Why Gravimetric Precipitates Must Be Washed and Dried

Two students precipitate sulfate ions from identical water samples using barium chloride. Both collect a white solid. One weighs the precipitate straight after filtration. The other washes it, dries it, and obtains a smaller mass.

Which result should be used to calculate the sulfate concentration?

The second result is more reliable, provided the precipitate has not been lost or chemically changed. A balance measures everything placed on it, including water and dissolved salts. It cannot distinguish the substance you want from unwanted material.

That is why a gravimetric precipitate must be both washed and dried:

• Washing removes soluble impurities trapped in the liquid surrounding the precipitate.
• Drying removes water remaining between and on the precipitate particles.

These steps solve different problems. Drying an unwashed precipitate can leave dissolved salts behind as solid crystals. Washing a precipitate without drying it leaves water that increases its measured mass.

01Follow the analyte from the original sample to the balance

To understand why washing and drying matter, follow the substance being measured through the experiment.

The substance being measured is called the analyte. For this example, the analyte is sulfate ions, \(\mathrm{SO_{4}}^{2-}\), dissolved in a water sample.

1. Take a representative sample

First, the chemist takes a measured portion of the original solution.

This portion is called an aliquot. It must represent the original sample accurately.

For example, if a water sample contains suspended material or has not been mixed properly, taking a small portion from the top might give a different concentration from taking one near the bottom.

Washing and drying cannot correct a sampling error. If the aliquot does not represent the original sample, the final calculation will not represent it either.

2. Convert the analyte into a precipitate

The chemist adds a solution containing barium ions, usually barium chloride.

Barium ions react with sulfate ions to form insoluble barium sulfate:

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

The barium sulfate forms a solid called a precipitate.

Notice what has happened: the sulfate ions from the original sample are now part of a new compound, \(\ce{BaSO4}\).

This is important because the chemist can collect and weigh the solid rather than trying to weigh dissolved sulfate ions directly.

3. Understand what remains in the beaker

Here is where the washing problem begins.

Imagine the original sample contained sodium sulfate. The overall reaction is:

\[ \ce{BaCl2(aq) + Na2SO4(aq) -> BaSO4(s) + 2NaCl(aq)} \]

The barium sulfate becomes a solid. However, sodium chloride remains dissolved in the water.

There may also be excess barium chloride because the chemist added slightly more precipitating reagent than was theoretically required.

After precipitation, the beaker therefore contains:

The liquid remaining after precipitation is called the mother liquor.

The chemist wants to collect the barium sulfate, not the mother liquor.

4. Filter the precipitate

The mixture is passed through a filter.

A funnel containing filter paper, or a suitable filtering crucible, holds the solid particles while the liquid passes through into a receiving flask.

The collected solid is called a filter cake.

However, the filter cake is not yet pure, dry barium sulfate.

Some mother liquor remains trapped between the particles and clings to their surfaces. This liquid still contains dissolved sodium chloride, excess barium chloride, and other soluble substances.

Think of wet sand after you have poured water over it. The sand grains are solid, but liquid remains in the spaces between them.

The same physical problem occurs with a precipitate. The difference is that the trapped liquid contains dissolved substances that can contaminate the final measurement.

There are also two other possible problems at this stage:

• If some analyte remains dissolved because precipitation was incomplete, the precipitate mass will be too small.
• If solid particles are lost during filtration or transfer, the recovered mass will also be too small.

Neither problem can be fixed simply by washing or drying.

02Why must the precipitate be washed?

Washing removes soluble substances that remain attached to or trapped within the collected precipitate.

Consider the barium sulfate example.

After filtration, the filter cake contains the desired \(\ce{BaSO4}\), but it also contains small amounts of mother liquor. That liquid contains dissolved sodium chloride and potentially excess barium chloride.

If the chemist skips washing, these dissolved substances remain with the precipitate.

What happens if you skip washing?

Suppose the chemist places the unwashed filter cake in a drying oven.

The water evaporates, but the dissolved salts do not evaporate with it.

Instead, the sodium chloride and other non-volatile dissolved substances remain behind as solid residue.

The final sample now contains:

• Barium sulfate, which is the desired precipitate.
• Sodium chloride crystals, which are unwanted.
• Potentially other solid residues from dissolved impurities.

The balance measures all of them.

The measured mass is therefore greater than the actual mass of pure barium sulfate.

This produces a positive error in the gravimetric result. If the chemist assumes all the measured mass is \(\ce{BaSO4}\), the calculated sulfate concentration will be too high.

How does washing remove these impurities?

The chemist passes a suitable washing liquid through the filter cake.

For barium sulfate, water is generally suitable because barium sulfate has very low solubility in water.

The water dissolves the soluble impurities and carries them through the filter into the receiving flask.

The desired barium sulfate remains in the filter.

A suitable washing liquid should:

• Dissolve the unwanted soluble substances.
• Not dissolve a significant amount of the desired precipitate.
• Not react with the precipitate.
• Not leave additional unwanted residue after drying.

This is why you cannot automatically assume that using more washing liquid is better. If the washing liquid dissolves some of the precipitate, some analyte-containing solid will be lost.

Why use several small washes instead of one large wash?

Using several small portions of washing liquid is generally more effective than using one large portion of the same total volume.

Each time the liquid drains away, it carries dissolved impurities with it. Adding fresh washing liquid then reduces the concentration of the impurities remaining in the filter cake.

This can be demonstrated mathematically.

How can a chemist check whether washing has removed chloride?

One useful test is to collect a small amount of the final washing liquid and add silver nitrate solution.

Silver ions react with chloride ions to form white silver chloride:

\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]

If a white precipitate forms, chloride ions are still present in the washing liquid.

If no visible precipitate forms, the chloride concentration is below the detection limit of that test.

This provides evidence that chloride has been removed sufficiently. However, it does not prove that every possible impurity has been removed. The test only checks for chloride.

03Why must the precipitate be dried?

Washing removes dissolved impurities, but it does not remove all the water.

After washing, the filter cake is still wet. Water remains between the solid particles and on their surfaces.

If the chemist weighs the precipitate immediately, the balance measures both the solid and the remaining water.

This gives:

\[ m_{\text{measured}} = m_{\text{precipitate}} + m_{\text{water}} \]

The extra water increases the measured mass.

Since the chemist uses this mass to calculate the amount of analyte, the calculated amount will also be too high.

What does drying actually achieve?

The washed precipitate is placed in a suitable drying arrangement, such as a drying oven.

The heat removes water from the collected solid.

However, the temperature must be appropriate for the compound being measured. Excessive heating could decompose the precipitate or change its chemical composition.

After drying, the sample is cooled before weighing. A desiccator is useful because it provides a dry environment while the sample returns to room temperature.

A desiccator is a sealed container containing a drying agent that reduces moisture in the surrounding air.

The cooled sample is then weighed. The chemist repeats the drying, cooling, and weighing process until successive measurements show no significant change in mass.

This is called reaching constant mass.

The distinction between washing and drying is important:

• Washing removes soluble contaminants from the precipitate.
• Drying removes water from the washed precipitate.
• Drying does not remove non-volatile salts left behind by insufficient washing.
• Washing does not guarantee that the precipitate contains no remaining water.

For a more detailed explanation of the drying and weighing procedure, see How to Determine Whether a Precipitate Has Reached Constant Mass.

Why is constant mass not necessarily the same as purity?

This is a particularly important distinction.

Suppose a precipitate has been dried repeatedly, and its mass no longer changes.

That tells the chemist that no significant additional mass is being lost under the drying conditions.

It does not prove that the solid is chemically pure.

For example, if sodium chloride remains mixed with barium sulfate, both substances can remain in the weighing vessel after drying. The measured mass can be perfectly stable while still being too high.

Constant mass provides evidence that removable moisture has been eliminated. Washing and appropriate precipitation conditions are still needed to obtain a sufficiently pure solid.

04What happens to the measured mass when something goes wrong?

The direction of the error depends on what remains in the sample or what is lost.

The key is to distinguish between contamination and loss.

Contamination adds mass that does not belong to the desired precipitate. Loss removes some of the precipitate that should have been measured.

Both affect the final calculation, but in opposite directions.

05An important limitation: Washing and drying cannot fix every impurity

There is one more complication worth understanding.

Washing is effective at removing soluble impurities. However, it cannot necessarily remove other substances that have also formed insoluble precipitates.

Imagine the original solution contains both sulfate ions and carbonate ions.

When barium chloride is added, both can react with barium ions:

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

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

Both barium sulfate and barium carbonate form insoluble solids.

The filter collects both substances.

Washing with water will remove soluble impurities, but it will not selectively remove the insoluble barium carbonate. Drying will remove water, but both solid compounds remain.

The final measured mass is therefore greater than the mass of pure barium sulfate.

This is called an interference in gravimetric analysis. The chemist must use a suitable selective method or remove the interfering ion before precipitation.

This gives us an important distinction:

A precipitate can be washed, dried, and brought to constant mass without being pure.

The washing and drying stages are necessary for accurate gravimetric analysis, but the precipitation reaction must also be sufficiently selective.

06Questions and solutions

The following questions test the difference between washing and drying, the effect of contamination on measured mass, and the limitations of both processes.

Question 1

A student analyses a sodium sulfate solution by adding excess barium chloride.

The resulting barium sulfate precipitate is collected by filtration.

Explain:

a) Why the precipitate must be washed.

b) Why the precipitate must be dried before weighing.

c) What would happen to the calculated sulfate concentration if the precipitate were dried without being washed.

Solution 1

Washing removes soluble impurities, while drying removes water. Skipping either step can cause the calculated sulfate concentration to be too high.

a) Washing removes dissolved substances such as sodium chloride and excess barium chloride that remain in the mother liquor trapped around the precipitate.

Without washing, these substances could remain in the final sample.

b) Drying removes water trapped between the precipitate particles and on their surfaces.

If the precipitate is weighed while wet, the balance includes the mass of the water.

c) The calculated sulfate concentration would be too high.

Even if the precipitate is dried, dissolved salts remaining from the mother liquor will be left behind as solid residue when the water evaporates.

The balance measures these salts as well as the barium sulfate. If all the measured mass is treated as pure \(\ce{BaSO4}\), the amount of sulfate will be overestimated.

Question 2

A student precipitates sulfate ions as barium sulfate using excess barium chloride.

After filtration, the student collects three successive portions of washing liquid. Each portion is tested separately by adding silver nitrate solution.

The following observations are recorded:

a) Identify the ion responsible for the white precipitate.

b) Explain what the observations suggest about the removal of soluble impurities.

c) Does the third observation prove that the barium sulfate precipitate is completely pure? Explain.

d) Explain why drying the precipitate immediately after the first wash would not solve the contamination problem.

Solution 2

The white precipitate is silver chloride, and the observations indicate that the amount of dissolved chloride decreases with successive washes. The final observation supports adequate chloride removal to the test’s detection limit, but it does not prove complete chemical purity.

a) The white precipitate is silver chloride, \(\ce{AgCl}\).

Silver ions react with chloride ions:

\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]

b) The first washing liquid contains a relatively high concentration of chloride ions.

The second contains less chloride, as indicated by the weaker white cloudiness.

The third produces no visible precipitate, suggesting that the chloride concentration has fallen below the detection limit of the test.

This provides evidence that successive washing has removed the soluble chloride contamination.

c) No. A negative chloride test does not prove that the precipitate is completely pure.

It only provides evidence that chloride is no longer detectable in the tested washing liquid.

Other dissolved impurities might still be present. Insoluble contaminants or substances trapped within the precipitate may also remain.

d) Drying removes water but does not remove non-volatile dissolved salts.

If chloride-containing solution remains after the first wash, evaporating its water leaves the dissolved salts behind.

The precipitate would therefore still have an artificially high mass.

Question 3

A 150.0 mL sulfate solution is analysed using barium chloride.

After filtration, the precipitate is dried to constant mass without being washed. The measured precipitate mass is 0.526 g.

The precipitate is then washed with a suitable washing liquid and dried again to constant mass. Its final mass is 0.498 g.

Assume:

• Precipitation was complete.
• Washing removed only soluble impurities.
• No barium sulfate dissolved or was lost during washing.
• The final precipitate is pure and dry barium sulfate.

Calculate:

a) The mass of soluble impurities removed during washing.

b) The correct sulfate concentration in the original sample.

c) The percentage by which the unwashed precipitate mass would overestimate the sulfate concentration.

Use \(M(\ce{BaSO4})=233.4\text{ g mol}^{-1}\).

Solution 3

The washing removed 0.028 g of soluble impurities, and the correct sulfate concentration is \(0.0142\text{ mol L}^{-1}\). Using the unwashed mass would overestimate the concentration by approximately 5.62%.

a) Calculate the mass of soluble impurities removed.

The mass before washing was 0.526 g.

The mass after washing and drying was 0.498 g.

Therefore:

\[ m_{\text{impurities}}=0.526-0.498 \]

\[ \boxed{m_{\text{impurities}}=0.028\text{ g}} \]

This mass was not barium sulfate. It was soluble material removed during washing.

b) Calculate the amount of pure barium sulfate.

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

\[ n(\ce{BaSO4})=\frac{0.498}{233.4} \]

\[ n(\ce{BaSO4})=2.13\times10^{-3}\text{ mol} \]

From the balanced ionic equation, the mole ratio of sulfate to barium sulfate is 1:1.

Therefore:

\[ n(\mathrm{SO_{4}}^{2-})=2.13\times10^{-3}\text{ mol} \]

Convert the sample volume:

\[ 150.0\text{ mL}=0.1500\text{ L} \]

Calculate the concentration:

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

\[ c=\frac{2.13\times10^{-3}}{0.1500} \]

\[ \boxed{c=0.0142\text{ mol L}^{-1}} \]

c) Calculate the percentage overestimate.

Because the same sample volume and mole ratio are used, the percentage error in concentration is the same as the percentage error in precipitate mass.

\[ \%\text{ overestimate} = \frac{0.526-0.498}{0.498}\times100 \]

\[ =\frac{0.028}{0.498}\times100 \]

\[ \boxed{5.62\%} \]

The unwashed precipitate produces an artificially high result because the measured mass includes soluble impurities that do not represent sulfate.

Notice that the first precipitate had already been dried to constant mass. This did not remove the contamination because the dissolved salts remained as solid residue.

Question 4

A chemist determines the sulfate concentration of a solution by precipitating sulfate ions with excess barium chloride.

However, the original solution also contains carbonate ions.

Both barium sulfate and barium carbonate are insoluble under the experimental conditions.

The chemist filters the precipitate, washes it thoroughly, and dries it to constant mass. The calculated sulfate concentration is still too high.

Explain:

a) Why the measured mass is too high.

b) Why further washing will not solve the problem.

c) Why further drying will not solve the problem.

d) Suggest one way the chemist could improve the analysis.

Solution 4

The result is too high because barium carbonate has precipitated alongside barium sulfate, adding unwanted solid mass. Neither ordinary washing nor further drying will selectively remove the barium carbonate.

a) Both sulfate and carbonate react with barium ions:

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

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

The chemist intends to measure the mass of \(\ce{BaSO4}\), but the filter collects both solid compounds.

The measured mass therefore includes barium carbonate, which does not represent the sulfate analyte.

If the entire mass is assumed to be barium sulfate, the calculated sulfate amount will be too high.

b) Washing removes soluble impurities.

Barium carbonate is an insoluble solid under the stated conditions, so it remains in the filter with barium sulfate.

More washing will not selectively remove it.

c) Drying removes water.

Both barium sulfate and barium carbonate remain as solids after drying. Further drying therefore cannot separate their contributions to the measured mass.

d) The chemist could remove the carbonate ions before adding barium chloride, or use a suitably selective precipitation method that prevents carbonate from forming a precipitate alongside sulfate.

The key is to address the interference before weighing the final solid.

This example shows why a dry, washed precipitate is not automatically a pure precipitate. The precipitation reaction itself must be selective enough for the measured mass to represent the analyte-containing compound.

Question 5

Two students analyse separate 50.0 mL aliquots of the same thoroughly mixed sulfate solution.

Both use barium chloride to precipitate sulfate as barium sulfate.

Their measured precipitate masses are identical: 0.272 g.

However, their experimental procedures differ.

Student A:

• Washes the precipitate correctly.
• Does not dry it before weighing.
• The measured sample contains exactly 0.010 g of water.
• No other contamination or precipitate loss occurs.

Student B:

• Dries the precipitate to constant mass.
• Does not wash it.
• The collected solid contains 0.018 g of soluble salt residue.
• During transfer, 0.008 g of barium sulfate is lost.

Assume precipitation is complete in both experiments, both aliquots have the same sulfate concentration, and there are no other errors.

a) Calculate the actual mass of barium sulfate recovered from each aliquot before any transfer loss.

b) Calculate the sulfate concentration of the original solution.

c) Calculate the concentration that either student would report if they incorrectly treated 0.272 g as pure, dry barium sulfate.

d) Explain why the identical measured masses do not demonstrate that either student’s procedure is accurate.

Use \(M(\ce{BaSO4})=233.4\text{ g mol}^{-1}\).

Solution 5

Both aliquots originally produced 0.262 g of barium sulfate, giving a sulfate concentration of \(0.0225\text{ mol L}^{-1}\). The identical measured masses are misleading because different experimental errors produce the same positive mass error.

This question combines three effects: retained water, soluble contamination, and loss of precipitate.

a) Determine the actual barium sulfate mass for each student.

For Student A, the sample contains 0.010 g of water.

The balance measures:

\[ m_{\text{measured}} = m_{\ce{BaSO4}}+m_{\text{water}} \]

Therefore:

\[ m_{\ce{BaSO4}}=0.272-0.010 \]

\[ \boxed{m_{\ce{BaSO4}}=0.262\text{ g}} \]

For Student B, two errors occur.

The student has:

• 0.018 g of unwanted soluble salt residue, which increases the mass.
• 0.008 g of lost barium sulfate, which decreases the mass.

Therefore:

\[ m_{\ce{BaSO4}} = 0.272-0.018+0.008 \]

\[ \boxed{m_{\ce{BaSO4}}=0.262\text{ g}} \]

Both aliquots originally contained the same amount of sulfate, as expected for equal volumes of a thoroughly mixed solution.

b) Calculate the sulfate concentration.

Use the correct precipitate mass:

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

\[ n(\ce{BaSO4})=\frac{0.262}{233.4} \]

\[ n(\ce{BaSO4})=1.12\times10^{-3}\text{ mol} \]

The reaction has a 1:1 mole ratio between sulfate and barium sulfate.

Therefore:

\[ n(\mathrm{SO_{4}}^{2-})=1.12\times10^{-3}\text{ mol} \]

Convert the aliquot volume:

\[ 50.0\text{ mL}=0.0500\text{ L} \]

Calculate the concentration:

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

\[ c=\frac{1.12\times10^{-3}}{0.0500} \]

\[ \boxed{c=0.0225\text{ mol L}^{-1}} \]

c) Calculate the concentration using the incorrect measured mass.

If either student assumes that 0.272 g is pure, dry barium sulfate:

\[ n=\frac{0.272}{233.4} \]

\[ n=1.17\times10^{-3}\text{ mol} \]

Therefore:

\[ c=\frac{1.17\times10^{-3}}{0.0500} \]

\[ \boxed{c=0.0233\text{ mol L}^{-1}} \]

The incorrect result is higher than the actual concentration.

The percentage overestimate is:

\[ \%\text{ error} = \frac{0.272-0.262}{0.262}\times100 \]

\[ \boxed{3.82\%} \]

d) Identical measured masses do not demonstrate accuracy because different errors can produce the same final reading.

Student A has extra water in the sample. Student B has extra soluble salt, partly offset by losing some barium sulfate during transfer.

Student B has also reached constant mass, but this does not remove the non-volatile salt residue.

The tempting conclusion is that both students obtained the same mass, so their methods must be reliable. However, agreement between measurements does not guarantee that the measurements represent the correct quantity.

This is the difference between precision and accuracy:

• Precision describes how closely measurements agree with one another.
• Accuracy describes how close a measurement is to the true value.

Both students report the same mass, but both results are too high.

The correct procedure must remove soluble impurities through washing, remove water through drying, and minimise the loss of the desired precipitate.

07What to remember when answering an HSC Chemistry question

When a question asks why a gravimetric precipitate must be washed and dried, keep the two purposes separate.

A strong written response should explain that washing removes soluble impurities without significantly dissolving the precipitate, while drying removes residual water without changing the chemical composition of the solid.

It is also worth remembering that constant mass does not automatically establish purity. Insoluble contaminants can remain even after repeated washing and drying.

Once a suitable pure, dry precipitate has been obtained, its mass can be converted into moles using \(n=m/M\). The balanced chemical equation then provides the mole ratio needed to calculate the original amount of analyte.

For the complete calculation process, see Gravimetric analysis: calculate analyte amounts from precipitates.

The important decision is not simply whether the precipitate looks dry or whether its mass has stopped changing. It is whether the mass being measured genuinely represents the intended chemical compound.