When aqueous solutions of nickel(II) bromide and sodium phosphate are combined, solid nickel(II) phosphate and a solution of sodium bromide are formed. The net ionic equation for this reaction is:

Answer 1

Since there are aqueous solutions of Nickel (II) Bromide and sodium phosphate combined, the state we need to put in the chemical equation is (aq) for each. Solid nickel (II) phosphate is formed, meaning it is called a precipitate.

First, write the chemical reaction for this situation:

NiBr₂(aq) + Na₃PO₄(aq)  -------> Ni₃(PO₄)₂ (s) + NaBr (aq)

This equation is not balanced as there are not the same quantity of atoms in both sides of equation. Therefore, it must be balanced before writing the ionic equation.

Balanced chemical equation:

3NiBr₂(aq) + 2Na₃PO₄(aq)  -------> Ni₃(PO₄)₂ (s) + 6NaBr (aq)

Now to write the ionic equation:

In these aqueous species, they typically form ions in solution except when forming a precipitate.

Br^- is soluble except with certain metals like silver, lead, and mercury.

Na⁺ is always a soluble ion, phosphate is typically insoluble unless the molecule also had sodium on in, which it does in sodium phosphate.

Equation: 3Ni²⁺(aq) + 6Br^-(aq) + 6Na⁺(aq) + 2PO₄^3-(aq)  -------> Ni₃(PO₄)₂ (s) + 6Na⁺(aq) + 6Br^- (aq)

Net ionic equation means that same ions located in both left and right of chemical equation get cancelled out. The common ions are the 6Br^-(aq) and 6Na⁺(aq). So cancel like terms and get the net ionic equation:

3Ni²⁺(aq) + 2PO₄^3-(aq)  -------> Ni₃(PO₄)₂ (s)