Using Ksp to Monitor Lead and Phosphate in Water

Suppose you add phosphate to a water sample containing dissolved lead. After mixing and settling, a laboratory reports that filtered lead has fallen from 2.00 to 0.40 µg/L.

Has the treatment removed 80% of the lead?

Not necessarily. The lead may have moved from dissolved ions into solid particles that are still sitting in the water. The treatment might have reduced the dissolved lead concentration without removing the same amount of lead from the sample.

To interpret the result, you need to answer two different questions:

• Does the water chemistry favour the formation of an insoluble lead-containing solid?
• Did the monitoring method actually establish where the lead went?

Solubility equilibria help answer the first question. Careful experimental design and analytical measurements answer the second.

01The monitoring experiment: following lead through water

Start with a simple prediction.

If phosphate is added to water containing dissolved lead(II) ions, what should happen?

You might predict that the dissolved lead concentration decreases because lead ions combine with phosphate ions to form an insoluble solid.

That is a reasonable prediction. However, there is an important distinction between precipitation and removal.

Precipitation changes the chemical form of lead. Removal takes that lead-containing material out of the water being monitored.

Think about sugar dissolved in tea. A strainer cannot remove sugar while it remains dissolved. If crystals form, however, the solid can be separated. The comparison is useful because it highlights the change from dissolved material to particles. The difference is that lead phosphate is a new compound formed through a chemical reaction, rather than simply crystallised sugar.

Designing a monitoring experiment

Imagine investigating whether adding phosphate reduces dissolved lead in a contaminated water sample.

A sensible experiment starts with one large, well-mixed water sample. Divide it between two identical treatment vessels.

• The control vessel receives no additional phosphate.
• The treatment vessel receives a measured amount of phosphate.

Keep the temperature, mixing time, sample volume, and initial pH as similar as possible.

After adding phosphate, the treatment vessel is mixed to distribute the ions. It is then left undisturbed so that any newly formed particles can settle.

A sample of the clearer water above the sediment is collected from a consistent depth.

The sample is then divided into separate portions for analysis.

1. Measuring dissolved lead

Pass one portion through a 0.45 µm filter.

This removes most suspended particles, although very small colloids can still pass through.

After filtration, acidify the filtrate using nitric acid and analyse its lead concentration using inductively coupled plasma mass spectrometry (ICP-MS).

The result represents operationally dissolved lead: lead that passed through the filter before the filtrate was acidified.

Importantly, ICP-MS does not distinguish free \(\mathrm{Pb}^{2+}\) ions from dissolved lead complexes. It measures the lead present in the prepared sample.

2. Measuring phosphate

Use a separate filtered portion to measure dissolved reactive orthophosphate.

A common approach is molybdenum-blue colorimetry, in which phosphate reacts to produce a coloured compound. The colour intensity is related to the phosphate concentration.

This measurement is usually reported as mg P/L, meaning milligrams of phosphorus per litre, rather than mg of the entire phosphate ion per litre.

The phosphate sample must be handled according to the analytical method. It should not simply be acidified in the same way as the lead sample.

3. Measuring total recoverable lead

Take another portion without filtering it.

Digest the sample using an appropriate acid digestion procedure, then measure the lead concentration.

This gives an operational measure of total recoverable lead, including dissolved lead and acid-recoverable suspended material.

Finally, collect and digest the settled solids separately. This helps determine how much lead has accumulated in the sediment.

Why the first experimental design is incomplete

A simple experiment might measure only filtered lead before and after phosphate addition.

Suppose the result falls from 2.00 to 0.40 µg/L.

You know that less lead passed through the filter after treatment. You do not yet know why.

Possible explanations include:

• Lead phosphate precipitation.
• Lead adsorption onto existing suspended particles.
• Formation of another lead-containing solid.
• Lead sticking to the treatment vessel.
• Changes in filtration or sample handling.

A stronger experimental design includes a phosphate-free control, filtered and unfiltered measurements, and analysis of the recovered solids.

It also uses laboratory quality controls such as blanks, duplicate samples, and matrix-spike recovery tests.

The important idea is that a chemical prediction and an analytical measurement provide different kinds of evidence. Neither should be used as a substitute for the other.

02Using Ksp to predict lead phosphate precipitation

Now consider the chemistry responsible for the predicted precipitation.

Lead(II) phosphate is represented by the formula \(\ce{Pb3(PO4)2}\).

Its dissolution equilibrium is:

\[ \ce{Pb3(PO4)2(s) <=> 3Pb^2+(aq) + 2PO4^3-(aq)} \]

Notice the coefficients.

One formula unit of solid produces three lead(II) ions and two phosphate ions.

The solubility product expression must therefore be:

\[ K_{sp}=[\mathrm{Pb}^{2+}]^3 [\mathrm{PO_4}^{3-}]^2 \]

The exponents come directly from the coefficients in the balanced dissolution equation.

This is a common examination trap. Writing \(K_{sp}=[\mathrm{Pb}^{2+}][\mathrm{PO_4}^{3-}]\) ignores the stoichiometry and gives the wrong expression.

For the illustrative calculations in this article, use:

\[ K_{sp}=1.0\times10^{-54} \]

at 25°C for the specified \(\ce{Pb3(PO4)2}\) solid phase.

This is a teaching-model value, not a universal constant for every lead-containing solid in natural water. Different solid phases and environmental conditions can have different equilibrium constants.

Comparing Q with Ksp

The ion product, \(Q\), has the same mathematical form as \(K_{sp}\):

\[ Q=[\mathrm{Pb}^{2+}]^3 [\mathrm{PO_4}^{3-}]^2 \]

The difference is that \(K_{sp}\) describes the equilibrium condition, whereas \(Q\) uses the current ion concentrations.

Compare the two values:

If \(Q>K_{sp}\), precipitation can reduce the concentrations of the dissolved ions.

However, supersaturation does not guarantee that precipitation happens immediately. The formation of solid particles may be slowed by nucleation barriers, a lack of suitable surfaces, or competing reactions.

Also, the Ksp expression only tests the specified solid phase. It does not establish which solid actually formed in a real water sample.

03The complication: phosphate exists in different forms

There is a problem with using the phosphate concentration measured by the laboratory directly in the Ksp expression.

The expression requires the concentration of \(\mathrm{PO_4^{3-}}\), but a laboratory measuring total dissolved orthophosphate does not directly measure only that ion.

Phosphate is a weak acid system. Depending on pH, it exists in several forms.

The two important equilibria in ordinary environmental water are:

\[ \ce{H2PO4^- <=> H+ + HPO4^2-} \]

\[ \ce{HPO4^2- <=> H+ + PO4^3-} \]

At 25°C, approximate values are:

\[ pK_{a2}=7.21 \]

\[ pK_{a3}=12.32 \]

The relationship between pH and the relative concentrations is described by:

\[ \frac{[\mathrm{HPO_4^{2-}}]} {[\mathrm{H_2PO_4^-}]} =10^{pH-pK_{a2}} \]

and

\[ \frac{[\mathrm{PO_4^{3-}}]} {[\mathrm{HPO_4^{2-}}]} =10^{pH-pK_{a3}} \]

At pH 7.20, for example:

\[ \frac{[\mathrm{HPO_4^{2-}}]} {[\mathrm{H_2PO_4^-}]} =10^{-0.01}=0.977 \]

But:

\[ \frac{[\mathrm{PO_4^{3-}}]} {[\mathrm{HPO_4^{2-}}]} =10^{-5.12} =7.59\times10^{-6} \]

The third form is present in a very small proportion.

This means that even when a water sample contains a measurable amount of total phosphate, its actual \(\mathrm{PO_4^{3-}}\) concentration can be tiny.

Why pH changes the precipitation prediction

Imagine two water samples with identical total phosphate concentrations.

One has a pH of 6.5. The other has a pH of 7.5.

The higher-pH sample has a greater proportion of its phosphate in the \(\mathrm{PO_4^{3-}}\) form.

Since the ion product contains the square of the phosphate concentration,

\[ Q=[\mathrm{Pb}^{2+}]^3[\mathrm{PO_4}^{3-}]^2 \]

even a moderate increase in free phosphate can produce a substantial increase in \(Q\).

This is why measuring pH is essential when using phosphate data to predict lead phosphate precipitation.

A total phosphate measurement alone cannot tell you whether the solution is saturated.

04The environmental monitoring dataset

Consider the following illustrative dataset from a water-treatment investigation.

The values are constructed for teaching purposes. They are not official drinking-water limits, environmental guidelines, or compliance measurements.

The measurements were obtained using the following procedures:

• Lead was measured by ICP-MS after 0.45 µm filtration and acidification of the filtrate.
• Dissolved reactive orthophosphate was measured on a separate filtered sample.
• The laboratory reporting limit for lead was 0.05 µg/L.
• For this exercise, the free lead concentration is estimated as 70% of the filtered dissolved lead concentration.

The 70% factor is a simplified speciation assumption for the calculations. It is not a universal proportion of free lead in natural water.

For the calculations, also assume:

• Temperature is 25°C.
• The only phosphate species considered significant are \(\mathrm{H_2PO_4^-}\), \(\mathrm{HPO_4^{2-}}\), and \(\mathrm{PO_4^{3-}}\).
• Activities can be approximated by concentrations.
• The specified solid phase is pure \(\ce{Pb3(PO4)2}\).

These assumptions make the calculations manageable. Real environmental samples require more detailed speciation and activity corrections.

07Precipitation is not the same as removal

Return to the original monitoring problem.

The filtered dissolved lead concentration decreased from 2.00 to 0.40 µg/L.

The percentage decrease is:

\[ \frac{2.00-0.40}{2.00}\times100 \]

\[ \boxed{80\%} \]

That is an 80% reduction in the measured filtered dissolved lead concentration.

It is not automatically an 80% reduction in total lead.

Consider the additional measurements from the treatment experiment.

The treated supernatant contains 0.55 µg/L total recoverable lead, including the 0.40 µg/L measured in its filtered fraction.

The settled solids contain the equivalent of 1.50 µg/L lead relative to the original water volume.

Adding the two fractions:

\[ 0.55+1.50=2.05\text{ µg/L} \]

The resuspended whole slurry also gives 2.05 µg/L.

The raw sample contained 2.10 µg/L.

The difference is only 0.05 µg/L, or approximately 2.4% of the original total. For this illustrative dataset, assume the combined analytical uncertainty is approximately ±0.10 µg/L.

The measurements are therefore consistent with a transfer of lead into the settled solids, rather than destruction of lead.

This distinction matters in environmental monitoring.

If the sediment remains in the water system, lead has not been removed from that system. If the sediment is collected and safely managed, lead has been removed from the treated water stream, although it still exists in the recovered waste.

A successful treatment process therefore needs both chemical conversion and physical separation.

08Detection limits: what does a result below the limit mean?

Look at the downstream sample in the dataset.

Its reported filtered dissolved lead concentration is:

\[ [\mathrm{Pb}]_{\text{filtered}}<0.05\text{ µg/L} \]

This does not mean that the water contains zero lead.

It means the measured concentration is below the laboratory’s stated reporting limit.

A common mistake is to replace a result reported as less than the detection or quantification limit with zero.

That can distort both concentration comparisons and Ksp calculations.

Instead, use the reporting limit to establish an upper bound.

Estimating an upper bound for the downstream sample

The downstream sample has:

• pH = 6.70
• Dissolved reactive phosphate = 0.010 mg P/L
• Filtered lead = less than 0.05 µg/L

At pH 6.70:

\[ \frac{[\mathrm{HPO_4^{2-}}]} {[\mathrm{H_2PO_4^-}]} =10^{6.70-7.21} =0.309 \]

And:

\[ \frac{[\mathrm{PO_4^{3-}}]} {[\mathrm{HPO_4^{2-}}]} =10^{6.70-12.32} =2.40\times10^{-6} \]

Following the same speciation method gives:

\[ \alpha_{\mathrm{PO_4}} =5.67\times10^{-7} \]

The total phosphate concentration is:

\[ C_P=\frac{1.0\times10^{-5}}{30.97} \]

\[ C_P=3.23\times10^{-7}\text{ mol/L} \]

Therefore:

\[ [\mathrm{PO_4^{3-}}] =(3.23\times10^{-7})(5.67\times10^{-7}) \]

\[ =1.83\times10^{-13}\text{ mol/L} \]

Now convert the lead reporting limit:

\[ [\mathrm{Pb}]_{\max} =\frac{5.0\times10^{-8}}{207.2} \]

\[ =2.41\times10^{-10}\text{ mol/L} \]

For a conservative upper-bound calculation, assume that all filtered lead is free \(\mathrm{Pb}^{2+}\).

Then:

\[ Q_{\max} =(2.41\times10^{-10})^3 (1.83\times10^{-13})^2 \]

\[ Q_{\max}=4.7\times10^{-55} \]

Comparing this with the model Ksp:

\[ \frac{Q_{\max}}{K_{sp}} = \frac{4.7\times10^{-55}}{1.0\times10^{-54}} \]

\[ \boxed{\frac{Q_{\max}}{K_{sp}}<0.47} \]

Under the stated assumptions, the downstream sample is undersaturated with respect to the specified lead phosphate solid.

This is a useful conclusion, but it has limits.

It does not prove that no lead remains. It also does not establish whether lead phosphate precipitated earlier in the treatment process.

It tells us that, at the measured downstream conditions, precipitation of this particular solid is not thermodynamically favoured.

If the pH has fallen, the proportion of phosphate present as \(\mathrm{PO_4^{3-}}\) also decreases. If lead phosphate solids are already present, an undersaturated solution can instead favour their dissolution.

The current equilibrium condition and the history of the sample are two different questions.

09Competing ions and other environmental complications

So far, the calculations have treated lead and phosphate as if they were the only chemically important ions.

Real water is more complicated.

Other dissolved substances can change the concentration of free lead or free phosphate, even when the laboratory’s total dissolved measurements remain unchanged.

Why competing ions matter

Consider carbonate.

If carbonate forms a separate lead-containing solid, the dissolved lead concentration can decrease even if lead phosphate is not the solid responsible.

Similarly, natural organic matter can bind lead into dissolved complexes.

ICP-MS may still measure that lead after the sample is acidified, even though the original free \(\mathrm{Pb}^{2+}\) concentration was much lower.

Calcium creates a different problem.

Adding phosphate to water does not guarantee that all of it remains available to react with lead. Some phosphate may form calcium phosphate solids or interact with mineral surfaces.

The amount of free phosphate can therefore be lower than expected from the total phosphate measurement.

These effects are why environmental Ksp calculations need clearly stated speciation assumptions.

Activities versus concentrations

There is another limitation to the simple calculations.

Strictly, thermodynamic solubility products are written using ion activities:

\[ K_{sp} =a_{\mathrm{Pb}^{2+}}^3 a_{\mathrm{PO_4^{3-}}}^2 \]

Activity describes the effective chemical concentration of an ion.

In sufficiently dilute solutions, activity can often be approximated by concentration. However, in water with substantial dissolved salts, this approximation can become inaccurate.

Activity coefficients are then needed to account for ionic interactions.

For HSC calculations, use concentrations when the question assumes ideal dilute conditions. For real environmental modelling, check whether activity corrections are necessary.

10Making the analytical results reliable

A good equilibrium calculation is only as useful as the measurements supplied to it.

Several practical details can change the interpretation of an environmental dataset.

Filter before acidifying when measuring dissolved lead.

If an unfiltered sample is acidified first, acid can dissolve suspended lead-containing particles.

Filtering afterwards does not restore the original distinction between dissolved and particulate lead.

The result may include lead that was originally present in solids.

Use separate sample portions for different measurements.

The lead and phosphate measurements require different preparation procedures.

Keep a separate unacidified portion for phosphate analysis. Measure pH as close as possible to the sampling location and time.

Measure total recoverable lead as well as filtered lead.

This helps establish whether lead has moved into suspended or settled material.

Include laboratory quality controls.

Blanks help identify contamination. Duplicate samples test repeatability. Matrix spikes help identify interference from the sample itself. Calibration standards establish the relationship between instrument response and concentration.

Report detection and quantification limits.

A result below the reporting limit must be treated as an upper bound, not as zero.

The most reliable environmental conclusion comes from combining the chemical model with measurements of dissolved lead, total recoverable lead, phosphate, pH, and recovered solids.

11Questions and solutions

Question 1

A water sample contains:

\[ [\mathrm{Pb}^{2+}]=2.0\times10^{-6}\text{ M} \]

\[ [\mathrm{PO_4^{3-}}]=3.0\times10^{-10}\text{ M} \]

At 25°C, use:

\[ K_{sp}=1.0\times10^{-54} \]

for \(\ce{Pb3(PO4)2}\).

Calculate the ion product, \(Q\), and determine whether precipitation is thermodynamically favoured.

Solution 1

The solution is strongly supersaturated, so precipitation of the specified lead phosphate solid is thermodynamically favoured.

Step 1: Write the balanced dissolution equation.

\[ \ce{Pb3(PO4)2(s) <=> 3Pb^2+ + 2PO4^3-} \]

Therefore:

\[ Q=[\mathrm{Pb}^{2+}]^3 [\mathrm{PO_4^{3-}}]^2 \]

Step 2: Substitute the concentrations.

\[ Q=(2.0\times10^{-6})^3 (3.0\times10^{-10})^2 \]

\[ Q=(8.0\times10^{-18}) (9.0\times10^{-20}) \]

\[ Q=7.2\times10^{-37} \]

Step 3: Compare with Ksp.

\[ \frac{Q}{K_{sp}} = \frac{7.2\times10^{-37}} {1.0\times10^{-54}} \]

\[ =7.2\times10^{17} \]

Since \(Q>K_{sp}\), the solution is supersaturated.

This predicts a thermodynamic tendency to form solid lead phosphate. It does not establish how quickly precipitation will occur or whether the actual environmental solid has this composition.

Question 2

Two water samples have identical total dissolved orthophosphate concentrations:

\[ C_P=1.00\times10^{-5}\text{ mol/L} \]

Both also have the same free lead concentration:

\[ [\mathrm{Pb}^{2+}]=1.00\times10^{-11}\text{ mol/L} \]

However, their pH values differ.

• Sample A: pH 6.50.
• Sample B: pH 7.50.

At 25°C, use:

\[ pK_{a2}=7.21 \]

\[ pK_{a3}=12.32 \]

Assume that \(\mathrm{H_3PO_4}\) is negligible.

Calculate the \(\mathrm{PO_4^{3-}}\) concentration and ion product for each sample.

Determine whether the pH difference changes the saturation classification.

Solution 2

Sample A is undersaturated, while Sample B is supersaturated. Raising the pH has increased the free phosphate concentration enough to change the predicted precipitation condition.

Step 1: Calculate phosphate speciation for Sample A.

At pH 6.50:

\[ \frac{[\mathrm{HPO_4^{2-}}]} {[\mathrm{H_2PO_4^-}]} =10^{6.50-7.21} \]

\[ =0.195 \]

Also:

\[ \frac{[\mathrm{PO_4^{3-}}]} {[\mathrm{HPO_4^{2-}}]} =10^{6.50-12.32} \]

\[ =1.51\times10^{-6} \]

The fraction of total phosphate present as \(\mathrm{PO_4^{3-}}\) is approximately:

\[ \alpha_{\mathrm{PO_4}} =2.47\times10^{-7} \]

Therefore:

\[ [\mathrm{PO_4^{3-}}] =(1.00\times10^{-5})(2.47\times10^{-7}) \]

\[ =2.47\times10^{-12}\text{ M} \]

Step 2: Calculate the ion product for Sample A.

\[ Q_A=(1.00\times10^{-11})^3 (2.47\times10^{-12})^2 \]

\[ Q_A=6.10\times10^{-57} \]

Compare with Ksp:

\[ \frac{Q_A}{K_{sp}} = \frac{6.10\times10^{-57}} {1.00\times10^{-54}} \]

\[ =0.00610 \]

Since \(Q_A<K_{sp}\), Sample A is undersaturated.

Step 3: Calculate phosphate speciation for Sample B.

At pH 7.50:

\[ \frac{[\mathrm{HPO_4^{2-}}]} {[\mathrm{H_2PO_4^-}]} =10^{7.50-7.21} \]

\[ =1.95 \]

And:

\[ \frac{[\mathrm{PO_4^{3-}}]} {[\mathrm{HPO_4^{2-}}]} =10^{7.50-12.32} \]

\[ =1.51\times10^{-5} \]

The resulting fraction of total phosphate present as \(\mathrm{PO_4^{3-}}\) is approximately:

\[ \alpha_{\mathrm{PO_4}} =1.00\times10^{-5} \]

Therefore:

\[ [\mathrm{PO_4^{3-}}] =(1.00\times10^{-5})(1.00\times10^{-5}) \]

\[ =1.00\times10^{-10}\text{ M} \]

Step 4: Calculate the ion product for Sample B.

\[ Q_B=(1.00\times10^{-11})^3 (1.00\times10^{-10})^2 \]

\[ Q_B=1.00\times10^{-53} \]

Comparing with Ksp:

\[ \frac{Q_B}{K_{sp}} = \frac{1.00\times10^{-53}} {1.00\times10^{-54}} \]

\[ =10.0 \]

Sample B is supersaturated.

The change in pH has increased the free phosphate concentration by approximately a factor of 40.5.

Since phosphate is squared in the ion-product expression, this increases \(Q\) by approximately:

\[ (40.5)^2\approx1.64\times10^3 \]

The important point is that total phosphate has not changed. Its distribution between different phosphate species has changed.

Question 3

An environmental laboratory measures the following concentrations in a filtered water sample:

\[ [\mathrm{Pb}]_{\text{filtered}} =0.083\text{ µg/L} \]

A phosphate speciation calculation gives:

\[ [\mathrm{PO_4^{3-}}] =1.25\times10^{-12}\text{ M} \]

Use:

\[ K_{sp}=1.0\times10^{-54} \]

for \(\ce{Pb3(PO4)2}\).

Two different models are proposed:

• Model A assumes that 100% of the filtered dissolved lead is free \(\mathrm{Pb}^{2+}\).
• Model B assumes that only 10% of the filtered dissolved lead is free \(\mathrm{Pb}^{2+}\), with the remainder present in dissolved complexes.

Calculate \(Q/K_{sp}\) for both models.

Explain why the laboratory’s filtered lead result alone is insufficient to determine whether the sample is supersaturated.

Solution 3

The filtered lead result alone cannot establish the saturation condition. The two reasonable speciation assumptions produce opposite conclusions.

Step 1: Convert the filtered lead concentration into mol/L.

\[ 0.083\text{ µg/L} =8.3\times10^{-8}\text{ g/L} \]

Using the molar mass of lead:

\[ [\mathrm{Pb}]_{\text{filtered}} = \frac{8.3\times10^{-8}}{207.2} \]

\[ =4.00\times10^{-10}\text{ M} \]

Step 2: Calculate Model A.

If all filtered lead is free:

\[ [\mathrm{Pb}^{2+}] =4.00\times10^{-10}\text{ M} \]

Then:

\[ Q_A=(4.00\times10^{-10})^3 (1.25\times10^{-12})^2 \]

\[ Q_A=1.00\times10^{-52} \]

Therefore:

\[ \frac{Q_A}{K_{sp}} = \frac{1.00\times10^{-52}} {1.00\times10^{-54}} \]

\[ \boxed{\frac{Q_A}{K_{sp}}=100} \]

Model A predicts supersaturation.

Step 3: Calculate Model B.

Only 10% of the filtered lead is assumed to be free:

\[ [\mathrm{Pb}^{2+}] =0.10(4.00\times10^{-10}) \]

\[ =4.00\times10^{-11}\text{ M} \]

Therefore:

\[ Q_B=(4.00\times10^{-11})^3 (1.25\times10^{-12})^2 \]

\[ Q_B=1.00\times10^{-55} \]

Comparing with Ksp:

\[ \frac{Q_B}{K_{sp}} = \frac{1.00\times10^{-55}} {1.00\times10^{-54}} \]

\[ \boxed{\frac{Q_B}{K_{sp}}=0.10} \]

Model B predicts an undersaturated solution.

Step 4: Interpret the difference.

The free lead concentration differs by a factor of ten between the models.

However, because lead has an exponent of three in the Ksp expression, the resulting ion products differ by a factor of:

\[ 10^3=1000 \]

ICP-MS measures total lead in the prepared filtered sample. It does not directly determine how much lead was originally present as free \(\mathrm{Pb}^{2+}\).

A reliable environmental prediction therefore needs an appropriate speciation estimate, including the effects of complexing ions and dissolved organic matter.

Question 4

A phosphate-treatment experiment produces the following results.

Assume the combined analytical uncertainty for the mass balance is approximately ±0.10 µg/L.

(a) Calculate the percentage decrease in filtered dissolved lead after treatment.

(b) Check whether the treated supernatant and recovered solids give a reasonable lead mass balance.

(c) Explain whether the results demonstrate that lead has been destroyed, precipitated, or removed from the water system.

Solution 4

The filtered dissolved lead concentration decreases by 80%, but the mass balance indicates that most of the missing lead has been transferred into settled solids rather than destroyed.

(a) Calculate the percentage decrease.

Use:

\[ \%\text{ decrease} = \frac{\text{initial}-\text{final}} {\text{initial}}\times100 \]

Substitute:

\[ =\frac{2.00-0.40}{2.00}\times100 \]

\[ =\boxed{80\%} \]

This is the reduction in the filtered dissolved fraction.

It does not establish an equivalent reduction in total lead.

(b) Check the mass balance.

Add the total recoverable lead in the treated supernatant to the lead recovered in the settled solids:

\[ 0.55+1.50=2.05\text{ µg/L} \]

Compare this with the original total recoverable concentration:

\[ 2.10\text{ µg/L} \]

The difference is:

\[ 2.10-2.05=0.05\text{ µg/L} \]

This is within the stated analytical uncertainty.

The resuspended whole slurry also contains 2.05 µg/L, which independently supports the same approximate total.

The mass balance is therefore consistent with the lead being redistributed between the water and solid fractions.

(c) Interpret the environmental result.

Lead has not been destroyed. Chemical reactions do not eliminate the lead atoms.

The measurements support a transfer of lead into suspended and settled solids.

The decrease in filtered dissolved lead is consistent with precipitation or adsorption, but the results do not identify the exact solid or surface responsible.

If the settled material is physically collected and removed, lead has been removed from the treated water stream.

However, it remains in the recovered waste and must be managed appropriately.

Question 5

A water-treatment operator makes the following claims after a phosphate-treatment trial.

The treated reactor has a pH of 7.20 and a dissolved reactive phosphate concentration of 0.080 mg P/L.

Filtered dissolved lead is measured at 0.40 µg/L. A simplified speciation calculation estimates that 70% of this lead is present as free \(\mathrm{Pb}^{2+}\).

The resulting ion-product calculation gives:

\[ \frac{Q}{K_{sp}}=2.3\times10^5 \]

A second aliquot from the same reactor gives a lead concentration of 1.90 µg/L when it is acidified before filtration.

At a downstream monitoring location, the following measurements are obtained:

• pH = 6.70
• Dissolved reactive phosphate = 0.010 mg P/L
• Filtered dissolved lead = less than 0.05 µg/L

The laboratory reporting limit is 0.05 µg/L.

No phosphate-free control was used in the treatment experiment, and the composition of the recovered solids has not been analysed.

The operator concludes:

1. The large value of \(Q/K_{sp}\) proves that the treatment formed pure \(\ce{Pb3(PO4)2}\).
2. The 1.90 µg/L result proves that the dissolved lead concentration increased when the second aliquot was tested.
3. The downstream result proves that no lead remains in the water and that lead phosphate precipitation is still removing lead.

Evaluate all three claims.

For the downstream sample, calculate an upper bound for \(Q/K_{sp}\), assuming all filtered lead is free \(\mathrm{Pb}^{2+}\). Use:

\[ pK_{a2}=7.21,\qquad pK_{a3}=12.32 \]

\[ K_{sp}=1.0\times10^{-54} \]

Finally, propose a follow-up experimental design that could establish whether phosphate specifically caused lead to move into a separable solid.

Solution 5

None of the three claims is fully justified.

The treatment reactor is supersaturated under the stated model, but this does not prove the identity of the solid. The acidified sample does not measure the original dissolved fraction, and the downstream non-detect does not establish zero lead or ongoing precipitation.

Step 1: Evaluate the claim that pure lead phosphate has formed.

The treated reactor has:

\[ \frac{Q}{K_{sp}}=2.3\times10^5 \]

Since this is much greater than one, the solution is strongly supersaturated with respect to the specified \(\ce{Pb3(PO4)2}\) phase.

This supports a thermodynamic tendency to form that solid.

However, supersaturation does not prove that a particular solid has actually formed.

Lead may instead have:

• Adsorbed onto iron or aluminium oxide particles.
• Formed another lead-containing precipitate.
• Associated with existing suspended material.
• Remained in a metastable solution because precipitation has not proceeded sufficiently.

The absence of a phosphate-free control also means that the operator cannot attribute the observed changes specifically to phosphate addition.

The correct conclusion is that the reactor chemistry is favourable for lead phosphate precipitation under the simplified model, not that pure lead phosphate has been experimentally identified.

Step 2: Evaluate the acidified sample result.

The 1.90 µg/L result does not demonstrate that the original dissolved lead concentration was 1.90 µg/L.

The order of sample preparation matters.

In the first analysis, the sample was filtered before acidification. Suspended particles were removed before the filtrate was acidified.

In the second analysis, acid was added before filtration.

The acid can dissolve suspended lead-containing particles. When the sample is subsequently filtered, some of that newly dissolved lead passes through the filter.

The two measurements therefore represent different operational fractions.

The difference between 1.90 and 0.40 µg/L is consistent with additional acid-recoverable particulate or colloidal lead in the second aliquot.

It is not evidence that the dissolved lead concentration increased between measurements.

Step 3: Calculate the downstream phosphate concentration.

At pH 6.70:

\[ \frac{[\mathrm{HPO_4^{2-}}]} {[\mathrm{H_2PO_4^-}]} =10^{6.70-7.21} \]

\[ =0.309 \]

And:

\[ \frac{[\mathrm{PO_4^{3-}}]} {[\mathrm{HPO_4^{2-}}]} =10^{6.70-12.32} \]

\[ =2.40\times10^{-6} \]

The fraction present as \(\mathrm{PO_4^{3-}}\) is:

\[ \alpha_{\mathrm{PO_4}} =5.67\times10^{-7} \]

Convert total phosphate into mol/L:

\[ C_P=\frac{0.010\times10^{-3}}{30.97} \]

\[ =3.23\times10^{-7}\text{ M} \]

Therefore:

\[ [\mathrm{PO_4^{3-}}] =(3.23\times10^{-7})(5.67\times10^{-7}) \]

\[ =1.83\times10^{-13}\text{ M} \]

Step 4: Calculate the maximum possible free lead concentration.

The laboratory reports:

\[ [\mathrm{Pb}]_{\text{filtered}}<0.05\text{ µg/L} \]

For a conservative upper bound, assume all of this lead is free.

Convert to mol/L:

\[ [\mathrm{Pb}^{2+}]_{\max} = \frac{5.0\times10^{-8}}{207.2} \]

\[ =2.41\times10^{-10}\text{ M} \]

Step 5: Calculate the maximum ion product.

\[ Q_{\max} =(2.41\times10^{-10})^3 (1.83\times10^{-13})^2 \]

\[ Q_{\max}=4.7\times10^{-55} \]

Compare with Ksp:

\[ \frac{Q_{\max}}{K_{sp}} = \frac{4.7\times10^{-55}} {1.0\times10^{-54}} \]

\[ \boxed{\frac{Q_{\max}}{K_{sp}}<0.47} \]

Even under the conservative assumption that all filtered lead is free, the downstream sample is undersaturated with respect to the specified lead phosphate solid.

Therefore, precipitation of this solid is not thermodynamically favoured at the measured downstream conditions.

The lower pH also means that a smaller proportion of total phosphate exists as \(\mathrm{PO_4^{3-}}\).

If lead phosphate solids are already present, the undersaturated conditions may instead favour their dissolution.

Step 6: Evaluate the claim that downstream lead is zero.

The result:

\[ [\mathrm{Pb}]_{\text{filtered}}<0.05\text{ µg/L} \]

does not mean zero lead.

It means the concentration is below the laboratory’s reporting limit.

There may still be dissolved lead below that concentration, as well as particulate lead that was removed during filtration.

The result also describes the sample at the time it was collected. It cannot establish whether lead phosphate precipitation occurred earlier in the treatment process.

Step 7: Design a stronger follow-up experiment.

A useful follow-up should establish both whether phosphate caused the change and where the lead ended up.

First, divide a homogenised water sample into two matched treatment vessels.

• Add a measured phosphate dose to the treatment vessel.
• Leave the control vessel without added phosphate.
• Keep pH, temperature, mixing, settling time, and sample volume consistent.

Then collect samples at the same stages from both vessels.

For each vessel:

1. Measure pH and dissolved reactive phosphate.
2. Filter a portion through 0.45 µm before acidifying it for dissolved-lead analysis.
3. Digest a separate unfiltered portion to measure total recoverable lead.
4. Collect and digest the settled solids to determine their lead content.
5. Use blanks, duplicates, and matrix spikes to check analytical reliability.

Finally, analyse the recovered solids using a suitable solid-phase characterisation method.

This is necessary to distinguish lead phosphate from lead carbonate, lead associated with mineral surfaces, or other lead-containing material.

The phosphate-free control establishes whether phosphate addition caused a difference under otherwise comparable conditions. The lead mass balance establishes whether lead moved into the solids. Solid-phase analysis provides evidence about the identity of those solids.

Together, these measurements provide much stronger evidence than a Ksp calculation alone.

The central lesson is that \(Q\) and \(K_{sp}\) predict the thermodynamic behaviour of a specified solid. They do not identify a solid, validate a laboratory measurement, or prove that a contaminant has been removed.

The next useful concept is therefore mass balance across dissolved, suspended, and settled fractions. It connects the equilibrium chemistry to the actual performance of an environmental treatment process.