What you'll learn
Acid–base chemistry is central to life, industry and the environment — from the acidity of soils used for Caribbean sugar cane and citrus, to the buffers that keep blood at a steady pH, to the treatment of acidic waste water. This guide covers the Brønsted–Lowry definitions of acids and bases, conjugate acid–base pairs, strong and weak acids and bases, the pH scale and calculations for strong acids, weak acids and strong bases, the ionic product of water, K_w, the acid dissociation constant, K_a, and pK_a, titration curves and the choice of indicators, and buffer solutions — how they work, how to calculate their pH, and where they are used.
CAPE acid–base questions are often calculation-heavy. You need confidence with logarithms, with deciding which formula applies to which situation, and with explaining buffer action using equilibrium ideas.
Key terms and definitions
Brønsted–Lowry acid — A proton (H⁺) donor.
Brønsted–Lowry base — A proton acceptor.
Conjugate acid–base pair — Two species that differ by one proton, such as CH₃COOH and CH₃COO⁻.
Strong acid — An acid that dissociates completely in water, such as HCl, HNO₃ and H₂SO₄.
Weak acid — An acid that dissociates only partially in water, such as ethanoic acid.
pH — −log₁₀[H⁺].
K_w — The ionic product of water: [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K.
K_a — The acid dissociation constant: [H⁺][A⁻] ÷ [HA].
pK_a — −log₁₀K_a; a smaller pK_a means a stronger acid.
Equivalence point — The point in a titration where the acid and base have reacted in exact stoichiometric amounts.
Indicator — A weak acid whose conjugate base has a different colour, used to show the end point.
Buffer solution — A solution that resists changes in pH when small amounts of acid or alkali are added.
Core concepts
Brønsted–Lowry theory
In the reaction HCl + H₂O → H₃O⁺ + Cl⁻, HCl donates a proton (acid) and water accepts it (base). Water can act as an acid or a base — it is amphoteric (amphiprotic).
In NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, ammonia accepts a proton (base) and water donates one (acid). The conjugate pairs are NH₃/NH₄⁺ and H₂O/OH⁻.
Strong and weak
Strength refers to the degree of dissociation, not concentration. A dilute strong acid is still fully dissociated; a concentrated weak acid is only partly dissociated.
- Strong acids: HCl, HNO₃, H₂SO₄ (first dissociation).
- Weak acids: ethanoic acid, carbonic acid, citric acid.
- Strong bases: NaOH, KOH.
- Weak bases: ammonia, amines.
pH of strong acids and strong bases
For a strong monoprotic acid, [H⁺] = concentration of the acid. pH = −log[H⁺].
For a strong base, find [OH⁻], then use K_w: [H⁺] = K_w ÷ [OH⁻], then pH = −log[H⁺]. (Alternatively, pOH = −log[OH⁻] and pH = 14 − pOH at 298 K.)
pH of weak acids
For a weak acid HA ⇌ H⁺ + A⁻:
K_a = [H⁺][A⁻] ÷ [HA]
Assuming [H⁺] = [A⁻] and that dissociation is so small that [HA] at equilibrium ≈ initial concentration:
[H⁺] = √(K_a × [HA])
K_w and neutrality
Water ionises slightly: H₂O ⇌ H⁺ + OH⁻. At 298 K, K_w = 1.0 × 10⁻¹⁴, so in pure water [H⁺] = 1.0 × 10⁻⁷ and pH = 7. Because the ionisation is endothermic, K_w increases with temperature, so the pH of pure water falls below 7 at higher temperatures — but it is still neutral because [H⁺] = [OH⁻].
Titration curves
| Titration | Starting pH | Equivalence point pH | Suitable indicator |
|---|---|---|---|
| Strong acid–strong base | ~1 | 7 | Phenolphthalein or methyl orange |
| Weak acid–strong base | ~3 | > 7 (about 8–9) | Phenolphthalein |
| Strong acid–weak base | ~1 | < 7 (about 5) | Methyl orange |
| Weak acid–weak base | — | no sharp change | No suitable indicator |
An indicator is suitable if its colour-change range lies entirely within the steep, vertical section of the titration curve. Methyl orange changes between about pH 3 and 4.5; phenolphthalein between about pH 8 and 10.
For weak acid–strong base titrations, the half-equivalence point has pH = pK_a, because [HA] = [A⁻] there. This allows K_a to be determined experimentally.
Buffer solutions
A buffer contains a weak acid and its conjugate base (such as ethanoic acid and sodium ethanoate), or a weak base and its conjugate acid (such as ammonia and ammonium chloride).
How an acidic buffer works: CH₃COOH ⇌ CH₃COO⁻ + H⁺
- When acid is added, the large reservoir of CH₃COO⁻ removes the extra H⁺, forming CH₃COOH; equilibrium shifts left and pH changes very little.
- When alkali is added, OH⁻ reacts with H⁺ (or directly with CH₃COOH); more CH₃COOH dissociates to replace H⁺, so pH changes very little.
Buffer pH:
[H⁺] = K_a × [HA] ÷ [A⁻], or pH = pK_a + log([A⁻] ÷ [HA])
Uses of buffers
- Blood is buffered at about pH 7.4, mainly by the carbonic acid–hydrogencarbonate system: H₂CO₃ ⇌ H⁺ + HCO₃⁻. Small changes in blood pH can be dangerous.
- Shampoos, cosmetics and foods contain buffers to keep pH stable.
- Fermentation and pharmaceutical processes need controlled pH.
- Calibrating pH meters uses buffer solutions of known pH.
Worked examples
Example 1: pH of a strong acid
Calculate the pH of 0.025 mol dm⁻³ HCl.
[H⁺] = 0.025. pH = −log(0.025) = 1.60.
Example 2: pH of a strong base
Calculate the pH of 0.010 mol dm⁻³ NaOH at 298 K.
[OH⁻] = 0.010. [H⁺] = 1.0 × 10⁻¹⁴ ÷ 0.010 = 1.0 × 10⁻¹². pH = 12.0.
Example 3: pH of a weak acid
Calculate the pH of 0.10 mol dm⁻³ ethanoic acid (K_a = 1.7 × 10⁻⁵ mol dm⁻³).
[H⁺] = √(1.7 × 10⁻⁵ × 0.10) = √(1.7 × 10⁻⁶) = 1.30 × 10⁻³. pH = 2.89.
Example 4: Buffer pH
A buffer contains 0.20 mol dm⁻³ ethanoic acid and 0.10 mol dm⁻³ sodium ethanoate (K_a = 1.7 × 10⁻⁵). Calculate its pH.
[H⁺] = 1.7 × 10⁻⁵ × 0.20 ÷ 0.10 = 3.4 × 10⁻⁵. pH = −log(3.4 × 10⁻⁵) = 4.47.
Example 5: Indicator choice
Which indicator should be used to titrate ethanoic acid with sodium hydroxide? Explain. (2 marks)
Phenolphthalein (1), because the equivalence point is above 7, and phenolphthalein's colour change range (about 8–10) falls within the steep part of the curve (1).
Example 6: pH after dilution
0.10 mol dm⁻³ hydrochloric acid is diluted 100 times with distilled water. Calculate the new pH.
The new concentration of H⁺ is 0.10 ÷ 100 = 1.0 × 10⁻³ mol dm⁻³, so pH = −log(1.0 × 10⁻³) = 3.00. Each tenfold dilution of a strong acid raises the pH by one unit. (Diluting a weak acid changes the pH less, because more of it dissociates as it is diluted.)
Example 7: pH of a mixture with excess strong acid
25.0 cm³ of 0.100 mol dm⁻³ HCl is mixed with 20.0 cm³ of 0.100 mol dm⁻³ NaOH. Calculate the pH of the mixture.
Moles H⁺ = 0.100 × 0.0250 = 2.50 × 10⁻³ mol. Moles OH⁻ = 0.100 × 0.0200 = 2.00 × 10⁻³ mol. Excess H⁺ = 5.0 × 10⁻⁴ mol in a total volume of 45.0 cm³ = 0.0450 dm³. [H⁺] = 5.0 × 10⁻⁴ ÷ 0.0450 = 0.0111 mol dm⁻³. pH = 1.95.
The key steps are to find which reagent is in excess, then divide by the total volume of the mixture.
Example 8: Determining K_a from pH
A 0.100 mol dm⁻³ solution of a weak monoprotic acid has a pH of 2.87. Calculate K_a.
[H⁺] = 10⁻²·⁸⁷ = 1.35 × 10⁻³ mol dm⁻³. Assuming [A⁻] = [H⁺] and [HA] ≈ 0.100: K_a = (1.35 × 10⁻³)² ÷ 0.100 = 1.82 × 10⁻⁵ mol dm⁻³ — close to the value for ethanoic acid.
Example 9: Buffer action in numbers
A buffer contains 0.20 mol dm⁻³ ethanoic acid and 0.10 mol dm⁻³ sodium ethanoate (K_a = 1.7 × 10⁻⁵ mol dm⁻³; pH 4.47). Enough HCl is added to provide 0.010 mol dm⁻³ of H⁺. Calculate the new pH.
The added H⁺ reacts with ethanoate ions: CH₃COO⁻ + H⁺ → CH₃COOH. New [CH₃COOH] = 0.20 + 0.010 = 0.21; new [CH₃COO⁻] = 0.10 − 0.010 = 0.09. [H⁺] = 1.7 × 10⁻⁵ × 0.21 ÷ 0.09 = 3.97 × 10⁻⁵. pH = 4.40.
The pH has fallen by only 0.07 units. Adding the same amount of acid to water would have changed the pH from 7 to 2 — a dramatic illustration of buffer action.
Choosing the right method
| Situation | Method |
|---|---|
| Strong acid | [H⁺] = concentration (× number of H⁺ for diprotic acids where fully dissociated) |
| Strong base | [OH⁻] = concentration; [H⁺] = K_w ÷ [OH⁻] |
| Weak acid alone | [H⁺] = √(K_a × [HA]) |
| Weak acid + its salt (buffer) | [H⁺] = K_a × [HA] ÷ [A⁻] |
| Strong acid + strong base mixture | find the excess, divide by total volume |
| Half-way through weak acid titration | pH = pK_a |
Deciding which row applies before starting is the most reliable way to avoid errors.
Common mistakes and how to avoid them
- Confusing strong with concentrated.
- Using the weak-acid formula for strong acids, or vice versa.
- Forgetting K_w when calculating the pH of bases.
- Saying pure water at 50 °C is acidic because its pH is below 7 — it is still neutral.
- Choosing an indicator whose range lies outside the vertical section.
- Explaining buffers without equilibrium shifts — say what the added H⁺ or OH⁻ reacts with.
Self-test questions
- Calculate the pH of 0.050 mol dm⁻³ sulfuric acid, assuming both protons dissociate fully. Answer: [H⁺] = 2 × 0.050 = 0.10 mol dm⁻³, so pH = 1.00.
- Calculate the pH of 0.0050 mol dm⁻³ barium hydroxide at 298 K. Answer: Each Ba(OH)₂ gives two OH⁻, so [OH⁻] = 0.010; [H⁺] = 1.0 × 10⁻¹⁴ ÷ 0.010 = 1.0 × 10⁻¹²; pH = 12.0.
- Identify the conjugate base of H₂PO₄⁻ and the conjugate acid of NH₃. Answer: HPO₄²⁻; NH₄⁺.
- Why does a mixture of hydrochloric acid and sodium chloride not act as a buffer? Answer: HCl is a strong acid, fully dissociated, so there is no reservoir of undissociated acid and no equilibrium to absorb added OH⁻; Cl⁻ is too weak a base to remove added H⁺.
How this links to your Internal Assessment
Acid–base titrations are among the most common CAPE SBAs. You might determine the concentration of ethanoic acid in vinegar, the purity of a commercial antacid, or K_a of a weak acid by measuring pH at the half-equivalence point. To score well, record burette readings to 0.05 cm³, obtain concordant titres, plot a titration curve if using a pH meter, and evaluate errors such as indicator judgement, parallax and the calibration of the pH meter. Investigating the buffering capacity of local drinks or soils can make an engaging, Caribbean-focused project.
Exam technique for Acids, bases, pH and buffers
- Decide first: strong acid, strong base, weak acid or buffer? Then choose the correct formula.
- Show logarithm steps and give pH to two decimal places unless told otherwise.
- For buffer explanations, write the equilibrium, then describe what happens when acid and alkali are added.
- When choosing indicators, refer explicitly to the pH range and the vertical section of the curve.
Quick revision summary
- Brønsted–Lowry: acids donate protons; bases accept them; conjugate pairs differ by one H⁺.
- Strong = fully dissociated; weak = partially dissociated.
- pH = −log[H⁺]; K_w = 1.0 × 10⁻¹⁴ at 298 K; bases: [H⁺] = K_w/[OH⁻].
- Weak acids: [H⁺] = √(K_a × [HA]); pK_a = −log K_a.
- Titration curves determine indicator choice; half-equivalence pH = pK_a.
- Buffers (weak acid + conjugate base) resist pH change; [H⁺] = K_a[HA]/[A⁻].
- Blood is buffered by H₂CO₃/HCO₃⁻ at about pH 7.4.