Question

part 2 of 2
write the net ionic equation for the reaction of aqueous nickel(ii) chloride with aqueous sodium phosphate. remember to include the states of matter.

Explanation:

Step1: Identify reactants, products, states

Reactants: $\text{NiCl}_2(aq)$, $\text{Na}_3\text{PO}_4(aq)$
Products: $\text{Ni}_3(\text{PO}_4)_2(s)$, $\text{NaCl}(aq)$

Step2: Write full balanced equation

$$3\text{NiCl}_2(aq) + 2\text{Na}_3\text{PO}_4(aq) ightarrow \text{Ni}_3(\text{PO}_4)_2(s) + 6\text{NaCl}(aq)$$

Step3: Split into ionic form

$$3\text{Ni}^{2+}(aq) + 6\text{Cl}^-(aq) + 6\text{Na}^+(aq) + 2\text{PO}_4^{3-}(aq) ightarrow \text{Ni}_3(\text{PO}_4)_2(s) + 6\text{Na}^+(aq) + 6\text{Cl}^-(aq)$$

Step4: Cancel spectator ions

Remove $\text{Na}^+(aq)$ and $\text{Cl}^-(aq)$ from both sides.

Answer:

$$3\text{Ni}^{2+}(aq) + 2\text{PO}_4^{3-}(aq) ightarrow \text{Ni}_3(\text{PO}_4)_2(s)$$