ChemistryICSE Class 10 · Analytical Chemistry

Action of Ammonium Hydroxide on Salt Solutions of Fe, Cu, Zn and Pb (ICSE Class 10)

Updated 17 October 2026 · Fact-checked

Ammonium hydroxide added drop by drop gives hydroxide precipitates: dirty green (Fe²⁺), reddish-brown (Fe³⁺), pale blue (Cu²⁺), white (Zn²⁺) and chalky white (Pb²⁺). In excess, the pale blue copper precipitate dissolves to an inky blue solution and the white zinc precipitate dissolves to a colourless solution, while the iron and lead precipitates do not dissolve. Hence NH₄OH distinguishes zinc from lead, which NaOH cannot.

Understand Ammonium Hydroxide on Salt Solutions

1. Why a precipitate forms. Aqueous ammonia, written NH₄OH, is a weak alkali that supplies a small number of hydroxide ions. Metal ions in the salt solution combine with them to give the insoluble metal hydroxide, a coloured or white precipitate. General reaction: metal salt + ammonium hydroxide gives metal hydroxide (precipitate) + ammonium salt.

2. Colour of the salt and its solution first. Copper(II) salts are blue, iron(II) salts pale green, iron(III) salts yellowish-brown; zinc and lead salts are white solids that form colourless solutions.

3. Drop by drop, then in excess.

Fe²⁺ (iron(II) sulphate): dirty green precipitate of iron(II) hydroxide, insoluble in excess.

$F e S O_{4} + 2 N H_{4} O H \to F e (O H)_{2} + (N H_{4})_{2} S O_{4}$

Fe³⁺ (iron(III) chloride): reddish-brown precipitate of iron(III) hydroxide, insoluble in excess.

$F e C l_{3} + 3 N H_{4} O H \to F e (O H)_{3} + 3 N H_{4} C l$

Cu²⁺ (copper(II) sulphate): pale blue precipitate of copper(II) hydroxide, which dissolves in excess to give a deep (inky) blue solution of tetraamminecopper(II) sulphate.

$C u S O_{4} + 2 N H_{4} O H \to C u (O H)_{2} + (N H_{4})_{2} S O_{4}$
$C u (O H)_{2} + (N H_{4})_{2} S O_{4} + 2 N H_{4} O H \to [C u (N H_{3})_{4}] S O_{4} + 4 H_{2} O$

Zn²⁺ (zinc sulphate): white gelatinous precipitate of zinc hydroxide, which dissolves in excess to give a colourless solution.

$Z n S O_{4} + 2 N H_{4} O H \to Z n (O H)_{2} + (N H_{4})_{2} S O_{4}$
$Z n (O H)_{2} + (N H_{4})_{2} S O_{4} + 2 N H_{4} O H \to [Z n (N H_{3})_{4}] S O_{4} + 4 H_{2} O$

Pb²⁺ (lead nitrate): chalky white precipitate of lead hydroxide, insoluble in excess.

$P b (N O_{3})_{2} + 2 N H_{4} O H \to P b (O H)_{2} + 2 N H_{4} N O_{3}$

4. Key conclusions.

Excess NH₄OH dissolves the precipitates of copper and zinc (a complex ion forms) but not those of Fe²⁺, Fe³⁺ and Pb²⁺.
Copper gives a special inky blue solution; this is its identifying test with ammonium hydroxide.
Zinc and lead both give white precipitates with NaOH and both dissolve in excess NaOH, but with NH₄OH zinc dissolves and lead does not: NH₄OH distinguishes Zn²⁺ from Pb²⁺.
The ions with the same colours in the table below can only be separated by this solubility.

5. Summary table (NH₄OH).

Ion	Few drops	Excess
Fe²⁺	dirty green precipitate	no change
Fe³⁺	reddish-brown precipitate	no change
Cu²⁺	pale blue precipitate	dissolves, inky blue solution
Zn²⁺	white precipitate	dissolves, colourless solution
Pb²⁺	chalky white precipitate	no change

Ammonium Hydroxide on Salt Solutions: formulas and key facts

Iron(II) and iron(III): $F e S O_{4} + 2 N H_{4} O H \to F e (O H)_{2} + (N H_{4})_{2} S O_{4}$ ; $F e C l_{3} + 3 N H_{4} O H \to F e (O H)_{3} + 3 N H_{4} C l$
Copper(II): precipitate and complex: $C u S O_{4} + 2 N H_{4} O H \to C u (O H)_{2} + (N H_{4})_{2} S O_{4}$ ; $C u (O H)_{2} + (N H_{4})_{2} S O_{4} + 2 N H_{4} O H \to [C u (N H_{3})_{4}] S O_{4} + 4 H_{2} O$
Zinc: precipitate and complex: $Z n S O_{4} + 2 N H_{4} O H \to Z n (O H)_{2} + (N H_{4})_{2} S O_{4}$ ; $Z n (O H)_{2} + (N H_{4})_{2} S O_{4} + 2 N H_{4} O H \to [Z n (N H_{3})_{4}] S O_{4} + 4 H_{2} O$
Lead(II): $P b (N O_{3})_{2} + 2 N H_{4} O H \to P b (O H)_{2} + 2 N H_{4} N O_{3}$
Observations: Fe²⁺ dirty green, insoluble in excess
Fe³⁺ reddish-brown, insoluble in excess
Cu²⁺ pale blue, dissolves in excess to inky blue
Zn²⁺ white, dissolves in excess to colourless
Pb²⁺ chalky white, insoluble in excess

Fe²⁺	dirty green, insoluble in excess
Fe³⁺	reddish-brown, insoluble in excess
Cu²⁺	pale blue, dissolves in excess to inky blue
Zn²⁺	white, dissolves in excess to colourless
Pb²⁺	chalky white, insoluble in excess

How to solve Ammonium Hydroxide on Salt Solutions questions

Identify an unknown salt solution by two observations.

1Note the colour of the solution: blue means Cu²⁺, pale green means Fe²⁺, yellowish-brown means Fe³⁺, colourless means Zn²⁺ or Pb²⁺.
2Add NH₄OH drop by drop and note the colour of the precipitate.
3Add excess and note whether the precipitate dissolves and what colour the solution is.
4Link the observation to the metal: dissolves to inky blue = Cu; dissolves to colourless = Zn; insoluble white = Pb.
5Write the equation with the precipitate and the ammonium salt; for excess give the complex salt.

Ammonium Hydroxide on Salt Solutions in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Spotting a blue solution in a school lab

Suppose a lab assistant has two colourless bottles labelled zinc sulphate and lead nitrate, and a student adds ammonium hydroxide to each, a little at a time and then in excess. One bottle gives a white precipitate that vanishes in excess; the other gives a white precipitate that stays.

Which bottle is which? Why would sodium hydroxide not help?

Try it first, then show the solutionSolution

The precipitate that dissolves in excess ammonium hydroxide is zinc hydroxide, so that bottle is zinc sulphate.
The precipitate that stays is lead hydroxide, so the other bottle is lead nitrate.
Sodium hydroxide fails because both white precipitates dissolve in excess NaOH.

Answer: Soluble in excess NH₄OH: zinc sulphate; insoluble: lead nitrate. NaOH cannot separate them.

Memory hook: Zinc dissolves in NH₄OH; lead stays.

Tricks to remember Ammonium Hydroxide on Salt Solutions

Short memory aids. Say each one aloud once, then use it on a question the same day.

Two dissolvers: Cu and Zn
In excess NH₄OH only copper and zinc precipitates dissolve (they form ammine complexes). Fe and Pb stay.
Use it when: Reading the excess step.
Inky blue = copper
A deep blue solution after excess ammonia says copper every time.
Use it when: Copper identification.
NaOH twins, NH₄OH splits
Zinc and lead behave the same with NaOH (white precipitate, dissolves) and differ with NH₄OH.
Use it when: Choosing the reagent to tell Zn from Pb.

Ammonium Hydroxide on Salt Solutions: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Observation · 2 marks · Short answerBoard level

State the observation when ammonium hydroxide is added drop by drop to copper(II) sulphate solution and then in excess.

Try it first, then show the solutionSolution

A pale blue precipitate forms.
In excess it dissolves to give a deep (inky) blue solution.

Answer: Pale blue precipitate; dissolves in excess to an inky blue solution.

Example 2: Identify the metal · 3 marks · Short answerBoard level

Salt A forms a colourless solution. With NH₄OH a white precipitate forms that does not dissolve in excess. Name the cation and the precipitate, and write an equation using the nitrate.

Try it first, then show the solutionSolution

White precipitate insoluble in excess points to lead.
The precipitate is lead(II) hydroxide, Pb(OH)₂.
$P b (N O_{3})_{2} + 2 N H_{4} O H \to P b (O H)_{2} + 2 N H_{4} N O_{3}$ .

Answer: Pb²⁺; Pb(OH)₂; equation as above.

Example 3: Distinguishing Zn and Pb · 3 marks · Short answerHarder board level

Name a reagent that distinguishes zinc sulphate solution from lead nitrate solution and give the observations. Why can sodium hydroxide not be used?

Try it first, then show the solutionSolution

Use ammonium hydroxide: with zinc sulphate the white precipitate dissolves in excess to a colourless solution; with lead nitrate it does not dissolve.
Sodium hydroxide gives a white precipitate that dissolves in excess with both, so there is no difference.

Answer: NH₄OH; zinc: precipitate dissolves; lead: it stays. NaOH acts the same on both.

Example 4: Complex formation equation · 4 marks · Long answerHarder board level

Write the balanced equations for the two stages of the reaction of copper(II) sulphate with ammonium hydroxide, naming the final blue compound. How many moles of NH₄OH are used in total for 1 mole of CuSO₄ (in both stages)?

Try it first, then show the solutionSolution

Stage 1: $C u S O_{4} + 2 N H_{4} O H \to C u (O H)_{2} + (N H_{4})_{2} S O_{4}$ .
Stage 2: $C u (O H)_{2} + (N H_{4})_{2} S O_{4} + 2 N H_{4} O H \to [C u (N H_{3})_{4}] S O_{4} + 4 H_{2} O$ ; the compound is tetraamminecopper(II) sulphate.
NH₄OH used: 2 + 2 = 4 moles.

Answer: Two equations as above; 4 moles of NH₄OH per mole of CuSO₄.

Example 5: Unknown salts · 5 marks · Long answerHOTS / competency

Three solutions X, Y, Z are tested with ammonium hydroxide. X gives a reddish-brown precipitate insoluble in excess. Y gives a pale blue precipitate that dissolves in excess to an inky blue solution. Z gives a dirty green precipitate that darkens on standing. (a) Identify the cations. (b) Write the equation for Y with the sulphate. (c) Suggest why Z changes on standing. (d) Which one of X, Z can form from the other by oxidation?

Try it first, then show the solutionSolution

(a) X: Fe³⁺; Y: Cu²⁺; Z: Fe²⁺.
(b) $C u S O_{4} + 2 N H_{4} O H \to C u (O H)_{2} + (N H_{4})_{2} S O_{4}$ and then complex.
(c) Fe(OH)₂ is oxidised by air to brown Fe(OH)₃.
(d) X (iron(III)) forms from Z (iron(II)) by oxidation.

Answer: (a) Fe³⁺, Cu²⁺, Fe²⁺ (b) as above (c) Air oxidises Fe(OH)₂ (d) Z gives X.

Where students lose marks in Ammonium Hydroxide on Salt Solutions

The slips examiners see most often, and what to write instead.

Mistake: Writing "blue solution" for Cu(OH)₂ in excess
The colour of the final solution is deep or inky blue, not just blue.
Do this: Write "inky blue solution".
Usually costs: 1 mark.
Mistake: Saying the lead precipitate dissolves in excess NH₄OH
This mixes it up with excess NaOH.
Do this: Lead hydroxide dissolves in NaOH but not NH₄OH.
Usually costs: 1 mark.
Mistake: Not mentioning "excess"
The answer says only "a precipitate forms" and omits the effect of excess reagent.
Do this: Always describe both: drop by drop and in excess.
Usually costs: 1 mark.
Mistake: Writing wrong colours for the iron hydroxides
Iron(II) is dirty green and iron(III) is reddish-brown; they are swapped.
Do this: Remember the green-to-brown change on oxidation.
Usually costs: 1 mark.

Exam tips

Always state the observation (colour, precipitate, solubility in excess), not just the name of the ion.
Name the special complex for copper and say inky blue.
For Zn versus Pb, give NH₄OH as the reagent, with the result in one line each.
Remember the syllabus covers only Fe, Cu, Zn and Pb salts.

Practice questions on Ammonium Hydroxide on Salt Solutions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
On adding excess ammonium hydroxide, the precipitate that dissolves to give a colourless solution is of:
(a) Fe(OH)₂
(b) Pb(OH)₂
(c) Zn(OH)₂
(d) Fe(OH)₃
Hint
It forms an ammine complex.
Answer
(c) Zn(OH)₂.
Q2 · 1 mark · MCQBoard level
The colour of the precipitate of iron(III) hydroxide is:
(a) dirty green
(b) reddish-brown
(c) pale blue
(d) white
Hint
Rust.
Answer
(b) reddish-brown.
Q3 · 1 mark · Very short answerBoard level
Name the compound responsible for the inky blue colour in the reaction of copper(II) sulphate and excess ammonium hydroxide.
Hint
An ammine complex.
Answer
Tetraamminecopper(II) sulphate, [Cu(NH₃)₄]SO₄.
Q4 · 3 marks · Short answerHarder board level
A student adds ammonium hydroxide to a colourless solution of a zinc salt. Describe what is seen in each stage and write the equation for the first stage using zinc sulphate.
Hint
Few drops then excess.
Answer
A white gelatinous precipitate forms with a few drops; it dissolves in excess to a colourless solution. $Z n S O_{4} + 2 N H_{4} O H \to Z n (O H)_{2} + (N H_{4})_{2} S O_{4}$ .
Q5 · 2 marks · Short answerHarder board level
Why does a precipitate of iron(II) hydroxide turn brown on standing in air?
Hint
Oxidation of iron(II) to iron(III).
Answer
Atmospheric oxygen oxidises dirty green Fe(OH)₂ to reddish-brown Fe(OH)₃.
Q6 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Excess ammonium hydroxide dissolves the precipitates of copper(II) hydroxide and zinc hydroxide.
Reason (R): Both form soluble complex compounds with ammonia.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Why do they dissolve?
Answer
(a).
Q7 · 4 marks · Long answerHarder board level
A solution of a salt P is pale blue. (a) Name the cation. (b) What happens on adding ammonium hydroxide in excess? (c) Name the product that gives the final colour. (d) What would be seen if sodium hydroxide were added in excess instead?
Hint
Compare the two reagents.
Answer
(a) Cu²⁺. (b) Pale blue precipitate, then it dissolves to give an inky blue solution. (c) [Cu(NH₃)₄]SO₄. (d) The pale blue precipitate remains and does not dissolve.
Q8 · 5 marks · Long answerHOTS / competency
Two colourless solutions Q and R. On adding NH₄OH, both give white precipitates. In excess, the precipitate in Q dissolves and that in R does not. Then NaOH is added to both: both precipitates dissolve in excess. (a) Identify Q and R (as cations). (b) Explain why the NaOH results alone could not separate them. (c) Write the equation for the dissolving of the precipitate in R in excess NaOH.
Hint
Both are amphoteric.
Answer
(a) Q: Zn²⁺; R: Pb²⁺. (b) Both zinc and lead hydroxides are amphoteric and dissolve in excess NaOH. (c) $P b (O H)_{2} + 2 N a O H \to N a_{2} P b O_{2} + 2 H_{2} O$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Colour of the Cu(OH)₂ precipitate?Show answer
Pale blue.
2.Excess NH₄OH on Zn(OH)₂?Show answer
The precipitate dissolves to a colourless solution.
3.Excess NH₄OH on Pb(OH)₂?Show answer
No change; the precipitate remains.
4.Colour of the solution when Cu(OH)₂ dissolves in excess ammonia?Show answer
Inky blue.
5.Colour of Fe(OH)₂?Show answer
Dirty green.

Ammonium Hydroxide on Salt Solutions: frequently asked questions

How do you test for Cu²⁺ using ammonium hydroxide?

A pale blue precipitate forms and dissolves in excess to give an inky blue solution.

Why does NH₄OH distinguish zinc from lead?

Zinc hydroxide dissolves in excess NH₄OH by forming a complex; lead hydroxide does not.

Is the iron(III) hydroxide precipitate soluble in excess ammonia?

No, it stays as a reddish-brown precipitate.

Why is NH₄OH used drop by drop?

So that the first precipitate and its change on adding excess can both be observed.