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When you dilute a strong acid, you increase the volume of solution without changing the number of moles of acid. Since pH depends on [H⁺], dilution changes the pH.
Worked Example: 50.0 cm³ of 0.10 mol dm⁻³ HCl is diluted to 500.0 cm³. Calculate the new pH.
Moles of HCl = 0.10 × (50.0/1000) = 5.00 × 10⁻³ mol
New concentration = 5.00 × 10⁻³ / (500.0/1000) = 0.010 mol dm⁻³
Since HCl fully dissociates, [H⁺] = 0.010 mol dm⁻³.
pH = -log₁₀(0.010) = 2.00
The original pH was -log₁₀(0.10) = 1.00, so the tenfold dilution increased the pH by exactly 1 unit. This is always the case for strong acids: a tenfold dilution raises pH by 1.
Why exactly 1 unit? If the original concentration is c, the diluted concentration is c/10. pH changes from -log(c) to -log(c/10) = -log(c) + log(10) = -log(c) + 1. The difference is always exactly 1.
The same principle applies to bases, but you work through Kw.
Worked Example: 25.0 cm³ of 0.20 mol dm⁻³ NaOH is diluted to 250.0 cm³. Calculate the new pH at 25°C.
Moles of NaOH = 0.20 × (25.0/1000) = 5.00 × 10⁻³ mol
New [NaOH] = 5.00 × 10⁻³ / (250.0/1000) = 0.020 mol dm⁻³
[OH⁻] = 0.020 mol dm⁻³ (full dissociation)
[H⁺] = Kw / [OH⁻] = (1.00 × 10⁻¹⁴) / 0.020 = 5.00 × 10⁻¹³ mol dm⁻³
pH = -log₁₀(5.00 × 10⁻¹³) = 12.30
For strong bases, a tenfold dilution decreases the pH by exactly 1 unit (the pH moves towards 7).
Diluting a weak acid does NOT increase the pH by exactly 1 unit per tenfold dilution. This is because dilution changes the degree of dissociation.
Worked Example: 0.10 mol dm⁻³ ethanoic acid (Ka = 1.74 × 10⁻⁵) is diluted tenfold to 0.010 mol dm⁻³.
Before dilution: [H⁺] = √(1.74 × 10⁻⁵ × 0.10) = 1.32 × 10⁻³, pH = 2.88 After dilution: [H⁺] = √(1.74 × 10⁻⁵ × 0.010) = 4.17 × 10⁻⁴, pH = 3.38
The pH changed by only 0.50 units, not 1. For a weak acid, each tenfold dilution increases pH by approximately 0.5 units (because pH = ½(pKa + pc) and pc increases by 1, so pH increases by 0.5).
No matter how much you dilute an acid, the pH can never exceed 7 (at 25°C). Even if you dilute HCl so much that [H⁺] from the acid becomes negligible, the autoionisation of water still provides [H⁺] = 1.00 × 10⁻⁷ mol dm⁻³, giving pH = 7.00. You cannot make an acidic solution alkaline by adding water.
Similarly, diluting a base cannot produce a pH below 7 at 25°C.
This is one of the most common calculation types at A-Level. The strategy is always the same:
flowchart TD
A[Calculate moles of H⁺] --> C{Compare moles}
B[Calculate moles of OH⁻] --> C
C --> D[H⁺ in excess]
C --> E[OH⁻ in excess]
C --> F[Exact neutralisation]
D --> G["[H⁺] = excess moles / total volume"]
E --> H["[OH⁻] = excess moles / total volume"]
F --> I["pH = 7.00 at 25°C"]
G --> J["pH = -log[H⁺]"]
H --> K["[H⁺] = Kw / [OH⁻]"]
K --> J
Worked Example: 30.0 cm³ of 0.10 mol dm⁻³ HCl is mixed with 20.0 cm³ of 0.10 mol dm⁻³ NaOH. Calculate the pH.
Moles of H⁺ = 0.10 × (30.0/1000) = 3.00 × 10⁻³ mol
Moles of OH⁻ = 0.10 × (20.0/1000) = 2.00 × 10⁻³ mol
H⁺ is in excess. Excess moles of H⁺ = 3.00 × 10⁻³ - 2.00 × 10⁻³ = 1.00 × 10⁻³ mol
Total volume = 30.0 + 20.0 = 50.0 cm³ = 0.0500 dm³
[H⁺] = 1.00 × 10⁻³ / 0.0500 = 0.0200 mol dm⁻³
pH = -log₁₀(0.0200) = 1.70
Worked Example: 20.0 cm³ of 0.15 mol dm⁻³ NaOH is mixed with 25.0 cm³ of 0.080 mol dm⁻³ HCl. Calculate the pH at 25°C.
Moles of OH⁻ = 0.15 × (20.0/1000) = 3.00 × 10⁻³ mol
Moles of H⁺ = 0.080 × (25.0/1000) = 2.00 × 10⁻³ mol
OH⁻ is in excess. Excess moles of OH⁻ = 3.00 × 10⁻³ - 2.00 × 10⁻³ = 1.00 × 10⁻³ mol
Total volume = 20.0 + 25.0 = 45.0 cm³ = 0.0450 dm³
[OH⁻] = 1.00 × 10⁻³ / 0.0450 = 0.0222 mol dm⁻³
[H⁺] = Kw / [OH⁻] = (1.00 × 10⁻¹⁴) / 0.0222 = 4.50 × 10⁻¹³ mol dm⁻³
pH = -log₁₀(4.50 × 10⁻¹³) = 12.35
If you mix two strong acids, simply find the total moles of H⁺ and divide by total volume.
Worked Example: 10.0 cm³ of 0.30 mol dm⁻³ HCl is mixed with 20.0 cm³ of 0.15 mol dm⁻³ HNO₃.
Moles of H⁺ from HCl = 0.30 × (10.0/1000) = 3.00 × 10⁻³ mol Moles of H⁺ from HNO₃ = 0.15 × (20.0/1000) = 3.00 × 10⁻³ mol Total H⁺ = 6.00 × 10⁻³ mol
Total volume = 30.0 cm³ = 0.0300 dm³
[H⁺] = 6.00 × 10⁻³ / 0.0300 = 0.200 mol dm⁻³
pH = -log₁₀(0.200) = 0.70
When a strong base is added to a weak acid in less than stoichiometric amount, the result is a buffer solution. This connects to Lessons 5 and 6.
Worked Example: 50.0 cm³ of 0.20 mol dm⁻³ CH₃COOH is mixed with 25.0 cm³ of 0.10 mol dm⁻³ NaOH. Ka = 1.74 × 10⁻⁵.
Moles CH₃COOH = 0.20 × 0.0500 = 0.010 mol Moles NaOH = 0.10 × 0.0250 = 0.0025 mol
The NaOH reacts with some CH₃COOH: CH₃COOH + NaOH → CH₃COONa + H₂O
Remaining CH₃COOH = 0.010 - 0.0025 = 0.0075 mol Produced CH₃COO⁻ = 0.0025 mol
This is a buffer. Using Henderson-Hasselbalch: pH = pKa + log₁₀(0.0025 / 0.0075) = 4.76 + log₁₀(0.333) = 4.76 - 0.48 = 4.28
A common student mistake is to average pH values when mixing two acids. pH is a logarithmic scale, so averaging pH values gives the wrong answer.
Example: 25 cm³ of pH 1 acid mixed with 25 cm³ of pH 3 acid does NOT give pH 2.
Correct method:
The pH 1 acid dominates because it has 100 times more H⁺ ions.
The key to these problems is careful bookkeeping: track moles (not concentrations) through the reaction, then convert back to concentration using the total volume. Always ask: "Which species is in excess?" and work from there.
Common errors include:
| Solution type | Effect of 10x dilution on pH |
|---|---|
| Strong acid | pH increases by exactly 1.00 |
| Strong base | pH decreases by exactly 1.00 |
| Weak acid | pH increases by approximately 0.50 |
| Weak base | pH decreases by approximately 0.50 |
| Buffer | pH barely changes (ratio stays the same) |
Edexcel 9CH0 specification, Topic 12: Acid-base equilibria, sub-strand 12.3 requires you to calculate pH following dilution of a strong acid or base; following partial neutralisation of a strong acid by a strong base (excess of either reactant); following partial neutralisation of a weak acid by a strong base (forming a buffer mid-titration); and following the mixing of two strong acid or two strong base solutions of different concentrations (refer to the official specification document for exact wording). Stoichiometric reasoning (moles in / moles consumed / moles remaining) underpins every calculation; the species remaining determines whether the post-mix pH is set by [H⁺]_excess (strong acid case), [OH⁻]_excess (strong base case), or the Ka equilibrium (weak-acid + strong-base partial-neutralisation case). Examined in Paper 2 (Topic 12 numerical) and Paper 3 (synoptic + practical).
Question (8 marks):
(a) 25.0 cm³ of 0.10 mol dm⁻³ aqueous NaOH is added to 50.0 cm³ of 0.10 mol dm⁻³ aqueous HCl. Calculate the pH of the resulting solution at 298 K. (4)
(b) 50.0 cm³ of 0.10 mol dm⁻³ aqueous HCl is diluted to 5.00 dm³ with distilled water. Calculate the pH of the diluted solution. (2)
(c) 25.0 cm³ of 0.10 mol dm⁻³ aqueous NaOH is added to 50.0 cm³ of 0.10 mol dm⁻³ aqueous CH₃COOH (Ka = 1.74 × 10⁻⁵ mol dm⁻³). Calculate the pH of the resulting solution. (2)
Solution with mark scheme:
(a) Moles HCl = 0.050 × 0.10 = 5.00 × 10⁻³ mol.
Moles NaOH = 0.025 × 0.10 = 2.50 × 10⁻³ mol.
Reaction: HCl + NaOH → NaCl + H₂O. NaOH is the limiting reagent.
Moles HCl remaining = 5.00 × 10⁻³ − 2.50 × 10⁻³ = 2.50 × 10⁻³ mol.
Total volume = 75.0 cm³ = 0.0750 dm³.
[HCl]_remaining = 2.50 × 10⁻³ / 0.0750 = 0.0333 mol dm⁻³.
[H⁺] = 0.0333 mol dm⁻³ (strong acid, full dissociation).
pH = −log₁₀(0.0333) = 1.48.
M1 — moles of each reactant correct. M1 — limiting-reagent identification and moles of excess HCl. M1 — concentration in combined volume. A1 — pH = 1.48.
(b) Moles HCl = 5.00 × 10⁻³ mol; new volume = 5.00 dm³.
[HCl]_diluted = 5.00 × 10⁻³ / 5.00 = 1.00 × 10⁻³ mol dm⁻³.
pH = −log₁₀(1.00 × 10⁻³) = 3.00.
M1 — concentration after dilution. A1 — pH = 3.00.
(c) Moles CH₃COOH = 5.00 × 10⁻³; moles NaOH = 2.50 × 10⁻³.
NaOH neutralises half the ethanoic acid → 2.50 × 10⁻³ mol CH₃COO⁻ formed and 2.50 × 10⁻³ mol CH₃COOH remaining. Equal moles of weak acid and conjugate base in the same solution → buffer at half-equivalence, so pH = pKa = 4.76.
M1 — recognises half-equivalence buffer condition. A1 — pH = pKa = 4.76.
Total: 8 marks (M5 A3, split as shown).
Question (6 marks): A laboratory technician mixes 20.0 cm³ of 0.20 mol dm⁻³ HCl with 30.0 cm³ of 0.10 mol dm⁻³ HNO₃.
(a) Calculate the [H⁺] in the resulting mixture. (2)
(b) Calculate the pH. (1)
(c) The mixture is then diluted to 500 cm³ by adding distilled water. Calculate the new pH. (3)
Mark scheme decomposition by AO:
(a)
(b)
(c)
Total: 6 marks split AO1 = 2, AO2 = 4. A clean stoichiometry-then-pH structure typical of Paper 2.
Connects to:
Topic 5 — Stoichiometry: every mixing/dilution problem reduces to mole bookkeeping. Confidence with moles / volumes / concentrations from Year 1 is the prerequisite.
Topic 12.4 — Buffers (Lesson 5): a weak acid partially neutralised by a strong base creates a buffer. Lesson 4 sets up the half-equivalence pH = pKa observation that Lesson 5 extends to the full Henderson-Hasselbalch equation.
Topic 12.5 — Titration curves (Lesson 8): every point on a titration curve is a "mixture" calculation. Lesson 4's stoichiometry routine, repeated at multiple titrant volumes, generates the curve.
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