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

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

  • 1
    $\begingroup$ You might just as well start with $\ce{Ni^2+}$ and $\ce{OH-}$. $\endgroup$ Commented Mar 14, 2016 at 10:12
  • $\begingroup$ Or use this: $$\ce{Ni^{2+}_{(aq)} + 2NH4^+OH^- -> Ni(OH)2_{(s)} + 2NH4^+}$$ $\endgroup$
    – MaxW
    Commented Mar 14, 2016 at 14:08

1 Answer 1


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.

  • $\begingroup$ I'm not sure where $(2)$ is. $\endgroup$
    – A.K.
    Commented Sep 29, 2018 at 21:56

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