Write a balanced complete ionic equation for Na3PO4(aq) + CuCl2(aq) = Cu3(PO4)2(s) + 6NaCl(aq). Express your answer as a chemical equation. Identify all of the phases in your answer
The Correct Answer and Explanation is:
Balanced Complete Ionic Equation: 3Na+(aq)+PO43−(aq)+3Cu2+(aq)+6Cl−(aq)→Cu3(PO4)2(s)+6Na+(aq)+6Cl−(aq)3\text{Na}^+(aq) + \text{PO}_4^{3-}(aq) + 3\text{Cu}^{2+}(aq) + 6\text{Cl}^-(aq) \rightarrow \text{Cu}_3(\text{PO}_4)_2(s) + 6\text{Na}^+(aq) + 6\text{Cl}^-(aq)3Na+(aq)+PO43−(aq)+3Cu2+(aq)+6Cl−(aq)→Cu3(PO4)2(s)+6Na+(aq)+6Cl−(aq)
Textbook-Style Explanation
When sodium phosphate (Na3PO4\text{Na}_3\text{PO}_4Na3PO4) and copper(II) chloride (CuCl2\text{CuCl}_2CuCl2) are mixed in aqueous solution, a double displacement reaction occurs. This reaction involves the exchange of ions between two soluble ionic compounds, leading to the formation of a precipitate, copper(II) phosphate (Cu3(PO4)2\text{Cu}_3(\text{PO}_4)_2Cu3(PO4)2), and sodium chloride (NaCl\text{NaCl}NaCl), which remains dissolved in water.
First, we write the balanced molecular equation: 2Na3PO4(aq)+3CuCl2(aq)→Cu3(PO4)2(s)+6NaCl(aq)2\text{Na}_3\text{PO}_4(aq) + 3\text{CuCl}_2(aq) \rightarrow \text{Cu}_3(\text{PO}_4)_2(s) + 6\text{NaCl}(aq)2Na3PO4(aq)+3CuCl2(aq)→Cu3(PO4)2(s)+6NaCl(aq)
This shows that 2 moles of sodium phosphate react with 3 moles of copper(II) chloride to produce 1 mole of solid copper(II) phosphate and 6 moles of sodium chloride in solution.
Next, we express the reaction as a complete ionic equation by dissociating all strong electrolytes into their constituent ions: 6Na+(aq)+2PO43−(aq)+3Cu2+(aq)+6Cl−(aq)→Cu3(PO4)2(s)+6Na+(aq)+6Cl−(aq)6\text{Na}^+(aq) + 2\text{PO}_4^{3-}(aq) + 3\text{Cu}^{2+}(aq) + 6\text{Cl}^-(aq) \rightarrow \text{Cu}_3(\text{PO}_4)_2(s) + 6\text{Na}^+(aq) + 6\text{Cl}^-(aq)6Na+(aq)+2PO43−(aq)+3Cu2+(aq)+6Cl−(aq)→Cu3(PO4)2(s)+6Na+(aq)+6Cl−(aq)
Here, Na3PO4\text{Na}_3\text{PO}_4Na3PO4 and CuCl2\text{CuCl}_2CuCl2 fully dissociate in water. Copper(II) phosphate, however, is insoluble and precipitates out as a solid. Sodium ions and chloride ions remain in solution as spectator ions.
Finally, if we remove the spectator ions (species that appear unchanged on both sides), we get the net ionic equation: 3Cu2+(aq)+2PO43−(aq)→Cu3(PO4)2(s)3\text{Cu}^{2+}(aq) + 2\text{PO}_4^{3-}(aq) \rightarrow \text{Cu}_3(\text{PO}_4)_2(s)3Cu2+(aq)+2PO43−(aq)→Cu3(PO4)2(s)
This net ionic equation clearly shows the formation of solid copper(II) phosphate from its constituent ions in solution.
