Coaching Centre
Identifying anions
Tests for carbonate, hydroxide, sulfate, phosphate, chloride, bromide, iodide and ethanoate ions, with net ionic equations, and why the order of the tests matters when a sample holds more than one anion.
A bag of white powder is found in a farm shed with its label washed off. It could be washing soda, gypsum, a phosphate fertiliser or rock salt. Each of these does a very different job, and putting the wrong one on a paddock or into a water tank could be costly. All four are white solids that dissolve (at least partly) in water, so the clue has to come from the negative ion. This page shows how to identify it.
Three kinds of reaction
Anion tests use the same ideas as cation tests, plus one more:
- Gas-producing reactions: carbonate fizzes with acid, releasing carbon dioxide.
- Acid–base reactions: hydroxide, carbonate and phosphate are bases, so their solutions are alkaline.
- Precipitation reactions: Ba²⁺ and Ag⁺ form insoluble salts with particular anions.
Because the reagents for anions are usually cations (Ba²⁺, Ag⁺), you add them as nitrate salts. Nitrate never precipitates, so it can't give a false result, and nitric acid adds no extra anion that you might later test for.
The eight anions
| Anion | Test and observation | Net ionic equation |
|---|---|---|
| CO₃²⁻ | add dilute HNO₃: fizzing; the gas turns limewater milky | $\text{CO}_3^{2-}(aq) + 2\text{H}^+(aq) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l)$ |
| OH⁻ | universal indicator turns purple (pH 12–14); a few drops of Cu²⁺ give a pale blue precipitate; no fizzing with acid | $\text{Cu}^{2+}(aq) + 2\text{OH}^-(aq) \rightarrow \text{Cu(OH)}_2(s)$ |
| SO₄²⁻ | acidify with HNO₃, then add Ba(NO₃)₂: white precipitate that does not dissolve in acid | $\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s)$ |
| PO₄³⁻ | in neutral solution, Ba(NO₃)₂ gives a white precipitate that dissolves in acid; ammonium molybdate in HNO₃ gives a yellow precipitate on warming | $3\text{Ba}^{2+}(aq) + 2\text{PO}_4^{3-}(aq) \rightarrow \text{Ba}_3(\text{PO}_4)_2(s)$ |
| Cl⁻ | acidify with HNO₃, add AgNO₃: white precipitate, soluble in dilute NH₃ | $\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s)$ |
| Br⁻ | as above: cream precipitate, soluble only in concentrated NH₃ | $\text{Ag}^+(aq) + \text{Br}^-(aq) \rightarrow \text{AgBr}(s)$ |
| I⁻ | as above: yellow precipitate, insoluble in NH₃ | $\text{Ag}^+(aq) + \text{I}^-(aq) \rightarrow \text{AgI}(s)$ |
| CH₃COO⁻ | warm with ethanol and a few drops of concentrated H₂SO₄: a sweet, fruity smell of ethyl ethanoate. Dilute acid alone gives a faint vinegar smell | $\text{CH}_3\text{COO}^-(aq) + \text{H}^+(aq) \rightarrow \text{CH}_3\text{COOH}(aq)$, then esterification |
Why does acid stop barium carbonate and barium phosphate from forming but not barium sulfate? Carbonate and phosphate are the conjugate bases of weak acids, so H⁺ converts them into CO₂ and H₂PO₄⁻, which don't precipitate. Sulfuric acid is strong, so sulfate stays as SO₄²⁻ even in acid and BaSO₄ still forms.
Telling the halides apart
The colours of AgCl, AgBr and AgI are close enough to be misjudged, especially in a mixture. Ammonia gives a clearer answer: it forms the complex ion $[\text{Ag(NH}_3)_2]^+$, which pulls silver ions out of the solid. AgCl is the most soluble of the three, so dilute ammonia dissolves it:
$\text{AgCl}(s) + 2\text{NH}_3(aq) \rightarrow [\text{Ag(NH}_3)_2]^+(aq) + \text{Cl}^-(aq)$
AgBr needs concentrated ammonia, and AgI is so insoluble ($K_{sp} \approx 8 \times 10^{-17}$) that even concentrated ammonia can't dissolve it. All three darken in sunlight as silver ions are reduced to silver metal, which is the chemistry behind black-and-white photographic film.
Why the order matters
In a mixture, reagents meet more than one anion. Silver carbonate, silver phosphate and silver oxide are all insoluble, so if you added AgNO₃ to an untreated sample with carbonate in it, you would see a precipitate even if no halide were there. The fix is to plan the order:
- Acidify with dilute HNO₃ first. Carbonate is removed as CO₂ (and identified by the fizzing), hydroxide is neutralised, and phosphate is protonated so it no longer precipitates.
- Add Ba(NO₃)₂. Only sulfate now gives a precipitate. Filter it off.
- Add AgNO₃ to the filtrate. Only the halides now precipitate. Use ammonia to decide which.
- Test fresh, un-acidified portions for phosphate (neutral Ba²⁺ then acid, or molybdate), hydroxide (indicator) and ethanoate (ester smell), because step 1 destroys the evidence for these.
Using BaCl₂ instead of Ba(NO₃)₂ in step 2 is a classic error: it adds Cl⁻, so the silver nitrate test in step 3 will always give a white precipitate.
- Solve three mixtures. Before you start each one, write down the order you'll use.
- On a fresh portion that you have not acidified, add AgNO₃ straight away. Compare the colour with what you get after acidifying first. Which ions caused the difference?
- Use BaCl₂ instead of Ba(NO₃)₂, filter, then add AgNO₃. A silver precipitate appears every time, whatever the mixture. Explain why you can't trust it.
- Find a mixture with a silver halide precipitate. Add dilute NH₃, then concentrated NH₃, and use what is left to decide which halide is present.
Test yourself
A solution gives a white precipitate with Ba(NO₃)₂. The precipitate dissolves with fizzing when dilute HNO₃ is added. The anion is
BaCO₃ dissolves in acid and releases CO₂ (the fizzing). BaSO₄ would not dissolve; Ba₃(PO₄)₂ dissolves without fizzing.
Why is nitric acid, rather than hydrochloric acid, used to acidify a sample before testing for halides?
Any added Cl⁻ gives white AgCl whether or not the sample contained chloride.
A pale precipitate forms when AgNO₃ is added to an acidified sample. It does not dissolve in dilute ammonia but does dissolve in concentrated ammonia. The sample contains
AgBr is cream and only dissolves in concentrated NH₃. AgCl dissolves in dilute NH₃; AgI dissolves in neither.
Which net ionic equation describes the test for phosphate with barium ions?
Charges must balance: three Ba²⁺ (+6) for two PO₄³⁻ (−6). Option 3 is balanced but includes spectator ions.
Name each step in the sequence of anion tests.
Put the steps for testing a mixture that may contain CO₃²⁻, SO₄²⁻ and Cl⁻ in order.
Carbon dioxide turns milky. Silver iodide is a precipitate that does not dissolve in ammonia. Reagents are added as salts because these are always soluble. Warming an ethanoate with ethanol and concentrated sulfuric acid produces a fruity-smelling .
A student added barium chloride solution to an unknown solution and saw a white precipitate. They then added silver nitrate to a separate portion of the same solution and again saw a white precipitate. They concluded that the solution contained both sulfate and chloride ions. Evaluate this conclusion and describe a better procedure.
The conclusion is not justified. Neither test was carried out in acid, so carbonate (or phosphate) could have caused both precipitates: BaCO₃ and Ag₂CO₃ are white or pale. The tests also don't prove that the white silver precipitate was AgCl.
A better procedure: add dilute nitric acid to the sample. Fizzing, with a gas that turns limewater milky, would show carbonate: $\text{CO}_3^{2-}(aq) + 2\text{H}^+(aq) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l)$. Continue adding acid until fizzing stops, then add barium nitrate. A white precipitate now shows sulfate: $\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s)$. Filter, then add silver nitrate to the filtrate. A white precipitate that dissolves in dilute ammonia confirms chloride: $\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s)$.