How to Plan a Sequence of Tests for Two Unknown Ions
Learn how to systematically identify two unknown ions using carefully selected reagents, fresh sample aliquots, and confirmatory tests. Explore worked examples covering precipitate colours, reagent interference, blank tests, and ambiguous results.
A white precipitate appears when you add a reagent to an unknown solution. You might be tempted to identify the ion immediately. But what if two different ions produce the same white precipitate? Or what if the reagent itself has introduced the ion you think you have discovered?
The solution is to plan your tests before you start mixing chemicals. A reliable sequence uses carefully chosen reagents, separate portions of the original sample, and at least one confirmatory test for any ambiguous result.
The best way to identify two unknown ions is to test the anion and cation independently, eliminate competing explanations, and then check that all the observations support the same identification.
This guide works through how to design that sequence, choose appropriate reagents, avoid interference, and justify your final identification in an HSC Chemistry response.
01Start with the possible ions, not the reagents
Assume you are given one aqueous sample containing exactly:
- One unknown cation (positive ion).
- One unknown anion (negative ion).
The question will often provide a list of possible ions. Your first job is to organise that list into two groups.
For example:
| Possible cations | Possible anions |
|---|---|
| Copper(II), \(\mathrm{Cu}^{2+}\) | Chloride, \(\mathrm{Cl}^{-}\) |
| Iron(II), \(\mathrm{Fe}^{2+}\) | Bromide, \(\mathrm{Br}^{-}\) |
| Iron(III), \(\mathrm{Fe}^{3+}\) | Iodide, \(\mathrm{I}^{-}\) |
| Zinc, \(\mathrm{Zn}^{2+}\) | Sulfate, \(\mathrm{SO_4}^{2-}\) |
| Aluminium, \(\mathrm{Al}^{3+}\) | Carbonate, \(\mathrm{CO_3}^{2-}\) |
You don’t need to perform every possible chemical test. You need to select tests that distinguish between the ions in the question.
For example, if the possible anions are chloride and sulfate, there is little point performing a carbonate test unless the question suggests carbonate could be present.
The four decisions behind a good test sequence
Before choosing your reagents, work through these decisions:
- What are the possible ions? Write down the candidate cations and anions separately.
- Which reagent distinguishes them? Look for a characteristic precipitate, gas, colour change, or complex formation.
- Could another ion produce the same observation? Identify possible interference before interpreting the result.
- How can you confirm the identification? Choose another test that provides independent chemical evidence.
This is the difference between recognising a colour and actually identifying an ion.
02Divide the original sample into separate portions
This is one of the most important practical decisions.
An aliquot is a small portion of a larger sample. You should use fresh aliquots for tests that require different reagents.
Why?
Because adding a reagent changes the chemical composition of the solution.
Suppose you add barium chloride to an unknown solution to test for sulfate ions. You then decide to test for chloride ions by adding silver nitrate to the same test tube.
You observe a white precipitate.
It might look like silver chloride, but there is a problem: the barium chloride you added earlier introduced chloride ions into the solution.
You have created the evidence you are trying to measure.
The correct approach is to take a fresh portion of the original sample for the silver nitrate test.
A practical arrangement
For a sample containing unknown cations and anions, you could prepare the following portions:
| Portion | Purpose | Reagents |
|---|---|---|
| A | Test for carbonate | Dilute acid, then limewater to identify the gas |
| B | Test for sulfate | Dilute nitric acid, then barium nitrate |
| C | Test for halides | Dilute nitric acid, then silver nitrate |
| D | Screen for cations | Sodium hydroxide, dropwise and then in excess |
| E | Confirm the cation | Aqueous ammonia, dropwise and then in excess |
| Blank | Check contamination | Deionised water and the same reagents |
You might not need all six portions in an actual question. Choose the tests that match the candidate ions.
Notice that the sequence does not mean adding every reagent to the same test tube.
Within an individual test, you may need to add reagents in a particular order. For example, add dilute nitric acid before silver nitrate. But different tests should generally begin with fresh aliquots.
03Choose the correct tests for the anion
Let’s look at the three main types of anion tests that help distinguish common unknown ions.
Test A: Carbonate ions
Carbonate ions react with dilute acids to produce carbon dioxide gas.
The net ionic equation is:
\[ \ce{CO3^2-(aq) + 2H+(aq) -> CO2(g) + H2O(l)} \]
You should observe effervescence, which means bubbles of gas are produced.
However, bubbles alone do not prove that the gas is carbon dioxide.
To confirm the gas, pass it through limewater, which is aqueous calcium hydroxide.
If carbon dioxide is present, the limewater turns milky because insoluble calcium carbonate forms.
\[ \ce{CO2(g) + Ca(OH)2(aq) -> CaCO3(s) + H2O(l)} \]
The white calcium carbonate precipitate causes the cloudy appearance.
The important distinction: effervescence is an initial observation, while identifying the gas provides stronger evidence for carbonate.
If carbonate is a possible interferent in your other anion tests, acidification is also important before adding silver or barium ions.
Test B: Sulfate ions
To test for sulfate, use dilute nitric acid followed by barium nitrate solution.
A positive result is a white precipitate of barium sulfate that remains insoluble in dilute acid.
The net ionic equation is:
\[ \ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)} \]
The barium ions combine with sulfate ions to form solid barium sulfate.
Why add nitric acid first?
Carbonate ions can also produce a white precipitate with barium ions:
\[ \ce{Ba^2+(aq) + CO3^2-(aq) -> BaCO3(s)} \]
If carbonate is present, it could give you a misleading positive result.
Adding dilute nitric acid first removes carbonate by converting it into carbon dioxide and water.
Barium sulfate, however, remains insoluble in dilute acid.
Therefore, a persistent white precipitate after acidification supports the presence of sulfate ions.
Test C: Halide ions
The common halide ions are chloride, bromide, and iodide.
To identify them, acidify a fresh aliquot with dilute nitric acid and then add silver nitrate solution.
The silver ions react with halide ions to form insoluble silver halides.
| Halide ion | Precipitate | Observation |
|---|---|---|
| Chloride, \(\mathrm{Cl}^{-}\) | \(\ce{AgCl}\) | White |
| Bromide, \(\mathrm{Br}^{-}\) | \(\ce{AgBr}\) | Cream |
| Iodide, \(\mathrm{I}^{-}\) | \(\ce{AgI}\) | Yellow |
For chloride, the net ionic equation is:
\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]
The colour of the precipitate helps narrow down the possible halide.
However, don’t rely on colour alone. Consider whether other ions could produce a similar observation and whether the correct preparation steps were followed.
Why use nitric acid instead of hydrochloric acid?
This is a common HSC examination trap.
Hydrochloric acid contains chloride ions. If you add it before silver nitrate, you introduce the very ion you are trying to detect.
Silver nitrate will then produce silver chloride even if the original sample contained no chloride.
Nitric acid avoids this problem because nitrate ions do not produce an insoluble silver nitrate reaction under these test conditions.
Comparing the anion tests
| Test | Reagent sequence | Positive observation | Main concern |
|---|---|---|---|
| Carbonate | Dilute acid, then limewater | Effervescence and milky limewater | Other gases may interfere with the limewater test |
| Sulfate | Dilute \(\ce{HNO3}\), then \(\ce{Ba(NO3)2}\) | Persistent white precipitate | Carbonate can also form a white precipitate |
| Halide | Dilute \(\ce{HNO3}\), then \(\ce{AgNO3}\) | White, cream, or yellow precipitate | Hydrochloric acid introduces chloride |
The key is to choose a reagent that reacts with the ion you want to identify without introducing another possible explanation.
04Choose the correct tests for the cation
Once you have planned your anion tests, use a separate aliquot to investigate the cation.
A common starting reagent is sodium hydroxide solution, \(\ce{NaOH}\).
Add it dropwise and record the initial precipitate. Then add it in excess and observe whether the precipitate dissolves.
Both observations matter.
What does sodium hydroxide tell you?
The following table summarises common HSC-level cation observations.
| Cation | With NaOH, dropwise | With excess NaOH |
|---|---|---|
| \(\mathrm{Cu}^{2+}\) | Pale blue precipitate | Remains insoluble |
| \(\mathrm{Fe}^{2+}\) | Pale green precipitate | Remains insoluble |
| \(\mathrm{Fe}^{3+}\) | Reddish-brown precipitate | Remains insoluble |
| \(\mathrm{Zn}^{2+}\) | White precipitate | Dissolves |
| \(\mathrm{Al}^{3+}\) | White precipitate | Dissolves |
The reactions occur because hydroxide ions combine with metal ions to form metal hydroxides.
For example, copper(II) ions form copper(II) hydroxide:
\[ \ce{Cu^2+(aq) + 2OH-(aq) -> Cu(OH)2(s)} \]
The pale blue precipitate is evidence for copper(II) ions.
Zinc ions form zinc hydroxide:
\[ \ce{Zn^2+(aq) + 2OH-(aq) -> Zn(OH)2(s)} \]
But zinc hydroxide dissolves in excess sodium hydroxide:
\[ \ce{Zn(OH)2(s) + 2OH-(aq) -> [Zn(OH)4]^2-(aq)} \]
This produces a colourless solution.
The important problem: zinc and aluminium look similar
Both zinc hydroxide and aluminium hydroxide are white precipitates.
Both also dissolve in excess sodium hydroxide.
This means a white precipitate that dissolves in excess sodium hydroxide does not uniquely identify zinc.
Aluminium hydroxide is also amphoteric, meaning it can react with both acids and bases.
\[ \ce{Al(OH)3(s) + OH-(aq) -> [Al(OH)4]-(aq)} \]
So how do you distinguish the two?
Use aqueous ammonia as a confirmatory test
Take another fresh aliquot and add aqueous ammonia, \(\ce{NH3}\), dropwise and then in excess.
The observations are different:
| Cation | With excess aqueous ammonia |
|---|---|
| \(\mathrm{Cu}^{2+}\) | Deep blue solution |
| \(\mathrm{Fe}^{2+}\) | Green precipitate, which can turn brown in air |
| \(\mathrm{Fe}^{3+}\) | Reddish-brown precipitate remains |
| \(\mathrm{Zn}^{2+}\) | White precipitate dissolves |
| \(\mathrm{Al}^{3+}\) | White precipitate remains |
Zinc ions form a soluble ammonia complex:
\[ \ce{Zn(OH)2(s) + 4NH3(aq) <=> [Zn(NH3)4]^2+(aq) + 2OH-(aq)} \]
Aluminium hydroxide does not dissolve in excess aqueous ammonia under these conditions.
This gives you a second way to distinguish zinc from aluminium.
A flame test can also provide useful supporting evidence for certain cations, but it should not automatically be treated as definitive. Contamination, particularly from sodium, can affect the observed flame colour.
05Worked example: Identify both ions in an unknown solution
Consider an unknown solution containing exactly one cation and one anion.
The possible cations are:
- \(\mathrm{Cu}^{2+}\)
- \(\mathrm{Fe}^{3+}\)
- \(\mathrm{Zn}^{2+}\)
- \(\mathrm{Al}^{3+}\)
The possible anions are:
- \(\mathrm{Cl}^{-}\)
- \(\mathrm{SO_4}^{2-}\)
The original solution is clear and colourless.
The following results are illustrative data, not measurements from an actual experiment.
| Test | Procedure | Observation |
|---|---|---|
| 1 | Add dilute nitric acid, followed by barium nitrate | No precipitate |
| 2 | Add dilute nitric acid, followed by silver nitrate | White precipitate |
| 3 | Add sodium hydroxide dropwise, then in excess | White precipitate forms and dissolves in excess |
| 4 | Add ammonia dropwise, then in excess to a fresh aliquot | White precipitate forms and dissolves in excess |
| Blank | Repeat the tests using deionised water | No precipitates |
Identify both unknown ions.
Step 1Identify the anion using the sulfate test
The first test produces no precipitate after acidification and the addition of barium nitrate.
If sulfate ions were present, they would react with barium ions to form insoluble barium sulfate.
Since no precipitate forms, sulfate ions are unlikely to be present under the test conditions.
Step 2Confirm the alternative anion
The second test produces a white precipitate after adding silver nitrate to an acidified fresh aliquot.
This is consistent with silver chloride.
\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]
The white precipitate supports the presence of chloride ions.
The negative sulfate test also eliminates the main competing explanation in the candidate list.
Therefore, the unknown anion is \(\mathrm{Cl}^{-}\).
Step 3Identify the cation using sodium hydroxide
A white precipitate forms and dissolves in excess sodium hydroxide.
This leaves two possible cations:
- Zinc, \(\mathrm{Zn}^{2+}\).
- Aluminium, \(\mathrm{Al}^{3+}\).
Both form white hydroxide precipitates that dissolve in excess sodium hydroxide.
You cannot distinguish them using this observation alone.
Step 4Use ammonia to confirm the cation
The fresh aliquot also produces a white precipitate with ammonia.
However, the precipitate dissolves in excess ammonia.
This is consistent with zinc ions forming a soluble ammonia complex.
Aluminium hydroxide would remain insoluble in excess ammonia.
Therefore, the unknown cation is \(\mathrm{Zn}^{2+}\).
Final identification
The unknown solution contains:
- Zinc ions, \(\mathrm{Zn}^{2+}\).
- Chloride ions, \(\mathrm{Cl}^{-}\).
The solution is consistent with aqueous zinc chloride, \(\ce{ZnCl2(aq)}\).
Notice that the identification required more than one positive observation. The negative sulfate result eliminated a competing anion, while the ammonia test resolved the ambiguity between two white hydroxide precipitates.
06Worked example: When the first test is not enough
An unknown solution contains either zinc ions or aluminium ions.
A student adds sodium hydroxide and observes:
- A white gelatinous precipitate forms.
- The precipitate dissolves when excess sodium hydroxide is added.
The student concludes that zinc ions must be present.
Is this conclusion justified?
No. Both zinc hydroxide and aluminium hydroxide dissolve in excess sodium hydroxide.
A second test is needed.
Step 1
Do not use the test tube containing sodium hydroxide. Its chemical composition has already changed.
Step 2
A white precipitate forms.
Step 3
There are two possible outcomes:
- If the precipitate dissolves, zinc ions are supported.
- If the precipitate remains, aluminium ions are supported.
Step 4
Zinc hydroxide forms a soluble ammonia complex, while aluminium hydroxide does not.
The important lesson is that a reagent can narrow down the possible ions without identifying one uniquely.
A confirmatory test should resolve the remaining possibilities, rather than simply repeat the same observation.
07Common sources of interference
A test sequence can look chemically convincing and still produce an incorrect identification.
These are the mistakes worth checking before accepting a result.
| Mistake | Why it causes a problem | Better approach |
|---|---|---|
| Using HCl before AgNO3 | Introduces chloride ions | Use dilute HNO3 |
| Using BaCl2 and then AgNO3 in the same tube | Introduces chloride ions that can form AgCl | Use fresh aliquots and barium nitrate |
| Adding barium nitrate without removing carbonate | Carbonate can form BaCO3 | Acidify with dilute HNO3 first |
| Reusing the same test tube | Previous reagents can change the sample | Use fresh aliquots |
| Identifying an ion from one precipitate colour | Different ions can produce similar observations | Use a confirmatory test |
| Ignoring a positive blank | Contamination may be producing the result | Investigate the reagents and equipment |
Why a blank test matters
A blank test uses deionised water instead of the unknown sample.
Everything else should remain the same:
- The same reagents.
- The same reagent volumes.
- The same equipment.
- The same experimental conditions.
For example, if you are testing for sulfate, your blank should contain deionised water, dilute nitric acid, and barium nitrate.
If the blank produces a white precipitate, you cannot confidently attribute the precipitate in your unknown sample to sulfate.
The reagent or equipment may be contaminated.
There is an important distinction here:
A positive blank does not prove that the unknown lacks the ion. It means the test cannot reliably distinguish the sample’s contribution from the background contamination.
For a more detailed explanation of interpreting these results, see How to Identify Contamination Using a Blank Test.
What if two tests give conflicting results?
Don’t immediately choose whichever result looks more convincing.
Work through the following possibilities:
- Did both tests use fresh aliquots?
- Did either reagent introduce the ion being investigated?
- Was the correct acid used?
- Could another ion have produced the same precipitate?
- Did the blank remain clear?
- Does the question allow mixtures of ions, or does it specify exactly one cation and one anion?
That final point is particularly important.
If a question specifies exactly one anion, a confirmed positive result for chloride rules out sulfate from the stated candidate list.
If mixtures are allowed, a positive chloride test does not establish that sulfate is absent.
Your conclusion must respect the assumptions given in the question.
08Questions and solutions
The following questions build from straightforward identification towards more complicated experimental reasoning. The observations are illustrative data.
Question 1
An unknown aqueous solution contains exactly one anion from the following possibilities:
- Chloride, \(\mathrm{Cl}^{-}\).
- Sulfate, \(\mathrm{SO_4}^{2-}\).
A student performs two tests using fresh aliquots.
Test A: The student adds dilute nitric acid, followed by barium nitrate. No precipitate forms.
Test B: The student adds dilute nitric acid, followed by silver nitrate. A white precipitate forms.
Identify the unknown anion.
Write the relevant net ionic equation and explain why nitric acid is preferred to hydrochloric acid.
Solution 1
The unknown anion is chloride, \(\mathrm{Cl}^{-}\).
The reasoning comes from combining the negative and positive results.
First, the negative barium nitrate test indicates that sulfate ions are unlikely to be present.
If sulfate ions were present, they would react with barium ions to produce white barium sulfate:
\[ \ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)} \]
The second test produces a white precipitate with silver nitrate.
This is consistent with silver chloride:
\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]
Since the candidate list contains only chloride and sulfate, the results identify chloride.
Nitric acid is used because hydrochloric acid would introduce chloride ions and could produce a false positive with silver nitrate.
Question 2
A student is investigating an unknown metal cation.
The possible ions are:
- Iron(II), \(\mathrm{Fe}^{2+}\).
- Iron(III), \(\mathrm{Fe}^{3+}\).
The student adds sodium hydroxide to a fresh aliquot.
A pale green precipitate forms immediately. After approximately ten minutes, the precipitate becomes reddish-brown.
A second fresh aliquot also produces a pale green precipitate immediately after sodium hydroxide is added.
(a) Identify the most likely cation.
(b) Explain why the precipitate changes colour when left standing.
(c) Write a balanced chemical equation for the reaction responsible for the colour change.
Solution 2
The unknown cation is iron(II), \(\mathrm{Fe}^{2+}\).
The initial pale green precipitate is iron(II) hydroxide, \(\ce{Fe(OH)2}\).
The formation reaction is:
\[ \ce{Fe^2+(aq) + 2OH-(aq) -> Fe(OH)2(s)} \]
Iron(II) hydroxide is unstable in contact with oxygen in the air. It is gradually oxidised to iron(III) hydroxide, which has a reddish-brown appearance.
The balanced equation is:
\[ \ce{4Fe(OH)2(s) + O2(g) + 2H2O(l) -> 4Fe(OH)3(s)} \]
The important observation is the initial colour, not simply the final colour.
If the student only recorded the reddish-brown precipitate after ten minutes, they might incorrectly identify iron(III).
The second fresh aliquot supports iron(II) because it reproduces the initial pale green observation.
This question shows why recording the timing of an observation can be just as important as recording its colour.
Question 3
An unknown solution contains either carbonate or sulfate ions.
A student performs the following investigation:
- An unacidified aliquot is treated with barium nitrate. A white precipitate forms.
- A portion of the isolated precipitate is treated with dilute nitric acid. The solid dissolves, producing bubbles.
- The gas is passed through limewater, which turns milky.
- A fresh aliquot is acidified with dilute nitric acid before barium nitrate is added. No precipitate forms.
Identify the original anion.
Explain why the initial white precipitate was insufficient to identify sulfate.
Solution 3
The original anion is carbonate, \(\mathrm{CO_3}^{2-}\).
The first white precipitate is ambiguous because both barium carbonate and barium sulfate are white solids.
The possible reactions are:
\[ \ce{Ba^2+(aq) + CO3^2-(aq) -> BaCO3(s)} \]
\[ \ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)} \]
The acid treatment provides the distinguishing evidence.
Barium carbonate reacts with hydrogen ions to release carbon dioxide:
\[ \ce{BaCO3(s) + 2H+(aq) -> Ba^2+(aq) + CO2(g) + H2O(l)} \]
The carbon dioxide then reacts with limewater:
\[ \ce{CO2(g) + Ca(OH)2(aq) -> CaCO3(s) + H2O(l)} \]
The resulting calcium carbonate makes the limewater milky.
Barium sulfate does not dissolve in dilute nitric acid in this way.
The final test provides further confirmation. Acidifying a fresh aliquot before adding barium nitrate removes carbonate ions, so no barium carbonate precipitate forms.
Together, these observations identify carbonate rather than sulfate.
Question 4
An unknown aqueous solution contains exactly one anion from the following possibilities:
- Chloride, \(\mathrm{Cl}^{-}\).
- Sulfate, \(\mathrm{SO_4}^{2-}\).
A student performs the following sequence in one test tube:
- Adds barium chloride to the original sample.
- Observes a white precipitate.
- Filters the mixture to remove the precipitate.
- Adds silver nitrate to the remaining filtrate.
- Observes another white precipitate.
The student concludes that both chloride and sulfate ions are present.
(a) Explain why the student’s conclusion is incorrect.
(b) Identify which anion is supported by the first test.
(c) Design a corrected sequence of tests that would avoid the interference.
Solution 4
The evidence supports sulfate ions, but the second test does not establish the presence of chloride ions.
The first test produces a white precipitate because barium ions react with sulfate ions to form insoluble barium sulfate.
\[ \ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)} \]
Under the stated candidate list, chloride ions would remain dissolved when barium chloride is added.
The problem occurs during the second test.
Barium chloride contains chloride ions. When the student adds it to the original sample, chloride ions are introduced into the test tube.
Filtering removes the barium sulfate precipitate, but it does not remove dissolved chloride ions.
When silver nitrate is subsequently added, those introduced chloride ions react with silver ions:
\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]
The resulting white precipitate is therefore not evidence that the original unknown contained chloride.
A corrected procedure would be:
Test 1: Sulfate identification
- Take a fresh aliquot of the original sample.
- Add dilute nitric acid.
- Add barium nitrate.
- A persistent white precipitate supports sulfate ions.
Test 2: Chloride identification
- Take another fresh aliquot.
- Add dilute nitric acid.
- Add silver nitrate.
- A white precipitate supports chloride ions.
Using barium nitrate rather than barium chloride also avoids introducing chloride during the sulfate test.
Since the question specifies exactly one possible anion, a positive sulfate result in the first test is sufficient to eliminate chloride.
Question 5
This final problem combines reagent interference, blank testing, and confirmatory cation identification.
An unknown solution is clear and colourless. It contains exactly:
- One cation from \(\mathrm{Zn}^{2+}\) or \(\mathrm{Al}^{3+}\).
- One anion from \(\mathrm{Cl}^{-}\) or \(\mathrm{SO_4}^{2-}\).
A student performs four tests using separate aliquots.
The following illustrative results are obtained.
| Investigation | Sample observation | Blank observation |
|---|---|---|
| Dilute HNO3 followed by Ba(NO3)2 | Faint white cloud | Faint white cloud |
| Dilute HNO3 followed by AgNO3 | White precipitate | Clear solution |
| NaOH, dropwise then excess | White precipitate dissolves | Not applicable |
| NH3, dropwise then excess | White precipitate remains | Not applicable |
The blank contains deionised water instead of the unknown sample. It receives the same reagents in the same quantities.
Assume the solutions are sufficiently dilute that silver sulfate does not produce a visible precipitate under the conditions used.
A student makes the following claim:
“The unknown is zinc sulfate because the barium nitrate test produced a white precipitate, and the sodium hydroxide test produced a white precipitate that dissolved in excess.”
Evaluate this claim.
Identify the most likely cation and anion. Explain what the blank tells you, and discuss what could be concluded if the original sample were allowed to contain mixtures of anions.
Solution 5
Under the stated assumption that exactly one anion is present, the most likely ions are aluminium, \(\mathrm{Al}^{3+}\), and chloride, \(\mathrm{Cl}^{-}\). The student’s identification of zinc sulfate is not justified.
There are three separate pieces of reasoning needed.
Step 1: Evaluate the sulfate test using the blank.
The sample produces a faint white cloud after barium nitrate is added.
At first, this appears to support sulfate ions.
However, the blank produces an identical observation.
This means the experimental procedure produces a white cloud even when the unknown sample is absent.
The result could be caused by sulfate contamination in the reagents, water, or equipment.
Consequently, the sulfate test is inconclusive.
It does not establish sulfate ions, but it also does not independently prove their absence.
Step 2: Interpret the silver nitrate test.
The acidified sample produces a white precipitate after silver nitrate is added.
The blank remains clear.
This is consistent with silver chloride:
\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]
Because the blank remains clear, there is no evidence that the silver nitrate procedure itself is introducing chloride contamination.
Under the stated candidate list and dilute test conditions, the positive result supports chloride ions.
Since the question specifies exactly one anion, sulfate is eliminated.
Step 3: Resolve the cation ambiguity.
The sodium hydroxide test produces a white precipitate that dissolves in excess reagent.
This is consistent with either zinc or aluminium.
The student has therefore not identified the cation yet.
The ammonia test provides the additional evidence.
A white precipitate remains even after excess ammonia is added.
This supports aluminium ions because aluminium hydroxide remains insoluble in excess ammonia.
Zinc hydroxide, by contrast, dissolves through the formation of a soluble ammonia complex.
Therefore, the unknown cation is aluminium, \(\mathrm{Al}^{3+}\).
Step 4: Consider what changes if mixtures are allowed.
This is the most important reasoning step.
Suppose the question did not specify exactly one anion.
The silver nitrate test would still provide evidence for chloride ions.
However, the contaminated blank would prevent you from deciding whether sulfate ions were also present.
You could conclude that chloride is present, but you could not confidently conclude that sulfate is absent.
The cation identification would remain aluminium because the sodium hydroxide and ammonia tests provide independent evidence.
This demonstrates why experimental controls and the assumptions in a question must be considered together.
A positive result is not automatically valid, and an inconclusive test is not the same as a negative test.
08A reliable method to remember in the HSC
When an exam asks you to plan a sequence of tests for two unknown ions, organise your response around the following principles:
- List the possible cations and anions separately.
- Select reagents that produce distinguishable observations.
- Use fresh aliquots of the original sample for independent tests.
- Avoid reagents that introduce the ions you are trying to identify.
- Record both the initial precipitate and what happens in excess reagent.
- Use a second chemical test when the first observation is not unique.
- Include appropriate controls and consider possible contamination.
- Check that your final identification is consistent with every observation and the assumptions in the question.
A strong experimental explanation follows a clear chain:
Reagent → observation → chemical reaction → identification.
For example:
“The addition of silver nitrate to the nitric-acid-acidified sample produced a white precipitate. This is consistent with the formation of insoluble silver chloride, supporting the presence of chloride ions.”
That is much stronger than simply writing “white precipitate means chloride”.
For a broader explanation of precipitation reactions and the solubility rules behind these tests, see How to Identify Two Unknown Ions Using Precipitation Tests.
Once you can plan and interpret an identification sequence, the next useful step is predicting what happens when two known ionic solutions are mixed. This brings together solubility rules, precipitation reactions, and net ionic equations, which are essential for explaining why a particular solid forms.