Dilution

Dilution is the process of decreasing the concentration of a solution by adding more solvent.

For an aqueous solution, the solvent is water, so dilution usually means adding water to an existing solution.

During simple dilution:

• more solvent is added
• the total volume increases
• the amount of solute stays the same
• the concentration decreases

The key idea is:

Dilution changes the concentration, but it does not change the amount of solute.

What Happens During Dilution?

Suppose we have:

100 mL of salt solution

and add:

400 mL water

The total volume becomes approximately:

500 mL

The amount of salt has not changed.

The salt particles are simply spread throughout a larger volume.

Therefore, there are fewer solute particles in each unit of volume.

The solution becomes more dilute.

At the particle level:

same number of solute particles + larger volume = lower concentration

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Concentrated and Dilute Solutions

A concentrated solution contains relatively more solute per unit volume.

A dilute solution contains relatively less solute per unit volume.

Imagine:

Before dilution

100 solute particles distributed through 100 mL.

After dilution

The same 100 solute particles distributed through 500 mL.

The number of solute particles has not changed.

Their concentration has decreased because they occupy a larger volume.

Moles Are Conserved During Dilution

When solvent is added, no additional solute is added and none is removed.

Therefore:

moles of solute before dilution = moles of solute after dilution

Using:

n = cV

we can write:

c₁V₁ = c₂V₂

where:

• c₁ = initial concentration
• V₁ = initial volume
• c₂ = final concentration
• V₂ = final volume

This is the dilution equation.

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Understanding the Dilution Equation

The equation:

c₁V₁ = c₂V₂

comes directly from conservation of the solute.

Before dilution:

n₁ = c₁V₁

After dilution:

n₂ = c₂V₂

Since the amount of solute does not change:

n₁ = n₂

Therefore:

c₁V₁ = c₂V₂

This is not a separate chemical law to memorize. It follows from the fact that the same amount of solute is present before and after dilution.

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Initial and Final Values

It is useful to identify the variables carefully.

Usually:

V₂ > V₁

and:

c₂ < c₁

because dilution increases volume and decreases concentration.

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Worked Example: Basic Dilution

A student has:

100 mL of 2.0 mol/L NaCl

Water is added until the total volume becomes:

500 mL

Calculate the new concentration.

Use:

c₁V₁ = c₂V₂

Substitute:

2.0 × 100 = c₂ × 500

Rearrange:

c₂ = (2.0 × 100) / 500

c₂ = 0.40 mol/L

Answer

Final concentration = 0.40 mol/L

The concentration decreased because the same amount of NaCl is now distributed through five times the volume.

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Why We Can Sometimes Use mL

Normally, calculations involving:

n = cV

require volume in litres when concentration is measured in mol/L.

However, in:

c₁V₁ = c₂V₂

both volumes can be in mL if they use the same units.

For example:

2.0 × 100 mL = c₂ × 500 mL

The mL units cancel in the ratio.

You could also use litres:

2.0 × 0.100 = c₂ × 0.500

Both methods give:

c₂ = 0.40 mol/L

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Worked Example: Diluting Hydrochloric Acid

A chemist takes:

50 mL of 3.0 mol/L HCl

and dilutes it to:

250 mL

Calculate the new concentration.

Use:

c₁V₁ = c₂V₂

3.0 × 50 = c₂ × 250

c₂ = 150 / 250

c₂ = 0.60 mol/L

Answer

Final concentration = 0.60 mol/L

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Worked Example: A Tenfold Dilution

A solution has an initial concentration of:

5.0 mol/L

A:

20 mL

sample is diluted to:

200 mL

Calculate the final concentration.

c₁V₁ = c₂V₂

5.0 × 20 = c₂ × 200

c₂ = 0.50 mol/L

The volume increased by a factor of:

10

Therefore, the concentration decreased by a factor of:

10

This is called a tenfold dilution.

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Dilution Factor

The dilution factor tells us how much the solution has been diluted.

One common definition is:

dilution factor = final volume / initial volume

For example:

Initial volume:

50 mL

Final volume:

500 mL

Then:

dilution factor = 500 / 50

= 10

This is a tenfold dilution.

The concentration therefore becomes:

1/10 of its original value

If the original concentration were:

2.0 mol/L

the new concentration would be:

0.20 mol/L

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Doubling the Volume

Suppose a solution has:

c₁ = 1.0 mol/L

and:

V₁ = 200 mL

Water is added until:

V₂ = 400 mL

The volume has doubled.

Since the amount of solute remains constant:

the concentration is halved

Therefore:

c₂ = 0.50 mol/L

This proportional reasoning can often help you predict the answer before calculating.

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Tripling the Volume

Suppose:

c₁ = 0.90 mol/L

and the volume is tripled.

The concentration becomes:

0.90 / 3

= 0.30 mol/L

The relationship is inverse:

volume increases → concentration decreases

provided the amount of solute remains constant.

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Finding the Final Volume

Sometimes the desired final concentration is given.

For example:

A chemist has:

100 mL of 2.0 mol/L NaCl

They want to prepare a:

0.50 mol/L

solution.

What should the final volume be?

Use:

c₁V₁ = c₂V₂

Substitute:

2.0 × 100 = 0.50 × V₂

Rearrange:

V₂ = 200 / 0.50

V₂ = 400 mL

Answer

The solution should be diluted to a final volume of 400 mL.

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Final Volume Is Not the Amount of Solvent Added

This is one of the most important ideas in dilution calculations.

In the previous example:

Initial volume:

100 mL

Required final volume:

400 mL

This does not mean that 400 mL of water should be added.

The amount of water required is approximately:

400 − 100 = 300 mL

Therefore:

final volume = 400 mL

but:

solvent added ≈ 300 mL

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Finding the Volume of Solvent Required

The general relationship is:

volume of solvent added = final volume − initial volume

or:

V_solvent = V₂ − V₁

For example:

Initial solution volume:

150 mL

Required final volume:

600 mL

Water added:

600 − 150 = 450 mL

Answer

Approximately:

450 mL water

must be added.

In precise laboratory work, however, a chemist normally dilutes the solution to the final volume rather than simply measuring and adding a calculated solvent volume, because solution volumes are not always perfectly additive.

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Worked Example: Finding Water Required

A student has:

200 mL of 1.5 mol/L solution

They need a:

0.50 mol/L solution

Find:

1. the required final volume
2. the approximate volume of water added

Use:

c₁V₁ = c₂V₂

1.5 × 200 = 0.50 × V₂

V₂ = 600 mL

Now calculate water added:

600 − 200 = 400 mL

Answer

Final volume = 600 mL

Water added ≈ 400 mL

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Worked Example: Larger Dilution

A chemist has:

250 mL of 4.0 mol/L solution

They want to dilute it to:

1.0 mol/L

Calculate the final volume.

4.0 × 250 = 1.0 × V₂

V₂ = 1000 mL

Therefore:

V₂ = 1.0 L

Approximate water added:

1000 − 250 = 750 mL

Answer

Final volume = 1.0 L

Water added ≈ 750 mL

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Finding the Initial Volume Needed

Another common laboratory problem asks how much concentrated solution is needed to prepare a dilute solution.

Suppose a chemist wants:

500 mL of 0.20 mol/L NaCl

using a stock solution with concentration:

2.0 mol/L

Use:

c₁V₁ = c₂V₂

2.0 × V₁ = 0.20 × 500

2.0V₁ = 100

V₁ = 50 mL

Answer

The chemist needs:

50 mL of the 2.0 mol/L stock solution

and dilutes it to a final volume of:

500 mL

Stock Solutions

A stock solution is a concentrated solution of known concentration.

Scientists often prepare dilute solutions from stock solutions because it is:

• convenient
• accurate
• efficient
• reproducible

Instead of preparing every solution directly from a solid, a measured volume of stock solution can be diluted to the required concentration.

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Preparing a Dilution in the Laboratory

Suppose we need:

250 mL of 0.20 mol/L solution

from:

1.0 mol/L stock solution

First calculate the required stock volume:

1.0 × V₁ = 0.20 × 250

V₁ = 50 mL

A laboratory procedure might then be:

1. Measure 50 mL of stock solution accurately.
2. Transfer it to a 250 mL volumetric flask.
3. Add distilled or deionized water.
4. Approach the calibration line carefully.
5. Add the final water dropwise.
6. Adjust the bottom of the meniscus to the calibration line.
7. Stopper the flask.
8. Mix thoroughly.

The final solution contains the same amount of solute that was present in the original 50 mL sample.

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Why Volumetric Glassware Is Used

Precise dilution requires accurate measurement.

Common equipment includes:

• volumetric pipettes
• graduated pipettes
• burettes
• volumetric flasks

A beaker is useful for holding and mixing liquids but is generally not the best equipment for preparing a highly accurate final volume.

A volumetric flask is designed specifically for preparing solutions to a precise volume.

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Reading the Meniscus

When using a volumetric flask, the liquid surface usually forms a curved shape called a meniscus.

For many aqueous solutions, the volume is read from the bottom of the meniscus.

The eye should be level with the calibration mark.

Reading from above or below can cause parallax error.

Accurate volume measurement helps ensure that the final concentration is correct.

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Multi-Step Example: Preparing a Dilute Acid

A laboratory needs:

250 mL of 0.40 mol/L HCl

from:

2.0 mol/L HCl

Calculate the volume of stock solution required.

Use:

c₁V₁ = c₂V₂

2.0 × V₁ = 0.40 × 250

2.0V₁ = 100

V₁ = 50 mL

Therefore:

50 mL stock solution

is required.

The chemist transfers this amount and adds water until the total volume reaches 250 mL.

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Checking the Calculation Using Moles

We can verify the previous calculation.

Initial solution:

c = 2.0 mol/L

V = 50 mL = 0.050 L

Moles:

n = cV

n = 2.0 × 0.050

n = 0.100 mol

Final solution:

c = 0.40 mol/L

V = 250 mL = 0.250 L

Moles:

n = 0.40 × 0.250

n = 0.100 mol

Therefore:

moles before = moles after

The calculation is consistent.

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Dilution Using Mass Concentration

The same dilution principle can be used with mass concentration.

Suppose a solution has:

80 g/L

and:

100 mL

is diluted to:

400 mL

Use:

c₁V₁ = c₂V₂

80 × 100 = c₂ × 400

c₂ = 20 g/L

The concentration decreases from:

80 g/L → 20 g/L

The dilution equation works because the mass of solute remains unchanged.

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Multi-Step Example: Mass Concentration

A fruit drink concentrate contains:

120 g/L

of dissolved sugar.

A:

250 mL

sample is diluted to:

1.0 L

Calculate the final concentration.

Convert:

1.0 L = 1000 mL

Then:

120 × 250 = c₂ × 1000

c₂ = 30 g/L

Answer

Final concentration = 30 g/L

The final volume is four times larger, so the concentration becomes one-quarter of its original value.

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Dilution and Particle Models

At the particle level, dilution does not change the identity of the particles.

Before dilution:

• solute particles are relatively close together
• there are many solute particles per unit volume

After dilution:

• the same solute particles remain
• more solvent particles are present
• solute particles are spread through a larger volume
• fewer solute particles occur per unit volume

This is why concentration decreases.

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Dilution Does Not Mean Removing Solute

Suppose a solution contains:

0.20 mol NaCl

Adding water does not reduce this to:

0.10 mol NaCl

The solution still contains:

0.20 mol NaCl

Only the concentration changes.

For example:

Before:

0.20 mol in 0.20 L = 1.0 mol/L

After:

0.20 mol in 1.0 L = 0.20 mol/L

Same amount of solute.

Different concentration.

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Dilution vs. Removing Solution

These processes are different.

Suppose we have a well-mixed:

1.0 mol/L solution

If we simply pour half of it away, the remaining solution is still:

1.0 mol/L

Both solute and solvent were removed in the same proportion.

The concentration has not changed.

If instead we add water:

concentration decreases

This distinction is important.

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Dilution vs. Evaporation

Dilution:

add solvent → volume increases → concentration decreases

Evaporation:

remove solvent → volume decreases → concentration increases

They have opposite effects.

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Serial Dilution

Sometimes a solution must be diluted by a very large factor.

Instead of performing one enormous dilution, scientists may perform several smaller dilutions in sequence.

This is called a serial dilution.

For example:

Start:

1.0 mol/L

Perform a tenfold dilution:

0.10 mol/L

Dilute tenfold again:

0.010 mol/L

Dilute tenfold again:

0.0010 mol/L

Serial dilutions are common in:

• chemistry
• microbiology
• medicine
• biochemistry
• environmental testing

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Worked Example: Serial Dilution

A student starts with:

1.0 mol/L solution

They take:

10 mL

and dilute it to:

100 mL

First dilution:

1.0 × 10 = c₂ × 100

c₂ = 0.10 mol/L

They then take:

10 mL

of this new solution and dilute it again to:

100 mL

Second dilution:

0.10 × 10 = c₂ × 100

c₂ = 0.010 mol/L

The original solution has now undergone an overall:

100-fold dilution

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Dilution in Everyday Life

Dilution occurs in many everyday situations.

Examples include:

• adding water to juice concentrate
• mixing cleaning products according to instructions
• preparing fertilizers
• adding water to concentrated food products
• preparing laboratory reagents

In each case, adding solvent reduces the concentration of substances already present.

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Dilution in Medicine

Solutions used in medicine often need carefully controlled concentrations.

A concentrated preparation may need to be diluted before use.

The calculation must be accurate because concentration affects the amount of substance delivered in a particular volume.

This is one reason accurate measurements and careful laboratory procedures are essential.

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Dilution in Environmental Science

Environmental scientists may need to dilute samples before analysis.

For example, a water sample containing a high concentration of a dissolved substance may be too concentrated for an analytical instrument.

A known dilution can be performed.

The original concentration can then be calculated from:

• the measured concentration
• the dilution factor

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Working Backwards from a Diluted Sample

Suppose a sample is diluted by a factor of:

20

The diluted sample is measured as:

0.15 mg/L

Original concentration:

0.15 × 20

= 3.0 mg/L

Therefore:

original concentration = 3.0 mg/L

This technique is common in analytical chemistry.

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Laboratory Safety During Dilution

Some dilutions can release significant heat.

A particularly important example is diluting concentrated acids.

When diluting a strong concentrated acid, laboratory procedures typically require adding the acid carefully to water, rather than pouring water directly into concentrated acid.

This helps reduce the risk of rapid heating and splashing.

Appropriate:

• eye protection
• protective clothing
• laboratory procedures
• supervision

should always be used.

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Common Mistakes

Thinking Dilution Removes Solute

It does not.

The amount of solute remains the same during simple dilution.

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Confusing Final Volume with Solvent Added

If:

V₁ = 100 mL

and:

V₂ = 500 mL

the approximate amount of water added is:

500 − 100 = 400 mL

not 500 mL.

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Adding the Required Final Volume of Water

If the instructions say:

Dilute to 250 mL.

This means the total final solution volume should be 250 mL.

It does not mean:

Add 250 mL water.

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Mixing Up Initial and Final Values

Remember:

c₁ and V₁ = before dilution

c₂ and V₂ = after dilution

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Predicting That Concentration Increases

Dilution means adding solvent.

Therefore:

concentration must decrease

If your calculation gives a higher concentration after simple dilution, check your work.

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Assuming Moles Decrease

During simple dilution:

n₁ = n₂

Only concentration and volume change.

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Using Different Volume Units

In:

c₁V₁ = c₂V₂

both volumes must use the same units.

Do not use:

V₁ = 50 mL

and:

V₂ = 0.250 L

without first converting one of them.

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Confusing Dilution with Removing Half the Solution

Removing half of a well-mixed solution removes approximately half of both the solute and solvent.

Its concentration remains the same.

Adding solvent decreases concentration.

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Using a Beaker for Precise Dilution

Beakers are not designed for highly precise volume measurements.

Volumetric glassware is generally preferred when accurate concentrations are required.

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Key Terms

Dilution — The process of decreasing the concentration of a solution by adding solvent.

Dilute solution — A solution containing relatively little solute per unit volume.

Concentrated solution — A solution containing relatively more solute per unit volume.

Initial concentration (c₁) — The concentration before dilution.

Final concentration (c₂) — The concentration after dilution.

Initial volume (V₁) — The volume before dilution.

Final volume (V₂) — The total volume after dilution.

Dilution equation — The relationship c₁V₁ = c₂V₂.

Dilution factor — A measure of how many times a solution has been diluted, commonly calculated as final volume divided by initial volume.

Stock solution — A concentrated solution of known concentration used to prepare more dilute solutions.

Solute — The substance dissolved in a solution.

Solvent — The substance that dissolves the solute.

Volumetric flask — Laboratory glassware designed to contain an accurately specified volume.

Volumetric pipette — Laboratory equipment used to accurately transfer a specific volume of liquid.

Meniscus — The curved surface of a liquid in narrow laboratory glassware.

Serial dilution — A sequence of repeated dilution steps.

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Key Takeaways

• Dilution decreases the concentration of a solution by adding solvent.
• During simple dilution, the amount of solute remains unchanged.
• The total solution volume increases.
• Solute particles become distributed through a larger volume.
• The central dilution equation is:

c₁V₁ = c₂V₂

• The equation comes from conservation of the amount of solute.
• Initial concentration and volume are represented by c₁ and V₁.
• Final concentration and volume are represented by c₂ and V₂.
• During dilution:

V₂ > V₁

and normally:

c₂ < c₁

• The volume of solvent added is approximately:

V_solvent = V₂ − V₁

• Final volume is not the same as the amount of solvent added.
• A stock solution can be diluted to prepare solutions of lower concentration.
• Volumetric glassware improves the accuracy of laboratory dilutions.
• Serial dilution allows very low concentrations to be prepared accurately.
• Simply removing some well-mixed solution does not dilute what remains.
• Dilution and evaporation have opposite effects on concentration.

The central idea is:

SAME AMOUNT OF SOLUTE + MORE SOLVENT = LOWER CONCENTRATION

and the central calculation is:

c₁V₁ = c₂V₂

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Check Your Understanding

Understanding Dilution

1. Define dilution.

2. What happens to concentration when a solution is diluted?

3. What happens to the amount of solute during simple dilution?

4. What happens to the total volume during dilution?

5. Explain dilution using the particle model.

6. Explain why adding water to a salt solution does not remove any salt.

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Calculate Final Concentration

Use:

c₁V₁ = c₂V₂

7. 100 mL of 2.0 mol/L solution is diluted to 500 mL. Calculate the final concentration.

8. 50 mL of 3.0 mol/L solution is diluted to 300 mL. Calculate the final concentration.

9. 250 mL of 1.2 mol/L solution is diluted to 1.0 L. Calculate the final concentration.

10. 20 mL of 5.0 mol/L solution is diluted to 200 mL. Calculate the final concentration.

11. 400 mL of 0.80 mol/L solution is diluted to 800 mL. Calculate the final concentration.

12. 25 mL of 4.0 mol/L solution is diluted to 500 mL. Calculate the final concentration.

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Calculate Final Volume

13. 100 mL of 2.0 mol/L solution must be diluted to 0.50 mol/L. Calculate the required final volume.

14. 250 mL of 1.5 mol/L solution must be diluted to 0.50 mol/L. Calculate the required final volume.

15. 50 mL of 4.0 mol/L solution must be diluted to 0.40 mol/L. Calculate the final volume.

16. 200 mL of 0.90 mol/L solution must be diluted to 0.30 mol/L. Calculate the final volume.

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Calculate Solvent Required

17. A 100 mL solution must be diluted to a final volume of 500 mL. Approximately how much water must be added?

18. A 250 mL solution must be diluted to 1.0 L. Approximately how much water must be added?

19. 200 mL of 1.5 mol/L solution is diluted to 0.50 mol/L. Calculate the final volume and the approximate volume of water added.

20. 50 mL of 4.0 mol/L solution is diluted to 0.80 mol/L. Calculate the final volume and approximate amount of water added.

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Preparing Solutions from Stock Solutions

21. What volume of 2.0 mol/L stock solution is needed to prepare 500 mL of 0.20 mol/L solution?

22. What volume of 5.0 mol/L stock solution is required to prepare 250 mL of 1.0 mol/L solution?

23. What volume of 1.5 mol/L stock solution is needed to prepare 300 mL of 0.50 mol/L solution?

24. What volume of 4.0 mol/L stock solution is required to prepare 1.0 L of 0.20 mol/L solution?

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Laboratory Applications

25. Describe how you would prepare 250 mL of 0.20 mol/L solution from a 1.0 mol/L stock solution.

26. Why is a volumetric flask preferred over a beaker for preparing an accurately diluted solution?

27. Explain why the final volume should be read at eye level.

28. Explain the difference between "add 250 mL water" and "dilute to 250 mL."

29. Why should a solution be mixed thoroughly after dilution?

30. Explain why the amount of solute before and after dilution should be equal.

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Analysis and Problem Solving

31. A solution is diluted from 100 mL to 500 mL. By what factor has it been diluted?

32. A 3.0 mol/L solution undergoes a tenfold dilution. What is its new concentration?

33. A 0.80 mol/L solution has its volume doubled by adding solvent. Predict its new concentration without using the dilution equation.

34. A 1.2 mol/L solution has its volume tripled. Predict its new concentration.

35. A student removes half of a well-mixed 1.0 mol/L solution. What is the concentration of the solution remaining? Explain.

36. Another student adds an equal volume of water to a 1.0 mol/L solution. What happens to its concentration? Explain.

37. A solution is diluted from 2.0 mol/L to 0.25 mol/L. Determine the dilution factor.

38. A 10 mL sample of 1.0 mol/L solution is diluted to 100 mL. Then 10 mL of this new solution is diluted again to 100 mL. Calculate the final concentration.

39. An environmental sample is diluted by a factor of 25. The diluted sample has a measured concentration of 0.40 mg/L. Calculate the concentration of the original sample.

40. A chemist needs 500 mL of 0.10 mol/L solution and has a 2.0 mol/L stock solution. Calculate the volume of stock solution required, describe how the dilution should be carried out using appropriate laboratory glassware, and explain at the particle level why the concentration decreases.