Tests for Ions

What this lesson covers in the specification

Incandio is aligned to this specification. It is not published by, endorsed by or affiliated with Pearson, and it reproduces none of Pearson's wording — the statement numbers are given so you can check every lesson against your own copy.

• 2.47 — Tests for NH₄⁺, Cu²⁺, Fe²⁺ and Fe³⁺ using sodium hydroxide solution
• 2.48 — Tests for halide ions, sulfate ions and carbonate ions

Specification section 2(h) Chemical tests — also taught in Titrations, Preparing a Soluble Salt, Preparing an Insoluble Salt, Tests for Gases, and for Water.

1 · Understand it

No exam language yet. The only question this section answers is: do I actually understand what is happening?

Adding SODIUM HYDROXIDE SOLUTION to a solution of an unknown salt identifies four positive ions at once, and it does so by a single mechanism: most metal hydroxides are insoluble, so a coloured precipitate forms and its colour names the metal.

Statement 2.47 — the four results with sodium hydroxide

1. COPPER(II), Cu²⁺ — a BLUE precipitate of copper(II) hydroxide.
2. IRON(II), Fe²⁺ — a GREEN precipitate of iron(II) hydroxide.
3. IRON(III), Fe³⁺ — a RED-BROWN precipitate of iron(III) hydroxide.
4. AMMONIUM, NH₄⁺ — NO precipitate. Instead, on WARMING, a gas is given off which turns damp red litmus BLUE. That gas is ammonia.
5. The ammonium test is the odd one out and is worth noting as such: it produces a gas rather than a precipitate, and it requires WARMING.

The two iron results are the pair most often confused, and the way to keep them straight is that the higher charge gives the darker, browner precipitate. Iron(II) is green; iron(III) is red-brown. If a green precipitate is left to stand in air it slowly turns brown at the surface, because the iron(II) is oxidised to iron(III) — which is a useful confirmation that the green one was iron(II).

The ionic equation for the copper case is Cu²⁺ + 2OH⁻ → Cu(OH)₂, giving the blue precipitate. The same pattern applies to the other two: Fe²⁺ gives a green hydroxide, and Fe³⁺ needs three hydroxide ions and gives a red-brown one.

Now the anions, statement 2.48, and there are three tests to know. Each adds a reagent that forms an insoluble compound with the ion being sought, so once again a precipitate is the signal.

Statement 2.48 — the three anion tests

1. HALIDES — add dilute NITRIC ACID, then SILVER NITRATE solution. Chloride gives a WHITE precipitate, bromide a CREAM one, and iodide a YELLOW one.
2. SULFATE — add dilute HYDROCHLORIC ACID, then BARIUM CHLORIDE solution. A WHITE precipitate of barium sulfate confirms sulfate.
3. CARBONATE — add DILUTE ACID. It FIZZES, and the gas turns limewater MILKY, confirming carbon dioxide.
4. The acid is added FIRST in the halide and sulfate tests, and this is the step most often omitted.
5. Its purpose is to remove any CARBONATE present, which would otherwise form its own white precipitate of silver carbonate or barium carbonate and give a false positive.

Think of it like clearing the table before laying out the evidence

If you are looking for a particular white envelope on a desk covered in white papers, the first thing to do is sweep the desk clear of everything that is not envelopes. Otherwise you will find something white and be none the wiser. The acid in these tests is the sweep. Carbonate ions form white precipitates with both silver nitrate and barium chloride, so without removing them first a white precipitate would be ambiguous — it might be the chloride or sulfate you were looking for, or it might be the carbonate you were not. Adding acid destroys any carbonate as carbon dioxide, so anything white appearing afterwards means what you hoped it meant.

That is why the acids differ between the two tests, and the choice is deliberate rather than arbitrary. The halide test uses NITRIC acid because hydrochloric acid would introduce chloride ions and guarantee a white precipitate whatever the sample contained. The sulfate test uses HYDROCHLORIC acid because sulfuric acid would introduce sulfate ions and do the same. In each case the acid chosen is the one that cannot supply the ion being tested for.

A full identification, which needs two tests

A green solution is tested. Adding sodium hydroxide gives a green precipitate. Adding dilute hydrochloric acid then barium chloride gives a white precipitate. Identify the compound.

1. The green precipitate with sodium hydroxide identifies the cation as IRON(II), Fe²⁺.
2. The white precipitate with barium chloride, after acidifying, identifies the anion as SULFATE, SO₄²⁻.
3. The hydrochloric acid was added first to remove any carbonate, which would otherwise have given a white barium carbonate precipitate and a false positive.
4. Putting the two results together: the compound is IRON(II) SULFATE, FeSO₄.
5. Neither test alone would have identified it — this is why a full identification always needs one cation test and one anion test.

Answer: Iron(II) sulfate — and it took both tests, because each identifies only half the compound.

That is the principle running through all three of these pages and it is worth stating plainly at the end of the topic. NO SINGLE TEST IDENTIFIES A COMPOUND. Every compound has a positive part and a negative part, so a full identification requires one test for each — and an answer naming a compound on the strength of one result has claimed more than the evidence supports.

2 · Grade 9 Notes

A different job from the section above. You have already understood it; this is the precise set of things to LEARN — definitions to reproduce word for word, processes in order, equations with units, and the answers that score full marks.

Statement 2.47 — sodium hydroxide added to the solution

Statement 2.48 — the three anion tests

Why the acid comes first, and why THAT acid

Purpose of the acid

To remove any carbonate present, which would give its own white precipitate and a false positive.

Nitric acid for halides

Hydrochloric acid would add chloride ions and guarantee a white precipitate whatever the sample was.

Hydrochloric acid for sulfates

Sulfuric acid would add sulfate ions and guarantee a white precipitate whatever the sample was.

The rule

Use the acid that cannot supply the ion you are testing for.

Identifying an unknown salt

1. Test for the CATION: add sodium hydroxide solution and note any precipitate and its colour, warming if no precipitate forms.
2. A flame test on the solid can also identify the cation for lithium, sodium, potassium, calcium and copper.
3. Test for the ANION: acidify first, then add silver nitrate for halides or barium chloride for sulfate.
4. For a carbonate, simply add dilute acid and test the gas with limewater.
5. Combine the two results to name the compound — neither test alone is enough.

Ba²⁺ + SO₄²⁻ → BaSO₄

Conditions: The white precipitate confirming sulfate. Acidify with hydrochloric acid first to remove any carbonate.

Model answer [4 marks]

Describe how you would test a solution for sulfate ions, and explain why acid is added first. [4]

Add a few drops of dilute hydrochloric acid to the solution, then add barium chloride solution. A white precipitate of barium sulfate confirms that sulfate ions are present. The acid is added first to remove any carbonate ions in the solution, because carbonate would form a white precipitate of barium carbonate and would be mistaken for a positive result. Hydrochloric acid is used rather than sulfuric acid because sulfuric acid would itself add sulfate ions and would give a white precipitate whatever the sample contained.

Model answer [4 marks]

A solution gives a red-brown precipitate with sodium hydroxide, and a cream precipitate with nitric acid followed by silver nitrate. Identify the compound and explain your reasoning. [4]

The red-brown precipitate with sodium hydroxide shows that the cation is iron(III), since iron(III) hydroxide is red-brown. The cream precipitate with silver nitrate, after acidifying with nitric acid, shows that the anion is bromide, since silver bromide is cream. Putting the two results together, the compound is iron(III) bromide. Both tests are needed because the sodium hydroxide test identifies only the positive ion and the silver nitrate test identifies only the negative ion.

Telling iron(II) from iron(III)

A. IRON(II), Fe²⁺ — a GREEN precipitate with sodium hydroxide, which slowly turns brown at the surface on standing in air.

B. IRON(III), Fe³⁺ — a RED-BROWN precipitate immediately. The higher charge gives the darker, browner solid.

Not this: A white precipitate with silver nitrate proves that chloride ions are present.

This: Only if the solution was ACIDIFIED WITH NITRIC ACID first. Carbonate ions also give a white precipitate with silver nitrate, so without the acid a white precipitate is ambiguous — and hydrochloric acid must never be used, since it would add chloride itself.

Mark-losing trap. ACIDIFY FIRST in the halide and sulfate tests, to destroy any carbonate that would give a false positive.

Mark-losing trap. NITRIC acid for halides, HYDROCHLORIC acid for sulfate — never the acid containing the ion you seek.

Mark-losing trap. The AMMONIUM test gives a GAS on warming, not a precipitate. It is the odd one out.

Mark-losing trap. NO SINGLE TEST identifies a compound. One cation test plus one anion test, always.

3 · Prove it — the five questions

The five questions climb Grade 6 → Grade 7 → Grade 8 → Grade 9 → Grade 9 challenge, and are marked inside Incandio on your own device, by rule, with an authored diagnosis of the mistake you actually made. The mark schemes stay in the app so that the practice is worth doing; the questions themselves are here.

1. Grade 6 · State [1 mark] — What colour precipitate does copper(II) give with sodium hydroxide solution?
2. Grade 7 · Identify [2 marks] — A solution is acidified with dilute nitric acid and silver nitrate is added, giving a yellow precipitate. Which ion is present?
3. Grade 8 · Explain [3 marks] — Explain why hydrochloric acid must not be used to acidify a solution before testing it for chloride ions with silver nitrate.
4. Grade 9 · Describe [6 marks] — Select every statement that belongs in a full-mark description of how to identify the compound in a solution known to contain one cation and one anion from those on the specification.
5. 9+ · Analyse [6 marks] — A student tests a solution with silver nitrate without acidifying first, and obtains a white precipitate. They conclude chloride ions are present. A second student points out the error, and after acidifying with nitric acid and repeating, no precipitate forms at all. Select every statement that belongs in a full-mark analysis.

The people behind this science

Two ways into the same idea — the one who had to prove a substance was new rather than merely unfamiliar, and the one who insisted a substance be identified by tests rather than by reputation. Inside Incandio each of them answers knowing exactly which lesson you have just finished.

Humphry Davy — the one who had to prove a substance was new rather than merely unfamiliar

Every test on this page answers the question 'what is this?', and Davy faced that question in its hardest form: having produced something nobody had seen, he had to establish that it was a new element rather than a compound or a known substance in disguise. That required showing it could not be broken down further and that its properties matched nothing already recorded. He is the right figure to close a topic on identification, because he is the case where the tests had to be assembled from scratch.

• “How do you prove a substance is a new element?”
• “What would convince you that you had not simply made a compound?”
• “How did you record the properties of something nobody had described?”
• “What tests did you have available?”
• “Were you ever wrong about a substance being new?”

Robert Boyle — the one who insisted a substance be identified by tests rather than by reputation

This whole section rests on identifying substances by their reactions, and that approach is Boyle's programme. He argued against defining substances by origin or appearance and in favour of testing them, and he introduced several tests himself, including the use of plant extracts as indicators. His definition of an element — a substance that cannot be broken down further — is itself a test rather than a description, which is exactly the move this topic makes about ions.

• “Why identify a substance by what it does rather than where it came from?”
• “What made you define an element by what cannot be done to it?”
• “Which tests did you devise yourself?”
• “How do you know a test result means what you think?”
• “What would you make of a test that gives the same result for two substances?”

Carry on through the course

• Previous lesson: Flame Tests
• Next lesson: Exothermic and Endothermic Reactions
• Group 1 — the Alkali Metals — Metals soft enough to cut with a knife, which set fire to water — and get worse as you go down.
• Group 7 — the Halogens — A green gas, a brown liquid and a grey solid — and a trend that runs the opposite way to Group 1.
• The Gases in the Air — Four gases, one of which is doing almost all the chemistry — and how to measure it without ever touching it.
• Combustion and Carbonates — Three things burning in oxygen, one carbonate breaking apart, and the 0.04% gas that changes the climate.
• The Reactivity Series — Putting eleven metals in order by watching what each one will and will not do.
• All of Chemistry · Chemistry key terms · required practicals · equations · Incandio Science · The 4CH1 specification