Question

Write the net ionic equation for the reaction of aqueous copper(II) chloride with aqueous sodium phosphate. Remember to include the states of matter.

Answer 1

To find the net ionic equation for the reaction of copper(II) chloride with sodium phosphate, one must write the overall chemical equation, the complete ionic equation, and cancel the spectator ions to obtain the net ionic equation:
`3Cu2+(aq) + 2PO43−(aq) → Cu3(PO4)2(s).`

Step-by-step explanation:

To write the net ionic equation for the reaction of aqueous copper(II) chloride with aqueous sodium phosphate, we need to follow a few steps. First, we write the overall chemical equation and then the full ionic equation. Finally, we can write the net ionic equation by canceling out the spectator ions. Here is how these steps might look for this specific reaction:

Overall chemical equation:

`3CuCl2(aq) + 2Na3PO4(aq) → Cu3(PO4)2(s) + 6NaCl(aq)`

Complete ionic equation:

`3Cu2+(aq) + 6Cl−(aq) + 6Na+(aq) + 2PO43−(aq) → Cu3(PO4)2(s) + 6Na+(aq) + 6Cl−(aq)`

Net ionic equation (after canceling out the spectator ions):

`3Cu2+(aq) + 2PO43−(aq) → Cu3(PO4)2(s)`

Answer 2

3 Cu²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Cu₃(PO₄)₂(s)

Step-by-step explanation:

Let's consider the molecular equation between aqueous copper(II) chloride and aqueous sodium phosphate.

3 CuCl₂(aq) + 2 Na₃PO₄(aq) ⇒ 6 NaCl(aq) + Cu₃(PO₄)₂(s)

The complete ionic equation includes all the ions and insoluble species.

3 Cu²⁺(aq) + 6 Cl⁻(aq) + 6 Na⁺(aq) + 2 PO₄³⁻(aq) ⇒ 6 Na⁺(aq) + 6 Cl⁻(aq) + Cu₃(PO₄)₂(s)

The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and insoluble species.

3 Cu²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Cu₃(PO₄)₂(s)