Sodium hydroxide on metal salts
Hand a chemist an unlabelled bottle of metal salt solution, and the first reagent they usually reach for is sodium hydroxide. Add it drop by drop and a coloured precipitate often appears; keep adding it in excess and that precipitate either stays put or quietly dissolves again. That two-stage observation — colour on dropwise addition, then behaviour in excess — is one of the most dependable identification tests in qualitative analysis, and it is the standard cation test prescribed for ICSE Class 10. This article expands on the core reactions for six common cations (Pb²⁺, Cu²⁺, Fe²⁺, Fe³⁺, Zn²⁺, Ca²⁺), explains why only two of the six precipitates redissolve in excess alkali, and works through the reasoning an examiner expects when two unknown salts must be told apart.
Sodium hydroxide precipitates a coloured metal hydroxide from a salt solution, and only the amphoteric hydroxides — lead and zinc — dissolve back into solution when NaOH is added in excess.
The core idea
Sodium hydroxide is a strong alkali; in solution it is fully dissociated into Na⁺ and OH⁻ ions. When it is added to a solution containing a metal cation Mⁿ⁺, the hydroxide ions combine directly with the cation:
where: is the metal cation present in the unknown salt, is its charge (valency), and is the resulting metal hydroxide, which is insoluble in water for every cation on this list and therefore appears as a precipitate. Written as a full molecular equation with a chloride salt, the same reaction reads:
Two things must be recorded for each cation: the colour of the precipitate that appears on dropwise addition, and whether that precipitate redissolves once NaOH is added in large excess. These are independent pieces of information, and a correct identification needs both.
| Cation | Precipitate (dropwise NaOH) | Excess NaOH |
|---|---|---|
| Pb²⁺ | White | Dissolves — colourless solution |
| Cu²⁺ | Light/pale blue | Insoluble — stays blue |
| Fe²⁺ | Dirty/dull green | Insoluble — stays dirty green |
| Fe³⁺ | Reddish-brown | Insoluble — stays reddish-brown |
| Zn²⁺ | White | Dissolves — colourless solution |
| Ca²⁺ | White | Insoluble (sparingly soluble) — stays white |
The rule to memorise: Pb(OH)₂ and Zn(OH)₂ are amphoteric, so excess strong alkali attacks them and carries them back into solution as soluble plumbate and zincate ions. The hydroxides of copper, iron(II), iron(III), and calcium have no such acidic character towards NaOH; once precipitated, they simply sit there no matter how much more alkali is poured in.
Why it works
The behaviour above follows from one chemical fact: amphoteric hydroxides react with both acids and strong bases, while ordinary basic hydroxides react only with acids.
- Every metal hydroxide on this list is basic — it neutralises acids. That is why all six form as insoluble precipitates when OH⁻ is first supplied by dropwise NaOH; there is no acid present to redissolve them at that stage.
- Lead hydroxide and zinc hydroxide are additionally acidic in character — they can also react with a strong base. Once dropwise NaOH is exhausted and a large excess is added, the surplus OH⁻ ions attack the freshly formed Pb(OH)₂ or Zn(OH)₂ itself, converting it into a soluble complex anion (plumbate, , or zincate, ). The solid disappears and the solution clears.
- Copper, iron(II), iron(III), and calcium hydroxides have no such acidic side to their character. However much extra NaOH is added, there is no reaction left for the excess reagent to drive — the precipitate simply remains suspended in the flask.
This is why the table must be read as colour-plus-behaviour pairs, never as two separate facts to memorise. A pale blue precipitate that stays blue in excess NaOH can only be Cu²⁺ — no other cation on this list gives that exact combination. A white precipitate that dissolves completely in excess NaOH narrows the field to Pb²⁺ or Zn²⁺ — these two look identical here and are told apart using a different reagent (ammonium hydroxide, where they behave differently). A white precipitate that does not dissolve in excess points to Ca²⁺.
Worked example
Question. A student is given a solution of zinc sulphate. NaOH solution is added dropwise, and then in large excess. Describe what is observed at each stage, write balanced equations, and state what would be different if the salt had instead been calcium chloride.
Step 1 — dropwise NaOH. Zn²⁺ ions in solution combine with OH⁻ ions supplied by the alkali, forming an insoluble white precipitate of zinc hydroxide:
Step 2 — excess NaOH. Zn(OH)₂ is amphoteric, so it reacts further with the surplus alkali and redissolves, forming soluble sodium zincate:
Step 3 — the observation. A white precipitate forms first, then dissolves as more NaOH is added, leaving a clear, colourless solution once excess has been reached. This dissolving-in-excess behaviour is a distinguishing test for Zn²⁺ (and, identically at this stage, for Pb²⁺).
Step 4 — the calcium comparison. If the salt were calcium chloride instead, dropwise NaOH would still give a white precipitate of Ca(OH)₂, but since Ca(OH)₂ is not amphoteric, the precipitate would remain white and undissolved even in large excess — the solution would stay cloudy, not clear. This one difference — clears versus stays cloudy in excess NaOH — is exactly how a student would distinguish a zinc salt from a calcium salt in an unknown-solution question.
Competitive stretch
The same logic distinguishes FeSO₄ from Fe₂(SO₄)₃. Dropwise NaOH gives dirty/dull green Fe(OH)₂ from Fe²⁺, and reddish-brown Fe(OH)₃ from Fe³⁺:
Neither hydroxide is amphoteric, so excess NaOH leaves both precipitates unchanged — here the diagnostic feature is colour alone, decided at the dropwise stage, not solubility in excess.
Common traps
- Assuming every white precipitate behaves the same way in excess NaOH — Pb(OH)₂ and Zn(OH)₂ dissolve; Ca(OH)₂ does not, even though all three start out white.
- Writing "Cu(OH)₂ dissolves in excess NaOH" — it does not. Among this list, only Pb²⁺ and Zn²⁺ hydroxides are amphoteric towards NaOH.
- Mixing up Fe²⁺ (dirty/dull green) with Fe³⁺ (reddish-brown) — a very common exam slip, since both are "iron."
- Forgetting to balance equations for salts of different valency — Fe₂(SO₄)₃ needs 6 mol NaOH per formula unit, not 2, because there are two Fe³⁺ ions to satisfy.
Ask the tutor: Still mixing up which hydroxides redissolve in excess NaOH? Ask your AI tutor for a quick memory trick to lock in Pb²⁺, Zn²⁺, Cu²⁺, Fe²⁺, Fe³⁺, and Ca²⁺ colours before your exam.
If Pb(OH)₂ and Zn(OH)₂ both dissolve in excess NaOH, how do I actually tell Pb²⁺ and Zn²⁺ apart
NaOH alone cannot separate them, since both give a white precipitate that dissolves in excess. The standard follow-up is ammonium hydroxide (NH₄OH): Zn(OH)₂ also dissolves in excess NH₄OH, but Pb(OH)₂ does not redissolve in excess NH₄OH. So the two-reagent combination — NaOH then, if needed, NH₄OH — is what a full qualitative-analysis scheme uses to pin down which of the two is present.
Why does adding more NaOH ever dissolve a precipitate — isn't more reagent supposed to make more precipitate, not less
It seems backwards at first, but only amphoteric hydroxides behave this way. A basic hydroxide like Fe(OH)₃ can only react with acids, so once it precipitates from OH⁻ it has nothing left to react with — more NaOH just sits unreacted. An amphoteric hydroxide like Zn(OH)₂ can also react with a strong base, so once dropwise NaOH is used up, any further OH⁻ actually attacks the precipitate itself and converts it into a soluble zincate ion, clearing the solution. It is not "more reagent, more product" — it is a second, different reaction taking over once the first one is complete.
Do I need to memorise all six colours and behaviours, or is there a pattern
There is a pattern worth learning instead of six isolated facts. First sort by colour: pale blue means Cu²⁺, dirty green means Fe²⁺, reddish-brown means Fe³⁺ — these three are each uniquely identified by colour alone. The remaining three (Pb²⁺, Zn²⁺, Ca²⁺) are all white, so colour alone cannot separate them; you then need the excess-NaOH test, where Pb²⁺ and Zn²⁺ dissolve (amphoteric) and Ca²⁺ does not. Learn it as "three colours, then one solubility split within the whites" rather than six unconnected entries.