# Question 76dc3

Apr 8, 2015

We say that reactions occur in aqueous solution when you produce one of the following:

• Precipitate
• Water
• Gas

No reaction takes place between those two compounds because the products will both be soluble in aqueous solution.

Both nickel (II) chloride and sodium bromide are soluble in aqueous solution, so you'd get

$N {i}_{\left(a q\right)}^{2 +} + 2 C {l}_{\left(a q\right)}^{-} + N {a}_{\left(a q\right)}^{+} + B {r}_{\left(a q\right)}^{-} \to \textcolor{red}{\text{N.R.}}$

In theory, this could be a double replacement reaction in which the cations and anions exchange partners; the balanced chemical equation looks like this

$N {i}_{\left(a q\right)}^{2 +} + 2 C {l}_{\left(a q\right)}^{-} + 2 N {a}_{\left(a q\right)}^{+} + 2 B {r}_{\left(a q\right)}^{-} \to 2 N a C {l}_{\left(a q\right)} + N i B {r}_{2 \left(a q\right)}$

However, sodium chloride and nickel bromide are both soluble as well, so the net ionic equation will look like this

$N {i}_{\left(a q\right)}^{2 +} + 2 C {l}_{\left(a q\right)}^{-} + 2 N {a}_{\left(a q\right)}^{+} + 2 B {r}_{\left(a q\right)}^{-} \to 2 N {a}_{\left(a q\right)}^{+} + 2 C {l}_{\left(a q\right)}^{-} + N {i}_{\left(a q\right)}^{2 +} + 2 B {r}_{\left(a q\right)}^{-}$

(added note: net ionic equations really only apply to reactions that actually occur, so there is not need to write a net ionic equation for this question)

Every ion is present both on the reactants' and on the products' side, so they act as spectator ions. As a conclusion, no reaction takes place because both products are soluble in aqueous solution.

Here is an example of a reaction which will produce a precipitate and a discussion of how to write the net ionic equation.

video from: Noel Pauller

Apr 8, 2015

There is no net ionic reaction.

You get a net ionic reaction if one of the products is a precipitate, a gas, or a covalent substance.

The possible reaction is a double displacement:

NiCl₂ + 2NaBr → NiBr₂ + 2NaCl

According to the solubility rules, every substance is soluble, so the $\textcolor{red}{\text{ionic equation}}$ is

$\text{Ni²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) +2Br⁻(aq) → Ni²⁺(aq) + 2Br⁻(aq) + 2Na⁺(aq) +2Cl⁻(aq)}$

To get the $\textcolor{red}{\text{net ionic equation}}$, you cancel all the ions that are common to each side.

cancel("Ni²⁺(aq)") + cancel("2Cl⁻(aq)") + cancel("2Na⁺(aq)") +cancel("2Br⁻(aq)") →#

$\cancel{\text{Ni²⁺(aq)") + cancel("2Br⁻(aq)") + cancel("2Na⁺(aq)") +cancel("2Cl⁻(aq)}}$

We have cancelled everything!

There is no net ionic equation.

Here is an example of a reaction which will produce a precipitate and a discussion of how to write the net ionic equation.

video from: Noel Pauller