amazon.co.ukAcids, Bases and Buffers
A* revision for Brønsted-Lowry acids and bases, pH calculations, Ka and pKa, buffers and titration curves, with worked examples and practice questions for AQA, Edexcel, OCR, WJEC and CCEA.
🎯 Learning Objectives
- Apply Brønsted-Lowry theory, identify conjugate pairs and amphoteric species
- Calculate pH of strong acids and strong bases using Kw, including diprotic acids and Ba(OH)₂
- Calculate weak acid pH and Ka from pH data using the √(Ka·[HA]) approximation and state its validity
- Explain buffer action with equations and calculate buffer pH using [H⁺] = Ka([HA]/[A⁻])
- Describe blood buffering by the carbonic acid/hydrogencarbonate system
- Interpret pH titration curves, justify indicator choice and determine Ka from the half-equivalence point
📋 Key Definitions and Core Concepts
Brønsted-Lowry acid: A proton (H⁺) donor, e.g. HCl, CH₃COOH.
Brønsted-Lowry base: A proton (H⁺) acceptor, e.g. NH₃, OH⁻.
Conjugate pair: Two species differing by one H⁺, e.g. CH₃COOH/CH₃COO⁻; the stronger the acid, the weaker its conjugate base.
Kw (ionic product of water): Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K; increases with temperature as ionisation is endothermic.
Ka (acid dissociation constant): Ka = [H⁺][A⁻]/[HA] for a weak acid HA ⇌ H⁺ + A⁻; larger Ka = stronger acid.
pKa: −log₁₀ Ka; smaller pKa = stronger acid. At half-equivalence pH = pKa.
Buffer solution: A solution that maintains approximately constant pH when small quantities of acid or base are added; a weak acid plus its conjugate base.
Equivalence point: The volume at which acid and base have reacted exactly — the near-vertical section of the pH curve.
Amphoteric species: A species that can act as both an acid and a base, e.g. H₂O, HCO₃⁻.
Brønsted-Lowry Acids and Bases, pH and Kw
Brønsted-Lowry: an acid is a proton (H⁺) donor, a base is a proton acceptor. In the equilibrium CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺, ethanoic acid donates a proton and water accepts it. Conjugate pairs differ by one H⁺: CH₃COOH/CH₃COO⁻ and H₃O⁺/H₂O. An amphoteric species can act as both — water donates to bases (H₂O → OH⁻) and accepts from acids (H₂O → H₃O⁺). Strong acids (HCl, HNO₃, H₂SO₄) dissociate completely; weak acids (ethanoic, carbonic) dissociate partially — the ⇌ arrow distinguishes them.
pH = −log₁₀[H⁺]; [H⁺] = 10⁻ᵖʰ. Strong acid calculations: [H⁺] equals the acid concentration (for HCl) or twice it (H₂SO₄ gives 2H⁺ — check whether the question intends full dissociation of both protons). Strong bases: [OH⁻] equals concentration for NaOH, twice for Ba(OH)₂. Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K — the ionic product of water. Convert [OH⁻] to [H⁺] via Kw, then to pH. Worked example: pH of 0.050 mol dm⁻³ Ba(OH)₂ at 298 K: [OH⁻] = 2 × 0.050 = 0.100; [H⁺] = 1.0 × 10⁻¹⁴/0.100 = 1.0 × 10⁻¹³; pH = 13.00.
Water's ionisation (2H₂O ⇌ H₃O⁺ + OH⁻) is endothermic, so Kw increases with temperature and the pH of pure water falls below 7 at higher temperature — water stays neutral because [H⁺] = [OH⁻]; neutrality is defined by equality, not by pH 7. This distinction is a classic A* point: pH 7 is neutral only at 298 K.
Further pH calculation examples to rehearse: (1) pH of 0.010 mol dm⁻³ HCl: [H⁺] = 0.010, pH = 2.00. (2) pH of 0.0050 mol dm⁻³ H₂SO₄ (assume full dissociation): [H⁺] = 0.010, pH = 2.00 — the factor of 2 matters; check whether the question specifies both protons dissociate. (3) dilution: diluting 0.100 mol dm⁻³ HCl by ×100 gives 0.00100 mol dm⁻³, pH rises from 1.00 to 3.00 — each ×10 dilution raises pH by 1 for strong acids. (4) [H⁺] from pH: pH 4.20 → [H⁺] = 10⁻⁴·²⁰ = 6.3 × 10⁻⁵. For strong bases, the two-step Kw route is always required — writing pH = 14 + log[OH⁻] without the Kw step loses the working marks; state Kw = 1.0 × 10⁻¹⁴ explicitly and quote units (mol² dm⁻⁶). At temperatures other than 298 K the question gives the new Kw — use it as given; do not assume 1.0 × 10⁻¹⁴.
Weak Acids, Ka and pKa
The acid dissociation constant for HA ⇌ H⁺ + A⁻: Ka = [H⁺][A⁻]/[HA]; pKa = −log₁₀ Ka. Larger Ka (smaller pKa) = stronger acid. For a weak acid, the approximation [H⁺] = [A⁻] (dissociation is tiny compared with [HA]) gives [H⁺] ≈ √(Ka × [HA]), so pH = −log₁₀[H⁺] and Ka ≈ [H⁺]²/[HA].
Worked examples: (1) pH of 0.100 mol dm⁻³ ethanoic acid (Ka = 1.74 × 10⁻⁵): [H⁺] = √(1.74 × 10⁻⁵ × 0.100) = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³; pH = 2.88. (2) Ka from pH: 0.150 mol dm⁻³ propanoic acid has pH 2.82: [H⁺] = 10⁻²·⁸² = 1.51 × 10⁻³; Ka = (1.51 × 10⁻³)²/0.150 = 1.52 × 10⁻⁵.
Approximation validity — the standard 4-marker: two approximations are made. (1) [H⁺] from water is negligible (Kw tiny) — valid for all but very weak/very dilute acids. (2) [HA] at equilibrium ≈ [HA] initial — valid only when dissociation is small: if [H⁺] > 5% of [HA], the approximation fails and the quadratic must be solved properly. Very dilute weak acids (e.g. 0.001 mol dm⁻³ ethanoic) violate approximation (2) — state the check: (5 × [H⁺]/[HA]) < 5%. Exam answers that quote and check the approximation earn the evaluation marks that A* requires.
Buffer Solutions: Action and pH Calculations
A buffer solution maintains approximately constant pH when small quantities of acid or base are added. An acidic buffer is made from a weak acid plus its conjugate base — either a salt (e.g. CH₃COOH + CH₃COONa) or partial neutralisation (excess weak acid + NaOH). The buffer contains large, roughly equal reservoirs of HA and A⁻, with the equilibrium HA ⇌ H⁺ + A⁻ pushed far left by the added A⁻.
Buffer action — the mechanism in words: when acid is added, the H⁺ reacts with the conjugate base reservoir: CH₃COO⁻ + H⁺ → CH₃COOH — equilibrium shifts left, removing most added H⁺, so [H⁺] (and pH) barely change. When base is added, the OH⁻ reacts with the weak acid reservoir: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O — equilibrium shifts right, replacing H⁺. The pH changes only slightly as long as both reservoirs remain large; a buffer is exhausted when one reservoir is consumed.
Buffer pH calculation: [H⁺] = Ka × ([HA]/[A⁻]), so pH = pKa + log₁₀([A⁻]/[HA]) (Henderson-Hasselbalch). Worked example (from the specification style): calculate the pH of a buffer with 0.20 mol dm⁻³ ethanoic acid (Ka = 1.74 × 10⁻⁵) and 0.15 mol dm⁻³ sodium ethanoate: [H⁺] = 1.74 × 10⁻⁵ × (0.20/0.15) = 2.32 × 10⁻⁵; pH = 4.63. Made-by-neutralisation questions: work out moles of HA remaining (initial minus reacted) and moles of A⁻ formed, then substitute into [H⁺] = Ka × n(HA)/n(A⁻) — mole ratios replace concentrations (same volume cancels).
Blood buffering: blood pH is controlled at 7.35-7.45 by the carbonic acid/hydrogencarbonate buffer (H₂CO₃ ⇌ H⁺ + HCO₃⁻, carbonic acid formed from dissolved CO₂). Acidosis (excess H⁺) is countered by HCO₃⁻ removing it (H⁺ + HCO₃⁻ → H₂CO₃ → H₂O + CO₂ exhaled); alkalosis (excess OH⁻) is countered by H₂CO₃ neutralising it. The link between respiration rate and blood pH (CO₂ levels) is the standard applied question: faster breathing removes CO₂, shifting equilibrium left, lowering [H⁺] and raising pH.
Buffer capacity and dilution — the finer points examiners test: diluting a buffer with water scales [HA] and [A⁻] by the same factor, so the ratio [HA]/[A⁻] is unchanged and the pH stays the same — but the buffer capacity falls (smaller reservoirs of each species, so fewer moles of added acid/base can be absorbed before exhaustion). Strong acid/concentrated-base additions beyond the capacity overwhelm the buffer and the pH changes substantially — buffers resist small additions only. When a buffer is made by partial neutralisation, check that neither reservoir is fully consumed: if the NaOH moles exceed the HA moles, no buffer forms (excess strong base, pH from [OH⁻] via Kw instead). Comparing two buffers of the same pH but different concentrations: same pH (same ratio) but the more concentrated one has greater capacity — a standard A* discrimination. Writing buffer answers: state the reservoirs present, give the reaction with the added species, state the equilibrium shift and conclude 'pH changes only slightly' — the full chain earns all marks.
pH Curves, Indicators and Determining Ka Experimentally
A pH titration curve plots pH against titrant volume. The equivalence point is the near-vertical section; curve shapes depend on the acid-base pair: strong-strong (pH 1 → 13, huge vertical section centred at 7); weak acid-strong base (vertical section above 7 — salt hydrolysis: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻); strong acid-weak base (vertical below 7); weak-weak (shallow, no vertical section). The flat region before equivalence is the buffer zone — in a weak acid titration the HA/A⁻ pair formed in the flask buffers the solution around half-equivalence.
Indicator choice: the indicator's colour-change range must lie entirely within the vertical section — phenolphthalein (8.3-10.0) for weak acid-strong base; methyl orange (3.1-4.4) for strong acid-weak base; either for strong-strong; neither for weak-weak. Indicators are themselves weak acids whose conjugate pair has different colours (HIn ⇌ H⁺ + In⁻); the range is where both colours coexist.
Determining Ka experimentally (standard 6-marker): titrate a weak acid with a strong base, measuring pH with a calibrated pH meter. At half the equivalence volume, [HA] = [A⁻], so pH = pKa — read the pH at half-equivalence and take Ka = 10⁻ᵖᵏᵃ. Method: find the equivalence volume from the vertical section (or from a preliminary titration), halve it, read the pH there, convert. Improvements: pH meter calibrated with buffers, temperature controlled (Ka depends on T), small increments near equivalence. Cross-links: this is the experimental side of the titration page's curve analysis, and the buffer zone explanation mirrors the buffer section above.
Salt Hydrolysis and Multi-Step pH Calculations
Salt solutions are not all neutral: the ions can hydrolyse water. NaCl (from strong acid + strong base) is neutral — neither ion reacts with water. CH₃COONa (weak acid + strong base) is basic: the ethanoate ion accepts a proton — CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻ — so [OH⁻] rises and pH > 7. NH₄Cl (strong acid + weak base) is acidic: NH₄⁺ donates a proton — NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺ — so pH < 7. The rule: the ion from the weaker partner hydrolyses; the salt of a strong acid/strong base does not. This explains why the equivalence-point pH of a weak acid titration is above 7 and a weak base titration below 7 — the standard link between salt hydrolysis and titration curves.
Multi-step calculation style (AQA Paper 1 favourite): a buffer made by mixing solutions, or a titration part-way to equivalence. Method: (1) moles of each species from concentration × volume/1000; (2) apply the reaction (neutralisation) to find remaining HA and formed A⁻; (3) substitute mole ratios into [H⁺] = Ka × n(HA)/n(A⁻) — volumes cancel because both species share the flask volume; (4) convert to pH. Example: 25.0 cm³ of 0.100 mol dm⁻³ CH₃COOH mixed with 10.0 cm³ of 0.100 mol dm⁻³ NaOH: moles HA = 2.50 × 10⁻³, OH⁻ = 1.00 × 10⁻³; remaining HA = 1.50 × 10⁻³, A⁻ formed = 1.00 × 10⁻³; [H⁺] = 1.74 × 10⁻⁵ × (1.50/1.00) = 2.61 × 10⁻⁵; pH = 4.58.
Exam technique: read whether the question wants pH, pKa or Ka and answer exactly that; keep unrounded values through the calculation and round only at the end (quoting pH to 2 decimal places); state assumptions ([H⁺] from water negligible; [HA] unchanged — valid when dissociation < 5%). For explanation marks, use the causal chain: more A⁻ relative to HA → equilibrium pushed left → lower [H⁺] → higher pH. Weak acid vs strong acid comparison answers must mention partial vs complete dissociation — 'weak acids have higher pH at the same concentration because they dissociate only partially'.
💡 Worked Example
Exam-style: Calculate the pH of a buffer solution containing 0.20 mol dm⁻³ ethanoic acid (Ka = 1.74 × 10⁻⁵) and 0.15 mol dm⁻³ sodium ethanoate, and explain how the pH changes when a small amount of dilute HCl is added.
Mark-scheme logic: [H⁺] = Ka × ([HA]/[A⁻]) = (1.74 × 10⁻⁵) × (0.20/0.15) = 2.32 × 10⁻⁵ mol dm⁻³; pH = −log₁₀(2.32 × 10⁻⁵) = 4.63. On adding HCl: the added H⁺ reacts with the ethanoate reservoir — CH₃COO⁻ + H⁺ → CH₃COOH; the equilibrium CH₃COOH ⇌ H⁺ + CH₃COO⁻ shifts left, removing most of the added H⁺, so pH changes only slightly (as long as the CH₃COO⁻ reservoir remains large). Mark points: buffer equation; correct [H⁺]; pH with units-free value; buffer action equation; equilibrium shift.
❓ Practice Questions & Mark Schemes
Calculate the pH of 0.050 mol dm⁻³ Ba(OH)₂ at 298 K (Kw = 1.0 × 10⁻¹⁴).
Answer: [OH⁻] = 2 × 0.050 = 0.100 mol dm⁻³ (each Ba(OH)₂ gives 2 OH⁻). [H⁺] = 1.0 × 10⁻¹⁴/0.100 = 1.0 × 10⁻¹³; pH = 13.00.
Calculate the pH of 0.100 mol dm⁻³ ethanoic acid (Ka = 1.74 × 10⁻⁵).
Answer: Weak acid: [H⁺] ≈ √(Ka × [HA]) = √(1.74 × 10⁻⁵ × 0.100) = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³. pH = −log₁₀(1.32 × 10⁻³) = 2.88.
A 0.150 mol dm⁻³ solution of propanoic acid has pH 2.82. Calculate Ka.
Answer: [H⁺] = 10⁻²·⁸² = 1.51 × 10⁻³ mol dm⁻³. Ka = [H⁺]²/[HA] = (1.51 × 10⁻³)²/0.150 = 1.52 × 10⁻⁵ (units: mol dm⁻³). Approximation valid: [H⁺] is ~1% of [HA].
Explain why the pH of pure water falls below 7 at higher temperatures but the water remains neutral.
Answer: Water ionisation (2H₂O ⇌ H₃O⁺ + OH⁻) is endothermic, so Kw increases with T, raising [H⁺] — pH falls. It remains neutral because neutrality means [H⁺] = [OH⁻], which still holds; pH 7 is neutral only at 298 K.
A buffer is made by neutralising half of 0.20 mol of CH₃COOH (Ka = 1.74 × 10⁻⁵) with NaOH. Calculate the pH.
Answer: Moles reacted: 0.10 → CH₃COOH remaining 0.10, CH₃COO⁻ formed 0.10. [H⁺] = Ka × n(HA)/n(A⁻) = 1.74 × 10⁻⁵ × (0.10/0.10) = 1.74 × 10⁻⁵. pH = 4.76 (= pKa, since [HA] = [A⁻]).
Describe how a titration curve can be used to determine Ka of a weak acid experimentally.
Answer: Titrate with a strong base measuring pH with a calibrated pH meter. Find the equivalence volume from the vertical section; take half that volume. At half-equivalence [HA] = [A⁻], so pH = pKa; read the pH at that volume and calculate Ka = 10⁻ᵖᵏᵃ. Controls: calibrated meter, constant temperature (Ka is temperature-dependent).