{Na}^+ + 2\mathrm{PO}_4^{3-} + 3\mathrm{Ca}^{2+} + 6\mathrm{Cl}^- \longrightarrow 6\mathrm{Na}^+ + 6\mathrm{Cl}^- + \mathrm{Ca}_3(\mathrm{PO}_4)_2 \). After removing them, we are left with \( 2\mathrm{PO}_4^{3-} + 3\mathrm{Ca}^{2+} \longrightarrow \mathrm{Ca}_3(\mathrm{PO}_4)_2 \).

Answer:

\( 2\mathrm{PO}_4^{3-} + 3\mathrm{Ca}^{2+} \longrightarrow \mathrm{Ca}_3(\mathrm{PO}_4)_2 \)