What Is Dilution?

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

For an aqueous solution, the solvent being added is usually water.

During dilution:

• More solvent is added.
• The total volume of solution increases.
• The amount of solute remains the same.
• The concentration decreases.

The key idea is:

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

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Concentrated and Dilute Solutions

A concentrated solution contains a relatively large amount of solute per unit volume.

A dilute solution contains a relatively small amount of solute per unit volume.

Suppose 0.20 mol of salt is dissolved in:

0.50 dm³ of solution

The concentration is:

0.40 mol/dm³

If enough water is added to increase the total volume to:

1.00 dm³

the same 0.20 mol is now spread through twice the volume.

The new concentration is:

0.20 mol/dm³

The concentration has been halved.

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What Happens to the Particles?

The particle model helps explain dilution.

Before dilution, solute particles are relatively close together because they are distributed through a smaller volume.

After solvent is added:

• The number of solute particles stays the same.
• The number of solvent particles increases.
• The solute particles become more widely dispersed.
• There are fewer solute particles in each unit volume.

This is why concentration decreases.

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What Does Not Change During Dilution?

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

If no solute is added or removed:

moles of solute before dilution = moles of solute after dilution

For example, if a solution initially contains:

0.10 mol NaCl

then after adding water it still contains:

0.10 mol NaCl

The volume changes.

The concentration changes.

But the amount of NaCl does not.

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Concentration, Moles, and Volume

Recall the relationship:

moles = concentration × volume

or:

n = c × V

During dilution, the moles of solute remain constant.

Therefore:

initial moles = final moles

So:

initial concentration × initial volume = final concentration × final volume

This gives the dilution equation:

c₁V₁ = c₂V₂

where:

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

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Understanding the Dilution Equation

The equation:

c₁V₁ = c₂V₂

works because both sides represent the same amount of solute.

Before dilution:

n = c₁V₁

After dilution:

n = c₂V₂

Since no solute has been added or removed:

c₁V₁ = c₂V₂

This is not simply a formula to memorize. It follows directly from conservation of the amount of solute.

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Worked Example: Basic Dilution

100 cm³ of a 2.0 mol/dm³ solution is diluted to a final volume of 500 cm³.

Calculate the final concentration.

Use:

c₁V₁ = c₂V₂

Substitute:

2.0 × 100 = c₂ × 500

Rearrange:

c₂ = (2.0 × 100) ÷ 500

c₂ = 0.40 mol/dm³

Therefore, the diluted solution has a concentration of:

0.40 mol/dm³

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Do the Volumes Always Need to Be Converted?

For the dilution equation:

c₁V₁ = c₂V₂

the two volumes can both be in cm³ because the same volume units appear on both sides.

For example:

2.0 × 100 = c₂ × 500

works correctly.

However, if you use:

n = cV

and concentration is in mol/dm³, the volume must be in dm³.

This distinction is important.

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Worked Example Using Moles

Let's solve the previous problem another way.

Initial solution:

c = 2.0 mol/dm³

V = 100 cm³ = 0.100 dm³

Calculate moles:

n = c × V

n = 2.0 × 0.100

n = 0.20 mol

After dilution:

V = 500 cm³ = 0.500 dm³

The amount remains:

0.20 mol

Therefore:

c = n ÷ V

c = 0.20 ÷ 0.500

c = 0.40 mol/dm³

This gives the same answer.

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Rearranging the Dilution Equation

Starting with:

c₁V₁ = c₂V₂

we can rearrange it depending on what we need.

To find final concentration:

c₂ = (c₁ × V₁) ÷ V₂

To find initial concentration:

c₁ = (c₂ × V₂) ÷ V₁

To find final volume:

V₂ = (c₁ × V₁) ÷ c₂

To find the required initial volume:

V₁ = (c₂ × V₂) ÷ c₁

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Worked Example: Finding Final Volume

A student has 50 cm³ of a 4.0 mol/dm³ solution.

To what final volume must it be diluted to produce a 1.0 mol/dm³ solution?

Use:

c₁V₁ = c₂V₂

Substitute:

4.0 × 50 = 1.0 × V₂

Therefore:

V₂ = 200 cm³

The final solution volume must be:

200 cm³

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Final Volume Is Not Water Added

This is a very common source of mistakes.

In the previous example, the solution must be diluted to 200 cm³.

The student already has:

50 cm³

Therefore, approximately:

200 - 50 = 150 cm³

of additional water is required.

So:

Final volume = 200 cm³

but:

Water added = 150 cm³

These are not the same thing.

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Worked Example: Finding the Initial Volume

A chemist wants to prepare 250 cm³ of a 0.20 mol/dm³ solution from a 1.0 mol/dm³ stock solution.

Use:

c₁V₁ = c₂V₂

Substitute:

1.0 × V₁ = 0.20 × 250

Therefore:

V₁ = 50 cm³

The chemist needs:

50 cm³ of the stock solution

This is then diluted until the total volume reaches:

250 cm³

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Stock Solutions

A stock solution is a relatively concentrated solution used to prepare more dilute solutions.

For example, a laboratory might keep:

2.0 mol/dm³ NaCl

as a stock solution.

Smaller concentrations can then be prepared by taking measured volumes of the stock solution and adding water.

This is often easier and more accurate than preparing every solution separately from solid chemicals.

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Practical Dilution

A common laboratory dilution uses:

• A pipette.
• A volumetric flask.
• Distilled or deionized water.

A typical procedure is:

• Measure a precise volume of stock solution using a pipette.
• Transfer it to a volumetric flask.
• Add some distilled water.
• Mix.
• Add more water until close to the calibration line.
• Carefully add the final drops until the bottom of the meniscus reaches the line.
• Stopper the flask.
• Mix thoroughly.

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Reading the Meniscus

Liquid surfaces often form a curved surface called a meniscus.

For many aqueous solutions, volume should be read at the bottom of the meniscus.

The observer's eye should be level with the calibration mark.

Viewing from above or below can produce a parallax error.

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Worked Practical Example

A student needs to prepare:

250 cm³ of 0.10 mol/dm³ NaCl

from:

1.0 mol/dm³ NaCl

Calculate the required volume of stock solution.

Use:

c₁V₁ = c₂V₂

1.0 × V₁ = 0.10 × 250

V₁ = 25 cm³

The student should:

• Measure 25 cm³ of the stock solution.
• Transfer it to a 250 cm³ volumetric flask.
• Add distilled water.
• Fill carefully to the 250 cm³ mark.
• Stopper and mix thoroughly.

The final concentration is:

0.10 mol/dm³

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Dilution Factor

The dilution factor describes how many times more dilute the final solution is.

For example:

100 cm³ is diluted to 500 cm³.

The volume has increased by a factor of:

500 ÷ 100 = 5

Therefore, the solution has undergone a:

5-fold dilution

Its concentration becomes:

1/5 of the original concentration

If the original concentration was:

2.0 mol/dm³

the new concentration is:

2.0 ÷ 5 = 0.40 mol/dm³

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Worked Example: Dilution Factor

20 cm³ of solution is diluted to 200 cm³.

Dilution factor:

200 ÷ 20 = 10

Therefore, the concentration becomes one-tenth of its original value.

If:

initial concentration = 3.0 mol/dm³

then:

final concentration = 3.0 ÷ 10

final concentration = 0.30 mol/dm³

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Doubling the Volume

Suppose a solution has:

100 cm³

and its volume is increased to:

200 cm³

The volume has doubled.

Since the amount of solute remains constant:

the concentration is halved.

For example:

0.80 mol/dm³ → 0.40 mol/dm³

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Tripling the Volume

Suppose the volume changes from:

100 cm³ → 300 cm³

The volume has tripled.

Therefore, the concentration becomes one-third of its original value.

For example:

0.90 mol/dm³ → 0.30 mol/dm³

This illustrates the inverse relationship between concentration and volume during dilution.

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Concentration and Volume

During dilution:

more volume → lower concentration

provided the amount of solute remains constant.

If the volume doubles:

concentration halves

If the volume triples:

concentration becomes one-third

If the volume increases five times:

concentration becomes one-fifth

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Particle Model Example

Imagine 100 solute particles inside a container.

Before Dilution

Volume = 100 cm³

There are many solute particles within each small region.

After Dilution

Volume = 500 cm³

There are still exactly 100 solute particles.

However, they are distributed throughout a much larger volume.

Therefore, each unit volume contains fewer solute particles.

This is the particle-level explanation for the decrease in concentration.

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Dilution Does Not Destroy Solute

Suppose a solution contains:

0.25 mol glucose

Water is added.

After dilution, the solution still contains:

0.25 mol glucose

The glucose has not:

• Disappeared.
• Reacted away.
• Been destroyed.
• Changed into water.

It has simply become more widely dispersed through the solution.

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Dilution and Mass Concentration

The same principle can be applied when concentration is measured in:

g/dm³

Suppose a solution contains 10 g of solute.

Initially:

Volume = 0.20 dm³

concentration = 10 ÷ 0.20 = 50 g/dm³

After dilution:

Volume = 1.0 dm³

concentration = 10 ÷ 1.0 = 10 g/dm³

The mass of solute remains:

10 g

but the concentration decreases.

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Worked Example with Mass Concentration

100 cm³ of a 60 g/dm³ sugar solution is diluted to 300 cm³.

Use:

c₁V₁ = c₂V₂

60 × 100 = c₂ × 300

c₂ = 20 g/dm³

Therefore:

final concentration = 20 g/dm³

The same dilution relationship works because the amount of solute remains constant.

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Serial Dilution

Sometimes a very dilute solution is prepared through several dilution steps.

This is called a serial dilution.

For example:

Start with:

1.0 mol/dm³

Perform a tenfold dilution:

0.10 mol/dm³

Dilute tenfold again:

0.010 mol/dm³

Dilute tenfold again:

0.0010 mol/dm³

Serial dilution is widely used in chemistry, biology, microbiology, and medicine.

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Worked Serial Dilution

A student begins with a solution of:

2.0 mol/dm³

Each dilution decreases the concentration by a factor of 10.

After the first dilution:

0.20 mol/dm³

After the second:

0.020 mol/dm³

After the third:

0.0020 mol/dm³

The overall dilution factor is:

10 × 10 × 10 = 1000

Therefore:

2.0 ÷ 1000 = 0.0020 mol/dm³

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Dilution in Chemical Experiments

Dilution is important because many experiments require specific concentrations.

Concentration can affect:

• Reaction rate.
• Titration results.
• Chemical equilibrium.
• Biological responses.
• Color intensity.
• Conductivity.

Accurate dilution allows scientists to control concentration while keeping other experimental conditions consistent.

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Dilution and Reaction Rate

A more concentrated solution generally contains more reacting particles per unit volume.

This can increase the frequency of collisions between reactant particles.

Dilution reduces the number of solute particles per unit volume.

As a result, dilution can reduce the rate of some chemical reactions.

However, the exact effect depends on the reaction involved.

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Dilution and Color

Some colored solutions become visibly paler when diluted.

This occurs because there are fewer colored solute particles per unit volume.

The total number of solute particles has not necessarily decreased.

They are simply distributed through a larger volume.

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Multi-Step Problem

A student has 40 cm³ of a 1.5 mol/dm³ solution.

Water is added until the total volume is 300 cm³.

Calculate the final concentration.

Use:

c₁V₁ = c₂V₂

1.5 × 40 = c₂ × 300

60 = 300c₂

c₂ = 0.20 mol/dm³

Therefore:

final concentration = 0.20 mol/dm³

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Challenge Problem

A chemist needs 500 cm³ of a 0.15 mol/dm³ solution.

The available stock solution is 2.5 mol/dm³.

Calculate the volume of stock solution required.

Use:

c₁V₁ = c₂V₂

2.5 × V₁ = 0.15 × 500

2.5V₁ = 75

V₁ = 30 cm³

Therefore:

30 cm³ of stock solution is required.

The chemist transfers this to a 500 cm³ volumetric flask and adds water until the final volume reaches:

500 cm³

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How Much Water Is Added?

From the previous example:

Initial stock volume:

30 cm³

Final solution volume:

500 cm³

Approximate water added:

500 - 30 = 470 cm³

However, in accurate laboratory work, the chemist would normally not simply measure 470 cm³ of water separately.

Instead, water is added until the total solution volume reaches the calibration mark.

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Checking Whether an Answer Makes Sense

Dilution means concentration must decrease.

Suppose:

Initial concentration = 2.0 mol/dm³

After adding water, you calculate:

Final concentration = 5.0 mol/dm³

Something is wrong.

Adding solvent cannot increase concentration if no solute is added.

A useful check is:

After dilution: final concentration < initial concentration

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Common Mistakes

Thinking Dilution Removes Solute

Dilution adds solvent.

The amount of solute remains unchanged.

Thinking the Number of Solute Particles Decreases

The particles become more spread out, but their number remains the same.

Confusing Final Volume With Volume of Water Added

If 50 cm³ is diluted to 200 cm³, the final volume is 200 cm³.

Approximately 150 cm³ of water has been added.

Adding the Volumes in the Equation Incorrectly

In:

c₁V₁ = c₂V₂

V₂ is the final total volume, not the amount of solvent added.

Forgetting Unit Consistency

If V₁ is in cm³, V₂ should also be in cm³ when using the dilution equation directly.

Using c₁V₁ = c₂V₂ When Solute Has Reacted

The equation assumes the amount of solute remains unchanged.

If the solute reacts or is removed, the simple dilution equation may not apply.

Thinking a Paler Solution Contains No Solute

A diluted solution may appear much paler but still contain the same original amount of solute.

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Check Your Understanding

1. Define dilution.

2. What happens to concentration when solvent is added?

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

4. Explain dilution using the particle model.

5. Write the dilution equation.

6. What do c₁, V₁, c₂, and V₂ represent?

7. 100 cm³ of a 1.0 mol/dm³ solution is diluted to 400 cm³. Calculate the final concentration.

8. 50 cm³ of a 2.0 mol/dm³ solution is diluted to 250 cm³. Calculate the final concentration.

9. What final volume is required to dilute 100 cm³ of a 3.0 mol/dm³ solution to 0.50 mol/dm³?

10. How much water is approximately added in Question 9?

11. A chemist wants 500 cm³ of 0.20 mol/dm³ solution from a 2.0 mol/dm³ stock solution. Calculate the volume of stock required.

12. Explain the difference between final solution volume and volume of solvent added.

13. What is a stock solution?

14. Explain why a volumetric flask is useful when preparing diluted solutions.

15. What is a serial dilution?

16. A solution undergoes a fivefold dilution. What happens to its concentration?

17. A solution is diluted from 100 cm³ to 1000 cm³. What is the dilution factor?

18. Explain why a colored solution may become paler after dilution.

19. Explain why dilution can reduce the rate of some chemical reactions.

20. A student dilutes a 0.50 mol/dm³ solution with water and calculates a final concentration of 0.80 mol/dm³. Explain why the answer cannot be correct.

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Key Terms

• Dilution – process of decreasing concentration by adding solvent.
• Concentrated solution – solution containing a relatively large amount of solute per unit volume.
• Dilute solution – solution containing a relatively small amount of solute per unit volume.
• Stock solution – concentrated solution used to prepare more dilute solutions.
• Dilution factor – factor by which a solution's concentration has been reduced.
• Serial dilution – series of repeated dilution steps.
• Solute – substance dissolved in a solvent.
• Solvent – substance in which a solute dissolves.
• Molar concentration – amount of solute in moles per unit volume of solution.
• Volumetric flask – laboratory glassware designed to contain an accurate fixed volume.
• Pipette – laboratory equipment used to transfer an accurately measured volume.
• Meniscus – curved surface of a liquid in a container.

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Key Takeaways

• Dilution decreases the concentration of a solution by adding solvent.
• The amount of solute remains unchanged during simple dilution.
• The number of solute particles therefore remains unchanged.
• Added solvent increases the total solution volume.
• Solute particles become more widely dispersed.
• Fewer solute particles are present per unit volume.
• This explains why concentration decreases.
• The dilution equation is c₁V₁ = c₂V₂.
• The equation works because the amount of solute before and after dilution is the same.
• If volume increases by a certain factor, concentration decreases by the same factor.
• Doubling the volume halves the concentration.
• A fivefold increase in volume produces a fivefold decrease in concentration.
• Final solution volume is not the same as the volume of solvent added.
• Stock solutions can be diluted to prepare solutions of lower concentration.
• Pipettes and volumetric flasks allow accurate laboratory dilutions.
• Serial dilution produces progressively lower concentrations.
• Dilution can be understood mathematically and using the particle model.
• A correctly diluted solution must have a lower concentration than the original solution.