How to Distinguish Chloride, Bromide and Iodide Using Silver Nitrate

Three clear solutions are each mixed with silver nitrate. One produces a white precipitate, another produces a cream precipitate, and the third produces a yellow precipitate. Which contains chloride, bromide, or iodide?

The standard test uses two stages. First, acidify the sample with dilute nitric acid and add silver nitrate. Then filter and wash the precipitate before testing its solubility in dilute and concentrated aqueous ammonia.

The colours give you an initial identification. The ammonia test provides stronger confirmation, particularly when two halides are present together.

01The results: chloride, bromide, and iodide

Memorise this table, but understand what each observation tells you.

These observations provide the central identification method:

• White precipitate that dissolves in dilute ammonia: chloride.
• Cream precipitate that dissolves in concentrated ammonia, but not dilute ammonia: bromide.
• Yellow precipitate that remains insoluble in concentrated ammonia: iodide.

There is an important detail here.

Suppose you obtain a cream precipitate that does not dissolve in dilute ammonia. Can you immediately conclude that bromide is present?

No. Both silver bromide and silver iodide remain largely undissolved in dilute ammonia. You need the concentrated ammonia test to distinguish them.

02Why does silver nitrate produce three different precipitates?

Silver nitrate, \(\ce{AgNO3}\), dissolves in water to produce silver ions and nitrate ions:

\[ \ce{AgNO3(aq) -> Ag+(aq) + NO3-(aq)} \]

The silver ions react with dissolved halide ions to form insoluble silver halides.

The three net ionic equations are:

Chloride:

\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]

Bromide:

\[ \ce{Ag+(aq) + Br-(aq) -> AgBr(s)} \]

Iodide:

\[ \ce{Ag+(aq) + I-(aq) -> AgI(s)} \]

Notice that silver ions are involved in all three reactions. The difference is the halide ion they react with.

The products are solids, which means they separate from the solution as precipitates.

Why are the three silver halides different?

The three silver halides have different solubilities in water.

Their relative solubility follows this order:

\[ \ce{AgCl > AgBr > AgI} \]

Silver chloride is the most soluble of the three, while silver iodide is the least soluble.

This difference becomes particularly useful when ammonia is introduced.

Silver iodide is much more difficult to dissolve than silver chloride or silver bromide. That difference allows us to distinguish the precipitates even when their initial colours are difficult to tell apart.

03The experimental method: step by step

You need:

• The unknown aqueous solution.
• Dilute nitric acid, \(\ce{HNO3}\).
• Silver nitrate solution, \(\ce{AgNO3}\).
• Dilute aqueous ammonia, \(\ce{NH3}\).
• Concentrated aqueous ammonia.
• Clean test tubes, a test tube rack, and separate droppers.
• Deionised water for washing.
• A white background for observing precipitate colours.

Follow this sequence.

Step 1: Prepare the unknown solution.

Place a small volume of the unknown solution into a clean test tube.

Use a separate aliquot if you need to repeat the test or investigate possible interference.

Step 2: Add dilute nitric acid.

Add a few drops of dilute nitric acid and mix gently.

This acidifies the solution and reduces interference from ions such as carbonate.

Do not substitute hydrochloric acid. It introduces chloride ions, which would interfere directly with the test.

Step 3: Add silver nitrate.

Add a few drops of silver nitrate solution.

Mix gently and observe the solution against a white background.

Record whether a precipitate forms and whether it appears white, cream, or yellow.

Step 4: Separate and wash the precipitate.

Filter or centrifuge the mixture to separate the solid from the liquid.

Wash the precipitate with a small amount of deionised water.

This is important because residual nitric acid can interfere with the subsequent ammonia test.

Step 5: Add dilute ammonia.

Add excess dilute aqueous ammonia to a small portion of the washed precipitate.

Observe whether the solid dissolves.

• Complete dissolution supports the presence of AgCl.
• A remaining solid is consistent with AgBr or AgI.

Step 6: Add concentrated ammonia if necessary.

If a solid remains, separate it from the dilute ammonia solution and wash it.

Then add concentrated aqueous ammonia.

• Dissolution supports the presence of AgBr.
• A yellow solid remaining supports the presence of AgI.

Concentrated ammonia should be handled in a well-ventilated laboratory, following appropriate safety instructions.

04Why must you use nitric acid rather than hydrochloric acid?

This is one of the most important experimental details.

Imagine testing a solution that contains no halide ions.

You add hydrochloric acid, followed by silver nitrate, and a white precipitate appears.

You might conclude that chloride ions were present in the original sample.

But that conclusion is wrong.

Hydrochloric acid contains chloride ions. You have introduced the very ion you are trying to detect.

The reaction is:

\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]

The silver chloride could have formed entirely because of the hydrochloric acid you added.

This is a false positive.

Nitric acid avoids this problem because nitrate ions remain soluble when silver nitrate is added.

There is another reason to use dilute nitric acid rather than concentrated nitric acid: sufficiently concentrated nitric acid can oxidise bromide and iodide ions, potentially changing the sample before identification.

Removing carbonate interference

Carbonate ions are another important source of interference.

Silver ions can react with carbonate ions to form silver carbonate:

\[ \ce{2Ag+(aq) + CO3^2-(aq) -> Ag2CO3(s)} \]

Silver carbonate is pale yellow. It can therefore be confused with silver iodide.

This is exactly why the sample is acidified before adding silver nitrate.

The hydrogen ions react with carbonate ions:

\[ \ce{CO3^2-(aq) + 2H+(aq) -> CO2(g) + H2O(l)} \]

Carbon dioxide gas escapes, removing carbonate from the solution before the silver nitrate test.

If you see bubbling when dilute nitric acid is added, carbonate may have been present.

Other possible interferences

Acidification is useful, but it does not guarantee that every other interfering ion has been removed.

Some other ions can also form precipitates with silver ions.

Acidification reduces some of these interferences, particularly carbonate, and protonates phosphate ions. It does not automatically eliminate every possible interfering species.

Sulfide-containing samples also require particular care because acidification can release toxic hydrogen sulfide gas.

If the sample is known to contain other potentially interfering ions, you need additional evidence before identifying the precipitate as a silver halide.

For a related explanation of how different anions can be distinguished experimentally, see How to Distinguish Carbonate from Sulfate Ions in Solution.

05Why does ammonia dissolve silver chloride and silver bromide?

This is the chemical reasoning behind the second stage of the experiment.

Ammonia is not simply reacting with the precipitate as an acid or base.

Instead, ammonia molecules form a complex ion with silver ions.

The complex is called the diamminesilver(I) ion:

\[ \ce{[Ag(NH3)2]+} \]

Silver chloride dissolves in ammonia because this complex ion is soluble in water.

The equilibrium equation is:

\[ \ce{AgCl(s) + 2NH3(aq) <=> [Ag(NH3)2]+(aq) + Cl-(aq)} \]

Think of ammonia as a shuttle that binds silver ions and keeps them dissolved in the solution.

As ammonia removes free silver ions by forming the complex, more silver chloride can dissolve to replace them.

The analogy has a limit: ammonia is not physically carrying solid particles around. It forms a chemical coordination complex with silver ions.

Why does concentration matter?

Silver chloride dissolves in dilute ammonia.

Silver bromide requires a higher ammonia concentration to dissolve appreciably.

Silver iodide remains insoluble under the usual dilute and concentrated ammonia test conditions.

The difference comes from the relative stability of the silver halide solids and the formation of the soluble silver-ammonia complex.

This gives us a second identification property beyond colour.

An important experimental mistake: adding ammonia before washing

Suppose you have just acidified your sample with nitric acid and added silver nitrate.

A precipitate forms.

You then add ammonia directly to the acidic mixture.

Nothing dissolves.

Does this prove that the precipitate is silver iodide?

No.

The remaining acid reacts with ammonia:

\[ \ce{NH3(aq) + H+(aq) -> NH4+(aq)} \]

This converts ammonia into ammonium ions, reducing the amount of free ammonia available to form the silver-ammonia complex.

Consequently, even silver chloride may fail to dissolve under these conditions.

This is why filtering and washing the precipitate before the ammonia test is essential.

A lack of dissolution is only meaningful when the experimental conditions allow the expected complex formation to occur.

An additional confirmation: re-forming the precipitate

You can strengthen your identification by acidifying the ammonia extract with dilute nitric acid.

The acid protonates the ammonia molecules and breaks apart the silver-ammonia complex:

\[ \ce{[Ag(NH3)2]+ + 2H+ -> Ag+ + 2NH4+} \]

The released silver ions can then react with the halide ions in the solution to reform a precipitate:

\[ \ce{Ag+(aq) + X-(aq) -> AgX(s)} \]

Here, X⁻ represents chloride, bromide, or iodide.

For example, if a white precipitate reforms after acidifying the dilute ammonia extract, that provides additional evidence for silver chloride.

This is particularly useful when analysing mixtures, where the original precipitate colour may be ambiguous.

06What if the unknown contains more than one halide?

The colour test becomes less reliable when multiple halide ions are present.

Imagine a solution containing both bromide and iodide.

Silver nitrate produces both silver bromide and silver iodide.

The resulting solid may appear cream-yellow.

You cannot reliably determine the two ions from the overall colour, and you cannot calculate their relative concentrations from how yellow or cream the mixture looks.

Instead, use the different ammonia solubilities to separate the precipitates.

The sequence is:

First: add dilute ammonia.

AgCl dissolves.

AgBr and AgI remain largely undissolved.

Second: separate the remaining solid.

Filter or centrifuge the mixture to separate the dissolved silver complex from the remaining precipitates.

Third: add concentrated ammonia to the remaining solid.

AgBr dissolves.

AgI remains as a yellow solid.

Fourth: confirm the dissolved fractions.

Acidify the ammonia extracts with dilute nitric acid.

A white precipitate from the dilute ammonia extract supports chloride.

A cream precipitate from the concentrated ammonia extract supports bromide.

The final yellow residue supports iodide.

There are two important limitations.

First, you need sufficient silver nitrate to precipitate the available halide ions. If silver nitrate is the limiting reagent, some halide ions remain in the original solution and will not be identified by analysing the collected solid.

Second, this is a qualitative separation, not a reliable quantitative measurement. The visible size of each fraction does not directly tell you the concentration of each halide.

09Questions and solutions

Question 1

An unknown aqueous solution is acidified with dilute nitric acid and treated with silver nitrate.

A yellow precipitate forms.

The precipitate is filtered, washed, and treated with excess concentrated ammonia. It remains undissolved.

Identify the most likely halide ion and write the relevant net ionic equation.

Solution 1

The unknown most likely contains iodide ions, \(\mathrm{I}^{-}\), forming silver iodide.

The initial yellow colour is consistent with AgI:

\[ \ce{Ag+(aq) + I-(aq) -> AgI(s)} \]

The important confirmation is that the yellow precipitate remains undissolved in concentrated ammonia.

Silver chloride dissolves in dilute ammonia, while silver bromide dissolves in concentrated ammonia. Silver iodide remains insoluble in both.

The combination of the yellow colour and the ammonia result therefore supports iodide.

This conclusion assumes that other ions capable of producing an insoluble silver salt have been excluded.

Question 2

A student has a \(10.0\ \text{mL}\) solution containing bromide ions at a concentration of \(0.0150\ \text{mol L}^{-1}\).

The student adds \(12.0\ \text{mL}\) of \(0.0100\ \text{mol L}^{-1}\) silver nitrate.

Assume precipitation is complete for stoichiometric calculations.

The reaction is:

\[ \ce{Ag+(aq) + Br-(aq) -> AgBr(s)} \]

Use \(M(\ce{AgBr})=187.77\ \text{g mol}^{-1}\).

a. Determine the limiting reactant.

b. Calculate the mass of silver bromide formed.

c. Calculate the concentration of excess bromide ions remaining in the filtrate, assuming the final volume is additive.

d. Explain why excess silver nitrate is important when testing a mixture of halide ions.

Solution 2

Silver ions are the limiting reactant. The reaction produces \(0.0225\ \text{g}\) of AgBr, while \(1.36\times10^{-3}\ \text{mol L}^{-1}\) bromide remains in the filtrate.

Step 1: Calculate the initial amount of bromide ions.

Use:

\[ n=cV \]

where \(n\) is amount in moles, \(c\) is concentration in \(\text{mol L}^{-1}\), and \(V\) is volume in litres.

\[ \begin{aligned} n(\mathrm{Br}^{-}) &=0.0150\times0.0100\\ &=1.50\times10^{-4}\ \text{mol} \end{aligned} \]

Step 2: Calculate the initial amount of silver ions.

\[ \begin{aligned} n(\mathrm{Ag}^{+}) &=0.0100\times0.0120\\ &=1.20\times10^{-4}\ \text{mol} \end{aligned} \]

Step 3: Determine the limiting reactant.

The reaction has a \(1:1\) mole ratio.

There are:

• \(1.50\times10^{-4}\) mol of bromide ions.
• \(1.20\times10^{-4}\) mol of silver ions.

Silver ions are present in the smaller amount, so they are the limiting reactant.

Step 4: Calculate the amount of silver bromide formed.

\[ n(\ce{AgBr})=1.20\times10^{-4}\ \text{mol} \]

Using \(m=nM\):

\[ \begin{aligned} m(\ce{AgBr}) &=1.20\times10^{-4}\times187.77\\ &=0.0225324\ \text{g}\\ &=0.0225\ \text{g} \end{aligned} \]

Step 5: Calculate the excess bromide ions.

\[ \begin{aligned} n_{\text{excess}}(\mathrm{Br}^{-}) &=1.50\times10^{-4}-1.20\times10^{-4}\\ &=3.00\times10^{-5}\ \text{mol} \end{aligned} \]

The final volume is:

\[ V=10.0+12.0=22.0\ \text{mL} \]

Converting to litres:

\[ V=0.0220\ \text{L} \]

Therefore:

\[ \begin{aligned} c(\mathrm{Br}^{-}) &=\frac{3.00\times10^{-5}}{0.0220}\\ &=1.36\times10^{-3}\ \text{mol L}^{-1} \end{aligned} \]

Step 6: Explain the importance of excess silver nitrate.

If silver nitrate is limiting, some halide ions remain dissolved in the original solution.

When analysing a mixture, this could cause an incomplete identification because not all the halide ions have entered the collected precipitate.

Using sufficient silver nitrate helps ensure that the available halide ions precipitate before the ammonia separation begins.

The calculation assumes effectively complete precipitation for the stoichiometric treatment. In a real solution, small equilibrium concentrations of dissolved ions remain.

Question 3

An unknown solution is known to contain carbonate ions and may also contain iodide ions.

A student divides the sample into two portions.

The following observations are recorded:

Assume carbonate is the only interfering anion capable of forming a silver precipitate in this scenario.

a. Explain why Experiment A cannot establish the presence of iodide.

b. Explain the observation during acidification.

c. Identify the halide supported by Experiments B and C, and write its net ionic equation.

Solution 3

The experiments support the presence of iodide ions. The first experiment is inconclusive because carbonate can produce a pale-yellow silver precipitate that resembles silver iodide.

Step 1: Explain the first experiment.

Silver ions react with carbonate ions to form silver carbonate:

\[ \ce{2Ag+(aq) + CO3^2-(aq) -> Ag2CO3(s)} \]

Silver carbonate is pale yellow.

Therefore, the pale-yellow precipitate in Experiment A could be silver carbonate rather than silver iodide.

The colour alone cannot distinguish the two.

Step 2: Explain the bubbles in Experiment B.

Nitric acid reacts with carbonate ions to produce carbon dioxide and water:

\[ \ce{CO3^2-(aq) + 2H+(aq) -> CO2(g) + H2O(l)} \]

The bubbles are carbon dioxide gas.

The acidification removes carbonate interference before silver nitrate is introduced.

Step 3: Identify the yellow precipitate.

Silver nitrate now reacts with the iodide ions to produce silver iodide:

\[ \ce{Ag+(aq) + I-(aq) -> AgI(s)} \]

Step 4: Use the ammonia result.

The precipitate remains yellow and undissolved in concentrated ammonia.

This is consistent with AgI, which is insoluble under the test conditions.

The two experiments therefore illustrate why sample preparation matters. The same initial colour can have different chemical explanations depending on which ions are present and which reagents have already been added.

Question 4

A student tests an unknown solution using dilute nitric acid and silver nitrate.

The student also prepares a reagent blank containing deionised water and the same reagents.

The results are:

Assume no other interfering silver salts are present.

a. What does the blank result indicate?

b. What does the ammonia behaviour of the unknown suggest?

c. Can the student confidently report that bromide ions are present in the original unknown? Explain what should happen next.

Solution 4

The results suggest chloride contamination in the testing procedure and possible bromide in the unknown, but the experiment cannot yet establish the original sample’s composition with confidence.

Step 1: Interpret the blank.

A reagent blank should remain clear when silver nitrate is added.

The white precipitate indicates that something in the testing procedure is producing a background reaction.

Its dissolution in dilute ammonia is consistent with silver chloride.

Possible contamination sources include:

• Chloride-contaminated water.
• Contaminated nitric acid or silver nitrate.
• Residual chemicals in the test tube.
• Contaminated droppers or other equipment.

The blank establishes that the procedure is not producing a clean background.

Step 2: Interpret the unknown’s ammonia behaviour.

Some of the unknown’s precipitate dissolves in dilute ammonia.

This is consistent with AgCl.

The remaining precipitate dissolves in concentrated ammonia.

This is consistent with AgBr.

The observations therefore suggest that the unknown’s precipitate contains both silver chloride and silver bromide.

Step 3: Evaluate the conclusion.

The student cannot confidently attribute the entire result to the original unknown.

The blank has already demonstrated a chloride-containing background signal.

The chloride-like fraction in the unknown could therefore have come from contamination introduced during the experiment.

The concentrated-ammonia-soluble fraction provides evidence consistent with bromide, but the failed blank means the result needs to be repeated before a confident conclusion is reported.

Step 4: Improve the experiment.

Prepare fresh reagents and use thoroughly cleaned equipment.

Investigate possible contamination in the water, reagents, and droppers. Change one component at a time if the source needs to be identified.

Then repeat the blank and unknown tests.

A reliable result requires a clear blank and ammonia behaviour consistent with the proposed halide.

A positive blank does not prove that bromide is absent from the unknown. It means that the experiment cannot yet distinguish the sample’s contribution from the background reliably.

Question 5

An unknown aqueous solution may contain any combination of chloride, bromide, and iodide ions.

A student performs the following experiment:

1. The unknown is acidified with dilute nitric acid.
2. Silver nitrate is added, producing a cream-yellow precipitate.
3. Without filtering or washing the solid, the student adds dilute ammonia directly to the acidic mixture.
4. No visible dissolution occurs.
5. The student concludes that the sample contains only iodide ions.

A second experiment is performed using a fresh aliquot.

The precipitate is filtered and thoroughly washed before the ammonia tests.

The following observations are recorded:

The reagent blank remains clear, and the positive controls behave as expected.

Assume:

• The unknown contains no other ions that form insoluble silver precipitates under the conditions used.
• Silver nitrate was added in sufficient excess.
• The ammonia solutions were prepared correctly.
• All precipitates and extracts were handled without cross-contamination.

a. Identify which halide ions are demonstrated by the second experiment.

b. Explain why the student’s first experiment cannot establish that only iodide is present.

c. Explain the chemical significance of the two reprecipitation results.

d. Could the relative sizes of the three precipitate fractions be used to calculate the relative concentrations of the halide ions? Explain.

Solution 5

All three halides are present: chloride, bromide, and iodide. The second experiment demonstrates this through sequential dissolution and reprecipitation. The first experiment is invalid because residual nitric acid can consume ammonia before it forms the silver-ammonia complex.

Step 1: Identify the chloride fraction.

The first ammonia treatment dissolves part of the washed precipitate.

Silver chloride dissolves in dilute ammonia by forming the diamminesilver(I) complex:

\[ \ce{AgCl(s) + 2NH3(aq) <=> [Ag(NH3)2]+(aq) + Cl-(aq)} \]

When the ammonia extract is acidified, the complex breaks apart and silver ions are released.

These react with chloride ions to reform white silver chloride.

Therefore, the white reprecipitate confirms chloride ions.

Step 2: Identify the bromide fraction.

After the dilute ammonia extraction, some solid remains.

Concentrated ammonia dissolves part of this remaining precipitate.

This is consistent with silver bromide dissolving through the same complex-formation process.

When the second ammonia extract is acidified, a cream precipitate reforms.

This confirms bromide ions.

Step 3: Identify the final yellow residue.

The final yellow solid remains insoluble in concentrated ammonia.

Under the stated assumptions, this is silver iodide.

The relevant precipitation reaction is:

\[ \ce{Ag+(aq) + I-(aq) -> AgI(s)} \]

The iodide ions are therefore also present.

Step 4: Explain why the first experiment was invalid.

The student added ammonia directly to a mixture that still contained nitric acid.

The remaining hydrogen ions reacted with ammonia:

\[ \ce{NH3(aq) + H+(aq) -> NH4+(aq)} \]

This reduced the concentration of free ammonia available to form the diamminesilver(I) complex.

As a result, silver chloride and silver bromide could remain undissolved even though they would normally dissolve under suitable ammonia conditions.

The student interpreted a failure of the experimental conditions as evidence about the identity of the precipitate.

That is the central mistake.

The precipitate needed to be separated and washed before ammonia was added.

Step 5: Interpret the significance of reprecipitation.

The two reprecipitation results provide independent confirmation of the dissolved fractions.

The white precipitate identifies the fraction that behaved as silver chloride.

The cream precipitate identifies the fraction that behaved as silver bromide.

The remaining yellow solid identifies the fraction that behaved as silver iodide.

Together, these observations establish the presence of all three halides under the stated assumptions.

Step 6: Evaluate whether the fractions provide concentration information.

No. The relative sizes of the fractions cannot be used directly to calculate the original halide concentrations.

The experiment is qualitative.

The apparent amount of solid depends on precipitation completeness, particle size, washing losses, and the different molar masses and solubilities of the silver halides.

A quantitative determination would require an appropriate calibrated analytical method.

The important distinction is that the experiment establishes which ions are present, not how much of each ion is present.

10Connecting the test to solubility and complex ions

The silver nitrate test is a useful example of two chemical processes working together.

First, precipitation occurs because silver ions combine with halide ions to form poorly soluble silver halides.

Second, complex-ion formation allows some of those solids to dissolve again when ammonia is added.

Understanding both processes is more useful than memorising three precipitate colours.

For HSC Chemistry, the next useful connection is to solubility equilibria and complex ions. These explain why a precipitate forms in the first place, why ammonia can dissolve some silver halides, and why experimental conditions must be controlled before an observation can support a chemical identification.