This chemistry grades 11-12 lesson on acids pH buffers notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on acids pH buffers and check yourself with the quiz.
Quick answer: A Brønsted–Lowry acid donates a proton and a base accepts one. Strong acids ionise fully, so pH = −log[H⁺]; weak acids ionise partly, so [H⁺] ≈ √(Ka × c). A buffer, made of a weak acid and its conjugate base, resists pH change, and its pH is pKa + log([A⁻]/[HA]).
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB R3.1
Acids pH buffers notes: the key ideas
What do acid, base and conjugate pair mean?
Chemists describe acid–base reactions as proton transfer. Also, a proton here is an H⁺ ion, which in water is attached to a water molecule as H₃O⁺ and is usually written simply as H⁺(aq).
A Brønsted–Lowry acid is a proton donor. A Brønsted–Lowry base is a proton acceptor. When an acid loses H⁺ it becomes its conjugate base; when a base gains H⁺ it becomes its conjugate acid.
Strong acids such as HCl and HNO₃ ionise almost completely in water. Weak acids such as ethanoic acid ionise only slightly, so most of the acid stays as molecules in an equilibrium: CH₃COOH ⇌ CH₃COO⁻ + H⁺.
How do you calculate pH and pOH?
The pH scale turns very small concentrations into manageable numbers. Water itself ionises a little, and its equilibrium constant is the ionic product of water, Kw.
pH = −log₁₀[H⁺] and [H⁺] = 10⁻ᵖᴴ
Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol²/dm⁶ at 25 °C
pH + pOH = 14 at 25 °C
[H⁺] is the hydrogen ion concentration in mol/dm³.
Aarav in Pune dilutes hydrochloric acid until it is 0.0080 mol/dm³. Find the pH.
HCl is strong, so [H⁺] = 0.0080 mol/dm³.
pH = −log(0.0080) = 2.10
How do you find the pH of a weak acid?
A weak acid HA sets up the equilibrium HA ⇌ H⁺ + A⁻. Its strength is measured by the acid dissociation constant Ka.
Ka = [H⁺][A⁻] / [HA] and pKa = −log₁₀ Ka
For a weak acid of concentration c: [H⁺] ≈ √(Ka × c)
Ka is in mol/dm³; c is the starting concentration in mol/dm³.
The shortcut assumes two things: the acid ionises so little that [HA] ≈ c, and [H⁺] = [A⁻] because water’s own H⁺ is negligible. In addition, it is reliable when less than about 5 % of the acid has ionised. A smaller pKa means a stronger acid.
| Acid | Ka (mol/dm³) | pKa |
|---|---|---|
| Chloroethanoic acid | 1.4 × 10⁻³ | 2.85 |
| Methanoic acid | 1.6 × 10⁻⁴ | 3.80 |
| Benzoic acid | 6.3 × 10⁻⁵ | 4.20 |
| Ethanoic acid | 1.8 × 10⁻⁵ | 4.74 |
| Hydrocyanic acid | 4.9 × 10⁻¹⁰ | 9.31 |
Find the pH of 0.100 mol/dm³ ethanoic acid (Ka = 1.8 × 10⁻⁵ mol/dm³).
[H⁺] = √(1.8 × 10⁻⁵ × 0.100) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ mol/dm³
pH = −log(1.34 × 10⁻³) = 2.87
Check: 1.34 × 10⁻³ ÷ 0.100 = 1.3 % ionised, so the approximation holds.
What is a buffer and how does it work?
A buffer is a solution that resists a change in pH when small amounts of acid or alkali are added. Moreover, it contains a weak acid HA together with a large amount of its conjugate base A⁻ (or a weak base with its conjugate acid).
Added H⁺ is removed by A⁻, forming HA. Added OH⁻ is removed by HA, forming A⁻ and water. Because both HA and A⁻ are plentiful, their ratio changes only slightly, so the pH hardly moves. The pH of a buffer follows the Henderson–Hasselbalch equation.
pH = pKa + log₁₀([A⁻] / [HA])
[A⁻] and [HA] are the concentrations of the conjugate base and the weak acid in mol/dm³. In one solution the mole ratio can be used instead.
A buffer holds 0.20 mol/dm³ ethanoic acid and 0.10 mol/dm³ sodium ethanoate.
pH = 4.74 + log(0.10 ÷ 0.20) = 4.74 − 0.30 = 4.44
What does a titration curve of a weak acid show?
When a weak acid is titrated with a strong alkali, the curve begins at a higher pH than a strong acid of the same concentration. A buffer region appears while HA and A⁻ coexist. Exactly half-way to the equivalence point, [HA] = [A⁻], so log(1) = 0 and pH = pKa. At equivalence only the weak base A⁻ remains, so the pH is above 7.
Do not use [H⁺] = √(Ka × c) for a strong acid, and do not trust it for very dilute solutions. Also, a buffer needs both HA and A⁻ present: adding exactly equal moles of a weak acid and a strong alkali gives only a salt solution, not a buffer.
Whenever a question says half-neutralised, equal moles of acid and conjugate base, or half-equivalence point, write pH = pKa straight away. Similarly, the logarithm term is zero.
- pH = −log₁₀[H⁺]; at 25 °C pH + pOH = 14 and Kw = 1.0 × 10⁻¹⁴.
- Weak acid: Ka = [H⁺][A⁻]/[HA]; [H⁺] ≈ √(Ka × c) when ionisation is under about 5 %.
- Buffer: pH = pKa + log([A⁻]/[HA]); A⁻ removes added H⁺ and HA removes added OH⁻.
- Half-equivalence point of a weak acid titration: pH = pKa.
- At the equivalence point of a weak acid with a strong base the pH is above 7.
Strong and weak acids, pH and buffer solutions: frequently asked questions
What is the difference between a strong and a weak acid?
A strong acid ionises completely in water, so [H⁺] equals its concentration. A weak acid ionises only partly and reaches an equilibrium described by Ka. Strength is about how much it ionises, not how concentrated it is.
How do you calculate the pH of a weak acid?
Write Ka = [H⁺][A⁻]/[HA], assume [H⁺] = [A⁻] and [HA] ≈ c, so [H⁺] = √(Ka × c). Then pH = −log[H⁺]. Check that less than 5 % of the acid has ionised.
How does a buffer resist pH change?
It contains a weak acid and its conjugate base. The base reacts with added H⁺ and the acid reacts with added OH⁻, so the ratio [A⁻]/[HA] changes only a little and the pH stays almost constant.
What is the Henderson–Hasselbalch equation?
It is pH = pKa + log([A⁻]/[HA]). Also, it gives the pH of a buffer from the pKa of the weak acid and the ratio of conjugate base to acid. It works best for ratios between 0.1 and 10.
Why is the equivalence point of a weak acid titration above pH 7?
At equivalence all the weak acid has become its conjugate base A⁻. A⁻ is a weak base that reacts with water to form OH⁻, so the solution is slightly alkaline.
