Question

Cu(OH)$_2$ is soluble in NH$_4$OH solution but not in NaOH solution. Why?

Hint:

Transition-metal hydroxides often dissolve in ligand bases like NH$_3$ by forming soluble complexes, but not necessarily in strong bases like NaOH unless they are amphoteric (e.g., Al(OH)$_3$).

Answer 1

Cu(OH)$_2$ dissolves in aqueous ammonia due to complex formation.
In excess NH$_3$, Cu$^{2+}$ forms the deep-blue ammine complex:
\[ \mathrm{Cu(OH)_2\;+\;4\,NH_3\;+\;2\,H_2O\;\longrightarrow\;[Cu(NH_3)_4(H_2O)_2]^{2+}\;+\;2\,OH^-} \]
This complex is soluble, so the precipitate dissolves.
In NaOH, no such soluble cuprate complex is formed (Cu(OH)$_2$ is not appreciably amphoteric),
hence Cu(OH)$_2$ remains as an insoluble blue precipitate.