Why Filtration Cannot Remove Dissolved Salt from Water

A student pours clear salt water through filter paper, expecting the salt to collect in the funnel. The water passes through, but no salt crystals appear on the paper. Where has the salt gone?

It has passed through with the water. Ordinary filtration cannot remove dissolved salt because the salt exists as dissolved ions, which pass through the pores of filter paper along with the water. Filtration separates insoluble solid particles from liquids, not dissolved substances from their solvents.

The difference becomes much easier to understand when you look at what is actually present in salt water.

01What happens when salt dissolves in water?

Before we look at the apparatus, predict what would happen if you added a spoonful of sand to a glass of salt water and filtered the mixture.

Would the sand and salt both collect on the filter paper?

Only the sand would be collected. The salt would remain dissolved in the water and pass through.

Here’s why.

When solid sodium chloride dissolves in water, its ionic structure separates into positively charged sodium ions and negatively charged chloride ions:

\[ \ce{NaCl(s) -> Na+(aq) + Cl-(aq)} \]

The symbols mean:

• \(\ce{NaCl(s)}\): solid sodium chloride.
• \(\ce{Na+(aq)}\): sodium ions dissolved in water.
• \(\ce{Cl-(aq)}\): chloride ions dissolved in water.

The notation \((aq)\) means aqueous, or dissolved in water.

Water molecules surround these ions. This process is called hydration.

The important point is that the salt is no longer present as separate, visible crystals. Its ions are distributed throughout the water.

This gives us our first important distinction:

• Sand in water is an insoluble solid. It remains as separate particles suspended in the liquid.
• Sodium chloride dissolved in water forms a homogeneous solution. Its ions are distributed throughout the liquid.

Filtration works well for the first mixture, but not the second.

02What does filter paper actually separate?

Imagine using a sieve to separate small stones from water.

The stones are too large to pass through the openings, so they remain in the sieve. The water flows through.

Filter paper works using a similar principle. It contains tiny pores that allow liquid to pass while retaining sufficiently large insoluble particles.

The analogy has a limit: filter paper isn’t identifying substances by their chemical identity. It isn’t designed to recognise salt, sand, or any particular chemical. It simply allows the liquid and dissolved substances to pass through its pores.

Understanding the filtration apparatus

A standard gravity filtration setup contains three main components:

1. A filter funnel lined with folded filter paper.
2. A conical flask or beaker positioned underneath the funnel.
3. A glass rod that guides the mixture into the funnel.

The mixture is poured down the glass rod and into the filter paper.

The liquid passes through the paper and collects in the container below. This collected liquid is called the filtrate.

Any insoluble solid particles retained by the filter paper are called the residue.

Now consider what happens when the mixture is salt water.

The water passes through the filter paper.

The dissolved sodium ions and chloride ions pass through with it.

There are no separate salt crystals for the filter paper to catch.

The result is a filtrate that still contains dissolved sodium chloride.

What happens to each component?

The following table compares different mixtures and their behaviour during ordinary filtration.

Notice the final row.

If salt is present as undissolved crystals, filtration can collect those crystals. However, any salt that has already dissolved will still pass through.

This distinction is about the physical state of the salt, not its chemical identity.

03Why does the filtrate still look like pure water?

This is one of the most common mistakes in separation-technique questions.

A student filters salt water and sees a clear liquid in the receiving flask. They conclude that the salt must have been removed.

That conclusion is incorrect.

A clear liquid is not necessarily a pure liquid.

The solution looks clear because dissolved ions are distributed throughout the water at a scale that cannot be seen with the naked eye. They don’t behave like suspended sand particles that scatter light and make water appear cloudy.

The filtrate can therefore look exactly like ordinary water while still containing a substantial amount of dissolved salt.

There are two useful ways to demonstrate this.

1. Test the electrical conductivity

Pure water has very low electrical conductivity.

Salt water conducts electricity because it contains mobile ions. These ions carry electrical charge through the solution.

If filtration removed the dissolved salt, the conductivity of the filtrate should decrease substantially.

However, ordinary filtration allows the ions to pass through. The filtrate should therefore retain a similar conductivity to the original salt solution, assuming the temperature and concentration are comparable.

2. Evaporate some of the filtrate

Take a small, measured volume of filtrate and evaporate the water.

If sodium chloride is still dissolved in the sample, solid salt will remain after the water has evaporated.

This is direct evidence that filtration did not remove the dissolved salt.

The key distinction is:

Clear does not mean pure. It only tells you that there are no visibly suspended particles making the liquid cloudy.

04Which separation technique should you use instead?

The correct technique depends on the physical properties of the substances and what you want to recover.

For salt water, the important properties are:

• Sodium chloride is effectively non-volatile under ordinary laboratory conditions.
• Water is volatile and can evaporate.
• Sodium chloride dissolves in water, so its dissolved ions pass through ordinary filter paper.

These properties suggest different methods for different objectives.

Evaporation: recovering the salt

If you only want to collect the sodium chloride, you can heat the salt solution in an evaporating basin.

As the water evaporates, the concentration of dissolved salt increases. Eventually, the water is removed and solid sodium chloride remains.

The salt doesn’t evaporate with the water under ordinary laboratory conditions.

Crystallisation: obtaining solid crystals

Crystallisation is useful when you want to obtain crystals rather than simply leave a dry solid residue.

The solution is concentrated, then allowed to cool under suitable conditions. If the solubility of the salt decreases as temperature decreases, some dissolved salt forms crystals.

These crystals can then be collected by filtration.

The important sequence is:

1. Dissolve the salt in water.
2. Concentrate the solution if necessary.
3. Allow crystals to form.
4. Filter to collect the crystals.

Filtration is useful here because the salt has already changed from dissolved ions into solid crystals.

For a more detailed comparison of these methods, see How to Choose Between Evaporation and Crystallisation.

Simple distillation: recovering both salt and water

What if you want the salt and the water?

Evaporation alone won’t achieve this because the water vapour escapes into the surroundings.

Simple distillation solves the problem by directing the water vapour through a condenser. The vapour cools, changes back into liquid water, and collects in a separate receiving container.

The sodium chloride remains in the original distillation flask.

You can explore the complete apparatus and separation process in How to Recover Both Salt and Water by Simple Distillation.

One final distinction: specialised membrane processes such as reverse osmosis can remove dissolved ions from water. These use membranes designed to reject dissolved substances, usually with applied pressure. They are not the same as ordinary filtration through laboratory filter paper.

05Worked examples

06Questions and solutions

Question 1

A student filters 25.0 mL of clear sodium chloride solution through ordinary filter paper.

Identify:

(a) The residue.

(b) The filtrate.

(c) Whether the filtrate is pure water.

Explain each answer using the physical properties of the substances.

Solution 1

(a) There is no solid residue, assuming the solution contains no suspended impurities.

The sodium chloride is completely dissolved, so there are no separate salt crystals to be retained by the filter paper.

(b) The filtrate contains water, dissolved sodium ions, and dissolved chloride ions.

These ions pass through the pores of the filter paper with the liquid.

(c) The filtrate is not pure water.

It is still a sodium chloride solution. Filtration separates insoluble solid particles from a liquid, but it cannot remove dissolved ions using ordinary filter paper.

Question 2

A laboratory technician prepares 200.0 g of sodium chloride solution containing 8.00% sodium chloride by mass.

The solution is passed through a filter to remove suspended dust.

Before filtration, the filter paper is wetted using 5.00 g of distilled water. During filtration, 4.00 g of the original salt solution remains trapped in the filter paper and funnel. Assume all 5.00 g of pre-wetting water enters the collected filtrate.

Assume no evaporation or selective adsorption of salt occurs.

Calculate:

(a) The initial mass of sodium chloride.

(b) The mass of sodium chloride in the filtrate.

(c) The total mass of the filtrate.

(d) The final mass percentage of sodium chloride in the filtrate.

Solution 2

(a) The initial mass of sodium chloride is 16.0 g.

The solution contains 8.00% sodium chloride by mass:

\[ m(\ce{NaCl})=200.0\times0.0800 \]

\[ \boxed{m(\ce{NaCl})=16.0\text{ g}} \]

(b) The filtrate contains 15.68 g of sodium chloride.

First, calculate the salt retained in the 4.00 g of solution held by the apparatus.

\[ m(\ce{NaCl})_{\text{retained}} =4.00\times0.0800 \]

\[ =0.320\text{ g} \]

Therefore:

\[ m(\ce{NaCl})_{\text{filtrate}} =16.0-0.320 \]

\[ \boxed{=15.68\text{ g}} \]

The 5.00 g of pre-wetting water contains no salt, so it does not change the amount of sodium chloride.

(c) The total mass of filtrate is 201.0 g.

The original solution contributes:

\[ 200.0-4.00=196.0\text{ g} \]

The pre-wetting water adds another 5.00 g:

\[ m(\text{filtrate})=196.0+5.00 \]

\[ \boxed{=201.0\text{ g}} \]

(d) The final mass percentage of sodium chloride is 7.80%.

Use:

\[ \text{Mass percentage} =\frac{\text{mass of solute}} {\text{mass of solution}}\times100 \]

Substituting:

\[ =\frac{15.68}{201.0}\times100 \]

\[ \boxed{=7.80\%} \]

The concentration has decreased slightly because the filter paper was pre-wetted with pure water.

The important distinction is that the reduction in salt concentration is caused by dilution, not filtration removing dissolved sodium chloride.

Question 3

A student claims that filtering a salt solution twice should remove more dissolved salt than filtering it once.

Another student tests this claim using an electrical conductivity meter.

The following illustrative measurements are recorded at a constant temperature of \(25^\circ\mathrm{C}\).

The student then evaporates 10.0 mL of the twice-filtered solution to dryness. A mass of 0.73 g of solid sodium chloride remains.

(a) Calculate the mass concentration of sodium chloride in the twice-filtered solution in g L\(^{-1}\).

(b) Explain what the conductivity measurements suggest.

(c) Evaluate the original student’s claim.

Solution 3

(a) The mass concentration of sodium chloride is 73 g L\(^{-1}\).

Mass concentration is calculated using:

\[ \text{Mass concentration} =\frac{\text{mass of solute}} {\text{volume of solution}} \]

Convert the volume into litres:

\[ 10.0\text{ mL}=0.0100\text{ L} \]

Substitute the measured values:

\[ c=\frac{0.73}{0.0100} \]

\[ \boxed{c=73\text{ g L}^{-1}} \]

This confirms that the twice-filtered sample still contains a substantial amount of dissolved sodium chloride.

(b) The conductivity measurements suggest that dissolved ions remain in the solution after filtration.

The conductivity is almost unchanged after the first filtration and remains similar after the second.

This is expected because sodium and chloride ions pass through ordinary filter paper with the water.

Conductivity depends on factors such as ion concentration, ion mobility, and temperature, so it is not an exact measurement of salt concentration without suitable calibration. However, the measurements are consistent with dissolved ions remaining in the filtrate.

(c) The original claim is incorrect.

Repeating ordinary filtration does not remove dissolved sodium chloride.

The filter paper may remove additional suspended impurities, but the dissolved ions continue to pass through.

The evaporation result provides further evidence: removing the water leaves 0.73 g of solid sodium chloride from just 10.0 mL of the twice-filtered solution.

The student has confused the removal of suspended particles with the removal of dissolved substances.

Question 4

A beaker contains the following mixture at room temperature:

• 40.0 g of water.
• 14.0 g of sodium chloride dissolved in the water.
• 5.0 g of undissolved sodium chloride crystals.
• 3.0 g of insoluble sand.

The mixture is filtered.

The solid residue is then transferred to a second beaker. A further 20.0 g of distilled water is added and stirred until all the remaining sodium chloride crystals dissolve.

The new mixture is filtered again.

Assume the first residue is drained of bulk liquid before washing, all 5.0 g of salt crystals dissolve in the added water, and liquid retained by the filter paper can be ignored.

(a) State the composition and mass of the first residue.

(b) State the composition of the first filtrate.

(c) State the composition and mass of the second residue.

(d) Explain why this experiment does not contradict the statement that filtration cannot remove dissolved salt.

Solution 4

(a) The first residue has a total mass of 8.0 g, consisting of 5.0 g of undissolved sodium chloride crystals and 3.0 g of sand.

The crystals are already in the solid state, so they are large enough to be retained by the filter paper.

The sand is also insoluble and remains as solid particles.

Therefore:

\[ m(\text{residue})=5.0+3.0 \]

\[ \boxed{=8.0\text{ g}} \]

(b) The first filtrate contains 40.0 g of water and 14.0 g of dissolved sodium chloride.

The dissolved salt exists as aqueous ions. These ions pass through the filter paper along with the water.

(c) The second residue contains 3.0 g of sand.

After adding 20.0 g of water, all 5.0 g of sodium chloride crystals dissolve.

The resulting mixture contains dissolved sodium chloride and sand.

During the second filtration, the sand is retained, while the dissolved ions pass through with the water.

Therefore:

\[ \boxed{m(\text{second residue})=3.0\text{ g}} \]

The second filtrate contains 20.0 g of water and 5.0 g of dissolved sodium chloride.

(d) The experiment does not contradict the statement because the first filtration collects undissolved salt crystals, not dissolved sodium chloride.

The original mixture contains two different physical forms of the same substance:

• Dissolved sodium chloride, which passes through the filter paper.
• Solid sodium chloride crystals, which are retained.

When the crystals are dissolved in additional water, they become aqueous ions and pass through the second filter.

The important factor is whether the salt is dissolved or present as a separate solid.

Question 5

A student prepares a hot solution containing 80.0 g of potassium nitrate, \(\ce{KNO3}\), dissolved in 100.0 g of water at \(60^\circ\mathrm{C}\).

The mixture also contains 1.50 g of insoluble dust.

The following illustrative solubility data are supplied:

Assume the solution reaches equilibrium at each temperature, all crystals formed can be collected, and no material is lost during either separation.

The student considers two possible procedures.

Procedure A

1. Filter the mixture while maintaining it at \(60^\circ\mathrm{C}\).
2. Cool the filtrate to \(20^\circ\mathrm{C}\).
3. Filter the cooled solution again.

Procedure B

1. Cool the original mixture to \(20^\circ\mathrm{C}\).
2. Filter the mixture once.

Answer the following:

(a) Calculate the mass of potassium nitrate that crystallises when the solution reaches \(20^\circ\mathrm{C}\).

(b) For Procedure A, describe the contents of the residue from each filtration and the final filtrate.

(c) For Procedure B, calculate the total mass of solid residue and identify its components.

(d) Which procedure produces the purer potassium nitrate crystal crop?

(e) A student argues that Procedure B proves filtration can remove dissolved salt because the filter paper collects a large mass of solid potassium nitrate. Evaluate this argument carefully.

Solution 5

(a) A mass of 48.0 g of potassium nitrate crystallises when the solution cools to \(20^\circ\mathrm{C}\).

At \(60^\circ\mathrm{C}\), the solubility is 110 g per 100 g of water.

The original solution contains only 80.0 g of potassium nitrate per 100.0 g of water, so all the potassium nitrate is initially dissolved.

At \(20^\circ\mathrm{C}\), only 32.0 g can remain dissolved in 100.0 g of water.

Therefore, the mass that crystallises is:

\[ m(\text{crystals})=80.0-32.0 \]

\[ \boxed{=48.0\text{ g}} \]

(b) In Procedure A, the first residue contains 1.50 g of insoluble dust.

At \(60^\circ\mathrm{C}\), all 80.0 g of potassium nitrate remains dissolved because the solution is below its saturation limit.

Therefore, the first filtration removes the dust but allows all the dissolved potassium nitrate to pass through.

After cooling the filtrate to \(20^\circ\mathrm{C}\), 48.0 g of potassium nitrate crystallises.

The second filtration collects these crystals.

The final filtrate contains:

• 100.0 g of water.
• 32.0 g of dissolved potassium nitrate.

The second residue contains 48.0 g of potassium nitrate crystals.

(c) In Procedure B, the total solid residue has a mass of 49.50 g.

Cooling the original mixture to \(20^\circ\mathrm{C}\) causes 48.0 g of potassium nitrate to crystallise.

The 1.50 g of insoluble dust is still present.

Both substances are retained during filtration:

\[ m(\text{residue})=48.0+1.50 \]

\[ \boxed{=49.50\text{ g}} \]

The residue contains:

• 48.0 g of potassium nitrate crystals.
• 1.50 g of insoluble dust.

The filtrate contains 100.0 g of water and 32.0 g of dissolved potassium nitrate.

(d) Procedure A produces the purer potassium nitrate crystal crop.

The first filtration removes the insoluble dust while the potassium nitrate is still completely dissolved.

Cooling then causes the potassium nitrate to crystallise in the absence of the dust.

The second filtration therefore collects the potassium nitrate crystals without mixing them with the original insoluble impurity.

In Procedure B, the dust and newly formed potassium nitrate crystals are collected together.

Although the total residue is heavier, the additional mass is contamination rather than extra potassium nitrate.

(e) The student’s argument is incorrect because the potassium nitrate was no longer entirely dissolved when Procedure B was performed.

Cooling changed the solubility of potassium nitrate. The solution could no longer keep all 80.0 g dissolved, so 48.0 g formed solid crystals.

Filtration then collected those crystals because they were already in the solid state.

The remaining 32.0 g of potassium nitrate stayed dissolved and passed through the filter paper.

This is the important distinction:

Cooling caused crystallisation. Filtration collected the crystals.

The filter paper did not remove dissolved potassium nitrate ions.

Both procedures recover the same 48.0 g crystal crop, but Procedure A produces a cleaner product because the insoluble dust is removed before crystallisation.

If the student wanted to recover the remaining 32.0 g of dissolved potassium nitrate as well, the final filtrate would need further concentration or evaporation.

07The key decision to remember

When deciding whether filtration will work, don’t start by asking whether a substance is a salt, a metal, or a particular chemical.

Start by asking what physical form it is in.

• If it is an insoluble solid suspended in a liquid, filtration can usually separate it.
• If it is dissolved in the liquid, ordinary filtration will not remove it.
• If it has crystallised or precipitated into a solid, filtration can collect it.

For salt water, dissolved sodium chloride passes through filter paper with the water because its ions are distributed throughout the solution.

Once you understand that distinction, the next decision becomes much easier: use crystallisation or evaporation when you want to recover the dissolved salt, and simple distillation when you also need to collect the water.