How to Compare Solubility Predictions with Experimental Results

Suppose your solubility rules predict that mixing two clear solutions will produce a white precipitate. You mix them, but the solution stays clear. Has the prediction failed?

Not necessarily. The solutions might be too dilute, the wrong reagent might have been used, or another ion might be interfering with the test. Equally, seeing a precipitate does not automatically prove that the compound you predicted has formed.

To compare a solubility-rule prediction with experimental results, you need to do three things: predict the chemical change, record the actual observation, and decide whether the observation provides convincing evidence for your prediction. If the two disagree, investigate the experimental conditions before rejecting the solubility rule.

01Separate the prediction from the observation

A solubility rule predicts whether an ionic compound is expected to remain dissolved or form a solid when two solutions are mixed.

Consider lead(II) nitrate and potassium iodide.

Both compounds dissolve in water, producing these ions:

• Lead(II) nitrate provides \(\mathrm{Pb}^{2+}\) and \(\ce{NO3-}\).
• Potassium iodide provides \(\mathrm{K}^{+}\) and \(\mathrm{I}^{-}\).

The possible new ionic compounds are lead(II) iodide, \(\ce{PbI2}\), and potassium nitrate, \(\ce{KNO3}\).

Potassium nitrate is soluble, but lead(II) iodide is insoluble. Therefore, the prediction is that yellow solid lead(II) iodide will form.

The balanced equation is:

\[
\ce{Pb(NO3)2(aq) + 2KI(aq) ->}
\]

\[
\ce{PbI2(s) + 2KNO3(aq)}
\]

The net ionic equation shows the actual precipitation reaction:

\[
\ce{Pb^2+(aq) + 2I-(aq) -> PbI2(s)}
\]

Now imagine carrying out the experiment and observing a yellow precipitate.

The prediction and observation agree. However, it is important to distinguish three separate statements:

The last statement is an interpretation, not a direct observation.

This distinction is important in HSC Chemistry. Writing that you observed lead(II) iodide is less scientifically precise than writing that you observed a yellow precipitate consistent with lead(II) iodide.

A colour can support an identification, but it cannot establish the identity of a compound on its own.

If you need to revise the underlying rules for predicting precipitates, see Solubility Rules and Precipitation Tests for HSC Chemistry.

02Set up an experiment that tests the prediction fairly

A good comparison requires a controlled experiment. Otherwise, you might be measuring contamination or an existing precipitate rather than the reaction you intended to investigate.

Let’s use the lead(II) iodide reaction.

Choosing the reagents

You need:

• Aqueous lead(II) nitrate, \(\ce{Pb(NO3)2}\).
• Aqueous potassium iodide, \(\ce{KI}\).
• Deionised water for reagent checks.
• Clean test tubes and a test-tube rack.
• Separate droppers or pipettes for each solution.
• A white background to make cloudiness and precipitates easier to observe.

Use three labelled test tubes.

Place the tubes in a rack against a white background.

Add the measured solutions using separate droppers. Observe each tube before mixing, then gently swirl the reaction tube and record any changes.

The two reagent checks help establish that the stock solutions were initially clear. For an unknown-sample investigation, you should also use a proper negative blank containing all the testing reagents but no unknown sample.

A positive control is useful too. This contains a known solution of the target ion and should produce the expected result.

If the blank produces a precipitate, contamination may be affecting the experiment. If the positive control fails, the reagents or procedure may not be working correctly.

Record what actually happens

A useful experimental record includes:

1. The reagents and their concentrations.
2. The volumes mixed.
3. The appearance of each solution before mixing.
4. The appearance immediately after mixing.
5. Whether a precipitate forms, its appearance, and whether it settles.
6. The results of any controls or confirmatory tests.

For example:

This is much more useful than simply writing that a reaction occurred.

Safety: Lead compounds are toxic. Lead nitrate experiments should only be conducted under appropriate school laboratory supervision, using suitable protective equipment and designated heavy-metal waste collection.

04What if the prediction and observation disagree?

A mismatch does not automatically mean the solubility rule is incorrect.

You should investigate the experimental conditions and consider other possible explanations.

Why might an insoluble compound fail to produce a visible precipitate?

The word insoluble can be misleading.

It does not mean that absolutely no particles of the compound can remain dissolved in water. It means that the compound has sufficiently low solubility for a solid to form under the relevant conditions.

Suppose you mix two extremely dilute solutions containing ions that normally form an insoluble salt.

The resulting concentration of those ions may be too low to produce an easily visible quantity of solid.

The solubility-rule prediction is still useful, but the experimental observation may be difficult to detect.

This is why concentrations and volumes matter.

Why might an unexpected precipitate form?

Consider an unknown solution that contains more than one type of anion.

You add a reagent intended to detect sulfate ions. A white precipitate forms.

You might immediately conclude that sulfate ions are present.

But another anion could have produced the same observation.

This is called interference. An interfering ion produces an observation that can be mistaken for evidence of the ion being investigated.

The solution is not simply to repeat the same test. You need to choose a reagent or procedure that distinguishes the competing explanations.

05A contrasting case: Why a white precipitate does not always confirm sulfate

Consider a solution that might contain sulfate ions, carbonate ions, or both.

You add barium chloride solution and observe a white precipitate.

The result appears to support the prediction of insoluble barium sulfate:

\[
\ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)}
\]

However, carbonate ions also produce a white precipitate:

\[
\ce{Ba^2+(aq) + CO3^2-(aq) -> BaCO3(s)}
\]

Both possible products are white.

Therefore, the original observation cannot distinguish between sulfate and carbonate.

This is a particularly important example of why an experimental result must be interpreted using more than colour.

Improve the experiment using dilute acid

You can distinguish the two possibilities by acidifying a fresh portion of the unknown solution before adding barium chloride.

Carbonate ions react with hydrogen ions to produce carbon dioxide and water:

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

The carbon dioxide escapes as bubbles.

Sulfate ions do not undergo this gas-producing reaction and remain available to form barium sulfate.

The improved procedure is:

1. Place a fresh portion of the unknown solution into a clean test tube.
2. Add dilute nitric acid.
3. Observe whether effervescence occurs.
4. If gas is produced, collect a sample and test it with limewater.
5. After the effervescence stops, add barium chloride to the acidified solution.
6. Observe whether a white precipitate forms.

Carbon dioxide turns limewater milky because it produces insoluble calcium carbonate:

\[
\ce{CO2(g) + Ca(OH)2(aq) -> CaCO3(s) + H2O(l)}
\]

The possible results can be interpreted as follows:

Notice that the experiment has been improved by changing the procedure, not by changing the solubility rule.

Choose reagents that do not introduce the ion being tested

This is another common source of false positive results.

The acid used in a test must not introduce an ion that produces the precipitate you are trying to detect.

Dilute nitric acid is suitable for these acidification steps because nitrate ions generally remain soluble with the relevant metal ions.

For the complete carbonate-versus-sulfate identification method, see How to Distinguish Carbonate from Sulfate Ions in Solution.

06What makes a confirmatory test convincing?

A confirmatory test should help distinguish the predicted compound from other plausible explanations.

Repeating the same test with the same reagent does not necessarily provide stronger evidence. If the reagent reacts with several possible ions, repeating it may simply reproduce the same ambiguity.

Consider the lead iodide example.

If the starting solutions are known to contain only lead(II) ions and iodide ions, the reaction equation and solubility rules provide a strong basis for identifying the precipitate.

However, suppose you are given an unknown solution and asked to establish whether it contains iodide ions.

A yellow precipitate with lead(II) nitrate is not sufficient on its own. Other anions can also form insoluble lead compounds.

An independent test can provide additional evidence.

For example, use a fresh portion of the unknown solution, acidify with dilute nitric acid, and add silver nitrate.

Iodide ions form pale-yellow silver iodide:

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

The silver iodide precipitate remains insoluble in aqueous ammonia. This behaviour helps distinguish it from silver chloride, which dissolves in dilute ammonia, and silver bromide, which is less soluble and requires more concentrated ammonia to dissolve.

Within a candidate set of common halide ions, the combination of the precipitation result and the ammonia behaviour provides stronger evidence for iodide than colour alone.

The key idea is to make your conclusion match the strength of your evidence.

For example:

• Weak conclusion: A yellow precipitate formed.
• Better conclusion: The yellow precipitate is consistent with lead(II) iodide.
• Stronger conclusion: The known starting ions, balanced ionic equation, observed precipitate, and independent iodide test collectively support the formation of lead(II) iodide.

You should also distinguish between a negative result and proof of absence. If no precipitate forms, the correct statement is that the ion was not detected under the experimental conditions.

07Questions and solutions

The following questions progress from straightforward solubility-rule predictions to experimental interpretation, interference, and more demanding quantitative reasoning.

Question 1

Equal volumes of aqueous sodium sulfate and barium nitrate are mixed in a clean test tube.

Predict whether a precipitate forms. Identify the precipitate, write the net ionic equation, and state the expected experimental observation.

Solution 1

A white precipitate of barium sulfate forms.

Sodium sulfate provides sodium ions and sulfate ions. Barium nitrate provides barium ions and nitrate ions.

The possible new compounds are barium sulfate and sodium nitrate.

Sodium nitrate is soluble, but barium sulfate is insoluble. Therefore, barium sulfate forms a solid.

The net ionic equation is:

\[
\ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)}
\]

The expected observation is that a white precipitate appears when the two initially clear solutions are mixed.

This observation would support the solubility-rule prediction, provided the reagents are uncontaminated and the concentrations are sufficient to produce a visible solid.

Question 2

An unknown solution is known to contain carbonate ions, sulfate ions, or both.

A student performs the following tests on separate portions of the unknown:

• Adding barium chloride directly produces a white precipitate.
• Adding dilute nitric acid to a fresh portion produces effervescence.
• The gas turns limewater milky.
• After the effervescence stops, adding barium chloride to the acidified solution produces a persistent white precipitate.

A water blank remains clear, and a known sulfate control produces a white precipitate after acidification and treatment with barium chloride.

Identify the ions present and explain why the original barium chloride test was insufficient.

Solution 2

Both carbonate and sulfate ions are present in the unknown solution.

The original white precipitate does not identify sulfate uniquely because both carbonate and sulfate ions form insoluble white compounds with barium ions.

The evidence from the improved tests distinguishes the two.

Step 1: Identify the evidence for carbonate.

The effervescence indicates gas production when acid is added.

The gas turns limewater milky, confirming the presence of carbon dioxide.

The relevant reaction is:

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

This supports the presence of carbonate ions.

Step 2: Identify the evidence for sulfate.

After acidification, the solution still produces a white precipitate when barium chloride is added.

Carbonate ions have been removed from the solution through their reaction with acid.

Sulfate ions remain available to react with barium ions:

\[
\ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)}
\]

The persistent white precipitate therefore supports the presence of sulfate ions.

Step 3: Evaluate the controls.

The clear water blank indicates that the testing procedure did not produce a detectable background precipitate.

The positive sulfate control demonstrates that the reagents and procedure can produce the expected sulfate result.

Together, these controls strengthen the interpretation.

The final conclusion is that both carbonate and sulfate ions are present. The original direct barium chloride test was ambiguous because it could not distinguish between the two white precipitates.

Question 3

A student mixes \(2.50\ \text{mL}\) of \(0.0150\ \text{mol L}^{-1}\) lead(II) nitrate with \(1.50\ \text{mL}\) of \(0.0200\ \text{mol L}^{-1}\) potassium iodide.

Calculate:

a) The amount of lead(II) ions supplied.

b) The amount of iodide ions supplied.

c) The limiting reactant.

d) The theoretical mass of lead(II) iodide formed.

Explain why using equal volumes of the two solutions would not necessarily produce complete precipitation.

Use \(M(\ce{PbI2})=461.0\ \text{g mol}^{-1}\).

Solution 3

The theoretical mass of lead(II) iodide is \(6.92\ \text{mg}\), and iodide ions are the limiting reactant.

The important detail is that the reaction requires two moles of iodide ions for every mole of lead(II) ions.

Step 1: Calculate the amount of lead(II) ions.

Use:

\[
n=cV
\]

Convert the volume into litres:

\[
2.50\ \text{mL}=0.00250\ \text{L}
\]

Therefore:

\[
n(\mathrm{Pb}^{2+})=0.0150\times0.00250
\]

\[
=3.75\times10^{-5}\ \text{mol}
\]

Step 2: Calculate the amount of iodide ions.

Convert the volume:

\[
1.50\ \text{mL}=0.00150\ \text{L}
\]

Therefore:

\[
n(\mathrm{I}^{-})=0.0200\times0.00150
\]

\[
=3.00\times10^{-5}\ \text{mol}
\]

Step 3: Determine the limiting reactant.

The net ionic equation is:

\[
\ce{Pb^2+ + 2I- -> PbI2(s)}
\]

The amount of iodide required to react with all the lead(II) ions would be:

\[
2(3.75\times10^{-5})
\]

\[
=7.50\times10^{-5}\ \text{mol}
\]

However, only \(3.00\times10^{-5}\ \text{mol}\) of iodide ions are available.

Therefore, iodide ions are the limiting reactant.

Step 4: Calculate the amount of lead(II) iodide.

The mole ratio between iodide ions and lead(II) iodide is \(2:1\).

Therefore:

\[
n(\ce{PbI2})=
\frac{3.00\times10^{-5}}{2}
\]

\[
=1.50\times10^{-5}\ \text{mol}
\]

Step 5: Calculate the theoretical mass.

\[
m=nM
\]

\[
m=(1.50\times10^{-5})(461.0)
\]

\[
=0.006915\ \text{g}
\]

Therefore:

\[
\boxed{m=6.92\ \text{mg}}
\]

The experiment should produce a yellow precipitate, assuming the reaction proceeds sufficiently for visible precipitation.

Equal volumes are not automatically required for complete precipitation. What matters is the mole ratio, which depends on both concentration and volume.

Even if the two solutions had equal volumes, different concentrations could produce an excess of either reactant.

Question 4

A student wants to test an unknown solution for sulfate ions.

They add dilute sulfuric acid to the unknown solution, followed by barium chloride. A white precipitate forms.

The student concludes that sulfate ions have been confirmed.

Evaluate this conclusion and explain how the experiment should be redesigned.

Solution 4

The conclusion is not justified because the sulfuric acid introduces sulfate ions into the experiment, creating a possible false positive result.

The student has not established that the sulfate in the precipitate originally came from the unknown sample.

Step 1: Identify the experimental error.

Sulfuric acid provides sulfate ions:

\[
\ce{H2SO4(aq) -> 2H+(aq) + SO4^2-(aq)}
\]

When barium chloride is added, the introduced sulfate ions react with barium ions:

\[
\ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)}
\]

A white precipitate can therefore form even if the original sample contains no sulfate.

Step 2: Choose a suitable replacement acid.

Use dilute nitric acid instead.

Nitric acid supplies hydrogen ions without introducing sulfate ions:

\[
\ce{HNO3(aq) -> H+(aq) + NO3-(aq)}
\]

The nitrate ions remain soluble with barium ions.

Step 3: Redesign the experiment.

Use a fresh portion of the unknown solution.

1. Add dilute nitric acid.
2. Add barium chloride.
3. Observe whether a persistent white precipitate forms.
4. Run a water blank using the same reagents.
5. Run a positive control containing a known sulfate solution.

Step 4: State the appropriate conclusion.

If the blank remains clear, the positive control produces a precipitate, and the acidified unknown produces a persistent white precipitate, the results support the presence of sulfate ions.

This conclusion assumes that other ions capable of producing an interfering precipitate have been excluded.

The key lesson is that the reagent itself must not introduce the substance being investigated.

Question 5

An unknown solution contains carbonate ions, sulfate ions, or both. No other anions that form precipitates with barium ions are present.

The solution is thoroughly mixed and divided into two equal portions, each measuring \(5.00\ \text{mL}\).

Two experiments are performed.

Experiment A

Excess barium chloride is added directly to the first portion. A white precipitate forms. After collection and drying, its mass is \(0.430\ \text{g}\).

Experiment B

The second portion is treated with excess dilute nitric acid. A gas is released and collected.

The gas volume is \(24.8\ \text{mL}\) at \(25^\circ\text{C}\) and \(100\ \text{kPa}\). It turns limewater milky.

After effervescence stops, excess barium chloride is added to the remaining acidified solution. A white precipitate forms. Its dried mass is \(0.233\ \text{g}\).

Use the following information:

• Molar volume of gas at the stated conditions: \(24.8\ \text{L mol}^{-1}\).
• Molar mass of \(\ce{BaCO3}\): \(197.3\ \text{g mol}^{-1}\).
• Molar mass of \(\ce{BaSO4}\): \(233.4\ \text{g mol}^{-1}\).

Assume the precipitation reactions are effectively complete, there are no significant experimental losses, and the two portions have identical compositions.

Determine:

a) Whether carbonate ions are present.

b) Whether sulfate ions are present.

c) The amount of each ion in one \(5.00\ \text{mL}\) portion.

d) Whether the calculated precipitate masses are consistent with the direct precipitation result.

Explain why the white precipitate in Experiment A cannot be identified as pure barium sulfate from its appearance alone.

Solution 5

Both carbonate and sulfate ions are present. Each \(5.00\ \text{mL}\) portion contains approximately \(1.00\times10^{-3}\ \text{mol}\) of carbonate ions and \(1.00\times10^{-3}\ \text{mol}\) of sulfate ions.

The crucial insight is that Experiment A produces two different white precipitates, whereas Experiment B selectively removes carbonate before testing for sulfate.

Step 1: Identify carbonate using the gas produced in Experiment B.

The gas turns limewater milky, confirming that it is carbon dioxide.

Carbonate ions react with hydrogen ions to produce carbon dioxide and water:

\[
\ce{CO3^2- + 2H+ -> CO2 + H2O}
\]

The mole ratio between carbonate ions and carbon dioxide is \(1:1\).

Calculate the amount of carbon dioxide:

\[
n=\frac{V}{V_m}
\]

Convert the gas volume into litres:

\[
24.8\ \text{mL}=0.0248\ \text{L}
\]

Substitute:

\[
n(\ce{CO2})=\frac{0.0248}{24.8}
\]

\[
=1.00\times10^{-3}\ \text{mol}
\]

Therefore:

\[
\boxed{n(\mathrm{CO_{3}}^{2-})=1.00\times10^{-3}\ \text{mol}}
\]

Carbonate ions are present.

Step 2: Identify sulfate using the precipitate from Experiment B.

The nitric acid has reacted with carbonate ions, releasing carbon dioxide.

Sulfate ions remain in solution because they do not undergo the same gas-producing reaction.

When barium chloride is added, sulfate ions form barium sulfate:

\[
\ce{Ba^2+ + SO4^2- -> BaSO4(s)}
\]

The measured mass of barium sulfate is \(0.233\ \text{g}\).

Calculate its amount:

\[
n=\frac{m}{M}
\]

\[
n(\ce{BaSO4})=\frac{0.233}{233.4}
\]

\[
=9.98\times10^{-4}\ \text{mol}
\]

The mole ratio between barium ions and sulfate ions is \(1:1\).

Therefore:

\[
\boxed{n(\mathrm{SO_{4}}^{2-})\approx1.00\times10^{-3}\ \text{mol}}
\]

Sulfate ions are also present.

Step 3: Calculate the expected mass of barium carbonate.

In Experiment A, carbonate ions react with excess barium ions to form barium carbonate.

\[
\ce{Ba^2+ + CO3^2- -> BaCO3(s)}
\]

The mole ratio is \(1:1\), so:

\[
n(\ce{BaCO3})=1.00\times10^{-3}\ \text{mol}
\]

Calculate the mass:

\[
m=nM
\]

\[
m=(1.00\times10^{-3})(197.3)
\]

\[
\boxed{m(\ce{BaCO3})=0.1973\ \text{g}}
\]

Step 4: Check the total mass against Experiment A.

Experiment A contains both barium carbonate and barium sulfate.

Therefore, the predicted total precipitate mass is:

\[
m_{\text{total}}=
m(\ce{BaCO3})+m(\ce{BaSO4})
\]

\[
=0.1973+0.233
\]

\[
\boxed{m_{\text{total}}=0.4303\ \text{g}}
\]

The experimentally measured mass is \(0.430\ \text{g}\).

These values agree within the precision of the supplied measurements.

Step 5: Explain what the comparison establishes.

The results provide consistent evidence for both carbonate and sulfate ions.

The gas measurement identifies carbonate through its reaction with acid. The remaining acidified solution produces barium sulfate, providing evidence for sulfate.

The direct precipitation experiment produces both barium carbonate and barium sulfate. Both are white, so the appearance of the original precipitate cannot distinguish them.

The mass calculation provides an additional consistency check.

A student who confidently identifies the entire \(0.430\ \text{g}\) precipitate as barium sulfate has overlooked the carbonate reaction. The two separate experiments are needed to resolve the ambiguity.

This is why comparing a solubility-rule prediction with experimental evidence involves more than checking whether a solid appeared.

08Where to go next

Once you can compare predicted and observed precipitates, the next useful skill is choosing an appropriate confirmatory test when different ions produce similar observations.

The distinction between an expected result, an actual observation, and a justified chemical conclusion is central to interpreting qualitative analysis experiments and writing strong HSC Chemistry responses.