1. Brønsted-Lowry Theory, pH & Conjugate Pairs
In quantitative chemistry, acid-base behaviour is governed by proton transfer:
- Acid: A proton (H⁺) donor.
- Base: A proton (H⁺) acceptor.
- Conjugate Acid-Base Pair: Two species that transform into each other by the gain or loss of a single proton.
Example: In CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺, CH₃COOH is the acid and CH₃COO⁻ is its conjugate base; H₂O is the base and H₃O⁺ is its conjugate acid.
pH = −log₁₀[H⁺]
[H⁺] = 10−pH
2. The Ionic Product of Water (Kw) & High-Temperature Neutrality
Water undergoes a slight self-ionisation (autoprotolysis):
H₂O(l) ⇌ H⁺(aq) + OH⁻(aq) [ΔH > 0, endothermic]
Because the concentration of undissociated water is immense (~55.5 mol dm⁻³) and effectively constant, it is incorporated into the equilibrium constant to give the ionic product of water (Kw):
Kw = [H⁺][OH⁻]
At standard temperature (298 K / 25 °C):
Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ => pKw = −log₁₀(Kw) = 14.00
Because the dissociation of water is endothermic (ΔH > 0), increasing temperature shifts the equilibrium to the right according to Le Chatelier's principle. Consequently, Kw increases:
- At 25 °C: Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶; [H⁺] = 1.00 × 10⁻⁷ mol dm⁻³; pH = 7.00.
- At 50 °C: Kw rises to 5.48 × 10⁻¹⁴ mol² dm⁻⁶; [H⁺] = √(Kw) = 2.34 × 10⁻⁷ mol dm⁻³; pH = 6.63.
Crucial Examination Point: Even though the pH drops below 7 at 50 °C, water remains strictly neutral because [H⁺] is strictly equal to [OH⁻]. Neutrality does NOT mean pH = 7; neutrality means [H⁺] = [OH⁻]!
pH of a Strong Base Using Kw
A strong base such as NaOH dissociates completely, so [OH⁻] = [base]. Use Kw to find [H⁺], then the pH.
Step 1: [OH⁻] = 0.0500 mol dm⁻³
Step 2: [H⁺] = Kw / [OH⁻] = 1.00 × 10⁻¹⁴ / 0.0500 = 2.00 × 10⁻¹³ mol dm⁻³
Step 3: pH = −log₁₀(2.00 × 10⁻¹³) = 12.70
Kw changes with temperature, so at a different temperature the same NaOH solution has a different pH. Always use the Kw value given for that temperature.
3. Weak Acids, Ka & pKa Calculations
Strong monoprotic acids (HCl, HNO₃) dissociate completely in aqueous solution, so [H⁺] = [acid]. Sulfuric acid is diprotic: for a simple A-level answer, take [H⁺] = 2 × [H₂SO₄]. For example, 0.0500 mol dm⁻³ H₂SO₄ gives [H⁺] = 0.100 mol dm⁻³ and pH = 1.00.
Weak acids (CH₃COOH, HCOOH, HCN) dissociate only partially:
HA(aq) ⇌ H⁺(aq) + A⁻(aq)
The acid dissociation constant (Ka) is:
Ka = [H⁺][A⁻] / [HA] and pKa = −log₁₀(Ka)
A higher Ka (or lower pKa) indicates a stronger weak acid that dissociates to a greater extent.
Converting between Ka and pKa: pKa = −log₁₀Ka and Ka = 10−pKa. For ethanoic acid, Ka = 1.74 × 10⁻⁵ mol dm⁻³ gives pKa = 4.76. Going back, 10−4.76 = 1.74 × 10⁻⁵ mol dm⁻³.
Standard Approximations for Weak Acid pH Calculations
- Equimolar dissociation: We assume all H⁺ ions come solely from the weak acid (ignoring the negligible contribution from water dissociation): [H⁺] = [A⁻].
- Negligible dissociation: Because the acid is weak, the amount dissociated is tiny compared to starting concentration: [HA]eq = [HA]initial.
Substituting these approximations gives:
Ka = [H⁺]² / [HA] => [H⁺] = √(Ka × [HA])
4. Buffer Solutions & Mechanism of Action
A buffer solution is a chemical system that resists changes in pH when small amounts of acid (H⁺) or base (OH⁻) are added.
- Acidic Buffer (pH < 7): Formed from a weak acid and a salt of its conjugate base (e.g. ethanoic acid, CH₃COOH, and sodium ethanoate, CH₃COONa).
- Basic Buffer (pH > 7): Formed from a weak base and a salt of its conjugate acid (e.g. ammonia, NH₃, and ammonium chloride, NH₄Cl).
Mechanism of Acidic Buffer Action
An acidic buffer contains a large reservoir of undissociated weak acid (HA) and a large reservoir of conjugate base ions (A⁻):
- When a small amount of H⁺ is added: The added protons react with the large reservoir of conjugate base ions: A⁻ + H⁺ → HA. The position of equilibrium shifts to the left, removing added H⁺ and keeping pH almost constant.
- When a small amount of OH⁻ is added: The added hydroxide reacts with the large reservoir of undissociated acid: HA + OH⁻ → A⁻ + H₂O. Protons are donated to neutralise OH⁻, keeping pH almost constant.
In both cases, the amounts of HA and A⁻ are large compared with the small amount of acid or alkali added. So the ratio [HA]/[A⁻] hardly changes, and the pH hardly changes. For ethanoic acid / sodium ethanoate: added H⁺ reacts with CH₃COO⁻, and added OH⁻ reacts with CH₃COOH.
Mechanism of Basic Buffer Action
A basic buffer, such as ammonia and ammonium chloride (NH₃ / NH₄Cl), contains a large amount of NH₃ and a large amount of NH₄⁺:
- When a small amount of H⁺ is added: it reacts with ammonia. NH₃ + H⁺ → NH₄⁺
- When a small amount of OH⁻ is added: it reacts with ammonium ions. NH₄⁺ + OH⁻ → NH₃ + H₂O
The ratio [NH₃]/[NH₄⁺] hardly changes, so the pH stays almost constant. You only need to explain basic buffers. You do not need to calculate their pH.
Effect of Dilution
Diluting a buffer with water lowers [HA] and [A⁻] by the same factor. The ratio [HA]/[A⁻] does not change, so [H⁺] = Ka × [HA]/[A⁻] stays the same. The pH stays about the same.
Applications of Buffers
- Blood: a carbonic acid / hydrogencarbonate buffer (H₂CO₃ / HCO₃⁻) keeps blood pH close to 7.4.
- Shampoos: buffered so they are not too alkaline for hair and skin.
- Food: buffers keep the pH of foods and drinks steady, which helps preservation and taste.
- Biological and industrial processes: enzymes and many reactions, such as fermentation, work only in a narrow pH range.
The Henderson-Hasselbalch Equation
Beyond the OxfordAQA specification: the Henderson-Hasselbalch equation is a shortcut for the pH of a buffer.
pH = pKa + log₁₀([A⁻] / [HA]) = pKa + log₁₀(moles of salt / moles of acid)
This equation is fine to use, but the method OxfordAQA expects is to rearrange Ka = [H⁺][A⁻] / [HA]:
[H⁺] = Ka × [HA] / [A⁻], then pH = −log₁₀[H⁺]
Both methods give the same answer. Worked Example 2 in Section 7 shows both.
5. pH Titration Curves
A pH curve plots the pH against the volume of base added. The four curves below are for 25.0 cm³ of 0.100 mol dm⁻³ acid with 0.100 mol dm⁻³ base, so the equivalence point is at 25.0 cm³ each time. The shaded bands show the indicator ranges.
- Strong acid + strong base: long vertical section (about pH 3 to 11). Equivalence point about pH 7.
- Weak acid + strong base: starts higher (about pH 3) with a buffer region. Vertical section about pH 7 to 11. Equivalence point above 7.
- Strong acid + weak base: vertical section about pH 3 to 7. Equivalence point below 7.
- Weak acid + weak base: no sharp vertical section, only a gradual change of slope near pH 7.
- Adding acid to base: if the base is in the flask and acid is added from the burette, each curve is flipped. It starts at high pH and falls.
| Titration Combination | Initial pH | Vertical Inflection Range | pH at Equivalence | Suitable Indicator |
|---|---|---|---|---|
| Strong Acid - Strong Base (HCl + NaOH) | ~1 | pH 3.0 to 11.0 (long vertical section) | 7.0 | Both Methyl Orange & Phenolphthalein |
| Weak Acid - Strong Base (CH₃COOH + NaOH) | ~3 | pH 7.0 to 11.0 (vertical in alkaline region) | > 7.0 (~8.7) | Phenolphthalein ONLY |
| Strong Acid - Weak Base (HCl + NH₃) | ~1 | pH 3.0 to 7.0 (vertical in acidic region) | < 7.0 (~5.3) | Methyl Orange ONLY |
| Weak Acid - Weak Base (CH₃COOH + NH₃) | ~3 | No sharp vertical section (gradual change of slope only) | ~7.0 | No indicator is suitable (use a pH meter) |
6. Indicator Selection & Half-Neutralisation Determination of Ka
- Phenolphthalein (range 8.3 to 10.0): use for a weak acid with a strong base.
- Methyl orange (range 3.1 to 4.4): use for a strong acid with a weak base.
- Strong acid with strong base: either indicator works, because the vertical section covers both ranges.
- Weak acid with weak base: no indicator is suitable, because there is no sharp vertical section.
RP7 covers both halves of this practical: a weak acid titrated with a strong base, and a strong acid titrated with a weak base. You record the pH after each addition, plot the curve and find the equivalence point. See the method and practice questions on the RP7 page.
Determining Ka from the Half-Neutralisation Point
During a weak acid titration, at the half-neutralisation point (halfway to equivalence volume, e.g. at 12.5 cm³ if equivalence is 25.0 cm³):
[HA] = [A⁻]
Substituting into Ka = ([H⁺][A⁻]) / [HA]:
Ka = [H⁺] => pH = pKa
Therefore, reading the pH at exactly half the equivalence volume directly gives the pKa of the weak acid!
7. Worked Calculations
Step 1: Apply weak acid approximation:
[H⁺] = √(Ka × [HA])
[H⁺] = √((1.74 × 10⁻⁵) × 0.150) = √(2.61 × 10⁻⁶) = 1.616 × 10⁻³ mol dm⁻³
Step 2: Calculate pH:
pH = −log₁₀(1.616 × 10⁻³) = 2.79
Calculate the pH of this buffer solution.
Step 1: Calculate moles of acid and salt:
Moles of HCOOH = 0.500 dm³ × 0.200 mol dm⁻³ = 0.100 mol
Moles of HCOO⁻ = 0.500 dm³ × 0.100 mol dm⁻³ = 0.050 mol
Step 2: Calculate pKa:
pKa = −log₁₀(1.78 × 10⁻⁴) = 3.75
Step 3: Apply the Henderson-Hasselbalch equation (beyond the OxfordAQA specification):
pH = pKa + log₁₀(moles of salt / moles of acid)
pH = 3.75 + log₁₀(0.050 / 0.100) = 3.75 + log₁₀(0.50) = 3.75 + (−0.301) = 3.45
The Ka method (expected by OxfordAQA): both are in the same total volume, so the mole ratio equals the concentration ratio.
[H⁺] = Ka × [HA] / [A⁻] = 1.78 × 10⁻⁴ × 0.100 / 0.050 = 3.56 × 10⁻⁴ mol dm⁻³
pH = −log₁₀(3.56 × 10⁻⁴) = 3.45
Step 1: Moles of each:
Moles of H⁺ = 0.0250 dm³ × 0.100 = 0.00250 mol
Moles of OH⁻ = 0.0200 dm³ × 0.100 = 0.00200 mol
Step 2: Excess H⁺ (H⁺ + OH⁻ → H₂O):
0.00250 − 0.00200 = 5.00 × 10⁻⁴ mol
Step 3: Divide by the total volume (25.0 + 20.0 = 45.0 cm³ = 0.0450 dm³):
[H⁺] = 5.00 × 10⁻⁴ / 0.0450 = 0.0111 mol dm⁻³
Step 4: pH = −log₁₀(0.0111) = 1.95
The most common mistake is to forget the total volume. If the base is in excess instead, find [OH⁻] the same way, then use Kw to get [H⁺].
Go deeper
These UK AQA pages cover the same ideas in more detail. They may include content that is not in 9620.
Exam-Style Practice Questions
Test your understanding of these core syllabus concepts with targeted questions and detailed explanations.
Question 1: At 60 °C, Kw is 9.60 × 10⁻¹⁴ mol² dm⁻⁶. What is the pH of pure water at this temperature, and is the water acidic, basic, or neutral?
Show Answer & Explanation
Correct Answer: C
[H⁺] = √(Kw) = √(9.60 × 10⁻¹⁴) = 3.10 × 10⁻⁷ mol dm⁻³. pH = −log₁₀(3.10 × 10⁻⁷) = 6.51. The water remains neutral because [H⁺] = [OH⁻].
Question 2: Which indicator is most suitable for titrating ethanoic acid (weak acid) with sodium hydroxide (strong base)?
Show Answer & Explanation
Correct Answer: C
A weak acid - strong base titration curve has its vertical inflection in the alkaline region (pH 7 to 11). Phenolphthalein's working range (8.3 to 10.0) lies entirely within this vertical section.
Question 3: In a titration of 25.0 cm³ of a weak monoprotic acid with 0.10 mol dm⁻³ NaOH, equivalence is reached at 30.0 cm³. At what volume of added NaOH does the solution pH equal the pKa of the acid?
Show Answer & Explanation
Correct Answer: C
pH equals pKa at the half-neutralisation point. Halfway to the 30.0 cm³ equivalence point is 15.0 cm³.
Question 4: What happens to the pH of an acidic buffer when a small volume of hydrochloric acid is added?
Show Answer & Explanation
Correct Answer: C
The large reservoir of conjugate base ions (A⁻) reacts with added protons (A⁻ + H⁺ → HA), preventing any major change in [H⁺] and keeping pH virtually constant.
Question 5 (2 marks): Calculate the pH of 0.0200 mol dm⁻³ potassium hydroxide at 298 K. Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶.
Show Answer & Mark Scheme
Answer
[OH⁻] = 0.0200 mol dm⁻³
[H⁺] = 1.00 × 10⁻¹⁴ / 0.0200 = 5.00 × 10⁻¹³ mol dm⁻³
pH = −log₁₀(5.00 × 10⁻¹³) = 12.30
Mark points
- [H⁺] = Kw / [OH⁻] = 5.00 × 10⁻¹³ mol dm⁻³
- pH = 12.30 (2 d.p.)
Question 6 (4 marks): 25.0 cm³ of 0.100 mol dm⁻³ HCl is mixed with 20.0 cm³ of 0.150 mol dm⁻³ NaOH. Calculate the pH of the mixture at 298 K. Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶.
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Answer
Moles of H⁺ = 0.0250 × 0.100 = 0.00250 mol; moles of OH⁻ = 0.0200 × 0.150 = 0.00300 mol.
OH⁻ is in excess: 0.00300 − 0.00250 = 5.00 × 10⁻⁴ mol.
[OH⁻] = 5.00 × 10⁻⁴ / 0.0450 = 0.0111 mol dm⁻³
[H⁺] = 1.00 × 10⁻¹⁴ / 0.0111 = 9.00 × 10⁻¹³ mol dm⁻³
pH = 12.05
Mark points
- Excess OH⁻ = 5.00 × 10⁻⁴ mol
- [OH⁻] = 0.0111 mol dm⁻³ (total volume 45.0 cm³ used)
- [H⁺] = Kw / [OH⁻] = 9.00 × 10⁻¹³ mol dm⁻³
- pH = 12.05
Question 7 (3 marks): A buffer contains 0.100 mol of ethanoic acid and 0.0500 mol of sodium ethanoate in 1.00 dm³ of solution. Ka = 1.74 × 10⁻⁵ mol dm⁻³. Calculate the pH of the buffer.
Show Answer & Mark Scheme
Answer
Ka = [H⁺][A⁻] / [HA], so [H⁺] = Ka × [HA] / [A⁻]
[H⁺] = 1.74 × 10⁻⁵ × 0.100 / 0.0500 = 3.48 × 10⁻⁵ mol dm⁻³
pH = −log₁₀(3.48 × 10⁻⁵) = 4.46
Mark points
- Rearranged Ka expression (or [HA] and [A⁻] used correctly)
- [H⁺] = 3.48 × 10⁻⁵ mol dm⁻³
- pH = 4.46
Question 8 (4 marks): A buffer is made from ammonia and ammonium chloride. Explain, with equations, how this buffer keeps its pH almost constant when a small amount of acid or a small amount of alkali is added.
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Answer
The buffer contains large amounts of NH₃ and NH₄⁺.
Added acid: NH₃ + H⁺ → NH₄⁺, so the added H⁺ is removed.
Added alkali: NH₄⁺ + OH⁻ → NH₃ + H₂O, so the added OH⁻ is removed.
The ratio [NH₃]/[NH₄⁺] hardly changes, so the pH stays almost constant.
Mark points
- Large amounts of NH₃ and NH₄⁺ present
- NH₃ + H⁺ → NH₄⁺
- NH₄⁺ + OH⁻ → NH₃ + H₂O
- Ratio of NH₃ to NH₄⁺ hardly changes, so pH hardly changes
Question 9 (2 marks): Hydrochloric acid is titrated with aqueous ammonia. Explain why methyl orange (range 3.1 to 4.4) is a suitable indicator but phenolphthalein (range 8.3 to 10.0) is not.
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Answer
For a strong acid with a weak base, the vertical section of the pH curve is about pH 3 to 7. Methyl orange changes colour within this range, so it changes at the equivalence point. Phenolphthalein would change only at pH 8.3 to 10.0, which is after the equivalence point, where the curve is no longer steep.
Mark points
- Vertical section is in the acidic region (about pH 3 to 7)
- Methyl orange range lies inside it; phenolphthalein range lies outside it (above the vertical section)