# Net ionic equations for an aqueous solution

Given the reaction of $$\ce{Ni^{2+}}$$ with aqueous ammonia to give $$\ce{Ni(OH)_2}$$ Would this be the correct reaction?

$$\ce{Ni^{2+} (aq) + 2NH3 (aq) + 2H2O(l) -> Ni(OH)2 (s) + 2 NH4+ (aq)}$$

• You might just as well start with $\ce{Ni^2+}$ and $\ce{OH-}$. – Ivan Neretin Mar 14 '16 at 10:12
• Or use this: $$\ce{Ni^{2+}_{(aq)} + 2NH4^+OH^- -> Ni(OH)2_{(s)} + 2NH4^+}$$ – MaxW Mar 14 '16 at 14:08

Your answer is probably correct. There are some extra details one might wish to consider. If a short yes/no answer was enough, you must not read any further.

First, let us remind ourselves that in aqueous solutions metal ions coordinate with water molecules. Octahedral complexes are most common.

$$\ce{Ni^2+(aq) + 6 H2O(l) <=> [Ni(OH2)6]^{2+} (aq)}$$

Then $$\ce{NH4OH}$$, better described as a hydrate $$\ce{NH3.H2O}$$, is added. $$\ce{NH3 (aq) + H2O (l)<=>NH3.H2O (aq)<=>NH4+ (aq) +OH-(aq)}$$

Now either $$\ce{NH3}$$ or $$\ce{OH-}$$ steals a proton from $$\ce{[Ni(OH2)6]^{2+}}$$. Note that the resulting complexes are equivalent.

$$\ce{[Ni(OH2)6]^{2+}(aq)+NH3.H2O(aq) <=> [Ni(OH)(OH2)5]+(aq) + NH4+(aq) +H2O(l)}$$

$$\ce{[Ni(OH2)6]^{2+}(aq) + OH-(aq)<=> [Ni(OH)(OH2)5]+(aq) + H2O(l)}$$

If another proton is taken, we reach the required precipitate $$\ce{Ni(OH)2}$$. (Due to the equivalency mentioned, I will write only one option.)

$$\ce{[Ni(OH)(OH2)5]+ (aq) + OH-(aq)<=> [Ni(OH)2(OH2)4](s) + H2O (l) \tag 1}$$

Often, however, the $$\ce{Ni(OH)2}$$ precipitate is not visible. The whole point of this excercise was to give background to the explanation that follows. When $$\ce{NH3.H2O}$$ is in excess, the following processes occur. The precipitate could directly react, in which case we might end up with

$$\ce{[Ni(OH)2(OH2)4](s) + 6 NH3.H2O (aq)<=> [Ni(NH3)6](OH)2 (aq) + 10H2O(l)}$$

Generally, the preferred explanation is through Le Chatelier's principle. $$\ce{[Ni(OH2)6]^{2+}}$$ ions react via $$(2)$$

\begin{align} \ce{[Ni(OH2)6]^{2+} + 4 NH3.H2O (aq)} & \ce{<=> [Ni(NH3)4(OH2)2]^{2+}(aq) + 8H2O(l)}\\ \ce{[Ni(OH2)6]^{2+} + 6 NH3.H2O(aq) } & \ce{<=> [Ni(NH3)6]^{2+}(aq) + 10 H2O(l)}\\ \ce{[Ni(NH3)6]^2+(aq) +2 OH-(aq) } & \ce{<=> [Ni(NH3)6](OH)2(aq)} \end{align}

Since the $$\ce{[Ni(OH2)6]^{2+}}$$ ions are used up, the equilibrium of $$(1)$$ tilts to the left, thus the precipitate disappears.

Once more, this was simply to explain why showing $$\ce{Ni(OH)2}$$ as a precipitate might not correspond with reality when there is (excess) dissolved ammonia.

• I'm not sure where $(2)$ is. – A.K. Sep 29 '18 at 21:56