Why a White Precipitate Does Not Identify an Ion
A white precipitate can form from several different ions, so colour alone is not enough for identification. Learn how to use specific reagents, control interference, and apply confirmatory tests to identify unknown ions reliably.
A clear solution produces a white precipitate when you add a few drops of barium chloride. Does that prove sulfate ions are present?
Not by itself. Carbonate ions can also produce a white precipitate with barium chloride. So can sulfite ions. In fact, several different positive ions and negative ions form white, insoluble compounds in common HSC Chemistry tests.
The problem is that a precipitate’s colour tells you about the solid that formed, not uniquely about the ion that was originally in solution. To identify an unknown ion, you need to consider the reagent, the reaction conditions, and what happens when you perform a second test.
01A white precipitate is evidence, not an identification
Imagine three separate test tubes:
- One contains sulfate ions, and you add barium chloride.
- One contains chloride ions, and you add silver nitrate.
- One contains magnesium ions, and you add sodium hydroxide.
All three can produce a white precipitate. However, the solids are different:
| Ions reacting | White precipitate formed |
|---|---|
| \(\mathrm{Ba^{2+}}\) and \(\mathrm{SO_4^{2-}}\) | \(\mathrm{BaSO_4}\), barium sulfate |
| \(\mathrm{Ag^+}\) and \(\mathrm{Cl^-}\) | \(\mathrm{AgCl}\), silver chloride |
| \(\mathrm{Mg^{2+}}\) and \(\mathrm{OH^-}\) | \(\mathrm{Mg(OH)_2}\), magnesium hydroxide |
The same observation can therefore have several explanations. You cannot work backwards from the colour alone and uniquely identify the original ion.
A precipitation reaction occurs when dissolved ions combine to form a solid with sufficiently low solubility in water. The solid is called a precipitate, and its formation is evidence that particular ions have reacted.
For example, the net ionic equation for the reaction between barium ions and sulfate ions is:
\[ \mathrm{Ba^{2+}(aq)+SO_4^{2-}(aq)} \rightarrow \mathrm{BaSO_4(s)} \]
This equation tells us that barium sulfate forms. It does not tell us that every white solid produced by adding a barium-containing reagent must be barium sulfate.
The reagent matters just as much as the observation.
02Which reagents produce white precipitates?
Different reagents can help identify different ions, but several tests share the same visible result. You need to record the complete observation, including whether the precipitate dissolves in excess reagent.
1. Barium chloride: testing for sulfate ions
Adding barium chloride solution to a solution containing sulfate ions produces a white precipitate of barium sulfate.
\[ \mathrm{Ba^{2+}(aq)+SO_4^{2-}(aq)} \rightarrow \mathrm{BaSO_4(s)} \]
However, carbonate ions can also react with barium ions:
\[ \mathrm{Ba^{2+}(aq)+CO_3^{2-}(aq)} \rightarrow \mathrm{BaCO_3(s)} \]
Barium carbonate is also white. A white precipitate after adding barium chloride is therefore not enough to distinguish sulfate from carbonate.
The solution is to acidify the sample before adding the barium reagent.
Add dilute hydrochloric acid to a fresh sample, then add barium chloride solution. If sulfate ions are present, the white barium sulfate precipitate remains.
Carbonate ions behave differently. They react with the acid to produce carbon dioxide gas:
\[ \mathrm{CO_3^{2-}(aq)+2H^+(aq)} \rightarrow \mathrm{CO_2(g)+H_2O(l)} \]
The gas can be passed through limewater. If it is carbon dioxide, the limewater turns milky due to the formation of calcium carbonate.
| Observation | Interpretation |
|---|---|
| White precipitate after adding \(\mathrm{BaCl_2}\) to an untreated sample | Could be sulfate, carbonate, or another ion that forms an insoluble barium salt |
| Bubbles after adding dilute acid, with gas turning limewater milky | Evidence for carbonate ions |
| White precipitate remains after acidification and addition of \(\mathrm{BaCl_2}\) | Supports the presence of sulfate ions |
The acidification step is important because it removes carbonate as carbon dioxide before the barium test.
Sulfate ions remain in solution during this treatment, allowing them to form barium sulfate when the barium chloride is added.
For this test, dilute hydrochloric acid is suitable because chloride ions do not produce an insoluble precipitate with barium ions. Do not use sulfuric acid: it introduces sulfate ions and can create a false positive.
For a more detailed comparison of the two anions, see How to Distinguish Carbonate from Sulfate Ions in Solution.
2. Silver nitrate: testing for halide ions
Silver nitrate is commonly used to test for halide ions.
When silver ions react with halide ions, an insoluble silver halide may form.
\[ \mathrm{Ag^+(aq)+Cl^-(aq)} \rightarrow \mathrm{AgCl(s)} \]
Silver chloride is white. However, other halide ions produce different colours:
| Halide ion | Precipitate | Appearance |
|---|---|---|
| Chloride, \(\mathrm{Cl^-}\) | \(\mathrm{AgCl}\) | White |
| Bromide, \(\mathrm{Br^-}\) | \(\mathrm{AgBr}\) | Cream |
| Iodide, \(\mathrm{I^-}\) | \(\mathrm{AgI}\) | Yellow |
The colours are useful, but they should not be treated as infallible identification tests. Concentration, lighting, and contamination can affect what you observe.
There is another important complication: silver ions can also form white precipitates with some other anions, including carbonate ions.
This is why the sample should be acidified with dilute nitric acid before adding silver nitrate.
Nitric acid removes carbonate ions by converting them into carbon dioxide and water. It does not introduce chloride ions into the sample.
Never use hydrochloric acid to acidify a sample before a silver nitrate test.
Hydrochloric acid contains chloride ions. Adding it can produce silver chloride even when the original unknown solution contained no chloride ions.
\[ \mathrm{Ag^+(aq)+Cl^-(aq)} \rightarrow \mathrm{AgCl(s)} \]
You would be detecting the chloride ions from your own reagent rather than identifying the unknown.
Confirming a suspected chloride ion
Silver chloride has another useful property: it dissolves in dilute aqueous ammonia.
\[ \mathrm{AgCl(s)+2NH_3(aq)} \rightleftharpoons \mathrm{[Ag(NH_3)_2]^+(aq)+Cl^-(aq)} \]
This provides additional evidence for chloride ions.
Silver bromide behaves differently. It dissolves only in concentrated ammonia, while silver iodide remains insoluble.
The ammonia test is useful because it checks a property of the precipitate beyond its initial colour.
For more detail, see How to Distinguish Chloride, Bromide and Iodide Using Silver Nitrate.
3. Sodium hydroxide: testing for metal cations
White precipitates are also common in tests for positive metal ions.
When sodium hydroxide is added to a solution containing certain metal cations, hydroxide ions combine with the metal ions to form insoluble metal hydroxides.
For example:
\[ \mathrm{Mg^{2+}(aq)+2OH^-(aq)} \rightarrow \mathrm{Mg(OH)_2(s)} \]
The magnesium hydroxide is white.
But magnesium is not the only metal ion that produces a white hydroxide.
| Metal ion | Precipitate with \(\mathrm{NaOH}\) | Behaviour in excess \(\mathrm{NaOH}\) |
|---|---|---|
| \(\mathrm{Mg^{2+}}\) | White | Remains insoluble |
| \(\mathrm{Al^{3+}}\) | White | Dissolves |
| \(\mathrm{Zn^{2+}}\) | White | Dissolves |
| \(\mathrm{Pb^{2+}}\) | White | Dissolves |
The behaviour in excess sodium hydroxide is often more informative than the initial colour.
Why?
Some metal hydroxides are amphoteric. This means they can react with both acids and bases. Aluminium hydroxide, zinc hydroxide, and lead(II) hydroxide are examples.
For instance, aluminium hydroxide dissolves in excess hydroxide ions to form a soluble complex ion:
\[ \mathrm{Al(OH)_3(s)+OH^-(aq)} \rightarrow \mathrm{[Al(OH)_4]^-(aq)} \]
Magnesium hydroxide does not behave this way under the usual qualitative test conditions. Its white precipitate remains.
This gives you a useful decision rule:
- White precipitate that remains in excess sodium hydroxide: magnesium is a possible candidate.
- White precipitate that dissolves in excess sodium hydroxide: aluminium, zinc, and lead(II) are possible candidates.
Notice the wording: possible candidates. You have narrowed the options, not necessarily identified one ion.
Use ammonia to narrow the possibilities further
A second reagent can help distinguish metal ions that behave similarly with sodium hydroxide.
Ammonia solution is a weak base. It produces hydroxide ions in water, so it can initially form white metal hydroxide precipitates too.
However, ammonia can also form soluble complex ions with some metal cations.
Zinc ions are a useful example.
Initially, zinc ions produce a white precipitate of zinc hydroxide:
\[ \mathrm{Zn^{2+}(aq)+2OH^-(aq)} \rightarrow \mathrm{Zn(OH)_2(s)} \]
In excess ammonia, the precipitate dissolves as a soluble zinc-ammonia complex forms:
\[ \mathrm{Zn(OH)_2(s)+4NH_3(aq)} \rightleftharpoons \mathrm{[Zn(NH_3)_4]^{2+}(aq)+2OH^-(aq)} \]
This behaviour helps distinguish zinc ions from aluminium and lead(II) ions under the usual qualitative test conditions.
| Metal ion | Initial observation with ammonia | In excess ammonia |
|---|---|---|
| \(\mathrm{Zn^{2+}}\) | White precipitate | Dissolves |
| \(\mathrm{Al^{3+}}\) | White precipitate | Remains insoluble |
| \(\mathrm{Pb^{2+}}\) | White precipitate | Remains insoluble |
| \(\mathrm{Mg^{2+}}\) | White precipitate | Remains insoluble |
The combination of sodium hydroxide and ammonia tests gives you more information than either test alone.
A further test can distinguish lead(II) from aluminium. Adding potassium iodide to a fresh portion of the unknown produces a bright yellow precipitate if lead(II) ions are present:
\[ \mathrm{Pb^{2+}(aq)+2I^-(aq)} \rightarrow \mathrm{PbI_2(s)} \]
Aluminium ions do not produce this precipitate under the same conditions.
This is the central idea of qualitative analysis: combine observations from different reactions until the remaining explanations are sufficiently restricted.
03Why interference can produce a false positive
An interfering ion is an ion that reacts during a test and produces an observation that could be mistaken for evidence of the ion you are trying to identify.
There are several common ways this happens.
1. Another ion forms the same coloured precipitate
The simplest problem is that different ionic compounds can have the same appearance.
For example, both barium sulfate and barium carbonate are white. Seeing a white precipitate after adding barium chloride does not tell you which one formed.
The acidification step resolves this particular ambiguity.
2. A reagent introduces the ion being tested for
This is one of the easiest mistakes to make.
Suppose you want to test an unknown solution for chloride ions. You add hydrochloric acid, followed by silver nitrate.
You observe a white precipitate.
The problem is that hydrochloric acid already contains chloride ions. The silver nitrate can react with those added ions, creating silver chloride.
The test has become unreliable because the reagent has introduced the very ion you are trying to detect.
Choose reagents that do not introduce the ion being tested for.
3. More than one ion is present
An unknown solution does not necessarily contain only one anion or one cation.
A sample containing both chloride and sulfate ions can produce a white precipitate in a silver nitrate test and a white precipitate in a barium nitrate test.
These observations do not contradict each other. They may indicate two different ions.
You should use separate portions of the original solution when performing independent tests. Otherwise, an earlier reagent may change the composition of the sample and interfere with later results.
4. The conditions affect whether a precipitate forms
The concentration of the ions and the pH of the solution can affect precipitation.
A precipitate may not form visibly if the relevant ion concentration is too low. Conversely, changing the pH can cause a hydroxide precipitate to form or dissolve.
Adding excess reagent can also change the result if a soluble complex forms.
Therefore, an observation such as “no precipitate formed” does not automatically prove that an ion is absent.
It means that no visible precipitate formed under the conditions of that particular test.
5. Contamination can imitate a positive result
Dirty glassware, contaminated droppers, or impure reagents can introduce ions that form precipitates.
A blank test helps identify this problem. You perform the same procedure using distilled water instead of the unknown sample.
If the blank produces the same precipitate, the result may be caused by contamination in the reagents or apparatus rather than ions in the unknown.
See How to Identify Contamination Using a Blank Test.
04A reliable approach to identifying an unknown ion
When you encounter a white precipitate, avoid jumping straight to an identification.
Instead, build your conclusion from a sequence of observations.
- Identify the reagent. Record exactly what was added. A white precipitate after silver nitrate means something different from one after sodium hydroxide.
- Check the preparation. Was the sample acidified first? Was the correct acid used?
- Record the initial observation. Include the precipitate’s colour and whether the solution became cloudy or formed a substantial solid.
- Test its behaviour. Add excess reagent where appropriate. Does the precipitate dissolve or remain?
- Use an independent confirmatory test. Choose a different reaction that tests another property of the suspected ion.
- Check for interference. Consider other ions, contamination, and whether an earlier reagent introduced an interfering species.
A useful distinction is the difference between evidence that is consistent with an ion and evidence that confirms it within the stated set of possible ions.
For example, a white precipitate after adding barium chloride is consistent with sulfate ions. A white precipitate that persists after suitable acidification is stronger evidence for sulfate because carbonate and sulfite are removed by acid treatment.
An additional independent test can make the conclusion more convincing.
05Worked example: Is the white precipitate barium sulfate?
A student adds excess barium chloride to 10.0 mL of an unknown solution containing 0.0200 mol L\(^{-1}\) sulfate ions. Calculate the theoretical mass of white precipitate formed. Explain why the same calculation would not identify sulfate if the original observation were simply an unidentified white precipitate.
Step 1
\[ \mathrm{Ba^{2+}(aq)+SO_4^{2-}(aq)} \rightarrow \mathrm{BaSO_4(s)} \]
The mole ratio between sulfate ions and barium sulfate is 1:1.
Step 2
Use:
\[ n=cV \]
where \(n\) is amount in moles, \(c\) is concentration in mol L\(^{-1}\), and \(V\) is volume in litres.
Convert the volume:
\[ 10.0\text{ mL}=0.0100\text{ L} \]
Then:
\[ n(\mathrm{SO_4^{2-}}) =0.0200\times0.0100 \]
\[ =2.00\times10^{-4}\text{ mol} \]
Step 3
Because the ratio is 1:1:
\[ n(\mathrm{BaSO_4}) =2.00\times10^{-4}\text{ mol} \]
Step 4
\[ M(\mathrm{BaSO_4}) =137.33+32.06+4(16.00) \]
\[ =233.39\text{ g mol}^{-1} \]
Step 5
Use:
\[ m=nM \]
\[ m=2.00\times10^{-4}\times233.39 \]
\[ \boxed{m=0.0467\text{ g}=46.7\text{ mg}} \]
The theoretical mass of barium sulfate is 46.7 mg.
This calculation predicts how much solid forms when sulfate ions react completely with excess barium ions. It does not establish the identity of an unknown solid from its mass or colour alone.
The calculation assumes complete precipitation, no loss during collection, and a pure, dry precipitate. If another ion also forms an insoluble barium salt, the measured mass could include more than one compound.
06Worked example: Distinguishing aluminium, zinc, and lead(II)
An unknown solution produces a white precipitate when sodium hydroxide is added dropwise. The precipitate dissolves in excess sodium hydroxide.
When ammonia is added to a fresh sample, another white precipitate forms. This precipitate dissolves in excess ammonia.
Identify the most likely metal ion from the following candidates: \(\mathrm{Al^{3+}}\), \(\mathrm{Zn^{2+}}\), and \(\mathrm{Pb^{2+}}\).
Step 1
All three candidates can produce white hydroxide precipitates that dissolve in excess sodium hydroxide.
For example:
\[ \mathrm{Zn^{2+}+2OH^-\rightarrow Zn(OH)_2(s)} \]
The precipitate dissolves in excess hydroxide because zinc hydroxide is amphoteric.
The first test therefore narrows the candidates to aluminium, zinc, and lead(II), but it does not distinguish between them.
Step 2
Zinc hydroxide dissolves in excess ammonia because zinc ions form a soluble complex:
\[ \mathrm{Zn(OH)_2(s)+4NH_3(aq)} \rightleftharpoons \mathrm{[Zn(NH_3)_4]^{2+}(aq)+2OH^-(aq)} \]
Aluminium hydroxide and lead(II) hydroxide do not show the same dissolution behaviour in excess ammonia under the usual test conditions.
Step 3
\[ \boxed{\mathrm{Zn^{2+}}} \]
Zinc ions are the most likely candidate because they produce a white hydroxide precipitate that dissolves in excess sodium hydroxide and excess ammonia.
The important point is that the initial white precipitate was not enough. Its behaviour with a second reagent provided the distinguishing evidence.
07Questions and solutions
Question 1
A student adds dilute hydrochloric acid to an unknown solution. No visible gas is produced. The student then adds barium chloride and observes a white precipitate that remains after mixing.
Which anion is most strongly supported by these observations? Write the net ionic equation for the precipitation reaction.
Assume the sample contains no other ions that form acid-resistant, insoluble barium salts.
Solution 1
The observations support the presence of sulfate ions, \(\mathrm{SO_4^{2-}}\).
Carbonate ions would react with hydrochloric acid to produce carbon dioxide gas. The absence of gas makes carbonate less likely, while the persistent white precipitate after adding barium chloride is consistent with insoluble barium sulfate.
The net ionic equation is:
\[ \mathrm{Ba^{2+}(aq)+SO_4^{2-}(aq)} \rightarrow \mathrm{BaSO_4(s)} \]
The precipitate remains because barium sulfate is insoluble in dilute hydrochloric acid.
The conclusion depends on the stated assumption that no other acid-resistant barium precipitate-forming ions are present. In an actual unknown sample, additional confirmatory evidence would strengthen the identification.
Question 2
A student takes 20.0 mL of an unknown solution and adds excess barium nitrate. A white precipitate forms.
The concentration of the anion is known to be 0.0150 mol L\(^{-1}\). Assume the anion is either sulfate or carbonate, but not both.
(a) Calculate the theoretical mass of precipitate if the anion is sulfate.
(b) Calculate the theoretical mass of precipitate if the anion is carbonate.
(c) Explain why observing a white precipitate alone cannot identify which anion is present, even if the amount of precipitate is measured accurately.
Use:
- \(M(\mathrm{BaSO_4})=233.39\text{ g mol}^{-1}\)
- \(M(\mathrm{BaCO_3})=197.34\text{ g mol}^{-1}\)
Solution 2
(a) The theoretical mass of barium sulfate is 70.0 mg.
First, calculate the amount of anion:
\[ n=cV \]
\[ n=0.0150\times0.0200 \]
\[ n=3.00\times10^{-4}\text{ mol} \]
The reaction has a 1:1 mole ratio between sulfate ions and barium sulfate.
Therefore:
\[ m=nM \]
\[ m=3.00\times10^{-4}\times233.39 \]
\[ \boxed{m=0.0700\text{ g}=70.0\text{ mg}} \]
(b) The theoretical mass of barium carbonate is 59.2 mg.
The reaction between barium ions and carbonate ions also has a 1:1 mole ratio.
\[ m=3.00\times10^{-4}\times197.34 \]
\[ \boxed{m=0.0592\text{ g}=59.2\text{ mg}} \]
(c) Both reactions produce white precipitates, so colour alone cannot distinguish them.
If the anion concentration is known accurately and the precipitate is collected and weighed without loss, the measured mass can help distinguish these two possibilities.
However, that conclusion depends on the assumption that only one of the two anions is present and that precipitation is complete.
In a real unknown containing a mixture, the total mass could represent contributions from both barium salts. A chemical test is still needed to establish the identity of the anion.
Acidifying a fresh aliquot before adding barium nitrate provides a more direct distinction: carbonate is converted to carbon dioxide, while sulfate remains available to form barium sulfate.
Question 3
An unknown solution gives the following results:
| Test | Observation |
|---|---|
| A few drops of sodium hydroxide | White precipitate |
| Excess sodium hydroxide | Precipitate dissolves |
| A few drops of ammonia to a fresh sample | White precipitate |
| Excess ammonia | Precipitate remains |
The possible metal ions are \(\mathrm{Al^{3+}}\), \(\mathrm{Zn^{2+}}\), \(\mathrm{Pb^{2+}}\), and \(\mathrm{Mg^{2+}}\).
(a) Which ions remain possible after the sodium hydroxide test?
(b) Which ions remain possible after the ammonia test?
(c) Suggest a further test to distinguish between the remaining candidates, and state its expected observation.
Solution 3
(a) Aluminium, zinc, and lead(II) remain possible after the sodium hydroxide test.
All three form white hydroxide precipitates that dissolve in excess sodium hydroxide because their hydroxides are amphoteric.
Magnesium hydroxide is not soluble in excess sodium hydroxide under the usual test conditions. Magnesium ions can therefore be excluded from the candidate list.
(b) Aluminium and lead(II) remain possible after the ammonia test.
Zinc hydroxide dissolves in excess ammonia because zinc forms a soluble ammine complex.
Aluminium hydroxide and lead(II) hydroxide remain as white precipitates.
The ammonia result therefore eliminates zinc.
(c) Add potassium iodide to a fresh portion of the unknown solution.
If lead(II) ions are present, a bright yellow precipitate of lead(II) iodide forms:
\[ \mathrm{Pb^{2+}(aq)+2I^-(aq)} \rightarrow \mathrm{PbI_2(s)} \]
A yellow precipitate supports lead(II) ions. Aluminium ions do not produce this precipitate under the same conditions.
The sequence matters. Each test removes possible candidates based on a different chemical property rather than relying on the original white colour.
Question 4
A student wants to test an unknown solution for chloride ions.
They add dilute hydrochloric acid, followed by silver nitrate. A white precipitate forms, and the student concludes that chloride ions were present in the original unknown.
(a) Explain the error in the student’s method.
(b) Write the net ionic equation responsible for the misleading observation.
(c) Describe a corrected procedure.
(d) Explain how ammonia can provide further evidence for chloride ions.
Solution 4
(a) The student has introduced chloride ions through the hydrochloric acid reagent.
The white precipitate cannot be confidently attributed to the original unknown solution.
Hydrochloric acid contains chloride ions, so the test may produce a positive result even if the unknown contained no chloride ions at all.
(b) Silver ions react with the chloride ions introduced by the acid.
\[ \mathrm{Ag^+(aq)+Cl^-(aq)} \rightarrow \mathrm{AgCl(s)} \]
The resulting silver chloride is white.
(c) Use dilute nitric acid instead of hydrochloric acid.
The corrected procedure is:
- Take a fresh portion of the unknown solution.
- Add dilute nitric acid to acidify the sample and remove carbonate interference.
- Add silver nitrate solution.
- Observe whether a white precipitate forms.
Nitric acid does not introduce chloride ions, so it avoids the contamination caused by hydrochloric acid.
(d) Add dilute aqueous ammonia to the isolated white precipitate.
Silver chloride dissolves in dilute ammonia, forming the soluble complex ion \(\mathrm{[Ag(NH_3)_2]^+}\).
This behaviour provides further evidence that the precipitate is silver chloride.
The important distinction is that the confirmatory test examines the behaviour of the solid itself, while the initial silver nitrate test examines whether an insoluble silver compound forms.
Question 5
An unknown solution may contain any combination of carbonate, sulfate, and chloride ions. Assume there are no other ions that interfere with the tests, all reagents are pure, and the tests are sufficiently sensitive to detect the ions present.
A student performs the following investigation.
Test A: A fresh portion of the unknown is treated with dilute nitric acid. Bubbles form, and the gas turns limewater milky.
Test B: A second portion is treated with dilute nitric acid until bubbling stops. Excess barium nitrate is then added, producing a white precipitate. The mixture is filtered.
Test C: Silver nitrate is added to the filtrate from Test B. A white precipitate forms. This precipitate dissolves when dilute ammonia is added.
A second student repeats Test C using a fresh portion of the original unknown, but uses dilute hydrochloric acid instead of nitric acid before adding silver nitrate. They also obtain a white precipitate.
(a) Identify which ions are supported by Tests A, B, and C.
(b) Explain why the order of reactions in Tests B and C allows the student to obtain evidence for two different anions using the same original solution.
(c) The filtrate in Test B still contains barium ions from the excess barium nitrate. Explain why this does not invalidate Test C.
(d) Explain why the second student’s result does not provide independent confirmation of chloride ions.
(e) A student argues that because all three positive observations involve white or cloudy substances, they must be detecting the same ion. Explain precisely why this reasoning fails.
Solution 5
(a) The results support the presence of carbonate, sulfate, and chloride ions in the original unknown solution.
Test A supports carbonate ions.
Carbonate ions react with hydrogen ions to form carbon dioxide and water:
\[ \mathrm{CO_3^{2-}(aq)+2H^+(aq)} \rightarrow \mathrm{CO_2(g)+H_2O(l)} \]
The gas turns limewater milky because calcium carbonate forms.
Test B supports sulfate ions.
Nitric acid removes carbonate ions by converting them into carbon dioxide. Sulfate ions remain in solution and react with barium ions to form insoluble barium sulfate:
\[ \mathrm{Ba^{2+}(aq)+SO_4^{2-}(aq)} \rightarrow \mathrm{BaSO_4(s)} \]
The persistent white precipitate is therefore evidence for sulfate ions.
Test C supports chloride ions.
The white precipitate is consistent with silver chloride:
\[ \mathrm{Ag^+(aq)+Cl^-(aq)} \rightarrow \mathrm{AgCl(s)} \]
Its dissolution in dilute ammonia provides further evidence because silver chloride forms a soluble silver-ammonia complex.
(b) The order of reactions separates the ions according to their different chemical behaviours.
First, nitric acid removes carbonate ions as carbon dioxide. This prevents carbonate from forming another white precipitate with barium ions.
Next, barium nitrate reacts with sulfate ions to form solid barium sulfate. Filtering removes this precipitate from the mixture.
Chloride ions remain dissolved in the filtrate. They can then react with silver ions to form silver chloride.
The key is that each reaction has a different purpose:
- Acidification removes carbonate interference.
- Barium nitrate removes sulfate ions from the solution.
- Silver nitrate tests the remaining solution for chloride ions.
The original white precipitate from Test B and the white precipitate from Test C are different compounds formed by different ions.
(c) The remaining barium ions do not invalidate the silver nitrate test because barium nitrate and silver nitrate are both soluble in water.
After filtration, excess \(\mathrm{Ba^{2+}}\) ions may remain in the solution.
However, adding silver nitrate introduces \(\mathrm{Ag^+}\) and \(\mathrm{NO_3^-}\) ions. Nitrate ions remain dissolved, and the remaining barium ions do not form an insoluble precipitate with silver ions.
The chloride ions are therefore still available to form silver chloride.
This is an important detail: removing the barium sulfate solid does not require removing every dissolved ion from the filtrate. It only requires ensuring that the remaining ions do not interfere with the next test.
(d) The second student’s result is unreliable because hydrochloric acid introduces chloride ions.
The added hydrochloric acid contains \(\mathrm{Cl^-}\) ions, which react with silver ions to form white silver chloride.
\[ \mathrm{Ag^+(aq)+Cl^-(aq)} \rightarrow \mathrm{AgCl(s)} \]
The precipitate could therefore form even if the original unknown contained no chloride ions.
The first student’s Test C is more reliable because nitric acid was used, and barium nitrate does not introduce chloride ions.
The second test does not provide independent confirmation. It repeats the same visible observation using a reagent that can create the result by itself.
(e) The reasoning fails because the appearance of a precipitate does not uniquely identify the ions that produced it.
Test A produces cloudy limewater because calcium carbonate forms when carbon dioxide reacts with calcium hydroxide.
Test B produces solid barium sulfate.
Test C produces solid silver chloride.
All three observations involve white or cloudy substances, but the chemical species and reactions are different.
The identity of an ion must be inferred from the complete reaction sequence, including the reagents used, the ions they introduce, the solubility of the products, and the behaviour of the products in further tests.
This final investigation also shows why the order of qualitative tests matters. A reagent that is perfectly suitable in one test can contaminate a later test if it introduces an ion that the later test is designed to detect.
08The key distinction: observation versus identification
When you see a white precipitate, record exactly what happened rather than immediately naming an ion.
A strong chemical identification follows a chain of evidence:
- Which reagent produced the precipitate?
- Which ions could have reacted with that reagent?
- Could another ion produce the same colour?
- Did the precipitate dissolve in acid or excess reagent?
- Could the reagents themselves have introduced interfering ions?
- Does an independent confirmatory test support the same conclusion?
A white precipitate is the beginning of an investigation, not its conclusion.
Once you can distinguish an initial observation from a properly supported identification, the next useful step is learning how to plan the order of tests for an unknown containing multiple ions. This is where selective precipitation, interference, and confirmatory reactions come together.