A bottle labelled "unknown salt" is not identified by a single chemistry word. It is identified by a chain:

For this lesson, the useful observations are fizzing from a gas, a white precipitate, and alkaline ammonia gas. The equations are not decoration: they explain why a positive test is positive.

Observation opener

Three colourless solutions are tested.

Before naming the ions, ask what each observation proves. A gas that turns limewater milky is carbon dioxide; a white precipitate with barium ions in acidified solution is barium sulfate; an alkaline gas formed on warming with hydroxide ions is ammonia.

Common-error contrast

"A white precipitate formed" is not a full identification by itself. The reagent and conditions matter. A white precipitate after acidified barium chloride has a different meaning from a white precipitate in a halide test with silver nitrate, which belongs outside this lesson.

Carbonate ions, CO3^2-, and hydrogencarbonate ions, HCO3^-, both react with aqueous acid to produce carbon dioxide. The visible sign is effervescence; the confirmatory gas test is that carbon dioxide turns limewater milky.

The ionic equations show the difference in proton demand:

CO3^2-(aq) + 2H+(aq) -> CO2(g) + H2O(l)

HCO3^-(aq) + H+(aq) -> CO2(g) + H2O(l)

Carbonate needs two protons because its charge is 2-; hydrogencarbonate already has one proton attached, so it needs one more proton to form carbonic acid, which breaks down to carbon dioxide and water. In a qualitative test, both ions can give the same carbon dioxide observation, so the equation is where the distinction becomes clear.

The limewater confirmation can be represented as:

CO2(g) + Ca(OH)2(aq) -> CaCO3(s) + H2O(l)

The milkiness is a suspension of calcium carbonate.

Worked example

A student adds dilute hydrochloric acid to an aqueous sample. The mixture effervesces, and the gas turns limewater milky.

1. Deduce the gas produced.
2. Write an ionic equation if the ion in the sample is carbonate.
3. Write an ionic equation if the ion in the sample is hydrogencarbonate.

Route

The limewater result identifies the gas as carbon dioxide. For carbonate:

CO3^2-(aq) + 2H+(aq) -> CO2(g) + H2O(l)

For hydrogencarbonate:

HCO3^-(aq) + H+(aq) -> CO2(g) + H2O(l)

Chemical interpretation

The observation proves that an acid-reactive carbonate-type ion is present. It does not, by observation alone, distinguish carbonate from hydrogencarbonate; the question must give enough context, or ask for both possible equations.

Sulfate ions, SO4^2-, are identified using acidified barium chloride solution. A positive result is a white precipitate of barium sulfate.

Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)

The acidified condition matters. Carbonate ions can also form an insoluble white barium salt if barium ions are added directly, so acid is used to remove carbonate or hydrogencarbonate interference by converting it to carbon dioxide and water. Do not acidify a sulfate test with sulfuric acid, because that would add sulfate ions and could create the very precipitate being tested for.

Practical reasoning

Use a clean portion of the sample solution. If the sample is a solid, dissolve a small amount in deionised water first, then acidify the solution and add barium chloride solution. The conclusion should name both the observation and the ion: a white precipitate with acidified barium chloride indicates sulfate ions.

Feynman diagnostic

Explain the sulfate test to a younger student without using the word "specific".

A strong explanation says: barium ions make an insoluble solid with sulfate ions, so the mixture turns cloudy or forms a white precipitate. The acid is added first so carbonate ions are destroyed as carbon dioxide instead of making their own white barium precipitate. The trap is saying "acid makes the sulfate"; it does not. The sulfate ion is already in the sample if the test is positive.

Ammonium ions, NH4+, are identified by adding sodium hydroxide solution and warming. Hydroxide ions remove a proton from ammonium ions, producing ammonia gas.

NH4+(aq) + OH^-(aq) -> NH3(g) + H2O(l)

Ammonia is an alkaline gas. The usual positive observation is that damp red litmus paper held near the mouth of the warm test tube turns blue. The paper must be damp because ammonia needs to dissolve in water on the paper before it can show alkaline behaviour.

Method logic

Use a small amount of sample in a test tube, add sodium hydroxide solution, and warm gently. Hold damp red litmus paper at the mouth of the tube without touching the liquid. Keep the mouth of the tube pointed away from people, and do not rely on direct smelling as the main test.

Worked example

An unknown salt is warmed with aqueous sodium hydroxide. A gas is produced that turns damp red litmus paper blue.

1. Deduce the gas.
2. Deduce the ion in the original salt.
3. Write the ionic equation.

Route

Damp red litmus turning blue shows an alkaline gas. In this test, the alkaline gas is ammonia, NH3. Ammonia formed only after adding hydroxide ions and warming, so the original sample contained ammonium ions, NH4+.

NH4+(aq) + OH^-(aq) -> NH3(g) + H2O(l)

Common-error contrast

Sodium hydroxide does not "contain ammonia". It supplies OH^- ions. The ammonium ion in the sample supplies the nitrogen and most of the hydrogen atoms in the ammonia molecule.

Qualitative analysis is strongest when each test is done on a fresh small portion of the unknown. That avoids one reagent contaminating the next test and producing a misleading observation.

State symbols are often worth including when the question asks for an equation from a practical observation:

Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)
NH4+(aq) + OH^-(aq) -> NH3(g) + H2O(l)

Recap

For Edexcel 9CH0, do not stop at naming the ion. Link reagent, condition, observation and ionic equation: acid plus carbonate-type ions gives CO2; acidified barium chloride plus sulfate gives BaSO4(s); warm sodium hydroxide plus ammonium gives NH3(g). That chain is what turns a remembered test into a justified chemical conclusion.