Ammonium hydroxide on metal salts
Drip aqueous ammonia into a blue copper sulphate solution and a pale precipitate curdles out of nowhere. Keep adding, and the precipitate does something stranger still — it vanishes, leaving behind a solution of a deep, inky blue. That single tube captures the whole idea behind ammonium hydroxide (NH₄OH) as an analytical reagent: like sodium hydroxide it precipitates metal hydroxides from solution, but in excess it can behave completely differently, redissolving some precipitates through complex-ion formation while leaving others untouched. Learning exactly which metal ions dissolve back in — and, crucially, why — turns NH₄OH into a second, complementary tool alongside NaOH for identifying an unknown metal salt with confidence.
Ammonium hydroxide precipitates metal hydroxides from solution, and in excess it redissolves only those metals — chiefly copper and zinc — that can form a soluble ammine complex ion with excess NH₃, a chemically distinct reason from the amphoterism seen with excess NaOH.
The core idea
Ammonium hydroxide is a weak alkali: unlike NaOH, it does not ionise completely in water, so only a fraction of the dissolved gas is present as free hydroxide ions at any moment.
where: the double arrow signals a partial, reversible ionisation, so a solution of ammonium hydroxide always contains a mix of intact NH₄OH molecules, NH₄⁺ ions, and OH⁻ ions. Even though it is weak, the OH⁻ ions it does release are enough to precipitate most metal cations as insoluble hydroxides, exactly as concentrated NaOH does:
where: is the metal cation in solution, is its charge, and the downward arrow marks the hydroxide as a solid precipitate settling out of solution. The interesting chemistry begins once you keep adding NH₄OH past the point where the precipitate first appears. With NaOH, some of these hydroxides (like lead and zinc) dissolve back in because they are amphoteric — they react with the extra OH⁻ ions as an acid would. With NH₄OH there is no large excess of OH⁻ to exploit in the same way; instead, a different mechanism operates for a different set of ions. Copper and zinc hydroxides dissolve in excess NH₄OH because the metal ion latches onto four ammonia molecules to form a soluble ammine complex ion:
where: the square brackets denote a complex ion, in which four neutral NH₃ molecules bond directly to the central metal ion, and the whole assembly carries the metal's original charge, keeping it soluble in water. Lead, iron(II), and iron(III) hydroxides have no such soluble ammine complex available to them, so once precipitated by NH₄OH they simply stay as solids, however much more reagent you add. Calcium hydroxide is only sparingly soluble to begin with, so dilute NH₄OH typically gives no visible precipitate with Ca²⁺ at all — a useful negative result rather than an oversight.
| Cation | Dropwise NH₄OH | Excess NH₄OH |
|---|---|---|
| Pb²⁺ | White precipitate | Stays white, insoluble |
| Cu²⁺ | Pale/light blue precipitate | Dissolves — deep blue solution |
| Fe²⁺ | Dirty green precipitate | Stays dirty green, insoluble |
| Fe³⁺ | Reddish-brown precipitate | Stays reddish-brown, insoluble |
| Zn²⁺ | White precipitate | Dissolves — colourless solution |
| Ca²⁺ | No precipitate (or very faint) | — |
Why it works
This topic has no numerical formula to apply — the reasoning is entirely about which metal ions can and cannot form a stable, soluble ammine complex, and a student's job is to reproduce that reasoning correctly rather than compute anything.
Step 1. Start from what dropwise NH₄OH always does. Whatever the metal, the first few drops of NH₄OH supply enough OH⁻ ions to exceed the solubility limit of the metal hydroxide, so a precipitate forms exactly as it would with NaOH. The colour of this precipitate is diagnostic on its own and is identical whether NaOH or NH₄OH was used, because it depends only on the identity of M(OH)ₙ, not on which alkali produced it.
Step 2. Ask whether the metal ion can coordinate ammonia molecules. Copper(II) and zinc(II) ions have empty orbitals available that let four neutral NH₃ molecules bond directly to the metal centre, in place of the water molecules normally surrounding it in solution. This bonding is strong enough that, once enough NH₃ is present (from the excess NH₄OH), the equilibrium shifts entirely toward the complex ion, pulling the solid hydroxide back into solution. Lead, iron(II), and iron(III) do not form comparably stable ammine complexes under these conditions, so no matter how much NH₄OH is added, their hydroxides remain as solids.
Step 3. Do not confuse this with amphoterism. With excess NaOH, Pb(OH)₂ and Zn(OH)₂ redissolve because they are amphoteric — they act as a weak acid toward the flood of extra OH⁻ ions, forming plumbate and zincate ions. With excess NH₄OH, the mechanism is entirely different: it is complex-ion (ammine) formation with neutral NH₃ molecules, not an acid-base reaction with OH⁻. This is precisely why lead behaves differently in the two reagents — Pb²⁺ is amphoteric (redissolves in excess NaOH) but does not form a soluble ammine complex (stays precipitated in excess NH₄OH), while Zn²⁺ does both, redissolving in excess of either reagent for two unrelated chemical reasons.
Step 4. Use the contrast diagnostically. Because Pb²⁺ and Zn²⁺ both give a white precipitate with dropwise NaOH or NH₄OH, and both redissolve in excess NaOH, a single reagent cannot tell them apart. Testing a fresh sample with excess NH₄OH resolves the ambiguity instantly, since only the zinc precipitate disappears.
Worked example
Question: Dilute ammonium hydroxide is added dropwise, then in excess, to a solution of copper sulphate. Describe what is observed and write balanced equations for each stage.
Step 1 — Dropwise addition. The first drops of NH₄OH precipitate copper hydroxide as a pale blue solid:
Step 2 — Continued excess. As more NH₄OH is added past the precipitation point, the pale blue solid dissolves, because copper forms a soluble ammine complex with the surplus ammonia:
Step 3 — Observation. The solution passes from colourless, to a pale blue precipitate suspended in solution, to a clear, deep (inky) blue solution with no remaining solid.
The final answer: the pale blue precipitate of Cu(OH)₂ dissolves in excess NH₄OH to give a deep blue solution of the tetraamminecopper(II) complex — a highly reliable confirmatory test for Cu²⁺.
Competitive stretch
Question: Two fresh white precipitates are known to be Pb(OH)₂ and Zn(OH)₂, but the labels have been lost. How does excess NH₄OH identify which is which?
Step 1. Add excess NH₄OH to both test tubes.
Step 2. The tube containing Zn(OH)₂ clears to a colourless solution, since zinc forms the soluble complex.
Final answer: the tube whose white precipitate dissolves is Zn²⁺; the tube where the white precipitate persists unchanged is Pb²⁺, since lead forms no soluble ammine complex.
Common traps
- Assuming NH₄OH and NaOH give identical results in excess — they do not: Pb(OH)₂ dissolves in excess NaOH but stays undissolved in excess NH₄OH.
- Calling the dissolving of Cu(OH)₂ or Zn(OH)₂ in excess NH₄OH "amphoteric behaviour" — the correct term is complex-ion (ammine complex) formation, a different phenomenon from the acid-base amphoterism seen with NaOH.
- Describing the copper test as just "precipitate dissolves" without naming the diagnostic deep blue colour of the resulting solution — the colour change is the actual observation examiners look for.
- Expecting a clear precipitate with Ca²⁺ and NH₄OH — calcium hydroxide is only sparingly soluble, so little or no precipitate forms, and that absence is itself a meaningful (negative) result.
Ask the tutor: Still mixing up when a precipitate redissolves because it is amphoteric versus because it forms a complex ion? Ask your AI tutor to walk through the full NaOH vs NH₄OH comparison table for Pb²⁺, Cu²⁺, Fe²⁺, Fe³⁺, Zn²⁺ and Ca²⁺, and quiz you until it sticks.
Why does zinc redissolve in excess of both NaOH and NH4OH, but for different reasons
Zinc hydroxide is genuinely amphoteric, so in excess NaOH it reacts with the surplus hydroxide ions to form a soluble zincate ion — an acid-base reaction. In excess NH₄OH, however, there is no large surplus of free OH⁻ to exploit in the same way; instead zinc forms a soluble ammine complex, , by directly bonding four ammonia molecules to the zinc ion. Both reagents dissolve the precipitate, but through unrelated mechanisms, so you should never say "zinc hydroxide is soluble in excess NH₄OH because it is amphoteric" — amphoterism is specifically the NaOH story.
If Pb(OH)2 does not dissolve in excess NH4OH, is lead never detected using ammonium hydroxide
Lead is still detected, just via a different observation. Dropwise NH₄OH gives Pb²⁺ the same white precipitate as most other cations, and the diagnostic step is that this precipitate stays white and solid however much more NH₄OH you add — unlike copper or zinc. So the absence of dissolution, paired with the white colour, is itself the confirmatory evidence for Pb²⁺, especially useful once NaOH excess has already ruled out easy identification (since Pb(OH)₂ dissolves in excess NaOH too).
Do I need to memorise the ammine complex formula exactly, brackets and all
Yes, for full marks — examiners expect the complex ion written with the metal centre, four NH₃ ligands, and the correct overall charge or hydroxide counter-ions, for example . Writing simply "the precipitate dissolves in ammonia" without the balanced equation and complex-ion formula loses marks in ICSE analytical chemistry answers, since the equation is what demonstrates you understand why it dissolves rather than just describing what you saw.