Section 1 of 5Halide Ions
Halides are the negative ions of the Group 7 elements, the halogens, and each forms a characteristic precipitate with silver ions.
Chloride ion
- Chloride ion Cl⁻, from chlorine, gives a white precipitate of silver chloride when added to silver nitrate solution.
Bromide ion
- Bromide ion Br⁻, from bromine, gives a cream precipitate of silver bromide in the same test.
Iodide ion
- Iodide ion I⁻, from iodine, gives a yellow precipitate of silver iodide in the same test.
Halogens
- Halogens Chlorine, bromine and iodine are Group 7 non-metals that exist as diatomic molecules, such as Cl₂.
Section 2 of 5Carrying Out the Test
The method identifies halide ions by the colour of the precipitate that forms.
1
Place a small sample of the unknown solution into a test tube.
2
Add a few drops of dilute nitric acid to remove any carbonate or hydroxide ions that could also form precipitates.
3
Add a few drops of silver nitrate solution and observe whether a precipitate forms and what colour it is.
4
Compare the colour with the three known results, and use it to identify the halide ion present.
5
Write the ionic equation for the precipitation reaction, showing only the ions that react and the solid formed.
| Halide ion | Silver halide formed | Colour of precipitate | Ionic equation |
|---|---|---|---|
| Chloride, Cl⁻ | AgCl | White | Ag⁺(aq) + Cl⁻(aq) → AgCl(s) |
| Bromide, Br⁻ | AgBr | Cream | Ag⁺(aq) + Br⁻(aq) → AgBr(s) |
| Iodide, I⁻ | AgI | Yellow | Ag⁺(aq) + I⁻(aq) → AgI(s) |
Section 3 of 5Why the Method Works
Why the Method Works
Silver ions
- Silver ions Silver nitrate supplies Ag⁺ ions, which react with halide ions to form insoluble silver halides.
Dilute nitric acid
- Dilute nitric acid The acid is used because it contains no halide ions, so it cannot interfere with the result.
Insoluble solids
- Insoluble solids Silver halides are insoluble in water, so they separate as solid precipitates that can be seen.
Mixtures
- Mixtures A precipitate shows that a halide is present, but a mixture of halides gives a blended colour that is harder to interpret.
Section 4 of 5Worked Example
Question
- A colourless solution is acidified with dilute nitric acid and then treated with silver nitrate solution.
- A cream precipitate forms.
- Identify the halide ion and write the ionic equation.
- A cream precipitate matches silver bromide, AgBr
- So the halide ion present is bromide, Br⁻
- Ionic equationAg⁺(aq) + Br⁻(aq) → AgBr(s)
Answer Bromide ions are present, because the cream precipitate is silver bromide.
Section 5 of 5Exam Checklist & Tips
8.3.4 – Key Points for the Exam
- Halide ion A negative ion formed when a halogen atom gains one electron, such as Cl⁻ or I⁻.
- Precipitate An insoluble solid formed when two solutions are mixed.
- Ionic equation An equation that shows only the ions taking part in a reaction and any solid formed.
- Silver halide An insoluble compound of silver and a halogen, such as AgCl.
- Dilute nitric acid The acid used in the halide test to remove carbonate and hydroxide ions that could interfere.
Common Mistakes
- Using hydrochloric acid in the first step, which adds chloride ions and gives a false positive result for chloride.
- Omitting the acid, so carbonate ions could form a precipitate and wrongly suggest a halide is present.
- Describing silver halides as solutions rather than precipitates of solid particles.
- Confusing halogens with halide ions; chlorine is the element and chloride is the ion formed from it.