ChemistryICSE Class 10 · Sulphuric Acid

Dilute Sulphuric Acid: Acidic Reactions, Dibasic Nature and Tests (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

Dilute sulphuric acid behaves as a typical strong acid. It is dibasic, giving two H⁺ ions per molecule, so it forms acid salts (hydrogen sulphates) and normal salts (sulphates). It reacts with active metals, oxides, hydroxides, carbonates, sulphites and sulphides. It is identified by a white precipitate with barium chloride solution.

Dilute sulphuric acid acts as a typical dibasic acid, and a white precipitate with barium chloride is the test that identifies it.

Illustration drawn for QuestionBuddy.

Understand Dilute Sulphuric Acid and Tests

A strong, dibasic acid. In water sulphuric acid ionises in two steps:

$H_{2} S O_{4} + H_{2} O \to H_{3} O^{+} + H S O_{4}^{-}$ ; $H S O_{4}^{-} + H_{2} O \to H_{3} O^{+} + S O_{4}^{2 -}$

Because each molecule can give two hydronium ions, it is dibasic. It therefore forms two series of salts: an acid salt when only one hydrogen is replaced, and a normal salt when both are replaced:

$N a O H + H_{2} S O_{4} \to N a H S O_{4} + H_{2} O$ (sodium hydrogen sulphate, acid salt)

$2 N a O H + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H_{2} O$ (sodium sulphate, normal salt)

Acidic reactions of the dilute acid.

Active metals give hydrogen: $Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2}$ ; $F e + H_{2} S O_{4} \to F e S O_{4} + H_{2}$ . (Copper does not react.)
Metal oxides and hydroxides give salt and water: $C u O + H_{2} S O_{4} \to C u S O_{4} + H_{2} O$ (black solid dissolves to a blue solution).
Carbonates and hydrogen carbonates give CO₂ (brisk effervescence; turns lime water milky): $N a_{2} C O_{3} + H_{2} S O_{4} \to N a_{2} S O_{4} + H_{2} O + C O_{2}$ .
Sulphites and hydrogen sulphites give SO₂ (smell of burning sulphur; turns acidified dichromate green): $N a_{2} S O_{3} + H_{2} S O_{4} \to N a_{2} S O_{4} + H_{2} O + S O_{2}$ .
Sulphides give hydrogen sulphide (rotten-egg smell; blackens lead acetate paper): $F e S + H_{2} S O_{4} \to F e S O_{4} + H_{2} S$ .

Note: calcium carbonate and lead compounds react only briefly with dilute sulphuric acid, because insoluble CaSO₄ or PbSO₄ coats the solid.

Tests for sulphuric acid (sulphate ions).

Barium chloride solution: a white precipitate of barium sulphate, insoluble in dilute HCl and HNO₃: $B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l$ . Nitric acid and hydrochloric acid give no precipitate with barium chloride.
Lead nitrate solution: a white precipitate of lead sulphate, insoluble even on heating: $P b (N O_{3})_{2} + H_{2} S O_{4} \to P b S O_{4} + 2 H N O_{3}$ . (Hydrochloric acid also gives a white precipitate, PbCl₂, but it dissolves on heating.)
The dilute acid gives hydrogen with zinc; the concentrated acid chars sugar.

Dilute Sulphuric Acid and Tests: formulas and key facts

Dibasic: acid salt and normal salt: $N a O H + H_{2} S O_{4} \to N a H S O_{4} + H_{2} O; 2 N a O H + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H_{2} O$
With zinc: $Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2}$
With a sulphite and a sulphide: $N a_{2} S O_{3} + H_{2} S O_{4} \to N a_{2} S O_{4} + H_{2} O + S O_{2}; F e S + H_{2} S O_{4} \to F e S O_{4} + H_{2} S$
Barium chloride test: $B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l$
Lead nitrate test: $P b (N O_{3})_{2} + H_{2} S O_{4} \to P b S O_{4} + 2 H N O_{3}$
Gas given by dilute sulphuric acid: Carbonate or hydrogen carbonate carbon dioxide, turns lime water milky
Sulphite or hydrogen sulphite sulphur dioxide, turns acidified dichromate green
Sulphide hydrogen sulphide, rotten-egg smell
Zinc or iron hydrogen, pops with a burning splint
Telling the three acids apart: Barium chloride white precipitate with sulphuric acid only
Silver nitrate white precipitate with hydrochloric acid only
Copper turnings, warmed brown fumes with nitric acid only

Carbonate or hydrogen carbonate	carbon dioxide, turns lime water milky
Sulphite or hydrogen sulphite	sulphur dioxide, turns acidified dichromate green
Sulphide	hydrogen sulphide, rotten-egg smell
Zinc or iron	hydrogen, pops with a burning splint

Barium chloride	white precipitate with sulphuric acid only
Silver nitrate	white precipitate with hydrochloric acid only
Copper turnings, warmed	brown fumes with nitric acid only

How to solve Dilute Sulphuric Acid and Tests questions

Equations with dilute sulphuric acid follow the usual acid patterns. Tests rest on the insolubility of barium sulphate.

1Identify the type of reactant: metal, oxide/hydroxide, carbonate, sulphite or sulphide.
2Use the pattern: metal → salt + H₂; base → salt + water; carbonate → salt + water + CO₂; sulphite → salt + water + SO₂; sulphide → salt + H₂S.
3Balance, remembering the sulphate is SO₄²⁻ (so Na₂SO₄, but ZnSO₄).
4For acid salts, one mole of NaOH per mole of acid; for normal salts, two.
5For tests, name the reagent, the observation and the equation; mention what a different acid would do.

Dilute Sulphuric Acid and Tests 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.

Checking an unlabelled acid

Suppose a lab assistant has a bottle labelled only 'dilute acid' and needs to know whether it is sulphuric acid before using it for a zinc-hydrogen experiment.

Describe a quick test that confirms sulphuric acid, and the expected result if 4.9 g of the acid were completely precipitated. (Ba = 137, S = 32, O = 16, H = 1)

Try it first, then show the solutionSolution

Add barium chloride solution: a white precipitate insoluble in dilute HCl shows sulphate, so the acid is sulphuric acid.
$B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l$ ; moles of H₂SO₄ $= 4.9/98 = 0.05$ .
Mass of BaSO₄ $= 0.05 \times 233 = 11.65$ g.

Answer: A white, acid-insoluble precipitate with BaCl₂; 4.9 g of acid would give 11.65 g of BaSO₄.

Memory hook: Barium finds sulphate: white and stubborn, it will not dissolve in acid.

Tricks to remember Dilute Sulphuric Acid and Tests

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

Di = two H
Dibasic sulphuric acid has two replaceable hydrogens, so two salts: NaHSO₄ and Na₂SO₄.
Use it when: Acid and normal salts.
Ba for sulphate, Ag for chloride, Cu for nitric
Barium chloride spots sulphuric; silver nitrate spots hydrochloric; copper spots nitric.
Use it when: Distinguishing the three mineral acids.
Carbonate CO₂, sulphite SO₂, sulphide H₂S
The gas carries the name of the anion: carbon, sulphur, hydrogen sulphide.
Use it when: Predicting the gas evolved.

Dilute Sulphuric Acid and Tests: solved examples

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

Example 1: Acid salt · 1 mark · MCQBoard level

Which of these is an acid salt formed by sulphuric acid?
(a) Na₂SO₄
(b) NaHSO₄
(c) BaSO₄
(d) Na₂SO₃

Try it first, then show the solutionSolution

An acid salt still contains a replaceable hydrogen.
NaHSO₄ has one H left.

Answer: (b) NaHSO₄.

Example 2: Sulphuric versus hydrochloric acid with lead nitrate · 2 marks · Short answerBoard level

Lead nitrate solution gives a white precipitate with both dilute sulphuric acid and dilute hydrochloric acid. How can heating tell the two apart? Write the equation for the sulphuric acid reaction.

Try it first, then show the solutionSolution

Warm each mixture: lead chloride (from HCl) dissolves in hot water and reappears on cooling; lead sulphate (from H₂SO₄) stays undissolved.
$P b (N O_{3})_{2} + H_{2} S O_{4} \to P b S O_{4} + 2 H N O_{3}$ .

Answer: The precipitate that stays on heating is PbSO₄ (sulphuric acid); the one that dissolves is PbCl₂ (hydrochloric acid).

Example 3: Equations with salts · 3 marks · Short answerHarder board level

Write balanced equations for dilute sulphuric acid with (a) sodium carbonate, (b) sodium sulphite, (c) iron(II) sulphide. Name the gas in each.

Try it first, then show the solutionSolution

(a) $N a_{2} C O_{3} + H_{2} S O_{4} \to N a_{2} S O_{4} + H_{2} O + C O_{2}$ : carbon dioxide.
(b) $N a_{2} S O_{3} + H_{2} S O_{4} \to N a_{2} S O_{4} + H_{2} O + S O_{2}$ : sulphur dioxide.
(c) $F e S + H_{2} S O_{4} \to F e S O_{4} + H_{2} S$ : hydrogen sulphide.

Answer: CO₂, SO₂ and H₂S respectively.

Example 4: Dibasic in numbers · 3 marks · Short answerHarder board level

4.9 g of sulphuric acid is neutralised by sodium hydroxide. What mass of NaOH is needed to form (a) the acid salt, (b) the normal salt? Write both equations. (Na = 23, O = 16, H = 1, S = 32)

Try it first, then show the solutionSolution

Moles of H₂SO₄ $= 4.9/98 = 0.05$ .
(a) $N a O H + H_{2} S O_{4} \to N a H S O_{4} + H_{2} O$ : 0.05 mol NaOH $= 2$ g.
(b) $2 N a O H + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H_{2} O$ : 0.10 mol NaOH $= 4$ g.

Answer: (a) 2 g (b) 4 g.

Example 5: Hydrogen from zinc · 3 marks · Short answerHOTS / competency

3.25 g of zinc reacts completely with excess dilute sulphuric acid. Calculate (a) the volume of hydrogen at S.T.P., (b) the mass of zinc sulphate formed. (c) Why is copper not used for this preparation? (Zn = 65, S = 32, O = 16)

Try it first, then show the solutionSolution

Moles of Zn $= 3.25/65 = 0.05$ .
(a) H₂ $= 0.05 \times 22.4 = 1.12$ L. (b) ZnSO₄ (M = 161): $0.05 \times 161 = 8.05$ g.
(c) Copper is below hydrogen in the activity series and does not displace hydrogen from dilute acids.

Answer: (a) 1.12 L (b) 8.05 g (c) Cu is below hydrogen.

Example 6: Identifying three unlabelled acids · 4 marks · Long answerHOTS / competency

Three bottles contain dilute HCl, dilute HNO₃ and dilute H₂SO₄. Using only barium chloride solution, silver nitrate solution and copper turnings, describe how to identify each, giving the observation and one equation for each positive test.

Try it first, then show the solutionSolution

Barium chloride: only H₂SO₄ gives a white precipitate: $B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l$ .
Silver nitrate on the other two: HCl gives a white precipitate (AgCl): $A g N O_{3} + H C l \to A g C l + H N O_{3}$ .
The remaining acid warmed with copper turnings gives brown fumes and a blue solution: nitric acid. $C u + 4 H N O_{3} \to C u (N O_{3})_{2} + 2 N O_{2} + 2 H_{2} O$ (with the concentrated acid; the dilute acid gives colourless NO that browns in air).
Doing the BaCl₂ test first removes sulphuric acid from the choice, so each later test has only two acids to separate.

Answer: BaCl₂ → H₂SO₄; AgNO₃ → HCl; copper → HNO₃.

Where students lose marks in Dilute Sulphuric Acid and Tests

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

Mistake: No statement for the nitric acid result
In 2024 (Q5(ii)) examiners noted candidates failed to state that nitric acid gives no precipitate with BaCl₂, or claimed a brown precipitate.
Do this: In a 'distinguish' answer, state the result for both substances.
Usually costs: 1 mark.
Mistake: Wrong formula of the salt
Students write NaSO₄ or Zn₂SO₄.
Wrong working
2NaOH + H₂SO₄ → NaSO₄ + 2H₂O
Right working
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
Do this: Sulphate is SO₄²⁻: Na₂SO₄, ZnSO₄, FeSO₄.
Usually costs: 1 mark.
Mistake: Confusing SO₂ with CO₂
In 2025 (Q1(i)) candidates confused sulphur dioxide with carbon dioxide in the permanganate test.
Do this: SO₂ decolourises KMnO₄ and turns dichromate green; CO₂ does neither but turns lime water milky.
Usually costs: 1 mark.
Mistake: Writing the concentrated-acid reaction for the dilute acid
Students give SO₂ as the product of zinc with dilute sulphuric acid.
Do this: Dilute acid with zinc gives hydrogen; SO₂ forms only with the hot concentrated acid.
Usually costs: 1 mark.

Exam tips

Learn the general patterns; you can then write any dilute-acid equation in the syllabus.
In tests, say 'white precipitate insoluble in dilute HCl/HNO₃' for barium sulphate.
Use 'dilute' in every equation in this topic, and 'concentrated' in the next.
For the dibasic question, give both an acid salt and a normal salt with equations.

Practice questions on Dilute Sulphuric Acid and Tests

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
Sulphuric acid is called dibasic because it:
(a) has two oxygen atoms
(b) gives two hydronium ions per molecule in water
(c) reacts with two bases
(d) has a high boiling point
Hint
Basicity counts replaceable hydrogens.
Answer
(b).
Q2 · 1 mark · Very short answerBoard level
Name the white precipitate formed when barium chloride is added to dilute sulphuric acid.
Hint
Ba and SO₄.
Answer
Barium sulphate, BaSO₄.
Q3 · 2 marks · Short answerBoard level
Write balanced equations for dilute sulphuric acid with (a) copper(II) oxide, (b) zinc.
Hint
Salt + water; salt + hydrogen.
Answer
(a) $C u O + H_{2} S O_{4} \to C u S O_{4} + H_{2} O$ (b) $Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2}$ .
Q4 · 2 marks · Short answerHarder board level
Write the two-step ionisation of sulphuric acid in water.
Hint
HSO₄⁻ is the intermediate.
Answer
$H_{2} S O_{4} + H_{2} O \to H_{3} O^{+} + H S O_{4}^{-}$ ; $H S O_{4}^{-} + H_{2} O \to H_{3} O^{+} + S O_{4}^{2 -}$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Dilute hydrochloric acid gives a white precipitate with lead nitrate, just as dilute sulphuric acid does.
Reason (R): Lead chloride dissolves on heating but lead sulphate does not.
(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
R is true but explains how to tell them apart, not why both precipitate.
Answer
(b).
Q6 · 2 marks · Short answerHarder board level
10.6 g of sodium carbonate reacts with excess dilute sulphuric acid. Calculate the volume of CO₂ at S.T.P. (Na = 23, C = 12, O = 16)
Hint
1 : 1.
Answer
Moles $= 10.6/106 = 0.1$ ; CO₂ $= 0.1 \times 22.4 = 2.24$ L.
Q7 · 3 marks · Short answerHarder board level
6.3 g of sodium sulphite reacts with excess dilute sulphuric acid. Calculate the volume of SO₂ formed at S.T.P. and give a test for the gas. (Na = 23, S = 32, O = 16)
Hint
M of Na₂SO₃ = 126.
Answer
Moles $= 6.3/126 = 0.05$ ; SO₂ $= 0.05 \times 22400 = 1120$ cm³. Test: it turns acidified potassium dichromate from orange to green.
Q8 · 2 marks · Short answerHOTS / competency
Suppose dilute sulphuric acid is added to marble chips (calcium carbonate). The fizzing starts and then almost stops although plenty of acid and marble remain. Explain, and suggest which acid would work better.
Hint
Solubility of calcium sulphate.
Answer
Calcium sulphate formed is only slightly soluble and coats the marble, stopping further contact with the acid. Dilute hydrochloric acid works better because calcium chloride is soluble.

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.Gas from a sulphide and dilute H₂SO₄?Show answer
Hydrogen sulphide.
2.Basicity of sulphuric acid?Show answer
2 (dibasic).
3.Normal salt of sulphuric acid with KOH?Show answer
Potassium sulphate, K₂SO₄.
4.Reagent that gives a white precipitate with H₂SO₄ but not with HNO₃?Show answer
Barium chloride solution.
5.Does copper react with dilute H₂SO₄?Show answer
No.

Dilute Sulphuric Acid and Tests: frequently asked questions

Why is sulphuric acid called dibasic?

Each molecule releases two hydronium ions in water, so it can form an acid salt (NaHSO₄) and a normal salt (Na₂SO₄).

What is the test for sulphuric acid?

Add barium chloride solution: a white precipitate of barium sulphate, insoluble in dilute acids, confirms sulphate ions.

What gases does dilute sulphuric acid release from salts?

CO₂ from carbonates, SO₂ from sulphites and H₂S from sulphides; hydrogen from active metals.

How do I tell dilute H₂SO₄ from dilute HCl?

Barium chloride gives a white precipitate only with H₂SO₄; silver nitrate gives a white precipitate only with HCl.