Why Tap Water Interferes with HSC Chemistry Ion Tests
Learn how dissolved ions in tap water can cause false positives, hide ions, and mislead precipitation tests. Explore worked calculations, blank tests, confirmatory reactions, and HSC-style questions with solutions.
A student dissolves a supposedly chloride-free salt in tap water, adds silver nitrate, and sees a white precipitate. Does that mean the salt contains chloride ions?
Not necessarily. Tap water already contains dissolved ions, including chloride, calcium, magnesium, and sometimes enough sulfate to interfere with identification tests. These ions can react with your reagents, produce misleading precipitates, or remove the ion you wanted to identify before you test for it.
The problem is that a chemical test detects what is present in the test tube. It doesn’t automatically tell you whether an ion came from the unknown sample, the water used to dissolve it, or another reagent.
01Why a white precipitate can give the wrong conclusion
Suppose you’re identifying chloride ions, \(\mathrm{Cl}^{-}\), in an unknown solution.
The usual test involves adding dilute nitric acid, followed by aqueous silver nitrate, \(\ce{AgNO3}\).
Silver nitrate supplies silver ions, \(\mathrm{Ag}^{+}\). If chloride ions are present, the two ions combine to form silver chloride, an insoluble white solid.
\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]
The expected observation is a white precipitate.
Before testing the unknown, make a prediction: what happens if you add the same reagents to tap water containing dissolved chloride?
You get the same reaction.
The silver ions cannot distinguish chloride originating in the unknown from chloride already present in the water. Both produce \(\ce{AgCl}\).
That’s the first important limitation: a correct chemical observation can still lead to an incorrect identification.
Why add nitric acid first?
Dilute nitric acid helps remove carbonate interference. Carbonate ions can also form a poorly soluble silver compound, so acidifying the solution makes the silver nitrate test more selective for halides.
For carbonate:
\[ \ce{CO3^2-(aq) + 2H+(aq) -> CO2(g) + H2O(l)} \]
The carbonate is converted into carbon dioxide and water before silver nitrate is added.
Use nitric acid rather than hydrochloric acid for this test. Hydrochloric acid introduces chloride ions, creating the very interference you’re trying to avoid.
Even after acidification, a white precipitate isn’t a complete identification. Silver chloride is white, silver bromide is usually cream, and other interfering ions may affect the result.
A further test is the reaction with dilute aqueous ammonia:
\[ \ce{AgCl(s) + 2NH3(aq) -> [Ag(NH3)2]+(aq) + Cl-(aq)} \]
Silver chloride dissolves because the soluble diamminesilver(I) complex forms.
This supports identifying the precipitate as silver chloride. It still doesn’t establish where the chloride originally came from.
02How an improved experiment reveals the interference
Imagine the first design used only one test tube.
A student dissolves a sample in tap water, acidifies it, adds silver nitrate, and records a white precipitate.
That design has a flaw: there’s no way to tell whether the chloride was already in the sample or was introduced during preparation.
The revision is to test the water separately and repeat the sample preparation using deionised water.
Set up the comparison
Use five clean test tubes. Give every tube the same final volume before adding the testing reagents, then add the same amounts of dilute nitric acid and silver nitrate.
For this illustrative investigation, use a known chloride-free sodium nitrate sample, \(\ce{NaNO3}\), so the source of any interference can be investigated.
The following results are hypothetical teaching data, not measurements from an actual experiment.
| Tube | Contents before testing | Observation after acidified silver nitrate |
|---|---|---|
| A | Deionised-water blank | Remains clear |
| B | Tap-water blank | Faint white cloudiness |
| C | Sodium nitrate prepared in tap water | Faint white cloudiness |
| D | Same sodium nitrate prepared in deionised water | Remains clear |
| E | Known sodium chloride solution prepared in deionised water | White precipitate |
What do these results establish?
Start with tubes A and B.
The deionised-water blank remains clear, but the tap-water blank becomes cloudy. This indicates that something associated with the tap-water preparation can react with the silver nitrate.
Tube C also becomes cloudy, while tube D remains clear.
Because the sodium nitrate sample is the same in C and D, the difference is associated with the water used during preparation.
Under the assumption that chloride is the only significant silver-precipitating ion in the tap water after acidification, dissolved chloride explains the results.
Tube E provides another useful control. Its white precipitate shows that the silver nitrate reagent is capable of detecting chloride under these conditions.
The experiment now has three distinct pieces of evidence:
- The tap-water blank reveals a background reaction.
- The deionised-water preparation separates the sample from that background.
- The known chloride standard checks that the test can produce its expected positive result.
One qualification matters. A clear tube only means that no precipitate was detected under the conditions used. It doesn’t prove that the chloride concentration is exactly zero.
Why the two blanks are different
A deionised-water reagent blank checks whether the reagents, glassware, and deionised water introduce interference.
A tap-water blank checks what happens when tap water goes through the same procedure without the unknown sample.
If the preparation uses tap water, a clean deionised-water blank alone can miss the problem entirely.
For a wider explanation of matching controls to possible contamination sources, see How to Identify Contamination Using a Blank Test.
03Which dissolved ions in tap water cause problems?
Tap water isn’t pure \(\ce{H2O}\). It contains dissolved mineral ions, with concentrations depending on the water source, treatment, and distribution system.
The important question is not whether those ions are harmful to drink. It’s whether they can react with the reagents used in your practical.
| Ion that may be present | Reagent or condition | Possible interference |
|---|---|---|
| Chloride, \(\mathrm{Cl}^{-}\) | Acidified silver nitrate | Forms white \(\ce{AgCl}\), potentially giving a false positive for chloride in the unknown |
| Sulfate, \(\mathrm{SO_{4}}^{2-}\) | Acidified barium nitrate | Forms white \(\ce{BaSO4}\), potentially giving a false positive for sulfate |
| Calcium, \(\mathrm{Ca}^{2+}\) | Sodium carbonate | Forms white \(\ce{CaCO3}\), which may be incorrectly attributed to calcium in the unknown |
| Magnesium, \(\mathrm{Mg}^{2+}\) | Sodium hydroxide | Can form white \(\ce{Mg(OH)2}\), interfering with some metal-ion tests |
| Hydrogencarbonate, \(\ce{HCO3-}\) | Acid or strongly alkaline conditions | Consumes acid, releases carbon dioxide, or can be converted into carbonate |
Not every supply contains these ions at concentrations high enough to cause a visible result. A positive test depends on concentration, reagent conditions, and the sensitivity of the observation.
Interference in a sulfate test
For sulfate identification, a suitable procedure is to acidify a fresh portion of the sample with dilute nitric acid and then add aqueous barium nitrate.
\[ \ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)} \]
The expected observation is a white precipitate that remains insoluble in dilute nitric acid.
But sulfate ions already present in tap water can produce exactly the same solid.
Carbonate creates an additional complication. Without sufficient acidification, barium ions may also form white barium carbonate:
\[ \ce{Ba^2+(aq) + CO3^2-(aq) -> BaCO3(s)} \]
Barium carbonate dissolves in dilute acid, whereas barium sulfate remains.
This difference is useful, but it doesn’t solve the source problem. A precipitate identified as \(\ce{BaSO4}\) could still have formed from sulfate in the tap water.
There is also a sequencing problem: if you use barium chloride instead of barium nitrate, you’re adding chloride to the mixture. Testing that same mixture for chloride with silver nitrate would then be unreliable.
Use separate, fresh aliquots for tests involving different ions.
Tap water can also hide an ion
So far, tap water has added unwanted ions and produced false positives.
It can cause the opposite problem too.
Suppose your unknown contains carbonate ions, but you dissolve it in water containing calcium ions.
The calcium and carbonate can react immediately:
\[ \ce{Ca^2+(aq) + CO3^2-(aq) -> CaCO3(s)} \]
A white solid forms.
If you filter that solid away before testing the remaining solution for carbonate, much of the carbonate may already be gone.
A student who tests only the clear filtrate could incorrectly conclude that the original unknown contained no carbonate.
The key is to keep track of which substances were present originally and which were removed during preparation.
04Worked examples
Worked example: How much silver chloride could come from tap water alone?
A student uses \(5.00\ \mathrm{mL}\) of tap water to prepare \(10.0\ \mathrm{mL}\) of a solution intended for chloride testing.
For this illustrative calculation, the tap water contains \(35.5\ \mathrm{mg/L}\) of chloride ions.
Calculate the maximum mass of silver chloride that could form from the tap water alone when excess silver nitrate is added.
Assume all the chloride precipitates as silver chloride, no other silver precipitate forms, and no material is lost.
Step 1
The molar mass of chloride ions is approximately \(35.45\ \mathrm{g/mol}\).
First convert milligrams to grams:
\[ 35.5\ \mathrm{mg/L} =0.0355\ \mathrm{g/L} \]
Concentration in moles per litre is mass concentration divided by molar mass:
\[ \begin{aligned} c&=\frac{0.0355\ \mathrm{g/L}} {35.45\ \mathrm{g/mol}}\\ &=1.00\times10^{-3}\ \mathrm{mol/L} \end{aligned} \]
Here, \(c\) is the chloride concentration.
Step 2
Use:
\[ n=cV \]
Here, \(n\) is the amount in moles, and \(V\) is the volume in litres.
The volume of tap water is \(0.00500\ \mathrm{L}\).
\[ \begin{aligned} n&=(1.00\times10^{-3})(0.00500)\\ &=5.00\times10^{-6}\ \mathrm{mol} \end{aligned} \]
Notice that making the sample up to \(10.0\ \mathrm{mL}\) dilutes the chloride but doesn’t change the number of moles introduced.
Step 3
The precipitation equation is:
\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]
The mole ratio of chloride to silver chloride is \(1:1\).
Therefore:
\[ n(\ce{AgCl})=5.00\times10^{-6}\ \mathrm{mol} \]
The molar mass of silver chloride is \(143.32\ \mathrm{g/mol}\).
Using \(m=nM\), where \(m\) is mass and \(M\) is molar mass:
\[ \begin{aligned} m&=(5.00\times10^{-6})(143.32)\\ &=7.17\times10^{-4}\ \mathrm{g}\\ &=0.717\ \mathrm{mg} \end{aligned} \]
The maximum predicted mass is \(0.717\ \mathrm{mg}\) of silver chloride from the tap water alone.
That’s a small amount, but it’s chemically real. Whether the cloudiness is easily visible depends on the conditions, particle size, and observation method.
The calculation also shows why dilution doesn’t automatically eliminate contamination. Dilution lowers concentration, but the introduced ions remain in the mixture unless a process removes them.
Worked example: Can tap water make carbonate disappear?
A student mixes \(20.0\ \mathrm{mL}\) of an unknown solution containing \(1.00\times10^{-3}\ \mathrm{mol/L}\) carbonate ions with \(20.0\ \mathrm{mL}\) of tap water containing \(1.50\times10^{-3}\ \mathrm{mol/L}\) calcium ions.
A white solid forms and is removed by filtration.
Assuming calcium carbonate is the only precipitate and precipitation is effectively complete, calculate the mass formed and explain why testing the filtrate for carbonate might be misleading.
Step 1
\[ \ce{Ca^2+(aq) + CO3^2-(aq) -> CaCO3(s)} \]
Each mole of calcium ions reacts with one mole of carbonate ions.
Step 2
Using \(n=cV\):
For calcium:
\[ \begin{aligned} n_{\mathrm{Ca}}&=(1.50\times10^{-3})(0.0200)\\ &=3.00\times10^{-5}\ \mathrm{mol} \end{aligned} \]
For carbonate:
\[ \begin{aligned} n_{\mathrm{CO_3}}&=(1.00\times10^{-3})(0.0200)\\ &=2.00\times10^{-5}\ \mathrm{mol} \end{aligned} \]
Here, \(n_{\mathrm{Ca}}\) and \(n_{\mathrm{CO_3}}\) represent the initial amounts of calcium and carbonate ions.
Carbonate is the limiting reactant because fewer moles are available and the reaction ratio is \(1:1\).
Step 3
The maximum amount of calcium carbonate is:
\[ n(\ce{CaCO3})=2.00\times10^{-5}\ \mathrm{mol} \]
Using a molar mass of \(100.09\ \mathrm{g/mol}\):
\[ \begin{aligned} m&=(2.00\times10^{-5})(100.09)\\ &=2.00\times10^{-3}\ \mathrm{g}\\ &=2.00\ \mathrm{mg} \end{aligned} \]
The theoretical precipitate mass is \(2.00\ \mathrm{mg}\).
Step 4
Under the complete-precipitation approximation, all the original carbonate has been incorporated into the solid.
In practice, a small equilibrium concentration remains dissolved, but the clear filtrate may contain too little carbonate to produce detectable effervescence when acid is added.
The original sample still contained carbonate. It was removed during preparation.
To investigate properly, test a fresh portion prepared in deionised water. Acidification of a carbonate-containing solution releases carbon dioxide, which can be confirmed by passing the gas through limewater.
The useful distinction is between an ion being absent from the original sample and an ion being removed before measurement.
05How to interpret unexpected observations
When a practical gives an unexpected precipitate, identify exactly which stage might have changed the mixture.
| Observation | What it suggests | What to do next |
|---|---|---|
| Unknown and tap-water blank both give a precipitate | Background ions may be contributing | Repeat the unknown preparation using deionised water |
| Unknown gives a precipitate, but a matched blank stays clear | The unknown may contain the tested ion | Use a suitable confirmatory reaction and known positive control |
| Deionised-water blank gives a precipitate | Reagents, water, or apparatus may be contaminated | Investigate each preparation source and clean or replace materials |
| A solid appears before the intended test | Some ions may have reacted during preparation | Identify the solid and test a fresh, unaltered portion |
| A previously positive sample becomes negative after filtration | The analyte may have entered the removed solid | Examine the residue and repeat the preparation without unnecessary separation |
Two rules prevent many identification mistakes.
First, a precipitate is evidence of a chemical reaction, not automatic proof of a particular ion’s origin.
Second, a negative result only describes the sample at the point when it was tested. An ion might already have reacted, been diluted below the detection limit, or been removed.
In laboratory work, use clean apparatus, separate droppers, suitable eye protection, and teacher-approved reagent concentrations. Silver nitrate and soluble barium salts require appropriate handling and waste collection.
06Questions and solutions
Question 1
A student prepares an unknown solution using tap water and tests it with dilute nitric acid followed by silver nitrate.
A white precipitate forms.
The student then tests two water blanks:
- Deionised-water blank: no precipitate.
- Tap-water blank: white precipitate.
Explain why the student cannot confidently identify chloride in the original unknown, and suggest one improvement.
Solution 1
The student cannot attribute the white precipitate to chloride in the unknown because the tap-water blank also produces a precipitate. The tap water may have introduced the reacting ions.
Chloride reacts with silver ions according to:
\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]
The white precipitate is consistent with silver chloride, but the blank demonstrates that the water can produce a similar result without the unknown.
The improvement is to prepare a fresh portion of the unknown using deionised water and repeat the acidified silver nitrate test alongside a matched blank.
A common mistake is to treat the observed white colour as complete proof of chloride in the unknown. It is evidence consistent with chloride, but the source must also be established.
Question 2
A student wants to test an unknown solution for both sulfate and chloride ions.
The sample has been prepared in deionised water.
The student follows this sequence:
- Adds dilute nitric acid and barium chloride. A white precipitate forms and persists in the acidified mixture.
- Filters the precipitate and adds silver nitrate to the filtrate. Another white precipitate forms.
The student concludes that the unknown contains both sulfate and chloride.
However, a deionised-water blank treated with barium chloride and then silver nitrate also produces a white precipitate at the second stage.
A separate, fresh portion of the unknown gives no precipitate when tested directly with dilute nitric acid and silver nitrate.
Evaluate the student’s conclusion, and propose a more reliable testing sequence.
Solution 2
The observations support sulfate in the unknown, but they do not establish chloride in the original sample. The second white precipitate can be explained by chloride introduced with the barium chloride reagent.
The first reaction is consistent with sulfate:
\[ \ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)} \]
Barium sulfate forms a white precipitate that persists in dilute nitric acid.
However, barium chloride introduces both barium and chloride ions:
\[ \ce{BaCl2(aq) -> Ba^2+(aq) + 2Cl-(aq)} \]
The chloride ions remain in solution when barium sulfate precipitates. Filtering removes the solid, not the dissolved chloride.
When silver nitrate is added to the filtrate:
\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]
Silver chloride forms even if the original unknown contained no chloride.
The positive result in the blank confirms that the testing sequence itself can generate the second precipitate. The negative direct silver nitrate test on a fresh aliquot provides no evidence of detectable chloride in the unknown.
A better procedure uses two fresh portions:
- Sulfate test: Acidify one portion with dilute nitric acid, then add barium nitrate. An acid-resistant white precipitate supports sulfate.
- Chloride test: Acidify the other portion with dilute nitric acid, then add silver nitrate. Investigate any white precipitate using dilute ammonia and appropriate controls.
This prevents chloride from the sulfate-testing reagent entering the chloride test.
The trap is treating the filtrate as chemically unchanged simply because the first precipitate has been removed.
Question 3
A student prepares an unknown solution using \(8.00\ \mathrm{mL}\) of tap water.
The tap water contains \(35.5\ \mathrm{mg/L}\) chloride, equivalent to \(1.00\times10^{-3}\ \mathrm{mol/L}\).
The unknown contributes an additional \(3.00\times10^{-6}\ \mathrm{mol}\) of chloride.
The student adds \(0.200\ \mathrm{mL}\) of \(0.0200\ \mathrm{mol/L}\) silver nitrate to the sample.
A matched tap-water blank receives the same silver nitrate volume. Assume all available silver ions precipitate as silver chloride.
Both tubes produce the same mass of precipitate.
Calculate that mass and explain why the equal results do not prove the unknown lacks chloride.
Use \(M(\ce{AgCl})=143.32\ \mathrm{g/mol}\).
Solution 3
Both tubes can produce \(0.573\ \mathrm{mg}\) of silver chloride even though the unknown contains additional chloride. Silver nitrate, not chloride, is the limiting reagent in both tubes.
First calculate chloride from the tap water:
\[ \begin{aligned} n_{\mathrm{tap}} &=(1.00\times10^{-3})(0.00800)\\ &=8.00\times10^{-6}\ \mathrm{mol} \end{aligned} \]
The sample contains the tap-water chloride plus the extra chloride:
\[ \begin{aligned} n_{\mathrm{sample}} &=8.00\times10^{-6}+3.00\times10^{-6}\\ &=1.10\times10^{-5}\ \mathrm{mol} \end{aligned} \]
Here, \(n_{\mathrm{tap}}\) and \(n_{\mathrm{sample}}\) are the amounts of chloride in the two tubes.
Now calculate the silver ions added. Convert \(0.200\ \mathrm{mL}\) to \(0.000200\ \mathrm{L}\):
\[ \begin{aligned} n(\mathrm{Ag}^{+}) &=0.0200\times0.000200\\ &=4.00\times10^{-6}\ \mathrm{mol} \end{aligned} \]
The reaction has a \(1:1\) mole ratio:
\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]
Both tubes contain more chloride than the \(4.00\times10^{-6}\ \mathrm{mol}\) of available silver ions.
Therefore, both form the same maximum amount of precipitate:
\[ n(\ce{AgCl})=4.00\times10^{-6}\ \mathrm{mol} \]
Its mass is:
\[ \begin{aligned} m&=(4.00\times10^{-6})(143.32)\\ &=5.73\times10^{-4}\ \mathrm{g}\\ &=0.573\ \mathrm{mg} \end{aligned} \]
The blank retains \(4.00\times10^{-6}\ \mathrm{mol}\) of unprecipitated chloride, while the unknown retains \(7.00\times10^{-6}\ \mathrm{mol}\).
The extra chloride is present, but there aren’t enough silver ions to reveal the difference through additional precipitate formation.
A tempting approach is to compare the two precipitate masses and conclude that equal masses mean equal chloride concentrations. That reasoning would require suitable excess reagent and a quantitative method.
A better investigation would use sufficient silver nitrate, an appropriately matched blank, and a validated quantitative measurement method. Preparing another portion with deionised water would also help isolate the sample’s chloride from the tap-water background.
Question 4
An unknown salt is one of the following:
- Pure sodium carbonate, \(\ce{Na2CO3}\).
- Pure sodium sulfate, \(\ce{Na2SO4}\).
- A mixture containing both ions at concentrations detectable by the procedures used.
A student prepares the unknown once using tap water and again using deionised water.
Barium nitrate is added to separate portions before acidification. The student then adds sufficient dilute nitric acid and records whether the precipitate remains.
The following observations are illustrative.
| Preparation | Before acid | After adding dilute nitric acid |
|---|---|---|
| Tap-water blank + barium nitrate | Small white precipitate | Small amount remains |
| Unknown in tap water + barium nitrate | Large white precipitate | Small amount remains, similar to the blank |
| Unknown in deionised water + barium nitrate | Large white precipitate | Precipitate disappears |
| Deionised-water blank + barium nitrate | No precipitate | No change |
A fresh portion of the unknown prepared in deionised water releases a gas when dilute nitric acid is added. The gas turns limewater milky.
Determine the identity of the unknown under the stated assumptions. Explain the different observations and describe how to make the sulfate test more selective.
Solution 4
The unknown is sodium carbonate under the given choices and detection assumptions. The results do not support sulfate in the original salt.
The decisive evidence comes from the deionised-water preparation.
Before acidification, carbonate reacts with barium ions:
\[ \ce{Ba^2+(aq) + CO3^2-(aq) -> BaCO3(s)} \]
The white barium carbonate precipitate dissolves when sufficient acid is added:
\[ \ce{BaCO3(s) + 2H+(aq) -> Ba^2+(aq) + CO2(g) + H2O(l)} \]
The disappearance of the precipitate is consistent with carbonate rather than sulfate.
The gas test provides stronger supporting evidence. Carbon dioxide reacts with limewater:
\[ \ce{CO2(g) + Ca(OH)2(aq) -> CaCO3(s) + H2O(l)} \]
The formation of white calcium carbonate makes the limewater milky.
If sulfate were also present at a detectable concentration, a barium sulfate precipitate would remain after acidification:
\[ \ce{Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s)} \]
No such residue remains in the deionised-water preparation.
Now consider the tap-water results.
The tap-water blank already produces a small acid-resistant precipitate. This is consistent with sulfate or another relevant interference originating from the tap water.
The unknown prepared in tap water produces a larger initial precipitate because its carbonate forms barium carbonate, while the background may contribute another solid.
After acidification, the barium carbonate dissolves, leaving a small amount resembling the tap-water blank.
The tempting conclusion is that an acid-resistant precipitate anywhere in the unknown’s test must prove the unknown contains sulfate. But the same residue appears without the unknown, so it cannot be attributed confidently to the original salt.
For a more selective sulfate test, acidify a fresh aliquot with dilute nitric acid before adding barium nitrate, and compare it against the appropriate water blank.
The result illustrates why the order of reagent addition and the preparation water both affect an ion-identification conclusion.
Question 5
A student is given an unknown solution that contains either sodium hydroxide, \(\ce{NaOH}\), or sodium carbonate, \(\ce{Na2CO3}\).
Both possible solutions have a concentration of \(0.0100\ \mathrm{mol/L}\).
The student adds \(20.0\ \mathrm{mL}\) of the unknown to \(50.0\ \mathrm{mL}\) of model tap water.
The model tap water initially contains:
\[ [\ce{Ca^2}]=1.00\times10^{-3}\ \mathrm{mol/L} \]
\[ [\ce{HCO3-}]=2.00\times10^{-3}\ \mathrm{mol/L} \]
Here, square brackets indicate molar concentration. Assume all other dissolved ions are spectators, the original tap water is clear, and precipitation can be treated as effectively complete.
A white precipitate forms. After filtration, the student adds dilute nitric acid to the collected solid and passes the evolved gas into limewater. The limewater turns milky.
A tap-water blank with no unknown added produces no precipitate.
The student confidently identifies the unknown as sodium carbonate.
However, a new \(20.0\ \mathrm{mL}\) portion of the unknown is added to \(50.0\ \mathrm{mL}\) of clean deionised water and immediately treated with sufficient dilute nitric acid.
No carbon dioxide is detected. A known sodium carbonate control treated in the same way produces carbon dioxide.
Determine which compound the unknown actually contains.
In your explanation:
- Show how either candidate could produce the original white precipitate.
- Calculate the maximum mass of calcium carbonate possible from the model tap water.
- Explain why the acid test on the collected precipitate is not decisive.
- Identify the evidence that distinguishes the two candidates.
Use \(M(\ce{CaCO3})=100.09\ \mathrm{g/mol}\).
Solution 5
The unknown is sodium hydroxide, \(\ce{NaOH}\). Sodium carbonate would produce carbon dioxide when a fresh portion prepared in deionised water was acidified, whereas the unknown does not.
The difficult part is explaining why sodium hydroxide could produce calcium carbonate in the original tap-water mixture.
First, consider sodium carbonate as a possible unknown.
If the unknown were sodium carbonate, it would supply carbonate ions directly:
\[ \ce{Na2CO3(aq) -> 2Na+(aq) + CO3^2-(aq)} \]
These would react with calcium already present in the tap water:
\[ \ce{Ca^2+(aq) + CO3^2-(aq) -> CaCO3(s)} \]
A white precipitate would form.
Adding acid to that precipitate would produce carbon dioxide:
\[ \ce{CaCO3(s) + 2H+(aq) -> Ca^2+(aq) + CO2(g) + H2O(l)} \]
So the original observations are certainly consistent with sodium carbonate.
But they aren’t unique to it.
Now consider what sodium hydroxide does to hydrogencarbonate.
The tap water contains hydrogencarbonate, \(\ce{HCO3-}\), as well as calcium ions.
Sodium hydroxide supplies hydroxide ions, which react with hydrogencarbonate:
\[ \ce{HCO3-(aq) + OH-(aq) -> CO3^2-(aq) + H2O(l)} \]
This reaction creates carbonate ions from hydrogencarbonate already present in the water.
Those newly formed carbonate ions then precipitate calcium:
\[ \ce{Ca^2+(aq) + CO3^2-(aq) -> CaCO3(s)} \]
So sodium hydroxide can also produce white calcium carbonate, despite the original unknown containing no carbonate ions.
This is the central trap: identifying carbonate in the product doesn’t necessarily establish that carbonate was present in the starting sample.
Calculate whether the sodium hydroxide explanation is possible.
The initial calcium amount in \(50.0\ \mathrm{mL}\), or \(0.0500\ \mathrm{L}\), of tap water is:
\[ \begin{aligned} n_{\mathrm{Ca}} &=(1.00\times10^{-3})(0.0500)\\ &=5.00\times10^{-5}\ \mathrm{mol} \end{aligned} \]
The hydrogencarbonate amount is:
\[ \begin{aligned} n_{\mathrm{HCO_3}} &=(2.00\times10^{-3})(0.0500)\\ &=1.00\times10^{-4}\ \mathrm{mol} \end{aligned} \]
The unknown contributes:
\[ \begin{aligned} n_{\mathrm{unknown}} &=0.0100\times0.0200\\ &=2.00\times10^{-4}\ \mathrm{mol} \end{aligned} \]
If the unknown is sodium hydroxide, it provides \(2.00\times10^{-4}\ \mathrm{mol}\) of hydroxide ions.
That is enough to convert all \(1.00\times10^{-4}\ \mathrm{mol}\) of hydrogencarbonate into carbonate under the stated idealised conditions.
The resulting carbonate amount exceeds the available calcium amount.
Calcium therefore limits calcium carbonate formation.
The maximum precipitate mass is:
\[ \begin{aligned} m&=(5.00\times10^{-5})(100.09)\\ &=5.00\times10^{-3}\ \mathrm{g}\\ &=5.00\ \mathrm{mg} \end{aligned} \]
If the unknown were sodium carbonate, its \(2.00\times10^{-4}\ \mathrm{mol}\) of carbonate would also exceed the available calcium amount. The same theoretical maximum of \(5.00\ \mathrm{mg}\) would result.
Therefore, even an accurate measurement of the precipitate mass would not distinguish the two candidates in this setup.
Why does the fresh deionised-water test settle the identity?
When sodium carbonate is dissolved in deionised water, it supplies carbonate directly. Adding sufficient acid produces carbon dioxide.
When sodium hydroxide is dissolved in clean deionised water, it supplies hydroxide, but there is no hydrogencarbonate source in the stated system.
Adding acid simply neutralises the hydroxide:
\[ \ce{H+(aq) + OH-(aq) -> H2O(l)} \]
There is no carbonate available to release carbon dioxide.
The fresh unknown produces no detectable carbon dioxide, while the known sodium carbonate control produces the expected positive result.
Under the question’s assumptions, this identifies the unknown as sodium hydroxide.
The original precipitate test established that calcium carbonate formed. The revised test established whether carbonate came directly from the unknown.
This is a deeper form of tap-water interference: the water doesn’t merely add a background signal. Its ions can react with the unknown to create a different species before the identification test begins.
When planning a practical involving several ions, the next decision is therefore which tests must use fresh aliquots, which reagents change the sample’s composition, and which observations actually identify the original material. Those decisions are developed further in How to Plan a Sequence of Tests for Two Unknown Ions.