Revise the theory
This practical links to these notes pages:
- Group 2: The Alkaline Earth Metals (Unit 1 · Spec 3.2.2)
- Group 7: The Halogens (Unit 1 · Spec 3.2.3)
1. Practical Scope & Target Ions
Required Practical 3 requires candidates to perform systematic qualitative tests using dropper bottles, test tubes, and heating apparatus to identify unknown inorganic salts containing:
- Cations: Group 2 metal ions (Mg²⁺, Ca²⁺, Sr²⁺, Ba²⁺) and the ammonium ion (NH₄⁺).
- Anions: Halide ions (Cl⁻, Br⁻, I⁻), hydroxide ions (OH⁻), sulfate ions (SO₄²⁻), and carbonate ions (CO₃²⁻).
Safety and technique
- Wear eye protection.
- Add reagents drop by drop, with the dropper held above the mouth of the tube, never touching the solution.
- Swirl gently to mix.
- Look against a dark background for white precipitates.
- Warm the ammonium test gently in a water bath.
- Hold the litmus paper with tongs.
- Use concentrated ammonia in a fume cupboard.
2. Group 2 Cation Identification & Solubility Trends
The four alkaline earth metal cations are distinguished by exploiting the opposing solubility trends of their hydroxides and sulfates:
Hydroxide Solubility Trend (Down Group 2)
Solubility INCREASES down Group 2:
- Mg²⁺: Mg(OH)₂ is sparingly soluble → white precipitate.
- Ca²⁺: Ca(OH)₂ is slightly soluble → slight white precipitate.
- Sr²⁺: Sr(OH)₂ is fairly soluble → very slight white precipitate.
- Ba²⁺: Ba(OH)₂ is very soluble → no precipitate (colourless solution).
Sulfate Solubility Trend (Down Group 2)
Solubility DECREASES down Group 2:
- Mg²⁺: MgSO₄ is very soluble → no precipitate.
- Ca²⁺: CaSO₄ is slightly soluble → slight white precipitate.
- Sr²⁺: SrSO₄ is sparingly soluble → white precipitate.
- Ba²⁺: BaSO₄ is insoluble → thick white precipitate.
| Group 2 Cation | Addition of dilute NaOH(aq) | Addition of dilute H₂SO₄(aq) | Diagnostic Deduction |
|---|---|---|---|
| Magnesium (Mg²⁺) | White precipitate (Mg(OH)₂) | No precipitate | The only ion with a white precipitate with NaOH and none with H₂SO₄. |
| Calcium (Ca²⁺) | Slight white precipitate (Ca(OH)₂) | Slight white precipitate (CaSO₄) | A slight precipitate with both reagents. |
| Strontium (Sr²⁺) | Very slight white precipitate (Sr(OH)₂) | White precipitate (SrSO₄) | Only a very slight precipitate with NaOH, but a white precipitate with H₂SO₄. |
| Barium (Ba²⁺) | No precipitate (colourless solution) | Thick white precipitate (BaSO₄) | The only ion with no precipitate with NaOH and a thick white precipitate with H₂SO₄. |
Flame tests (Beyond the OxfordAQA specification)
Flame tests are not in the OxfordAQA 9620 specification, and you do not need them to identify these ions: the sodium hydroxide and sulfuric acid results above are enough. For reference, calcium gives a brick-red flame, strontium crimson and barium apple-green.
3. Test for the Ammonium Cation (NH₄⁺)
The ammonium ion does not form precipitates with most standard reagents. Its identification relies on generating volatile ammonia gas (NH₃):
2. Add approximately 1 cm³ of dilute sodium hydroxide (NaOH) solution.
3. Warm the test tube gently using a hot water bath (avoid boiling or spattering).
4. Hold a piece of damp red litmus paper over the mouth of the test tube (do not touch the glass walls).
5. Positive Observation: Damp red litmus paper turns blue due to alkaline ammonia gas:
NH₄⁺(aq) + OH⁻(aq) → NH₃(g) + H₂O(l).
In written examinations, stating "hold red litmus paper over the tube" loses the mark: the paper must be damp. Litmus responds to hydroxide ions in solution, so the ammonia gas must dissolve in the water on the paper: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. The hydroxide ions turn the litmus from red to blue. Do not let the paper touch the tube: any sodium hydroxide on the glass would turn it blue anyway.
4. Halide Anion Identification with Acidified AgNO₃ & Ammonia
Halide anions (Cl⁻, Br⁻, I⁻) are differentiated using silver nitrate followed by testing precipitate solubility in dilute and concentrated aqueous ammonia:
| Halide Ion | Addition of Acidified AgNO₃(aq) | Ionic Precipitation Equation | Effect of Dilute NH₃(aq) | Effect of Concentrated NH₃(aq) |
|---|---|---|---|---|
| Chloride (Cl⁻) | White precipitate | Ag⁺(aq) + Cl⁻(aq) → AgCl(s) |
Dissolves (forms colourless [Ag(NH₃)₂]⁺ complex) | Dissolves completely |
| Bromide (Br⁻) | Cream precipitate | Ag⁺(aq) + Br⁻(aq) → AgBr(s) |
Insoluble (precipitate remains) | Dissolves (forms [Ag(NH₃)₂]⁺ complex) |
| Iodide (I⁻) | Yellow precipitate | Ag⁺(aq) + I⁻(aq) → AgI(s) |
Insoluble | Insoluble (precipitate remains) |
Why these reagents?
- Silver nitrate is used because the silver halides are insoluble and each has its own colour.
- It is acidified with dilute nitric acid to remove carbonate and hydroxide ions, which would also form precipitates with silver ions. Hydrochloric acid must not be used because it adds chloride ions.
- Ammonia is added because the white, cream and yellow precipitates can be hard to tell apart; their solubility in dilute and concentrated ammonia confirms the halide. Solubility in ammonia: AgCl > AgBr > AgI.
A classic Unit 5 question asks: "Explain why the sample must be acidified with dilute nitric acid before adding silver nitrate."
Mark Scheme Answer: To react with and remove any carbonate (CO₃²⁻) or hydroxide (OH⁻) ions present as impurities. If not removed, silver carbonate (Ag₂CO₃) or silver oxide (Ag₂O) would form white/brown precipitates, giving a false positive test for chloride.
5. Tests for Sulfate (SO₄²⁻), Carbonate (CO₃²⁻) and Hydroxide (OH⁻) Anions
Sulfate Test (Acidified BaCl₂)
Protocol:
- Acidify the sample with dilute hydrochloric acid (HCl) to remove carbonate ions.
- Add barium chloride solution (BaCl₂).
- Observation: Formation of a thick white precipitate of barium sulfate:
Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s).
Why these reagents?
- Barium chloride is used because barium sulfate is insoluble, so sulfate ions give a white precipitate.
- It is acidified with dilute hydrochloric acid to remove carbonate ions, which would otherwise give a white precipitate of barium carbonate. Hydrochloric acid is fine because chloride ions give no precipitate with barium ions.
- Sulfuric acid must not be used because it adds sulfate ions.
Carbonate Test (Dilute Acid + Limewater)
Protocol:
- Add dilute hydrochloric acid (HCl) or nitric acid (HNO₃) to the solid or solution.
- Rapid effervescence (fizzing) occurs as CO₂ gas is released:
CO₃²⁻(aq) + 2H⁺(aq) → CO₂(g) + H₂O(l). - Bubble the gas through limewater (aqueous Ca(OH)₂).
- Observation: Limewater turns cloudy / milky due to fine white CaCO₃ precipitate:
Ca(OH)₂(aq) + CO₂(g) → CaCO₃(s) + H₂O(l).
Beyond the OxfordAQA specification: hydrogencarbonate ions (HCO₃⁻) also fizz with dilute acid and give carbon dioxide.
Hydroxide Test (Red Litmus Paper)
Hydroxide ions make the solution alkaline: red litmus paper turns blue (universal indicator turns blue or purple).
In the ammonium test the paper only meets the gas above the tube; here it touches the solution.
6. Worked Inorganic Deduction Problem
Problem: Solid X is a white crystalline inorganic salt. A student carries out three qualitative tests:
- Test 1: A portion of solid X is dissolved in water. Addition of dilute sodium hydroxide produces no visible precipitate. Warming this mixture produces a gas that turns damp red litmus paper blue.
- Test 2: A separate sample of solution X is acidified with dilute nitric acid. Addition of silver nitrate solution produces a cream precipitate. When dilute ammonia solution is added to this precipitate, it remains insoluble; however, upon adding concentrated ammonia, the precipitate completely dissolves to form a colourless solution.
- Test 3: Addition of acidified barium chloride solution to solution X gives no precipitate.
Deduce the identity of the cation and anion in Salt X. Write ionic equations for Test 1 and Test 2.
Deduction of Cation:
Warming with NaOH produces gas turning damp red litmus blue → NH₃ gas evolved.
Therefore, Cation is Ammonium (NH₄⁺).
Ionic equation: NH₄⁺(aq) + OH⁻(aq) → NH₃(g) + H₂O(l)
Deduction of Anion:
Acidified AgNO₃ gives a cream precipitate → silver bromide (AgBr).
Insoluble in dilute NH₃ but soluble in concentrated NH₃ confirms Bromide (Br⁻).
Ionic equation: Ag⁺(aq) + Br⁻(aq) → AgBr(s)
Final Formula of Salt X:
Salt X is Ammonium bromide (NH₄Br).
Conclusion: NH₄Br
7. Practice Exam Questions
Question 1: Which reagent can distinguish between aqueous solutions of magnesium chloride and barium chloride in a single test-tube step?
Show Answer & Explanation
Correct Answer: B
Adding sulfuric acid gives a thick white precipitate with barium ions (insoluble BaSO₄) but leaves magnesium ions in clear solution (soluble MgSO₄). Dilute NaOH could also distinguish them (giving white precipitate with Mg²⁺ but not Ba²⁺).
Question 2: Why is the solution acidified with dilute hydrochloric acid before barium chloride solution is added?
Show Answer & Explanation
Correct Answer: B
Carbonate ions would react with barium ions to give a white precipitate of barium carbonate, which looks like a positive sulfate result. The acid removes them as carbon dioxide and water: CO₃²⁻ + 2H⁺ → CO₂ + H₂O. Hydrochloric acid is used because chloride ions give no precipitate with barium ions; sulfuric acid must not be used because it adds sulfate ions.
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Exam-style question
3 marksDescribe tests you could use to distinguish between aqueous potassium chloride and aqueous potassium iodide. Give the results.
Show Mark Scheme
- Add dilute nitric acid, then silver nitrate solution, to each sample [1]
- Potassium chloride gives a white precipitate; potassium iodide gives a yellow precipitate [1]
- The white silver chloride precipitate dissolves in dilute ammonia; the yellow silver iodide does not dissolve, even in concentrated ammonia [1]