HookYour blood lives inside one tenth of a pH unit
Arterial blood holds its pH between 7.35 and 7.45 — one tenth of a unit of freedom. Drift much below 6.8 or above 7.8 for long and the outcome is fatal; hospitals treat a fall of 0.1 as an emergency. Yet your metabolism is an acid factory: every day you exhale well over ten moles of carbon dioxide, each molecule of which dissolves in water as carbonic acid. What holds the line is not luck but an equilibrium — the carbonic acid/hydrogencarbonate buffer, in which \(\text{HCO}_3^-\) outnumbers \(\text{H}_2\text{CO}_3\) roughly 20 to 1 at pH 7.40, with breathing as the release valve. Hyperventilate and you vent CO₂ faster than you make it: the equilibrium shifts, [H⁺] falls, blood pH climbs — and the dizziness of respiratory alkalosis is Le Chatelier arriving in person.
3.1.12 is the mathematics of that proton traffic, and it is the most calculation-dense section on Paper 1. Brønsted–Lowry supplies the language (donors and acceptors); pH supplies the logarithmic currency; \(K_w\) extends the sums to bases; \(K_a\) prices weak acids honestly; titration curves put the whole system on one graph; and buffers — blood's trick — close the loop. Every calculation in the section is the same three moves: write the expression, substitute, interpret the answer.
ModelBrønsted–Lowry — acids donate protons, bases accept them
A Brønsted–Lowry acid is a proton donor; a base is a proton acceptor; and every acid–base reaction is a single proton changing hands — which means an acid only acts when a base is present to receive. Hydrogen chloride donating to water, \(\text{HCl} + \text{H}_2\text{O} \rightarrow \text{H}_3\text{O}^+ + \text{Cl}^-\), casts water as the base; ammonia accepting from water, \(\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-\), casts water as the acid. Water plays both roles, and for brevity chemists write the hydrated proton as \(\text{H}^+\) even though \(\text{H}_3\text{O}^+\) is the honest species.
Two vocabulary distinctions carry marks all section long. Strong versus weak is about the extent of dissociation: strong acids — HCl, nitric, sulfuric — dissociate essentially completely in water; weak acids — ethanoic, carbonic — barely start, with 0.100 mol dm⁻³ ethanoic acid only about 1.3% dissociated. Concentrated versus dilute is about the amount dissolved per dm³. The axes are independent: a concentrated weak acid and a dilute strong acid are both perfectly coherent descriptions, and swapping the words is an instant AO1 penalty.
ModelpH — a logarithm that tames a million-fold range
\(\text{pH} = -\log_{10}[\text{H}^+]\), and back again \([\text{H}^+] = 10^{-\text{pH}}\). The logarithm is there because hydrogen-ion concentrations sprawl across fourteen powers of ten: stomach acid near pH 1.5 carries around a million times the \([\text{H}^+]\) of neutral water at pH 7. One pH unit is always a factor of ten — so 'the pH fell by 2' means a hundredfold rise in \([\text{H}^+]\), a sentence worth writing explicitly in data questions.
For a strong acid the arithmetic is immediate because dissociation is complete: \([\text{H}^+]\) equals the acid concentration for a monoprotic acid like HCl or nitric acid, and twice it for sulfuric acid, which AQA treats as releasing both protons. Convention to burn in now: AQA expects pH quoted to two decimal places.
(a) 0.0500 mol dm⁻³ HCl: \(\text{pH} = -\log_{10}(0.0500) = 1.30\). (b) 0.0250 mol dm⁻³ sulfuric acid: \([\text{H}^+] = 2 \times 0.0250 = 0.0500\ \text{mol dm}^{-3}\), so the pH is also 1.30 — half the concentration, same acidity, because each mole donates two protons. (c) Reversing: a solution of pH 2.70 has \([\text{H}^+] = 10^{-2.70} = 2.0 \times 10^{-3}\ \text{mol dm}^{-3}\). Keep every digit on the calculator and round once, at the end.
ModelKw — the ionic product that never sleeps
Water dissociates, slightly: \(\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-\). The equilibrium generates the ionic product of water, \(K_w = [\text{H}^+][\text{OH}^-]\), equal to \(1.0 \times 10^{-14}\ \text{mol}^2\,\text{dm}^{-6}\) at 298 K. The product holds in every aqueous solution — acidic, alkaline or neutral — so fixing either concentration fixes the other. In pure water \([\text{H}^+] = [\text{OH}^-] = 1.0 \times 10^{-7}\ \text{mol dm}^{-3}\), hence pH 7.00.
\(K_w\) is the bridge to strong bases: sodium hydroxide hands you \([\text{OH}^-]\) directly, and \(K_w\) converts it into \([\text{H}^+]\), thence pH. And \(K_w\) is temperature-dependent — the dissociation is endothermic, so heating pushes it up. Around 50 °C, \(K_w\) is roughly \(5.5 \times 10^{-14}\), making pure water's pH about 6.63. Is hot water acidic? No: \([\text{H}^+]\) still equals \([\text{OH}^-]\), and that is the definition of neutral. pH 7 is a fact about 298 K, not about neutrality — AQA's favourite discriminator in this whole section.
pH of 0.150 mol dm⁻³ NaOH at 298 K. Complete dissociation gives \([\text{OH}^-] = 0.150\ \text{mol dm}^{-3}\). Then \[ [\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{0.150} = 6.67 \times 10^{-14}\ \text{mol dm}^{-3} \] so \(\text{pH} = -\log_{10}(6.67 \times 10^{-14}) = 13.18\). Sense-check against the anchors: a moderately concentrated strong base should land near the top of the scale, and it does.
ModelWeak acids — Ka, pKa and two approximations you must confess
A weak acid HA sets up \(\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\), governed by the acid dissociation constant \[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \] with units \(\text{mol dm}^{-3}\). Bigger \(K_a\), stronger acid. Because the values are inconveniently tiny they are usually quoted as \(\text{p}K_a = -\log_{10} K_a\): ethanoic acid's \(K_a = 1.74 \times 10^{-5}\) becomes the friendlier \(\text{p}K_a = 4.76\) — and now smaller means stronger.
Calculating a weak acid's pH leans on two approximations, and AQA regularly asks you to state them. First, \([\text{H}^+] = [\text{A}^-]\): every dissociation event makes one of each, and water's own tiny contribution to \([\text{H}^+]\) is ignored. Second, \([\text{HA}]_{\text{eqm}} \approx [\text{HA}]_{\text{initial}}\): so little dissociates that the acid is effectively undepleted. Together they collapse the expression to \(K_a = [\text{H}^+]^2 / [\text{HA}]\), so \([\text{H}^+] = \sqrt{K_a \times c}\). The second approximation is the fragile one: it degrades for the stronger end of the weak acids and for very dilute solutions, where the fraction dissociated stops being negligible.
pH of 0.100 mol dm⁻³ ethanoic acid, \(K_a = 1.74 \times 10^{-5}\ \text{mol dm}^{-3}\): \[ [\text{H}^+] = \sqrt{1.74 \times 10^{-5} \times 0.100} = 1.32 \times 10^{-3}\ \text{mol dm}^{-3} \] so pH \(= 2.88\). Compare 0.100 mol dm⁻³ HCl at pH 1.00: identical concentration, nearly two pH units apart — a roughly 76-fold difference in \([\text{H}^+]\) — purely because ethanoic acid holds about 98.7% of its protons back. And the confession that earns the evaluation mark: only 1.3% dissociated, so treating [HA] as 0.100 was safe here.
DataTitration curves and RP9 — four shapes, one vertical truth
Plot pH against volume of base added to an acid and the story is in the vertical section — the near-instant leap where a fraction of a drop swings the pH by several units. Strong acid–strong base: starts near 1, leaps from roughly 3 to 11, passes pH 7 at equivalence. Weak acid–strong base (ethanoic against NaOH): starts near 3, climbs through a shallow buffer region, leaps from about 7 to 11 — equivalence sits above 7. Strong acid–weak base is the mirror image: vertical from about 3 to 7, equivalence below 7. Weak against weak: no vertical section at all.
An indicator is itself a weak acid whose two forms differ in colour, changing over a narrow pH range — and the choice rule is mechanical: the indicator's range must fall inside the curve's vertical section, so that the end point (the colour change) coincides with the equivalence point (exact stoichiometric neutralisation). Phenolphthalein (about 8.3–10) suits weak acid–strong base; methyl orange (about 3.1–4.4) suits strong acid–weak base; for strong–strong either works; for weak–weak nothing does — use a pH meter instead.
Which is Required Practical 9. Method: calibrate the pH probe against standard buffers (pH 4, 7 and 10); run base in from a burette in 1 cm³ steps, stirring and recording pH after each addition; shrink the additions to a drop or two as the pH starts to move; continue well past equivalence. Errors worth naming: an uncalibrated or drifting probe (systematic error), temperature changes (both the electrode response and \(K_w\) are temperature-dependent), and adding too fast near equivalence — which skips the very readings that define the vertical section. The finished curve pays out twice: the equivalence volume, read halfway up the leap, and — the elegant one — the pH at half-neutralisation, where half the acid has been converted so \([\text{HA}] = [\text{A}^-]\), the ratio cancels out of \(K_a\), and \(\text{pH} = \text{p}K_a\) can be read straight off the graph.
25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid is titrated with 0.100 mol dm⁻³ NaOH. Equivalence falls at 25.0 cm³, where the flask holds nothing but sodium ethanoate solution at a pH of about 8.7 — comfortably inside phenolphthalein's range, hopelessly above methyl orange's, which would change colour around pH 4, in the buffer region, long before equivalence. At 12.5 cm³ — half-neutralisation — the probe reads pH 4.76, so \(K_a = 10^{-4.76} = 1.7 \times 10^{-5}\ \text{mol dm}^{-3}\): the practical hands you the dissociation constant for free.
MechanismBuffers — spending a reservoir instead of moving the price
An acidic buffer is a weak acid plus a large supply of its conjugate base — mix ethanoic acid with sodium ethanoate, or make the salt in place by half-neutralising the acid with NaOH. The design principle is to keep both reservoirs large. Added acid is absorbed by the base reservoir, \(\text{A}^- + \text{H}^+ \rightarrow \text{HA}\); added alkali is absorbed by the acid reservoir, \(\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}\). Either way the ratio \([\text{HA}]/[\text{A}^-]\) barely moves — and since rearranging \(K_a\) gives \[ [\text{H}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]} \] a nearly fixed ratio means a nearly fixed pH. (Basic buffers — ammonia with ammonium chloride — mirror the same logic above pH 7.)
Blood runs the scheme with carbonic acid: \(\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-\), the 20:1 hydrogencarbonate reservoir soaking up metabolic acid while the lungs exhale the resulting CO₂ — a buffer fitted with a chimney. The reservoirs are finite, though, and that is the honest caveat: exhaust one and the buffer fails, which is exactly what untreated diabetic ketoacidosis does to blood.
Exam calculations arrive in two costumes: the pH of a made buffer (put the ratio straight in), and the pH after a small addition of strong acid or base (update the moles of HA and A⁻ first, then take the ratio).
A buffer is 0.500 mol dm⁻³ ethanoic acid with 0.250 mol dm⁻³ sodium ethanoate (\(K_a = 1.74 \times 10^{-5}\)): \([\text{H}^+] = 1.74 \times 10^{-5} \times (0.500/0.250) = 3.48 \times 10^{-5}\), so pH \(= 4.46\). Now add 0.0100 mol of NaOH to 1.00 dm³ of it: the alkali converts acid into base, leaving HA \(= 0.490\) mol and A⁻ \(= 0.260\) mol, so \([\text{H}^+] = 1.74 \times 10^{-5} \times (0.490/0.260) = 3.28 \times 10^{-5}\): pH 4.48. The same 0.0100 mol of NaOH dropped into a litre of pure water takes the pH from 7.00 to 12.00. Two hundredths of a unit against five whole units — that contrast is the definition of a buffer, and quoting it is the cleanest evaluation sentence in the section.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Acids and bases is Paper 1's calculation engine, and the marks follow the M1–M2–M3 rhythm: expression written, substitution shown, answer with units — and pH to two decimal places. Units are marks in their own right: \(K_a\) in \(\text{mol dm}^{-3}\), \(K_w\) in \(\text{mol}^2\,\text{dm}^{-6}\), pH unitless. A weak-acid calculation is not finished until you can state the two approximations on demand — \([\text{H}^+] = [\text{A}^-]\), and \([\text{HA}]\) effectively unchanged — because 'explain why your value is approximate' is a standard follow-up.
Language discipline earns AO1 marks across the section: 'strong' for dissociation, 'concentrated' for amount; 'equivalence point' for the stoichiometry, 'end point' for the indicator. Justify indicator choice by naming the vertical section and placing the indicator's range inside it — 'phenolphthalein, because the pH at equivalence leaps from about 7 to 11, which contains its 8.3–10 range' is the full-credit sentence. When sketching curves, make the start pH plausible (a 0.1 mol dm⁻³ weak acid starts near 3, not 1), fix the equivalence volume from the moles, and put the vertical section on the correct side of 7.
Buffer questions are moles-first, always: react any added acid or alkali against the reservoirs, tabulate the new moles of HA and A⁻, and only then take the ratio into \([\text{H}^+] = K_a[\text{HA}]/[\text{A}^-]\). The classic 6-mark extended response asks how a buffer resists both added acid and added alkali: give both equations, state that the ratio — and therefore the pH — is almost unchanged, and credit the large reservoirs. On RP9, name the calibration against standard buffers and the dropwise additions near equivalence; those two details are what separate level-3 method answers from the rest.