Neutralisation removes hydrogen ions by reacting them, and dilution only spreads them through more water. In neutralisation the amount of H⁺ in moles goes down. In dilution the amount stays the same and only the concentration changes. The difference decides which numbers you can use in a calculation.
This lesson builds on relating acidity to particles and belongs to acids, bases and salts.
How do the two changes differ?
| Neutralisation | Dilution | |
|---|---|---|
| What is added | A base or alkali | Water |
| Amount of H⁺ (mol) | Falls | Unchanged |
| Concentration of H⁺ | Falls | Falls |
| pH | Rises, can reach 7 or go above | Rises toward 7, never reaches it |
| New substance formed | Salt and water | None |
For an acid and an alkali the ionic equation is H⁺(aq) + OH⁻(aq) → H₂O(l). For hydrochloric acid and sodium hydroxide the full equation is HCl + NaOH → NaCl + H₂O.
The key formula for both
amount (mol) = concentration (mol/dm³) × volume (dm³)
Convert cm³ to dm³ by dividing by 1000. During dilution, the amount does not change, so you can find the new concentration from the new volume.
Worked example
Start with 25.0 cm³ of hydrochloric acid, 0.100 mol/dm³. The numbers are invented for teaching.
Part A: dilute to 250 cm³ in total.
Step 1: amount of HCl = 0.0250 × 0.100 = 0.00250 mol.
Step 2: the amount is unchanged, and the new volume is 250 cm³ = 0.250 dm³.
Step 3: new concentration = 0.00250 ÷ 0.250 = 0.0100 mol/dm³.
For a strong acid the pH moves from 1 to 2. The solution is still acidic.
Part B: neutralise the original 25.0 cm³ with sodium hydroxide, 0.100 mol/dm³.
Step 1: HCl + NaOH → NaCl + H₂O. The ratio is 1:1, so 0.00250 mol NaOH is needed.
Step 2: volume of NaOH = 0.00250 ÷ 0.100 = 0.0250 dm³ = 25.0 cm³.
Check: 25.0 cm³ × 0.100 mol/dm³ = 0.00250 mol, which equals the acid amount.
Part C: neutralise the diluted solution instead. Its amount of acid is still 0.00250 mol, so it still needs 25.0 cm³ of the same sodium hydroxide. Dilution did not change the amount of acid, so it did not change the amount of alkali needed.
The mistake to watch for
Mistaken answer: “After dilution to 250 cm³, the acid needs only a tenth as much alkali, because it is ten times weaker.”
The student used the concentration instead of the amount.
The correction is that a neutralisation needs the same moles of alkali as acid (for a 1:1 reaction), and dilution leaves the moles unchanged. The larger volume simply holds them at a lower concentration. Always convert to moles before comparing.
A second slip is to say that diluting an acid “neutralises it”. Only a base can do that.
Check yourself
1. 50.0 cm³ of 0.200 mol/dm³ HCl is diluted with water to 200 cm³. Find the new concentration.
Show answer
Amount = 0.0500 × 0.200 = 0.0100 mol. New concentration = 0.0100 ÷ 0.200 = 0.0500 mol/dm³.
2. A strong acid has pH 2. It is diluted 100 times by volume. What is the new pH in the simple model?
Show answer
Dilution by 100 lowers the H⁺ concentration by a factor of 100, which raises the pH by 2. The new pH is 4, and the solution is still acidic.
3. Find the volume of 0.100 mol/dm³ NaOH needed to neutralise 20.0 cm³ of 0.0500 mol/dm³ HCl.
Show answer
HCl amount = 0.0200 × 0.0500 = 0.00100 mol. The ratio is 1:1, so NaOH amount = 0.00100 mol. Volume = 0.00100 ÷ 0.100 = 0.0100 dm³ = 10.0 cm³.
Where this leads next
Once you can separate the two changes, see how the products of neutralisation are prepared in choosing a salt-preparation principle. The mole and equation-ratio tutor helps with the amounts, and the equation balance reasoning trainer checks the equations. Then test yourself on the practice set.
When the wording of a question changes and the method is no longer obvious, individual teaching can help. See online one-to-one Chemistry tuition.