Why One Water Sample Cannot Prove a River Is Unpolluted

A water sample collected from a river at 10 am produces a very small amount of white precipitate during chemical analysis. Does that mean the entire river is unpolluted?

No. It tells you something about the particular water collected, at the particular location and time of sampling. It cannot establish that the rest of the river contains no pollutants.

There are two separate problems to solve before making a reliable conclusion:

1. Sampling: Did the water sample actually represent the river?
2. Analysis: Did the experiment accurately measure the amount of the substance being investigated?

Understanding both problems is important when using precipitation reactions to analyse environmental water samples. Even a perfectly performed chemical analysis cannot compensate for a poorly collected sample.

01Why one water sample cannot represent an entire river

Imagine that a stormwater drain releases contaminated water into a river. The river carries this water downstream, but it does not necessarily mix evenly with the surrounding water straight away.

Now imagine collecting a sample from the opposite riverbank, upstream of the drain.

Would your sample detect the contamination?

Of course not. The contaminated water has never reached your sampling location.

Even if you collect your sample downstream, you might collect water from a part of the river that has not mixed completely with the discharge.

This is why a river cannot automatically be treated as one uniform solution.

A river changes in several ways:

• Location: Tributaries, industrial discharges, agricultural runoff and stormwater drains can introduce substances at particular points.
• Depth: Suspended particles and dissolved substances may be distributed differently through the water column.
• Weather: Rainfall can wash substances from surrounding land into the river. Concentrations may change significantly during and after rainfall.
• Time: A discharge might occur for only a few hours. A sample collected the following morning could miss it entirely.

Think about taking one spoonful from a pot of soup. If the soup has been stirred thoroughly, that spoonful might represent the whole pot.

A river is different. It is flowing, receiving new inputs, and constantly changing. There is no guarantee that the water at one sampling point has the same composition as the water kilometres upstream.

What does one sample actually tell you?

The distinction is important:

There is another limitation.

Suppose your analysis measures sulfate ions, \(\mathrm{SO_4^{2-}}\). A low sulfate concentration does not establish that the water is free from lead ions, harmful organic chemicals, excessive nutrients, or microorganisms.

You have measured one analyte, not every possible pollutant.

An analyte is the particular substance being measured in a chemical analysis.

Also, a high sulfate concentration does not automatically prove human pollution. Sulfate can occur naturally in water through contact with rocks and minerals. Results need to be interpreted against appropriate background measurements and environmental information.

The important conclusion is that a single sample cannot establish that an entire river is unpolluted, even when the laboratory analysis is accurate.

02Following sulfate from the river to a precipitate

Let’s follow one specific analyte through an environmental water analysis.

Suppose a monitoring team wants to measure the concentration of dissolved sulfate ions in river water.

The problem is that sulfate ions are dissolved in water. You cannot simply collect the water, put it on a balance, and measure the mass of sulfate.

Instead, you can convert the dissolved sulfate into an insoluble solid using a precipitation reaction.

Step 1: Collect a suitable water sample

The team collects water in a clean sample bottle.

For a measurement of dissolved sulfate, the sample is filtered through a suitable membrane to remove suspended particles. This must happen before the precipitation analysis.

This defines exactly what is being measured: sulfate present in the dissolved fraction.

If the aim is to measure total recoverable sulfate, including sulfate associated with suspended material, the sample must remain unfiltered and undergo an appropriate preparation or digestion procedure.

These are different measurements. Filtering a sample before analysis changes which fraction of the analyte is included.

A good sampling investigation also records the location, collection time, depth, weather conditions, and other relevant details.

If samples from different locations are going to be compared, their collection and preparation methods should be consistent.

Step 2: Add a reagent to form a precipitate

The filtered water sample is transferred into a beaker.

A small amount of dilute hydrochloric acid can be added to suppress carbonate interference. A solution of barium chloride is then added slowly while the mixture is stirred.

Barium chloride supplies barium ions, \(\mathrm{Ba^{2+}}\), which react with the dissolved sulfate ions.

The reaction is:

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

The product is barium sulfate, \(\mathrm{BaSO_4}\), an insoluble white solid.

What should happen if the original water contains sulfate ions?

A white precipitate should form as barium sulfate is produced.

That observation supports the presence of sulfate in the particular sample. However, the appearance of a white solid alone does not establish its exact mass or purity.

Other ions can also form insoluble barium compounds. The method must control potential interfering substances.

Step 3: Separate and wash the precipitate

The mixture is allowed to stand so that the precipitate can settle.

The solid is then separated from the remaining liquid by filtration.

A pre-weighed filter crucible can be used to collect the precipitate. The solid is washed with deionised water to remove soluble substances, including excess reagents and dissolved salts.

Why is washing necessary?

Because the solid may be surrounded by liquid containing substances that have not reacted. If these remain trapped in the collected material, they can contribute to the measured mass.

The aim is to isolate the barium sulfate rather than weigh a mixture of barium sulfate and leftover solution.

Step 4: Dry the precipitate before weighing

This is where an important experimental error can occur.

Immediately after filtration, the collected precipitate is wet. It contains barium sulfate mixed with water trapped between the solid particles.

If you weigh it immediately, the balance measures both the solid and the retained liquid.

You therefore need to dry the precipitate.

The filter crucible is heated or placed in a suitable drying oven, then cooled in a desiccator before weighing. This process is repeated until the mass remains constant within the precision of the balance.

This is called drying to constant mass.

The final mass is used to calculate the amount of barium sulfate produced.

However, remember an important distinction:

A dry precipitate is not automatically a pure precipitate.

Drying removes water, but it does not necessarily remove insoluble impurities that have formed alongside the barium sulfate.

For the measured mass to represent pure barium sulfate, the precipitation method must be sufficiently selective, the solid must be washed properly, and interfering substances must be controlled.

A reagent blank can also help identify contamination introduced by the reagents or laboratory procedure.

03Wet precipitate versus pure dry precipitate

These terms are not interchangeable.

The difference matters because the mass of the precipitate is used to calculate the amount of sulfate originally present.

If you mistakenly use the wet mass, you overestimate the amount of barium sulfate.

This leads to an overestimate of the sulfate concentration.

How does the precipitate mass relate to sulfate mass?

The precipitation reaction has a mole ratio of 1:1.

\[ 1\text{ mol BaSO}_4 = 1\text{ mol SO}_4^{2-} \]

However, the molar masses are different.

• Molar mass of \(\mathrm{BaSO_4}\) = 233.39 g mol\(^{-1}\)
• Molar mass of \(\mathrm{SO_4^{2-}}\) = 96.06 g mol\(^{-1}\)

Why aren’t they equal?

Because each mole of barium sulfate contains both barium and sulfate. The mass of the barium must not be included when reporting the mass of sulfate.

The calculation therefore has three stages:

1. Convert the dry barium sulfate mass into moles.
2. Use the 1:1 mole ratio to determine the moles of sulfate.
3. Convert the sulfate moles into mass, then calculate its concentration.

The relevant equations are:

\[ n(\mathrm{BaSO_4}) = \frac{m(\mathrm{BaSO_4})}{233.39} \]

\[ n(\mathrm{SO_4^{2-}}) = n(\mathrm{BaSO_4}) \]

\[ m(\mathrm{SO_4^{2-}}) = n(\mathrm{SO_4^{2-}})\times96.06 \]

Here, \(n\) represents amount in moles, and \(m\) represents mass in grams.

To calculate concentration in mg/L, convert the sulfate mass into milligrams and divide by the original sample volume in litres.

04Worked examples

05Questions and solutions

The following questions move from routine gravimetric calculations towards more demanding environmental analysis problems. Pay attention to what the experiment actually measures, not just which equation you need.

Question 1

A 200.0 mL filtered river-water sample is treated with excess barium chloride solution.

After filtration, washing, and drying to constant mass, 0.1867 g of pure barium sulfate is collected.

Calculate the concentration of dissolved sulfate ions in the original sample, in mg/L.

Assume precipitation is quantitative and there are no interfering ions.

Solution 1

The sulfate concentration is 384 mg/L.

The key is to convert the mass of barium sulfate into the mass of sulfate using the 1:1 mole ratio.

Step 1: Calculate the moles of barium sulfate.

\[ n(\mathrm{BaSO_4}) = \frac{0.1867}{233.39} \]

\[ =8.00\times10^{-4}\text{ mol} \]

Step 2: Calculate the mass of sulfate.

\[ n(\mathrm{SO_4^{2-}}) = 8.00\times10^{-4}\text{ mol} \]

\[ m=nM \]

\[ =(8.00\times10^{-4})(96.06) \]

\[ =0.07685\text{ g} \]

\[ =76.85\text{ mg} \]

Step 3: Calculate the concentration.

Convert 200.0 mL into litres:

\[ V=0.2000\text{ L} \]

\[ c=\frac{76.85}{0.2000} \]

\[ \boxed{c=384\text{ mg/L}} \]

The result describes the dissolved sulfate concentration in the collected sample. It does not establish the concentration at other locations in the river.

Question 2

A river receives water from a tributary.

Before the junction:

• The main river has a flow rate of 2.0 m³/s and a sulfate concentration of 120 mg/L.
• The tributary has a flow rate of 0.50 m³/s and a sulfate concentration of 600 mg/L.

Assume the two flows mix completely, there is no chemical reaction or removal of sulfate, and the total flow is the sum of the two flow rates.

(a) Calculate the expected sulfate concentration after complete mixing.

(b) A monitoring team collects two samples from a cross-section shortly below the junction, before complete mixing has occurred. A sample near the tributary bank measures 310 mg/L, while a sample from mid-channel measures 215 mg/L.

Explain why these results differ and whether either individual sample necessarily represents the fully mixed river.

Solution 2

(a) The fully mixed sulfate concentration is 216 mg/L. (b) The near-bank sample is likely measuring water containing a larger proportion of tributary water, while the mid-channel sample is closer to the eventual mixed concentration. Neither individual measurement guarantees a representative result.

Step 1: Calculate the total flow rate.

\[ Q_{\text{total}}=2.0+0.50 \]

\[ =2.50\text{ m}^3/\text{s} \]

Step 2: Calculate the sulfate entering the junction per second.

Because the flow rates are expressed in the same units, we can use a flow-weighted concentration.

\[ C_{\text{mixed}} = \frac{Q_1C_1+Q_2C_2}{Q_1+Q_2} \]

Substitute:

\[ C_{\text{mixed}} = \frac{(2.0)(120)+(0.50)(600)} {2.50} \]

\[ =\frac{240+300}{2.50} \]

\[ \boxed{C_{\text{mixed}}=216\text{ mg/L}} \]

Step 3: Explain the different measurements.

The tributary has a much higher sulfate concentration than the main river.

Immediately after the junction, the water may not be completely mixed. The tributary-rich plume can remain concentrated along one side of the river.

The near-bank sample therefore measures 310 mg/L, which is higher than the eventual mixed concentration.

The mid-channel sample measures 215 mg/L, close to the predicted fully mixed value of 216 mg/L.

This does not prove that the mid-channel sample represents the entire cross-section. It may simply happen to be close to the average.

The important experimental limitation is that the concentration varies across the river before mixing is complete.

A more reliable investigation would collect samples at several positions across the downstream cross-section, preferably after sufficient mixing has occurred.

Question 3

A student performs a gravimetric analysis on 250.0 mL of filtered river water.

The following masses are recorded:

Assume that the final dry precipitate is pure barium sulfate and that precipitation is quantitative.

(a) Calculate the mass of dry barium sulfate collected.

(b) Calculate the sulfate concentration in the original water sample, in mg/L.

(c) Calculate the sulfate concentration the student would report if they incorrectly used the wet precipitate mass.

(d) Explain why the two calculated concentrations differ.

Solution 3

(a) The dry barium sulfate mass is 0.233 g. (b) The correct sulfate concentration is 384 mg/L. (c) Using the wet mass would give approximately 453 mg/L, an overestimate of about 18%.

Step 1: Calculate the net dry precipitate mass.

Subtract the empty crucible mass from the final mass.

\[ m_{\text{dry}} = 32.075-31.842 \]

\[ \boxed{m_{\text{dry}}=0.233\text{ g}} \]

Step 2: Calculate the moles of dry barium sulfate.

\[ n(\mathrm{BaSO_4}) = \frac{0.233}{233.39} \]

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

The mole ratio between barium sulfate and sulfate is 1:1.

Step 3: Calculate the mass of sulfate.

\[ m(\mathrm{SO_4^{2-}}) = (9.98\times10^{-4})(96.06) \]

\[ =0.0959\text{ g} \]

\[ =95.9\text{ mg} \]

Step 4: Calculate the correct concentration.

\[ c=\frac{95.9}{0.2500} \]

\[ \boxed{c=384\text{ mg/L}} \]

Step 5: Calculate the concentration using the incorrect wet mass.

First, calculate the wet precipitate mass:

\[ m_{\text{wet}} = 32.117-31.842 \]

\[ =0.275\text{ g} \]

If this is incorrectly treated as pure barium sulfate:

\[ n(\mathrm{BaSO_4}) = \frac{0.275}{233.39} \]

\[ =1.18\times10^{-3}\text{ mol} \]

Convert to sulfate mass:

\[ m(\mathrm{SO_4^{2-}}) = (1.18\times10^{-3})(96.06) \]

\[ =0.1132\text{ g} \]

\[ =113.2\text{ mg} \]

Calculate the apparent concentration:

\[ c=\frac{113.2}{0.2500} \]

\[ \boxed{c\approx453\text{ mg/L}} \]

Step 6: Explain the difference.

The wet precipitate contains more than just barium sulfate.

Water remains trapped between the particles, and soluble substances may also remain in the collected liquid.

The difference between the wet and dry masses is:

\[ 0.275-0.233=0.042\text{ g} \]

This 0.042 g represents material removed during washing and drying. It should not automatically be interpreted as water alone.

Using the wet mass incorrectly treats all of this additional material as barium sulfate.

Consequently, the calculated number of moles of sulfate is too high, producing an overestimate of the concentration.

This is why gravimetric analysis requires a washed, dry precipitate of known purity.

Question 4

A student collects one 250.0 mL water sample from the downstream bank of a river at midday, during a period of dry weather.

The sample produces only a very small mass of barium sulfate during gravimetric analysis.

The student concludes:

“The river is unpolluted because the sulfate concentration is very low.”

Evaluate this conclusion.

Then describe how you would redesign the investigation to determine whether a nearby stormwater drain increases the concentration of dissolved sulfate in the river.

Your response should address:

• The limitations of the original sample.
• Appropriate sampling locations.
• The importance of sampling at different times.
• How the water samples should be prepared.
• How the precipitation method should be used to obtain reliable results.

Solution 4

The student’s conclusion is not justified. A small precipitate only indicates that the collected sample contained a small amount of sulfate under the conditions of the analysis. It cannot establish that the entire river is unpolluted.

There are several problems with the original investigation.

Step 1: Identify the sampling limitation.

The student collected water from only one location, at one time.

The sample was taken from the downstream bank, but the stormwater drain may introduce sulfate into a different part of the river.

The contaminant may not have mixed evenly across the river.

The sample was also collected during dry weather. It could therefore miss a temporary increase in sulfate caused by rainfall and stormwater runoff.

Step 2: Select suitable sampling locations.

A better investigation would include:

• An upstream reference site, before the stormwater drain enters the river.
• A sample from the stormwater discharge itself.
• Several sampling positions downstream, after the discharge has mixed with the river.

The upstream sample provides a comparison with water that has not received the drain’s discharge.

The stormwater sample helps identify the composition of the incoming water.

The downstream samples show whether the river’s sulfate concentration changes after receiving the discharge.

Step 3: Collect samples at different times.

The team should collect samples during dry weather and repeat the investigation during and after rainfall.

This allows the team to investigate whether runoff is associated with changes in sulfate concentration.

Repeated sampling is important because a single rainfall event may not represent typical conditions.

Replicate samples should also be collected to identify variation between individual measurements.

Step 4: Control the sample preparation.

If the aim is to measure dissolved sulfate, the samples should be filtered using a consistent procedure before precipitation.

The team should use clean sampling equipment and containers, label samples correctly, and include a suitable field blank to check for contamination introduced during collection and handling.

Step 5: Perform the precipitation analysis consistently.

Each sample should be treated with excess barium chloride under controlled conditions.

The resulting barium sulfate should be filtered, washed, dried to constant mass, and weighed.

The mass of dry barium sulfate can then be converted into sulfate concentration using the 1:1 mole ratio.

Sample volumes and any dilution factors must also be accounted for.

Step 6: Interpret the results carefully.

If the downstream samples repeatedly show higher sulfate concentrations than the upstream reference samples, particularly during or after rainfall, this would support the hypothesis that the stormwater discharge is associated with increased sulfate concentrations.

However, it would not automatically prove that the drain is the only source. Other inputs, natural geological sources, and changes in river conditions must also be considered.

Most importantly, even a well-designed sulfate investigation cannot establish that the river is free from every other pollutant.

The investigation has answered a specific question about sulfate, not a universal question about water quality.

Question 5

An environmental monitoring team investigates sulfate concentrations at the same fully mixed river cross-section on two different dates.

On each date, a 250.0 mL water sample is collected and homogenised before being divided into two equal analytical portions.

The first portion is filtered before precipitation to measure dissolved sulfate.

The second portion remains unfiltered and undergoes a validated digestion that quantitatively recovers sulfate associated with suspended material. This measures total recoverable sulfate.

Both portions are treated with excess barium chloride. All precipitates are washed, dried to constant mass, and assumed to be pure barium sulfate.

The following illustrative data are obtained:

Assume the measured concentrations represent the entire river cross-section, the flow measurements are accurate, and there are no additional inputs or losses between the sampling location and the cross-section being assessed.

Answer the following:

(a) Calculate the dissolved sulfate concentration on each date.

(b) Calculate the total recoverable sulfate concentration on each date.

(c) Determine the concentration of sulfate associated with suspended material on each date.

(d) Calculate the total recoverable sulfate mass passing the monitoring cross-section per second on each date. Express your answers in g/s.

(e) A student claims that the river is cleaner on Date 2 because its total recoverable sulfate concentration is lower. Evaluate this claim using your calculations.

Solution 5

Date 2 has a lower total recoverable sulfate concentration but a higher sulfate mass flow through the monitoring cross-section. Therefore, the student’s claim that the river is cleaner on Date 2 is not justified.

This question combines three ideas:

• The distinction between dissolved and particulate-associated sulfate.
• Gravimetric analysis using barium sulfate.
• The difference between concentration and the total mass of a substance transported per second.

Step 1: Calculate the dissolved sulfate concentration on each date.

The filtered portions measure dissolved sulfate.

For Date 1:

\[ n(\mathrm{BaSO_4}) = \frac{0.233}{233.39} \]

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

The sulfate mass is:

\[ m=(9.98\times10^{-4})(96.06) \]

\[ =0.0959\text{ g}=95.9\text{ mg} \]

The sample volume is 0.2500 L.

\[ c=\frac{95.9}{0.2500} \]

\[ \boxed{c_{\text{dissolved, Date 1}}=384\text{ mg/L}} \]

For Date 2:

\[ n(\mathrm{BaSO_4}) = \frac{0.1867}{233.39} \]

\[ =8.00\times10^{-4}\text{ mol} \]

The sulfate mass is:

\[ m=(8.00\times10^{-4})(96.06) \]

\[ =0.07685\text{ g}=76.85\text{ mg} \]

Therefore:

\[ c=\frac{76.85}{0.2500} \]

\[ \boxed{c_{\text{dissolved, Date 2}}=307\text{ mg/L}} \]

The dissolved sulfate concentration decreases between the two dates.

Step 2: Calculate the total recoverable sulfate concentration.

The unfiltered portions include sulfate recovered from both dissolved and suspended material.

For Date 1:

\[ n(\mathrm{BaSO_4}) = \frac{0.280}{233.39} \]

\[ =1.20\times10^{-3}\text{ mol} \]

The sulfate mass is:

\[ m=(1.20\times10^{-3})(96.06) \]

\[ \approx0.1152\text{ g} \]

\[ =115.2\text{ mg} \]

Therefore:

\[ c=\frac{115.2}{0.2500} \]

\[ \boxed{c_{\text{total, Date 1}}=461\text{ mg/L}} \]

For Date 2:

\[ n(\mathrm{BaSO_4}) = \frac{0.233}{233.39} \]

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

The sulfate mass is:

\[ m=(9.98\times10^{-4})(96.06) \]

\[ =0.0959\text{ g}=95.9\text{ mg} \]

Therefore:

\[ c=\frac{95.9}{0.2500} \]

\[ \boxed{c_{\text{total, Date 2}}=384\text{ mg/L}} \]

The total recoverable sulfate concentration also decreases between the two dates.

Step 3: Calculate the sulfate associated with suspended material.

Subtract the dissolved sulfate concentration from the total recoverable sulfate concentration.

For Date 1:

\[ c_{\text{suspended}} = 461-384 \]

\[ \boxed{77\text{ mg/L}} \]

For Date 2:

\[ c_{\text{suspended}} = 384-307 \]

\[ \boxed{77\text{ mg/L}} \]

The difference between the filtered and unfiltered measurements represents the sulfate recovered from suspended material under the stated method assumptions.

Notice that the particulate-associated sulfate concentration is approximately the same on both dates, even though the dissolved concentration changes.

This is why the sample preparation matters. Filtering before analysis would exclude this fraction.

Step 4: Calculate the sulfate mass flow on each date.

This is the important conceptual step.

A concentration in mg/L tells you how much sulfate is present in each litre of water.

It does not tell you how much sulfate passes through the river each second.

To calculate the mass flow:

\[ \text{Mass flow}=C\times Q \]

Where:

• \(C\) is the total recoverable sulfate concentration in mg/L.
• \(Q\) is the river flow rate in L/s.

First convert the flow rates.

Date 1:

\[ 1.50\text{ m}^3/\text{s} = 1500\text{ L/s} \]

Date 2:

\[ 2.00\text{ m}^3/\text{s} = 2000\text{ L/s} \]

For Date 1:

\[ \text{Mass flow} = 461\times1500 \]

\[ =691500\text{ mg/s} \]

Convert to grams:

\[ \boxed{\text{Mass flow}\approx691\text{ g/s}} \]

For Date 2:

\[ \text{Mass flow} = 384\times2000 \]

\[ =768000\text{ mg/s} \]

Using unrounded concentrations gives approximately:

\[ \boxed{\text{Mass flow}\approx767\text{ g/s}} \]

Step 5: Evaluate the student’s conclusion.

The student’s reasoning initially seems sensible.

The total recoverable sulfate concentration falls from approximately 461 mg/L to 384 mg/L.

However, the river is flowing faster on Date 2.

This means that more water passes through the monitoring cross-section each second.

Although each litre contains less sulfate, more litres pass through the river per second.

The result is that the sulfate mass flow increases from approximately 691 g/s to 767 g/s.

Therefore, a lower concentration does not necessarily mean that a smaller mass of sulfate is being transported.

The student’s conclusion is not supported by the concentration measurements alone.

There is also a final limitation: a higher or lower sulfate mass flow does not, by itself, establish the overall ecological condition of the river. That requires environmental context, appropriate reference measurements, and consideration of other relevant pollutants.

06The key distinction

A reliable environmental analysis depends on more than obtaining a precise mass from a laboratory balance.

You must first establish that the sample represents the water you want to investigate. You must then ensure that the precipitation reaction quantitatively converts the analyte into a suitable solid, and that the collected precipitate is washed, dried, and sufficiently pure before weighing.

For gravimetric analysis, remember:

• One sample describes one collected portion of water, not an entire river.
• One analyte does not represent every possible pollutant.
• Wet precipitate mass includes retained liquid and cannot be treated as pure solid mass.
• Drying to constant mass removes moisture, but purity must also be established.
• A lower concentration does not necessarily mean a lower mass of substance transported per second.

Once the sampling and gravimetric measurements are reliable, the next useful step is deciding whether you need a concentration in mg/L or a mass flow in g/s. Those measurements answer different environmental questions, and confusing them can lead to a confident but incorrect conclusion.