When you add water to a solution, the moles of solute stay the same and the volume increases, so the concentration decreases. The formula M₁V₁ = M₂V₂ is “moles before = moles after” written in short form.
This lesson is part of acid-base reasoning beyond memorised labels in SPM Chemistry. If units still cause trouble, read calculating acid-base quantities first.
Why is M₁V₁ = M₂V₂ true?
Moles of solute = concentration × volume. Before dilution, moles = M₁ × V₁. After dilution, the same moles are in a new volume, so moles = M₂ × V₂.
Since the moles are identical, M₁V₁ = M₂V₂. The volumes can stay in cm³ if both use the same unit, because the conversion cancels on both sides.
Worked example: tracking the moles
Question. 25.0 cm³ of 2.0 mol dm⁻³ hydrochloric acid is diluted to a total volume of 250 cm³. Find the new concentration and check the moles.
Moles before: 0.0250 dm³ × 2.0 mol dm⁻³ = 0.050 mol.
New concentration: 0.050 mol ÷ 0.250 dm³ = 0.20 mol dm⁻³. The volume increased tenfold, so the concentration fell to one tenth.
Moles after: 0.250 dm³ × 0.20 mol dm⁻³ = 0.050 mol. The moles match, so the answer is consistent.
How much water do I add?
Question. You need 500 cm³ of 0.10 mol dm⁻³ acid from a 1.0 mol dm⁻³ stock. Find the volume of stock to use.
Moles needed = 0.500 dm³ × 0.10 mol dm⁻³ = 0.050 mol. Volume of stock = 0.050 ÷ 1.0 = 0.050 dm³, which is 50 cm³.
In the laboratory, put the 50 cm³ in a 500 cm³ volumetric flask and add water to the mark. The water added is about 450 cm³, but the mark, not the subtraction, fixes the final volume.
What happens to pH when I dilute?
A strong acid keeps ionising completely, so a tenfold dilution cuts the hydrogen ion concentration to one tenth. The pH rises by one unit, for example from 1 to 2.
A weak acid also becomes more dilute, but it ionises a little more as it is diluted, so its pH rises by less than one unit. In both cases the acid strength does not change, only the concentration.
The mistake to avoid
The common mistake is to say “diluting halves the moles” when the volume doubles. The moles are constant, and the concentration is halved.
| Claim | Right or wrong |
|---|---|
| Doubling the volume halves the concentration | Right |
| Doubling the volume halves the moles of solute | Wrong, the moles are unchanged |
| Dilution makes the acid weaker | Wrong, it makes the acid more dilute |
The concentration and dilution tutor lets you see the moles stay fixed while the volume changes.
Check yourself
100 cm³ of 0.60 mol dm⁻³ sulfuric acid has 200 cm³ of water added. Find the new concentration and state the moles of acid before and after.
Answer
Moles before = 0.100 × 0.60 = 0.060 mol. The final volume is 100 + 200 = 300 cm³ = 0.300 dm³, assuming the volumes add.
New concentration = 0.060 ÷ 0.300 = 0.20 mol dm⁻³. Moles after = 0.300 × 0.20 = 0.060 mol, the same as before.
What to study next
Go on to reconciling a pH description with an explicitly stated classroom model. Then test all four ideas in the integrated practice set.
If you want a teacher to ask you to justify each step, see online one-to-one Chemistry tuition.