How to Distinguish Dissolved and Total Recoverable Metals

If a water sample contains dissolved copper ions and tiny copper-bearing particles, filtering the sample changes what you measure. Filter it before acid treatment, and you measure the copper that passes through the filter. Keep it unfiltered and treat it with acid, and you can also recover copper released from some of the suspended particles.

This is the key difference between dissolved pollutant and total recoverable pollutant in a water sample report. The distinction comes down to sample preparation, particularly when filtration happens and whether the sample undergoes acid digestion.

The important part is following the pollutant through the entire process. A filter used during sampling has a different job from a filter used later to collect a chemical precipitate. Mixing those two steps up can lead to completely incorrect conclusions.

01Start with what is actually in the water

Imagine collecting water from a creek downstream of an industrial site. The sample contains copper in two forms:

• Copper ions, such as \(\mathrm{Cu}^{2+}\), dispersed throughout the water.
• Tiny suspended particles containing copper, such as fragments of mineral matter with copper compounds attached.

The first group can pass through a suitable membrane filter. The second group may be retained.

What would happen if you filtered the sample before doing anything else?

The dissolved copper would pass through with the water, while larger copper-bearing particles would remain on the filter. You would have separated two different fractions of the original sample.

Think of the membrane as an extremely fine sieve. Larger particles are retained, while sufficiently small material passes through.

There is one important limitation to this analogy: a 0.45 µm membrane does not separate free metal ions from every possible colloid. Some very small colloidal material can pass through too.

This is why dissolved metal is an operational definition. It describes what passes through a specified filter, rather than guaranteeing that every measured metal atom exists as a freely dissolved ion.

02What the two report labels actually mean

For metal pollutants in water, the two measurements are distinguished by how the original sample is prepared.

Dissolved pollutant

The dissolved concentration represents the pollutant that passes through a 0.45 µm membrane filter in an unacidified sample.

For copper, this includes copper ions in solution and potentially some sufficiently small colloidal material.

The important sequence is:

Filter first, acidify second.

If you acidify the sample before filtering, you risk changing the amount of copper that remains dissolved. Some copper attached to suspended particles may be released into the solution.

That would affect the result.

Total recoverable pollutant

Total recoverable concentration is determined using an unfiltered sample.

The sample receives an appropriate acid treatment, and the analytical method may include heating or digestion to release metal associated with suspended material.

The result includes:

• Copper that was already dissolved in the original water.
• Additional copper released from acid-soluble suspended particles.

However, total recoverable does not necessarily mean every copper atom in every solid particle has been extracted. Some particularly resistant mineral material may remain undissolved.

The exact recovery depends on the analytical method.

Do not automatically treat total recoverable metals and total metals as identical terms. Some analytical procedures use more vigorous digestion to determine total metal content.

The Australian Drinking Water Guidelines distinguish filterable metals, total metals determined after vigorous digestion, and acid-extractable metals determined after hot mineral acid treatment. The preparation method therefore matters when interpreting a laboratory report.

Australian Drinking Water Guidelines: Metal fractions

03Follow the sample through the laboratory

The easiest way to understand the difference is to follow two portions of the same well-mixed water sample.

Pathway 1: Measuring dissolved metal

Suppose an analyst wants to determine the dissolved copper concentration.

The process is:

1. Collect a representative water sample in a clean container.
2. Pass an unacidified portion through a 0.45 µm membrane filter.
3. Collect the filtrate, which contains the copper that passed through the membrane.
4. Preserve the filtrate using the specified acid treatment, commonly nitric acid for metal analysis.
5. Analyse the prepared sample.

The copper retained on the original membrane is excluded from the dissolved measurement.

Pathway 2: Measuring total recoverable metal

Now consider the second portion of the same sample.

This time, the analyst wants to include copper associated with suspended particles.

The process is:

1. Collect a representative, well-mixed sample.
2. Transfer an unfiltered portion into an appropriate sample container.
3. Preserve the sample using the specified acid treatment.
4. In the laboratory, apply the prescribed acid digestion or extraction method.
5. If insoluble material remains after digestion, separate that residue as required by the method.
6. Analyse the resulting solution for copper.

The crucial difference is that the suspended particles were present during the acid treatment.

Some copper attached to those particles has now been released into the solution and can be measured.

Notice something important: the laboratory might filter the digested sample at the end of this process.

That does not turn it into a dissolved measurement.

The original suspended particles were exposed to the acid treatment before this second filtration. The copper released from them remains in the solution.

Why the order of filtration matters

Consider a sample containing copper attached to sediment.

If you acidify the sample first, some of that copper may enter the solution. Filtering afterwards cannot reverse that change.

The sample may now contain copper that was originally attached to particles but has become dissolved during preparation.

For this reason, the dissolved fraction must be separated before acid treatment changes the sample.

The National Measurement Institute’s Australian sample-submission guidance follows this distinction: dissolved metal samples are filtered through a 0.45 µm membrane, while total recoverable metal samples are submitted unfiltered in an acid-preserved container.

National Measurement Institute: Aqueous sample requirements

04Where precipitation fits into the analysis

There is another important step that can appear in a chemistry question: precipitation.

Precipitation is a chemical reaction in which dissolved ions form an insoluble solid.

It is different from the initial filtration of the water sample.

For example, imagine that the prepared copper solution is treated with sodium hydroxide.

The copper ions react with hydroxide ions to form solid copper(II) hydroxide:

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

The blue copper(II) hydroxide precipitate can then be separated from the solution using filtration.

The precipitate can then be heated to convert copper(II) hydroxide into copper(II) oxide:

\[ \ce{Cu(OH)2(s) ->[\Delta] CuO(s) + H2O(l)} \]

The copper remains in the solid, but the chemical compound has changed.

This gives us a useful way to demonstrate gravimetric analysis: determine the amount of copper by measuring the mass of a known compound containing copper.

The original dissolved or total recoverable classification is still determined by how the water sample was prepared before this precipitation step.

Wet precipitate versus pure dry mass

This is where students often make a costly calculation error.

Suppose you filter a blue copper hydroxide precipitate and immediately weigh it.

The material on the filter paper contains more than just copper hydroxide.

It can contain:

• The copper hydroxide solid.
• Water trapped between the particles.
• Small amounts of solution retained in the filter cake.
• Dissolved impurities that have not been washed away.

The measured mass is therefore the mass of a wet precipitate, not the mass of pure copper hydroxide.

You cannot use that wet mass directly in a stoichiometric calculation.

Instead, the precipitate must be washed and dried, or heated to a specified chemical form, according to the gravimetric method.

It is then cooled and weighed repeatedly until a constant mass is obtained.

This is called drying to constant mass. It provides confidence that further drying is no longer causing a measurable change in the sample’s mass.

There is one more distinction:

The dry precipitate is not necessarily the same substance as the original pollutant.

For example, if the final solid is copper(II) oxide, its mass includes both copper and oxygen.

You must use the molar masses to calculate the actual copper mass.

The two filtration steps now have different purposes:

1. The original 0.45 µm membrane determines which pollutant fraction is being measured.
2. The later filtration collects the newly formed precipitate so that its mass can be determined.

Never confuse them.

07Questions and solutions

Question 1

A water sample report contains the following results for zinc:

Both measurements were obtained from representative portions of the same well-mixed sample.

Calculate the additional zinc recovered by the total recoverable method and explain what this result represents.

Solution 1

The additional acid-recoverable zinc concentration is 0.33 mg/L.

The total recoverable result includes dissolved zinc and additional zinc released from suspended material.

Subtract the dissolved concentration:

\[ C_{\text{additional}}=0.49-0.16 \]

\[ \boxed{C_{\text{additional}}=0.33\text{ mg/L}} \]

This represents the difference between the two operationally defined fractions.

It does not represent 0.33 mg/L of suspended solids. Both results are concentrations of zinc, not concentrations of the particles carrying it.

Question 2

A 250.0 mL unfiltered water sample is prepared for total recoverable zinc analysis.

The zinc is precipitated and converted into pure zinc oxide, ZnO.

The wet filter cake has a mass of 15.2 mg. After washing and ignition, the dry zinc oxide has a mass of 1.22 mg.

Calculate the zinc concentration in the original sample.

Use:

• Molar mass of zinc = 65.38 g/mol.
• Molar mass of oxygen = 16.00 g/mol.

Assume complete recovery of zinc, a pure final precipitate, and no dilution.

Solution 2

The zinc concentration is 3.92 mg/L.

The wet filter cake mass cannot be used because it includes retained water and potentially dissolved impurities. We must use the dry zinc oxide mass and convert it into the mass of zinc.

Step 1: Identify the dry mass of zinc oxide.

\[ m(\mathrm{ZnO})=1.22\text{ mg} \]

Step 2: Calculate the molar mass of zinc oxide.

\[ M(\mathrm{ZnO})=65.38+16.00 \]

\[ M(\mathrm{ZnO})=81.38\text{ g/mol} \]

Step 3: Calculate the mass of zinc.

Each mole of ZnO contains one mole of Zn.

Therefore:

\[ m(\mathrm{Zn}) = 1.22\times\frac{65.38}{81.38} \]

\[ m(\mathrm{Zn})=0.980\text{ mg} \]

Step 4: Convert the sample volume into litres.

\[ 250.0\text{ mL}=0.2500\text{ L} \]

Step 5: Calculate the zinc concentration.

\[ c=\frac{0.980}{0.2500} \]

\[ \boxed{c=3.92\text{ mg/L}} \]

The original sample contained 3.92 mg of recoverable zinc per litre.

Using the wet mass of 15.2 mg would have substantially overestimated the zinc concentration.

Question 3

An analyst collects two portions of turbid water containing lead.

The following procedures are used:

Bottle A

• The unacidified sample is passed through a 0.45 µm membrane at the collection site.
• The filtrate is collected and acidified.
• The measured lead concentration is 0.21 mg/L.

Bottle B

• The sample is collected without filtration.
• Acid is added to preserve the sample.
• The sample undergoes heated acid digestion in the laboratory.
• Any remaining insoluble mineral residue is filtered off.
• The measured lead concentration is 0.64 mg/L.

A student argues that both results must represent dissolved lead because both samples were filtered at some stage.

Evaluate this argument.

Solution 3

The student’s argument is incorrect. Bottle A measures dissolved lead, while Bottle B measures total recoverable lead.

The key distinction is when filtration occurs relative to acid treatment.

For Bottle A, filtration happens before acidification. Lead attached to particles retained by the original membrane is excluded from the measurement.

For Bottle B, the unfiltered sample undergoes acid digestion first. Some lead attached to suspended particles is released into solution during this process.

The later filtration only removes remaining insoluble material. It does not remove lead that has already dissolved.

Therefore:

• Bottle A: dissolved lead = 0.21 mg/L.
• Bottle B: total recoverable lead = 0.64 mg/L.

The higher result from Bottle B is consistent with additional lead being recovered from the suspended fraction.

The important lesson is that the presence of a filtration step does not automatically make a result a dissolved measurement. Its position in the procedure matters.

Question 4

A laboratory reports the following nickel concentrations for a representative water sample:

• Dissolved nickel: 0.24 mg/L.
• Total recoverable nickel: 0.80 mg/L.

Calculate:

a) The additional acid-recoverable nickel concentration.

b) The percentage of the total recoverable nickel concentration represented by this additional fraction.

c) Explain why this percentage should not automatically be described as the percentage of all nickel atoms physically attached to suspended particles.

Solution 4

a) The additional acid-recoverable nickel concentration is 0.56 mg/L.

Subtract the dissolved result from the total recoverable result:

\[ C_{\text{additional}}=0.80-0.24 \]

\[ \boxed{C_{\text{additional}}=0.56\text{ mg/L}} \]

b) The additional fraction represents 70% of the reported total recoverable concentration.

Calculate the percentage:

\[ \%\text{ additional} = \frac{0.56}{0.80}\times100 \]

\[ \boxed{70\%} \]

c) This percentage describes the difference between two operationally defined analytical measurements.

It does not necessarily represent the exact physical distribution of every nickel atom in the original water.

The 0.45 µm membrane defines the filterable fraction, and the acid digestion defines how much additional nickel can be recovered from the unfiltered sample.

Some small colloids may pass through the membrane, while some particularly resistant particles may not release all their nickel during digestion.

The correct interpretation is that 70% of the measured total recoverable nickel concentration was additional to the dissolved measurement.

Question 5

A laboratory investigates copper contamination in a creek.

The sample is initially collected and thoroughly mixed. Three separate portions are then prepared.

Two students interpret these results differently.

Student A argues:

“Sample U has the same result as Sample D. Therefore, there was no copper attached to suspended particles in the original creek water.”

Student B argues:

“The difference between total recoverable and dissolved copper is 0.45 mg/L, so exactly 0.45 mg/L of copper must have been attached to suspended particles.”

Evaluate both explanations.

State the strongest conclusion supported by the data, and explain the limitation of using Sample U to investigate the original sample.

Solution 5

Neither student gives a completely correct interpretation.

The strongest conclusion is that the total recoverable method recovered an additional 0.45 mg/L of copper compared with the immediately filtered dissolved measurement. However, the data do not establish the exact physical distribution of every copper atom in the original sample.

Step 1: Interpret Samples D and T.

Sample D was filtered immediately before acidification.

Its result of 0.30 mg/L represents the dissolved, operationally defined copper fraction.

Sample T remained unfiltered during acid preservation and digestion.

Its result of 0.75 mg/L includes dissolved copper and additional copper released from suspended material.

The difference is:

\[ C_{\text{additional}}=0.75-0.30 \]

\[ \boxed{C_{\text{additional}}=0.45\text{ mg/L}} \]

This difference is meaningful because D and T were prepared from the same initially mixed sample using the appropriate procedures.

Step 2: Examine Student A’s reasoning.

Student A assumes that Sample U represents the original whole water sample.

It does not.

During the 30-minute settling period, suspended particles can move towards the bottom of the container.

When the analyst takes only the clear upper layer, much of this settled material is excluded before filtration even begins.

The membrane therefore receives a portion of water that has already been separated from some of its original suspended particles.

A result of 0.30 mg/L is entirely consistent with this process.

It does not prove that the original sample contained no particle-associated copper.

Step 3: Examine Student B’s reasoning.

Student B correctly calculates the difference between the dissolved and total recoverable results.

However, describing 0.45 mg/L as the exact amount of copper physically attached to suspended particles is too strong.

The difference is an operationally defined acid-recoverable fraction.

Some small colloidal material may have passed through the original membrane, and some copper in resistant particles may not have been recovered during digestion.

The method therefore does not provide a perfect measurement of the physical location of every copper atom.

Step 4: State the strongest justified conclusion.

The original sample contained:

• 0.30 mg/L of copper in the immediately filterable fraction.
• An additional 0.45 mg/L recovered when the unfiltered sample underwent acid treatment.
• 0.75 mg/L of total recoverable copper.

The additional fraction represents:

\[ \frac{0.45}{0.75}\times100=60\% \]

Thus, 60% of the reported total recoverable copper concentration was additional to the dissolved result.

To investigate the original sample properly, the analyst should use paired, representative portions: filter one immediately before acidification and keep the other unfiltered for the prescribed total recoverable procedure.

Sample U cannot replace either measurement because settling changed which material was present in the aliquot before analysis.

This is the important distinction: a laboratory result describes the fraction recovered by its method, not necessarily every physical form of the pollutant in the original environment.

08Taking the next step

When interpreting a sample report, always check three things:

1. Was the original sample filtered before or after acidification?
2. Did the unfiltered sample undergo an acid treatment or digestion?
3. Is the reported mass expressed as the pollutant itself or as a compound containing the pollutant?

Once you can identify the correct sample fraction, the next calculation is often converting a recovered precipitate mass into a concentration in mg/L.

For more practice with that calculation, see tutorgum’s How to Convert Gravimetric Results into mg/L.

Remember: the sample preparation determines which pollutant fraction you measure. The precipitation and stoichiometry determine how much of that pollutant is present.