Why Excess Reagent Can Dissolve a Precipitate in HSC Chemistry

Add a few drops of sodium hydroxide to a colourless solution containing aluminium ions. A white, gelatinous precipitate appears. Keep adding sodium hydroxide, swirl the test tube, and something unexpected happens: the white solid disappears, leaving a clear solution.

Why would adding more of the reagent that created a precipitate make that precipitate dissolve?

Excess reagent can dissolve a precipitate when it reacts with the solid to form a soluble complex ion. This can happen through two important processes: amphoteric hydroxides reacting with excess hydroxide ions, and metal ions forming stable complexes with ligands such as ammonia.

The metal hasn’t disappeared. It has simply returned to solution in a different chemical form.

However, this behaviour is not universal. Some precipitates become even less soluble when excess reagent is added. Understanding what the reagent does to the dissolved ions is the key to predicting which outcome occurs.

01Why does adding excess sodium hydroxide dissolve some precipitates?

Start with the aluminium example.

Imagine placing a small amount of aluminium nitrate solution in a clean test tube. Add aqueous sodium hydroxide, \(\ce{NaOH(aq)}\), one drop at a time.

Initially, a white precipitate forms. This is aluminium hydroxide, \(\ce{Al(OH)3}\).

The net ionic equation is:

\[ \ce{Al^3+(aq) + 3OH^-(aq) -> Al(OH)3(s)} \]

The aluminium ions react with hydroxide ions to produce an insoluble solid.

Now continue adding sodium hydroxide until it is in excess.

The precipitate dissolves because aluminium hydroxide reacts with additional hydroxide ions to produce a soluble complex ion:

\[ \ce{Al(OH)3(s) + OH^-(aq) <=> [Al(OH)4]^-(aq)} \]

The product, \(\ce{[Al(OH)4]^-}\), is a soluble aluminate ion. It remains dissolved in the solution.

This explains the observation:

• Initially, hydroxide ions cause a white precipitate to form.
• With sufficiently excess hydroxide ions, the precipitate reacts further and dissolves.
• The resulting solution is colourless because the dissolved aluminate ions do not produce a visible colour.

What makes aluminium hydroxide different?

Aluminium hydroxide is described as amphoteric.

An amphoteric substance can react with both acids and bases.

For aluminium hydroxide, this means:

• It reacts with acids, producing dissolved aluminium ions.
• It reacts with sufficiently concentrated excess hydroxide, producing soluble aluminate ions.

This second reaction explains why adding excess sodium hydroxide can reverse the initial precipitation.

Zinc hydroxide behaves similarly.

\[ \ce{Zn(OH)2(s) + 2OH^-(aq) <=> [Zn(OH)4]^2-(aq)} \]

The soluble product is the tetrahydroxozincate(II) ion.

Both aluminium hydroxide and zinc hydroxide can therefore dissolve in excess sodium hydroxide.

This is an important distinction in qualitative analysis: a white precipitate that dissolves in excess sodium hydroxide does not uniquely identify either aluminium ions or zinc ions.

02A different mechanism: forming complex ions with ammonia

Not every precipitate dissolves through amphoteric behaviour.

Another important mechanism involves a substance called a ligand.

A ligand is a molecule or ion that donates a lone pair of electrons to a metal ion, forming a coordinate bond. Several ligands can attach to one metal ion, producing a complex ion.

Aqueous ammonia, \(\ce{NH3(aq)}\), is a common ligand in qualitative analysis.

Consider silver chloride, \(\ce{AgCl}\), a white precipitate.

It forms when silver ions react with chloride ions:

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

Normally, silver chloride is only slightly soluble in water.

However, adding excess aqueous ammonia causes the precipitate to dissolve.

Why?

Ammonia molecules bond to the dissolved silver ions, forming the soluble diamminesilver(I) complex:

\[ \ce{Ag^+(aq) + 2NH3(aq) <=> [Ag(NH3)2]^+(aq)} \]

The ammonia removes free silver ions from the solution by converting them into complex ions.

This changes the dissolution equilibrium of silver chloride:

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

As the concentration of free \(\mathrm{Ag}^{+}\) ions decreases, more silver chloride dissolves to replace them.

The overall reaction is:

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

The white precipitate disappears because the silver is now present in a soluble complex.

What about copper(II) hydroxide?

Copper(II) hydroxide provides another useful example.

When a small amount of aqueous ammonia is added to a copper(II) solution, a pale blue precipitate forms.

This is copper(II) hydroxide:

\[ \ce{Cu^2+(aq) + 2OH^-(aq) -> Cu(OH)2(s)} \]

Ammonia is a weak base, so it reacts with water to produce some hydroxide ions:

\[ \ce{NH3(aq) + H2O(l) <=> NH4^+(aq) + OH^-(aq)} \]

Initially, these hydroxide ions cause the precipitation.

However, when ammonia is added in excess, it also acts as a ligand. The copper hydroxide dissolves, forming a deep blue copper(II)-ammine complex.

The simplified net ionic equation is:

\[ \ce{Cu(OH)2(s) + 4NH3(aq) <=> [Cu(NH3)4]^2+(aq) + 2OH^-(aq)} \]

The dissolved complex is often represented more fully in aqueous solution as \(\ce{[Cu(NH3)4(H2O)2]^2+}\).

Notice the two different roles of ammonia:

1. Initially, it acts as a weak base, supplying hydroxide ions that produce a precipitate.
2. In excess, it acts as a ligand, forming a soluble complex that dissolves the precipitate.

This is why the sequence of observations matters. A pale blue precipitate followed by a deep blue solution is much more informative than simply reporting that the final solution is blue.

03Comparing the different precipitation tests

The same visible change, a precipitate disappearing, can result from different chemical reactions.

The following table summarises several useful qualitative-analysis patterns.

These are the expected patterns under ordinary qualitative-test conditions, assuming suitable reagent concentrations and no interfering substances.

The important contrast: magnesium hydroxide

Consider magnesium hydroxide, \(\ce{Mg(OH)2}\).

It also forms a white precipitate when sodium hydroxide is added:

\[ \ce{Mg^2+(aq) + 2OH^-(aq) -> Mg(OH)2(s)} \]

However, adding excess sodium hydroxide does not dissolve this precipitate under ordinary qualitative-test conditions.

Why not?

Magnesium hydroxide does not form a sufficiently stable soluble hydroxo complex under these conditions.

This means the additional hydroxide ions do not produce the same dissolution reaction observed for aluminium hydroxide and zinc hydroxide.

The white precipitate remains.

This is a useful contrasting case because it exposes a common misconception:

Adding excess reagent does not automatically dissolve a precipitate.

The outcome depends on the chemical species that the additional reagent can form.

What if excess reagent makes a precipitate less soluble?

There is another important possibility.

Consider adding sodium chloride to a suspension of silver chloride.

The dissolution equilibrium is:

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

Adding sodium chloride increases the concentration of chloride ions.

The increased chloride concentration shifts the equilibrium towards the solid, suppressing the dissolution of silver chloride.

This is an example of the common-ion effect.

Compare the two situations:

The difference is what happens to the free silver-ion concentration.

Adding chloride pushes the dissolution equilibrium backwards. Adding ammonia removes free silver ions through complex formation, encouraging further dissolution.

If you need to revisit how a precipitate first forms and how solubility equilibria predict that point, see Which Ion Precipitates First? A Worked HSC Chemistry Example.

04How to use these reactions to identify unknown ions

Now consider how these observations are used in a laboratory identification.

Suppose you have a colourless unknown solution that contains one of the following ions:

• Aluminium, \(\mathrm{Al}^{3+}\)
• Zinc, \(\mathrm{Zn}^{2+}\)
• Magnesium, \(\mathrm{Mg}^{2+}\)
• Copper(II), \(\mathrm{Cu}^{2+}\)

A single white precipitate is not enough to identify the ion. Several of these ions form white hydroxides.

Instead, use the behaviour of the precipitate with two different reagents.

Step 1: Add sodium hydroxide dropwise, then in excess

Record the initial precipitate and whether it dissolves.

• A white precipitate that dissolves is consistent with aluminium or zinc.
• A white precipitate that remains is consistent with magnesium.
• A pale blue precipitate that remains is consistent with copper(II).

The sodium hydroxide test narrows down the possibilities, but it cannot distinguish aluminium from zinc.

Step 2: Test a fresh portion with ammonia

Use a separate aliquot of the original unknown solution.

Add ammonia dropwise, then continue until ammonia is in excess.

• Aluminium produces a white precipitate that remains in excess ammonia.
• Zinc produces a white precipitate that dissolves in excess ammonia.
• Magnesium produces a white precipitate that remains.
• Copper(II) produces a pale blue precipitate that dissolves in excess ammonia, giving a deep blue solution.

Using a fresh portion is important. If you have already added excess sodium hydroxide, the solution contains a high concentration of hydroxide ions. That changes the chemical conditions and can interfere with the ammonia test.

Step 3: Interpret the complete sequence

Do not identify an ion from one observation when several ions could produce it.

For example:

• White precipitate with sodium hydroxide, dissolving in excess: aluminium or zinc.
• White precipitate with ammonia, remaining in excess: aluminium.
• White precipitate with ammonia, dissolving in excess: zinc.

The second test resolves the ambiguity left by the first.

Interferences and limitations to remember

Qualitative tests become less reliable when several ions are present together.

Keep these points in mind:

• Use separate aliquots. Adding several reagents successively to one test tube can change the pH and complex-ion equilibria, making later observations difficult to interpret.
• Add the reagent gradually. You need to observe the initial precipitate before testing its behaviour in excess reagent.
• Allow time for mixing. A precipitate may not disappear immediately. Swirl the test tube and distinguish slow dissolution from a genuinely persistent solid.
• Consider mixtures. One ion may produce a coloured solution while another produces a precipitate. The final colour may not identify the solid.
• Check reagent concentration. Too little ammonia or sodium hydroxide may produce only partial dissolution.
• Do not rely on colour alone. Colour intensity depends on concentration, lighting, and the presence of other dissolved substances.

Silver chloride provides a particularly useful example of why confirmatory testing matters.

A white precipitate after adding silver nitrate is consistent with chloride ions, but it is not sufficient proof on its own. Other ions, including carbonate, can also produce precipitates with silver ions.

For a more reliable chloride test, acidify the original sample with dilute nitric acid before adding silver nitrate. Do not use hydrochloric acid, because it introduces additional chloride ions.

If white silver chloride forms, it should dissolve in aqueous ammonia. Acidifying the resulting solution with dilute nitric acid removes ammonia by converting it into ammonium ions. The silver complex breaks down, allowing silver chloride to precipitate again.

This dissolution-and-reprecipitation sequence provides stronger evidence than the original white precipitate alone.

The behaviour of silver bromide and silver iodide is different, so ammonia solubility should also be considered when other halide ions may be present. See How to Distinguish Chloride, Bromide and Iodide Using Silver Nitrate for the broader halide test.

05Worked examples

06Questions and solutions

The following questions progress from basic precipitation observations to more demanding problems involving equilibrium, complex formation, and interfering ions.

Question 1

A student adds sodium hydroxide dropwise to a solution containing \(\mathrm{Al}^{3+}\). A white precipitate forms.

The student continues adding sodium hydroxide until it is in excess.

(a) State the observation after excess sodium hydroxide is added.

(b) Write the balanced net ionic equation for the reaction responsible for this observation.

(c) Explain why this behaviour is described as amphoteric.

Solution 1

(a) The white precipitate dissolves, leaving a colourless solution.

The initial precipitate is aluminium hydroxide, \(\ce{Al(OH)3}\). Excess hydroxide ions react with this solid to produce soluble aluminate ions.

(b) The net ionic equation is:

\[ \ce{Al(OH)3(s) + OH^-(aq) <=> [Al(OH)4]^-(aq)} \]

The product is soluble, so the solid disappears from the test tube.

(c) Aluminium hydroxide is amphoteric because it reacts with both acids and bases.

It reacts with acids to form dissolved aluminium species, and it reacts with excess hydroxide ions to form soluble aluminate ions.

The important point is that aluminium hydroxide is not simply an insoluble solid under every set of conditions. Its solubility depends on the chemical environment.

Question 2

Two colourless unknown solutions, A and B, each contain one metal ion selected from:

\[ \ce{Al^3+,\ Zn^2+,\ Mg^2+} \]

The following observations are recorded.

(a) Identify the most likely ion in solution A.

(b) State the two possible ions in solution B.

(c) Describe an additional test using aqueous ammonia that would distinguish the two possibilities for solution B. Include the expected observations.

Solution 2

(a) Solution A contains \(\mathrm{Mg}^{2+}\).

Magnesium ions form a white precipitate of magnesium hydroxide:

\[ \ce{Mg^2+(aq) + 2OH^-(aq) -> Mg(OH)2(s)} \]

Magnesium hydroxide remains insoluble in excess sodium hydroxide under the stated conditions.

This matches solution A.

(b) Solution B contains either \(\mathrm{Al}^{3+}\) or \(\mathrm{Zn}^{2+}\).

Both ions form white hydroxide precipitates that dissolve in excess sodium hydroxide.

For aluminium:

\[ \ce{Al(OH)3(s) + OH^-(aq) <=> [Al(OH)4]^-(aq)} \]

For zinc:

\[ \ce{Zn(OH)2(s) + 2OH^-(aq) <=> [Zn(OH)4]^2-(aq)} \]

Therefore, the sodium hydroxide test alone cannot distinguish these ions.

(c) Add aqueous ammonia dropwise to a fresh portion of solution B, then continue adding ammonia in excess.

The expected results are:

• If the ion is \(\mathrm{Al}^{3+}\), a white precipitate forms and remains insoluble in excess ammonia.
• If the ion is \(\mathrm{Zn}^{2+}\), a white precipitate forms and dissolves in excess ammonia.

Zinc forms the soluble complex \(\ce{[Zn(NH3)4]^2+}\), whereas aluminium hydroxide remains insoluble.

This second test resolves the ambiguity left by the sodium hydroxide test.

Question 3

A student has \(0.00200\ \mathrm{mol}\) of \(\mathrm{Zn}^{2+}\) ions and adds \(0.00600\ \mathrm{mol}\) of hydroxide ions.

For this question, assume that all zinc is present either as solid \(\ce{Zn(OH)2}\) or dissolved \(\ce{[Zn(OH)4]^2-}\). Assume all supplied hydroxide ions are consumed by these species, with no free hydroxide remaining.

The relevant equations are:

\[ \ce{Zn^2+ + 2OH^- -> Zn(OH)2(s)} \]

\[ \ce{Zn(OH)2(s) + 2OH^- <=> [Zn(OH)4]^2-} \]

(a) Calculate the total mole ratio of hydroxide ions supplied per mole of zinc ions.

(b) Determine whether all the zinc can be converted into \(\ce{[Zn(OH)4]^2-}\).

(c) Calculate the maximum fraction of zinc that can exist as the dissolved complex under the stated assumptions.

Solution 3

(a) The hydroxide-to-zinc mole ratio is \(3:1\).

Calculate:

\[ \begin{aligned} \text{Ratio} &=\frac{n(\mathrm{OH}^{-})}{n(\mathrm{Zn}^{2+})}\\ &=\frac{0.00600}{0.00200}\\ &=3.00 \end{aligned} \]

There are three moles of hydroxide ions available for every mole of zinc ions.

(b) No, all the zinc cannot be converted into zincate ions.

To form solid zinc hydroxide, two hydroxide ions are needed per zinc ion.

To convert that solid into the soluble tetrahydroxozincate complex, another two hydroxide ions are needed.

The overall reaction is:

\[ \ce{Zn^2+ + 4OH^- <=> [Zn(OH)4]^2-} \]

Complete conversion therefore requires four moles of hydroxide ions per mole of zinc ions.

The available ratio is only three.

(c) The maximum fraction of zinc that can form the soluble complex is \(50.0\%\).

Let \(f\) represent the fraction of zinc converted into the soluble complex.

The remaining fraction, \(1-f\), exists as solid zinc hydroxide.

Each mole of complex requires four moles of hydroxide ions, while each mole of solid zinc hydroxide requires two.

Therefore, the average hydroxide requirement per mole of zinc is:

\[ 4f+2(1-f)=3 \]

Expand and simplify:

\[ 4f+2-2f=3 \]

\[ 2f=1 \]

\[ f=0.500 \]

Therefore:

\[ \boxed{f=50.0\%} \]

At most \(0.00100\ \mathrm{mol}\) of zinc can exist as the soluble complex, with the remaining \(0.00100\ \mathrm{mol}\) present as solid zinc hydroxide.

This is a stoichiometric upper bound under the assumptions in the question, not a prediction of the actual equilibrium composition.

The interesting point is that \(0.00600\ \mathrm{mol}\) of hydroxide is already in excess relative to the initial precipitation reaction. However, it is not enough to convert all the precipitate into zincate.

Being in excess relative to one reaction does not mean there is enough reagent for every subsequent reaction.

Question 4

An unknown solution contains both \(\mathrm{Cu}^{2+}\) and \(\mathrm{Cl}^{-}\) ions.

A student adds aqueous silver nitrate, \(\ce{AgNO3}\), to the unknown. A white precipitate forms.

The student then adds excess aqueous ammonia directly to the same test tube.

The white precipitate disappears, and the solution becomes deep blue.

(a) Identify the substance responsible for the initial white precipitate.

(b) Explain why the precipitate dissolves after excess ammonia is added.

(c) Explain the origin of the deep blue colour.

(d) Describe a more reliable experimental procedure to establish the presence of both ions. Include the reagents, observations, and any necessary separation steps.

Solution 4

(a) The initial white precipitate is silver chloride, \(\ce{AgCl}\).

The silver ions supplied by silver nitrate react with chloride ions in the unknown:

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

Silver chloride is a white solid.

(b) The precipitate dissolves because excess ammonia forms a soluble silver complex.

Ammonia molecules coordinate to silver ions, forming diamminesilver(I):

\[ \ce{Ag^+(aq) + 2NH3(aq) <=> [Ag(NH3)2]^+(aq)} \]

This removes free silver ions from the solution and encourages additional silver chloride to dissolve.

The overall reaction is:

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

The white solid disappears because the silver is now dissolved as a complex ion.

(c) The deep blue colour comes from copper(II)-ammine complex ions.

The original solution contained \(\mathrm{Cu}^{2+}\) ions.

When excess ammonia is added, these ions form a deep blue complex, commonly represented as:

\[ \ce{[Cu(NH3)4]^2+} \]

Therefore, two different reactions explain the two observations:

• Silver chloride dissolves because silver ions form a soluble ammonia complex.
• Copper(II) ions produce the deep blue colour because they form a copper(II)-ammine complex.

The deep blue colour does not mean that the original white precipitate was a copper compound.

This is a particularly important example of interference: two different substances contribute to the final observation.

(d) Use separate tests and physically separate the precipitate before testing it.

A more reliable procedure is:

1. Divide the original unknown into two fresh portions.
2. To the first portion, add dilute nitric acid, followed by aqueous silver nitrate.
3. Record the formation of a white precipitate.
4. Filter the mixture and wash the collected precipitate to remove dissolved copper ions.
5. Add aqueous ammonia to the washed solid. It should dissolve.
6. Acidify the resulting solution with dilute nitric acid. The white silver chloride precipitate should form again.

This sequence provides evidence for chloride ions.

For copper(II), use the second fresh portion:

1. Add aqueous ammonia dropwise.
2. Observe the formation of a pale blue precipitate.
3. Continue adding ammonia in excess.
4. Observe the precipitate dissolving to produce a deep blue solution.

This provides evidence for copper(II) ions.

The key improvement is separating the tests rather than interpreting a mixture of reactions in one test tube.

Filtering and washing the silver chloride also prevents dissolved copper ions from interfering with the ammonia test on the solid.

Question 5

An unknown solution may contain \(\mathrm{Al}^{3+}\), \(\mathrm{Zn}^{2+}\), or both ions. Assume there are no other metal ions or complexing substances present.

A student carries out the following experiment:

• Sodium hydroxide is added dropwise to the unknown, producing a white precipitate.
• Excess sodium hydroxide is added, and the precipitate completely disappears.
• The student then adds excess ammonia to the same strongly alkaline solution.
• No new precipitate forms.

The student concludes that only one metal ion was originally present, and that neither aluminium nor zinc can be confirmed.

(a) Explain why the initial sodium hydroxide observations cannot establish whether one or both metal ions are present.

(b) Explain why adding ammonia to the strongly alkaline solution is not a reliable way to distinguish the ions.

(c) Design an experiment using a fresh portion of the original solution that can establish whether aluminium ions, zinc ions, or both are present.

Your procedure must use excess ammonia, filtration, and a confirmatory test on each fraction.

(d) State the expected observations for each of the three possible original samples.

Solution 5

(a) The sodium hydroxide observations are consistent with aluminium ions, zinc ions, or a mixture of both.

Both metal ions initially form white hydroxide precipitates.

For aluminium:

\[ \ce{Al^3+ + 3OH^- -> Al(OH)3(s)} \]

For zinc:

\[ \ce{Zn^2+ + 2OH^- -> Zn(OH)2(s)} \]

Both precipitates dissolve in excess hydroxide:

\[ \ce{Al(OH)3(s) + OH^- <=> [Al(OH)4]^-} \]

\[ \ce{Zn(OH)2(s) + 2OH^- <=> [Zn(OH)4]^2-} \]

Therefore, a clear solution after adding excess sodium hydroxide does not establish that only one ion was present.

If both ions were originally present, both could have dissolved.

The student has mistaken the disappearance of a precipitate for the disappearance of a particular metal ion.

(b) Adding ammonia to the strongly alkaline solution does not provide a valid ammonia test.

The original addition of excess sodium hydroxide has already created a strongly alkaline environment.

Both aluminium and zinc can exist as soluble hydroxo complexes under these conditions.

Adding ammonia does not remove the excess hydroxide ions or restore the original conditions required for the usual ammonia precipitation test.

Consequently, the absence of a new precipitate cannot be used to conclude that aluminium or zinc is absent.

The mistake is the order of addition.

An ammonia test should begin with a fresh portion of the original solution, not a solution that has already been treated with excess sodium hydroxide.

(c) Use excess ammonia on a fresh sample, then filter to separate the two possible hydroxides.

The procedure is as follows.

Step 1: Take a fresh portion of the original unknown.

Do not use the solution that has already been treated with sodium hydroxide.

Step 2: Add aqueous ammonia dropwise.

Both aluminium and zinc ions can produce white hydroxide precipitates.

For aluminium:

\[ \ce{Al^3+ + 3OH^- -> Al(OH)3(s)} \]

For zinc:

\[ \ce{Zn^2+ + 2OH^- -> Zn(OH)2(s)} \]

The hydroxide ions are supplied through the weak-base equilibrium of ammonia with water.

Step 3: Continue adding ammonia until it is in excess.

The zinc hydroxide dissolves because it forms a soluble tetraamminezinc(II) complex:

\[ \ce{Zn(OH)2(s) + 4NH3(aq) <=> [Zn(NH3)4]^2+(aq) + 2OH^-(aq)} \]

Aluminium hydroxide remains insoluble in excess ammonia.

At this stage:

• Any aluminium present remains as solid \(\ce{Al(OH)3}\).
• Any zinc present is dissolved as \(\ce{[Zn(NH3)4]^2+}\).

Step 4: Filter the mixture.

The filter separates the remaining solid from the solution.

The two fractions now contain different possible metal ions.

Step 5: Confirm aluminium in the solid residue.

Wash the collected white residue and add excess sodium hydroxide.

If aluminium hydroxide is present, it dissolves to form soluble aluminate ions:

\[ \ce{Al(OH)3(s) + OH^-(aq) <=> [Al(OH)4]^-(aq)} \]

The disappearance of the white residue supports the presence of aluminium ions.

Step 6: Confirm zinc in the filtrate.

The filtrate contains dissolved zinc ammine complexes if zinc was originally present.

Add dilute nitric acid to destroy the ammonia complex. The ammonia is protonated to form ammonium ions, releasing the zinc ions into solution.

Then add sodium hydroxide dropwise.

A white zinc hydroxide precipitate should form:

\[ \ce{Zn^2+(aq) + 2OH^-(aq) -> Zn(OH)2(s)} \]

Continue adding sodium hydroxide in excess.

The white precipitate should dissolve, producing soluble zincate ions:

\[ \ce{Zn(OH)2(s) + 2OH^-(aq) <=> [Zn(OH)4]^2-(aq)} \]

This provides evidence for zinc ions in the original sample.

(d) The results distinguish all three possibilities.

The crucial insight is that the two ions behave differently in excess ammonia, even though they behave similarly in excess sodium hydroxide.

Filtering separates the aluminium hydroxide from the dissolved zinc complex, allowing each fraction to be tested independently.

A confident first answer might be that the original clear solution contained only one metal ion. The experiment shows why that conclusion is unjustified: two different metal ions can both form precipitates and then both dissolve in the same excess reagent.

The order of testing and the separation of chemical species are just as important as the observations themselves.

07Where this leads next

When a precipitate disappears, the next step is to identify what has replaced it in solution.

Ask three things:

1. Did excess hydroxide produce a soluble hydroxo complex?
2. Did a ligand such as ammonia bind the metal ion into a soluble complex?
3. Did the added reagent instead introduce a common ion and suppress dissolution?

These questions turn a simple observation into a chemical explanation.

They also connect qualitative precipitation tests to the broader study of chemical equilibrium: changing the concentration of one species can shift an equilibrium, change the dominant dissolved species, and determine whether a solid remains or dissolves.