ChemistryICSE Class 10 · Analytical Chemistry

Action of Sodium Hydroxide on Salt Solutions: Identifying Cations (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

Sodium hydroxide added drop by drop to a salt solution precipitates the metal hydroxide. Fe²⁺ gives dirty green, Fe³⁺ reddish-brown and Cu²⁺ pale blue precipitates, insoluble in excess. Zn²⁺ and Pb²⁺ give white precipitates that dissolve in excess NaOH. Ammonium salts give ammonia gas on warming.

Fe²⁺, Fe³⁺ and Cu²⁺ precipitates stay in excess NaOH, but the white Zn²⁺ and Pb²⁺ precipitates dissolve to a clear solution.

Illustration drawn for QuestionBuddy.

Understand Action of Sodium Hydroxide on Salt Solutions

Why a precipitate forms. Sodium hydroxide supplies OH⁻ ions. Hydroxides of most heavy metals are insoluble, so they come out of solution as precipitates whose colour identifies the metal ion. The general reaction is: metal salt + sodium hydroxide → metal hydroxide (precipitate) + sodium salt.

Colour of the salt first. Before adding any reagent, note the colour: blue solution suggests Cu²⁺, pale green Fe²⁺, yellowish-brown Fe³⁺; colourless suggests Zn²⁺ or Pb²⁺ (among the ions in the syllabus).

Drop by drop, then in excess.

Fe²⁺ (e.g. FeSO₄): dirty green precipitate of iron(II) hydroxide, insoluble in excess. $F e S O_{4} + 2 N a O H \to F e (O H)_{2} + N a_{2} S O_{4}$
Fe³⁺ (e.g. FeCl₃): reddish-brown precipitate of iron(III) hydroxide, insoluble in excess. $F e C l_{3} + 3 N a O H \to F e (O H)_{3} + 3 N a C l$
Cu²⁺ (e.g. CuSO₄): pale blue precipitate of copper(II) hydroxide, insoluble in excess. $C u S O_{4} + 2 N a O H \to C u (O H)_{2} + N a_{2} S O_{4}$
Zn²⁺ (e.g. ZnSO₄): white gelatinous precipitate that dissolves in excess to a colourless solution of sodium zincate. $Z n S O_{4} + 2 N a O H \to Z n (O H)_{2} + N a_{2} S O_{4}$ ; $Z n (O H)_{2} + 2 N a O H \to N a_{2} Z n O_{2} + 2 H_{2} O$
Pb²⁺ (e.g. Pb(NO₃)₂): chalky white precipitate that dissolves in excess to a colourless solution of sodium plumbite. $P b (N O_{3})_{2} + 2 N a O H \to P b (O H)_{2} + 2 N a N O_{3}$ ; $P b (O H)_{2} + 2 N a O H \to N a_{2} P b O_{2} + 2 H_{2} O$

Zinc and lead hydroxides dissolve in excess because they are amphoteric. Sodium hydroxide therefore cannot distinguish Zn²⁺ from Pb²⁺; ammonium hydroxide can (Zn(OH)₂ dissolves in excess NH₄OH, Pb(OH)₂ does not).

Ammonium salts. Warming an ammonium salt with sodium hydroxide releases ammonia gas and gives no precipitate: $N H_{4} C l + N a O H \to N a C l + H_{2} O + N H_{3}$ . The gas has a pungent smell, turns moist red litmus blue and gives dense white fumes with a rod dipped in concentrated HCl. This is the test for the ammonium ion; sodium chloride gives no gas.

On standing. The dirty green Fe(OH)₂ slowly turns brown at the surface as air oxidises it to Fe(OH)₃.

Action of Sodium Hydroxide on Salt Solutions: formulas and key facts

Iron(II) and iron(III): $F e S O_{4} + 2 N a O H \to F e (O H)_{2} + N a_{2} S O_{4}; F e C l_{3} + 3 N a O H \to F e (O H)_{3} + 3 N a C l$
Copper(II): $C u S O_{4} + 2 N a O H \to C u (O H)_{2} + N a_{2} S O_{4}$
Zinc: precipitate, then sodium zincate: $Z n S O_{4} + 2 N a O H \to Z n (O H)_{2} + N a_{2} S O_{4}; Z n (O H)_{2} + 2 N a O H \to N a_{2} Z n O_{2} + 2 H_{2} O$
Lead: precipitate, then sodium plumbite: $P b (N O_{3})_{2} + 2 N a O H \to P b (O H)_{2} + 2 N a N O_{3}; P b (O H)_{2} + 2 N a O H \to N a_{2} P b O_{2} + 2 H_{2} O$
Ammonium ion test: $N H_{4} C l + N a O H \to N a C l + H_{2} O + N H_{3}$
Sodium hydroxide: observations: Fe²⁺ dirty green precipitate, insoluble in excess
Fe³⁺ reddish-brown precipitate, insoluble in excess
Cu²⁺ pale blue precipitate, insoluble in excess
Zn²⁺ white gelatinous precipitate, soluble in excess
Pb²⁺ chalky white precipitate, soluble in excess
NH₄⁺ on warming no precipitate, ammonia gas
Colours of the salts in solution: Copper(II) salts blue
Iron(II) salts pale green
Iron(III) salts yellowish-brown
Zinc and lead salts colourless

Fe²⁺	dirty green precipitate, insoluble in excess
Fe³⁺	reddish-brown precipitate, insoluble in excess
Cu²⁺	pale blue precipitate, insoluble in excess
Zn²⁺	white gelatinous precipitate, soluble in excess
Pb²⁺	chalky white precipitate, soluble in excess
NH₄⁺ on warming	no precipitate, ammonia gas

Copper(II) salts	blue
Iron(II) salts	pale green
Iron(III) salts	yellowish-brown
Zinc and lead salts	colourless

How to solve Action of Sodium Hydroxide on Salt Solutions questions

Identification questions are solved by elimination. Combine the solution colour with the precipitate colour and its behaviour in excess.

1Note the colour of the solution (blue, pale green, yellowish-brown or colourless).
2Add NaOH drop by drop and note the precipitate colour.
3Add NaOH in excess: does the precipitate dissolve?
4Coloured and insoluble: Fe²⁺ (dirty green), Fe³⁺ (reddish-brown), Cu²⁺ (pale blue).
5White and soluble: Zn²⁺ or Pb²⁺; use NH₄OH to decide.
6No precipitate but a gas on warming that turns red litmus blue: NH₄⁺.
7Write the equation and describe the observation (colour, 'soluble/insoluble in excess').

Action of Sodium 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.

Is a fertiliser an ammonium salt?

Suppose a farmer is offered a white fertiliser powder said to contain ammonium chloride and wants a quick check with a small sample of 2.14 g. (N = 14, H = 1, Cl = 35.5)

Describe the test, and calculate the volume of ammonia at S.T.P. that 2.14 g of pure ammonium chloride would give.

Try it first, then show the solutionSolution

Warm a little powder with sodium hydroxide solution and hold moist red litmus at the mouth: it turns blue if ammonia is given off.
$N H_{4} C l + N a O H \to N a C l + H_{2} O + N H_{3}$ ; moles of NH₄Cl $= 2.14/53.5 = 0.04$ .
NH₃ $= 0.04 \times 22400 = 896$ cm³ at S.T.P.

Answer: Ammonia (red litmus turns blue) confirms NH₄⁺; 2.14 g would give 896 cm³ of ammonia.

Memory hook: Ammonium plus alkali, warmed, smells of ammonia.

Tricks to remember Action of Sodium Hydroxide on Salt Solutions

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

Green, Brown, Blue stay; White goes
Coloured hydroxides (Fe²⁺ green, Fe³⁺ brown, Cu²⁺ blue) stay in excess NaOH; white ones (Zn, Pb) go into solution.
Use it when: Predicting the excess-NaOH result.
Two-plus is green, three-plus is brown
Fe²⁺ dirty green, Fe³⁺ reddish-brown, like a leaf going brown with age (and air).
Use it when: Iron(II) versus iron(III).
Zinc and lead love NaOH
Zincate and plumbite: both amphoteric, both dissolve in excess caustic soda.
Use it when: Why NaOH cannot separate Zn²⁺ and Pb²⁺.

Action of Sodium Hydroxide on Salt Solutions: solved examples

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

Example 1: Reddish-brown precipitate · 1 mark · MCQBoard level

A solution gives a reddish-brown precipitate with sodium hydroxide that does not dissolve in excess. The cation is:
(a) Fe²⁺
(b) Fe³⁺
(c) Cu²⁺
(d) Zn²⁺

Try it first, then show the solutionSolution

Reddish-brown hydroxide is Fe(OH)₃.
So the ion is Fe³⁺.

Answer: (b) Fe³⁺.

Example 2: Ammonium sulphate versus sodium sulphate · 2 marks · Short answerBoard level

Two white powders are ammonium sulphate and sodium sulphate. Describe how warming each with sodium hydroxide solution tells them apart, and write the equation for the positive result.

Try it first, then show the solutionSolution

Warm each with NaOH solution: ammonium sulphate gives ammonia, which turns moist red litmus blue (or gives white fumes with conc. HCl); sodium sulphate gives no gas.
$(N H_{4})_{2} S O_{4} + 2 N a O H \to N a_{2} S O_{4} + 2 H_{2} O + 2 N H_{3}$ .

Answer: Only ammonium sulphate gives ammonia (red litmus turns blue); sodium sulphate gives no gas.

Example 3: Excess sodium hydroxide · 3 marks · Short answerHarder board level

Sodium hydroxide solution is added first drop by drop and then in excess to (a) zinc sulphate solution, (b) copper(II) sulphate solution. State the observations and write the equations.

Try it first, then show the solutionSolution

(a) White gelatinous precipitate which dissolves in excess: $Z n S O_{4} + 2 N a O H \to Z n (O H)_{2} + N a_{2} S O_{4}$ ; $Z n (O H)_{2} + 2 N a O H \to N a_{2} Z n O_{2} + 2 H_{2} O$ .
(b) Pale blue precipitate insoluble in excess: $C u S O_{4} + 2 N a O H \to C u (O H)_{2} + N a_{2} S O_{4}$ .

Answer: Zn: white, dissolves; Cu: pale blue, stays.

Example 4: Iron(II) or iron(III)? · 3 marks · Short answerHarder board level

Two solutions, P (pale green) and Q (yellowish-brown), both contain iron. Explain how sodium hydroxide distinguishes them, with equations. What happens to P's precipitate if left in air?

Try it first, then show the solutionSolution

P (Fe²⁺): dirty green precipitate. $F e S O_{4} + 2 N a O H \to F e (O H)_{2} + N a_{2} S O_{4}$ (taking P as iron(II) sulphate).
Q (Fe³⁺): reddish-brown precipitate. $F e C l_{3} + 3 N a O H \to F e (O H)_{3} + 3 N a C l$ (taking Q as iron(III) chloride).
P's dirty green precipitate slowly turns brown at the surface as air oxidises Fe(OH)₂ to Fe(OH)₃.

Answer: Dirty green (Fe²⁺) versus reddish-brown (Fe³⁺); the green one browns in air.

Example 5: Masses in a precipitation · 4 marks · Long answerHOTS / competency

3.2 g of anhydrous copper(II) sulphate is dissolved in water and sodium hydroxide is added until precipitation is complete. Calculate (a) the mass of NaOH used, (b) the mass of precipitate, and (c) state its colour and whether extra NaOH changes it. (Cu = 64, S = 32, O = 16, Na = 23, H = 1)

Try it first, then show the solutionSolution

Moles of CuSO₄ $= 3.2/160 = 0.02$ .
(a) NaOH $= 0.04$ mol $= 1.6$ g.
(b) Cu(OH)₂ (M = 98): $0.02 \times 98 = 1.96$ g.
(c) Pale blue; it does not dissolve in excess NaOH.

Answer: (a) 1.6 g (b) 1.96 g (c) pale blue, insoluble in excess.

Example 6: Sorting three solutions · 3 marks · Short answerHOTS / competency

Three colourless solutions are zinc sulphate, lead nitrate and ammonium sulphate. A student has only sodium hydroxide solution. (a) Which can she identify for certain, and how? (b) Why can she not tell the other two apart? (c) Suggest a second reagent to finish the job and the result.

Try it first, then show the solutionSolution

(a) Ammonium sulphate: no precipitate; on warming, ammonia is released (red litmus turns blue).
(b) Zinc sulphate and lead nitrate both give a white precipitate that dissolves in excess NaOH.
(c) Ammonium hydroxide: Zn(OH)₂ dissolves in excess NH₄OH; Pb(OH)₂ does not.

Answer: (a) Ammonium sulphate (ammonia on warming). (b) Both white, soluble in excess. (c) NH₄OH: zinc dissolves, lead does not.

Where students lose marks in Action of Sodium Hydroxide on Salt Solutions

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

Mistake: Testing for ammonia without saying how
In 2025 (Q8(ii)) examiners noted that candidates wrote only 'colourless gas' without naming ammonia or its test, or wrote reactions for both salts.
Do this: Name the gas (ammonia) and give a test: moist red litmus turns blue, or white fumes with conc. HCl. State that NaCl gives no gas.
Usually costs: 1 mark.
Mistake: Not stating both results in 'distinguish'
The same report found answers giving the result for only one salt.
Do this: Write one line for each substance.
Usually costs: 1 mark.
Mistake: Writing Fe²⁺ when the precipitate is reddish-brown
In 2024 (Q7(i)) examiners noted the cation written as Fe or Fe²⁺ instead of Fe³⁺ for a reddish-brown precipitate.
Do this: Reddish-brown hydroxide = Fe(OH)₃, so the ion is Fe³⁺, with its charge.
Usually costs: 1 mark.
Mistake: Zinc hydroxide instead of sodium zincate
In 2025 (Q2(iii)) candidates chose zinc hydroxide instead of sodium zincate as the salt formed from zinc oxide and hot concentrated NaOH; the same zincate forms when Zn(OH)₂ dissolves in excess NaOH.
Do this: In excess NaOH, Zn(OH)₂ becomes soluble sodium zincate, Na₂ZnO₂.
Usually costs: 1 mark.

Exam tips

Always mention 'drop by drop' and 'in excess' observations separately.
Give precipitate colours exactly: dirty green, reddish-brown, pale blue, white gelatinous, chalky white.
Ions in the 2027 syllabus are Fe²⁺, Fe³⁺, Cu²⁺, Zn²⁺, Pb²⁺ (and NH₄⁺); calcium salts were dropped.
Write the cation with its charge (Fe³⁺, not Fe).

Practice questions on Action of Sodium 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
The precipitate formed when sodium hydroxide is added to copper(II) sulphate solution is:
(a) white
(b) dirty green
(c) pale blue
(d) reddish-brown
Hint
Cu(OH)₂.
Answer
(c) pale blue.
Q2 · 1 mark · Very short answerBoard level
Name the salt formed when excess sodium hydroxide dissolves lead(II) hydroxide.
Hint
Plumbite.
Answer
Sodium plumbite, Na₂PbO₂.
Q3 · 2 marks · Short answerBoard level
Write the equation for the action of sodium hydroxide on iron(III) chloride solution and state the colour of the precipitate.
Hint
Three NaOH.
Answer
$F e C l_{3} + 3 N a O H \to F e (O H)_{3} + 3 N a C l$ ; reddish-brown.
Q4 · 2 marks · Short answerHarder board level
Why does the white precipitate from zinc sulphate dissolve in excess sodium hydroxide?
Hint
Amphoteric.
Answer
Zinc hydroxide is amphoteric; it reacts with excess NaOH to form soluble sodium zincate: $Z n (O H)_{2} + 2 N a O H \to N a_{2} Z n O_{2} + 2 H_{2} O$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Sodium hydroxide cannot be used to distinguish zinc nitrate from lead nitrate.
Reason (R): Both give white precipitates that dissolve in excess sodium hydroxide.
(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
Same behaviour, same result.
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
3.25 g of iron(III) chloride is completely precipitated with sodium hydroxide. Calculate the mass of NaOH needed and the mass of precipitate. (Fe = 56, Cl = 35.5, Na = 23, O = 16, H = 1)
Hint
M of FeCl₃ = 162.5; Fe(OH)₃ = 107.
Answer
Moles $= 3.25/162.5 = 0.02$ . NaOH $= 0.06$ mol $= 2.4$ g. Fe(OH)₃ $= 0.02 \times 107 = 2.14$ g.
Q7 · 2 marks · Short answerHarder board level
A pale green salt solution gives a dirty green precipitate with NaOH, insoluble in excess. With barium chloride solution it gives a white precipitate insoluble in dilute HCl. Name the salt and write the equation with NaOH.
Hint
Cation from NaOH, anion from BaCl₂.
Answer
Iron(II) sulphate, FeSO₄. $F e S O_{4} + 2 N a O H \to F e (O H)_{2} + N a_{2} S O_{4}$ .
Q8 · 3 marks · Short answerHOTS / competency
Suppose a solution contains both zinc sulphate and copper(II) sulphate. Excess sodium hydroxide is added and the mixture is filtered. What is left on the filter paper and what is in the filtrate? Explain.
Hint
Which hydroxide dissolves in excess?
Answer
On the filter: pale blue Cu(OH)₂, which is insoluble in excess NaOH. In the filtrate: sodium zincate (colourless), because Zn(OH)₂ dissolves in excess NaOH, plus sodium sulphate and unused NaOH.

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 Fe(OH)₂?Show answer
Dirty green.
2.Which two hydroxides dissolve in excess NaOH?Show answer
Zn(OH)₂ and Pb(OH)₂.
3.Product of Zn(OH)₂ with excess NaOH?Show answer
Sodium zincate, Na₂ZnO₂.
4.Gas from NH₄Cl warmed with NaOH?Show answer
Ammonia.
5.Colour of iron(III) chloride solution?Show answer
Yellowish-brown.

Action of Sodium Hydroxide on Salt Solutions: frequently asked questions

Why do metal salt solutions give precipitates with sodium hydroxide?

The OH⁻ ions combine with the metal ions to form insoluble metal hydroxides, which come out of solution.

How can you tell iron(II) from iron(III) salts?

With NaOH, Fe²⁺ gives a dirty green precipitate and Fe³⁺ a reddish-brown one; both are insoluble in excess.

Why can't NaOH distinguish zinc from lead salts?

Both give white precipitates that dissolve in excess NaOH because their hydroxides are amphoteric. Use ammonium hydroxide instead.

What is the test for an ammonium salt?

Warm with NaOH solution: ammonia is given off, which turns moist red litmus blue and gives white fumes with concentrated HCl.