You are viewing a free preview of this lesson.
Subscribe to unlock all 8 lessons in this course and every other course on LearningBro.
The pH scale is the chemist's compact way of expressing hydrogen-ion concentration across more than fourteen orders of magnitude. Defined as pH = −log₁₀[H⁺] with [H⁺] in mol dm⁻³, it converts the unwieldy numerical range of [H⁺] in real solutions (from roughly 10 mol dm⁻³ in concentrated mineral acids down to 10⁻¹⁵ mol dm⁻³ in concentrated alkali) into a familiar 0–14 scale. This lesson develops the quantitative skills required to manipulate that scale at A-Level. We start by computing the pH of strong monoprotic and diprotic acids of arbitrary concentration, then mirror the procedure for strong bases via the ionic product of water Kw = [H⁺][OH⁻], showing how Kw acts as the bridge between [H⁺] and [OH⁻] in any aqueous system. We examine the temperature dependence of Kw — autoionisation is endothermic, so Kw rises with T and the pH of pure water falls below 7 above 25 °C while remaining genuinely neutral. We end with dilution effects, the breakdown of "pH = [acid]" reasoning at very low concentrations where water autoionisation contributes, and a brief practical-skills box on pH measurement. Weak-acid Ka treatment, buffers and titration curves are previewed for later lessons.
Spec mapping (AQA 7405): This lesson maps to §3.1.12 (acids and bases) — specifically the definitions of pH and Kw, calculations of pH for strong acids and strong bases, and the temperature dependence of Kw. Brønsted-Lowry theory (conjugate pairs, proton transfer) is the focus of lesson 0 of this course; Ka, pKa and the quantitative treatment of weak acids appear in lesson 2; pH of weak acids and bases in lesson 3; buffer calculations and titration curves in lesson 4; indicator selection in lesson 5. Kw is also explicitly an equilibrium constant — students should cross-reference §3.1.6 (equilibria) and recognise that everything in this lesson is an application of Kc to the water autoionisation equilibrium. Refer to the official AQA specification document for the exact wording of each section.
Assessment objectives: AO1 recall items include the definitions of pH and pOH, the value and units of Kw at 298 K (1.00 × 10⁻¹⁴ mol² dm⁻⁶), and the pH ranges associated with "acidic", "neutral" and "alkaline" at 25 °C. AO2 application skills include computing pH from a given [H⁺] (and vice versa via [H⁺] = 10⁻^pH), computing the pH of strong monoprotic and diprotic acids of any concentration, and computing the pH of strong monoprotic and diprotic bases via the Kw bridge. AO3 reasoning items include predicting the effect of temperature on Kw and on the pH of pure water (qualitatively and quantitatively given data), reasoning about why very dilute strong acid does not have pH > 7, and explaining why pH 7 is not synonymous with "neutral" except at 25 °C.
The pH of an aqueous solution is defined by:
pH = −log₁₀[H⁺]
where [H⁺] is the hydrogen-ion concentration in mol dm⁻³. The choice of base-10 logarithm and the negative sign are pure convention — they convert a very small positive number (often 10⁻¹ to 10⁻¹⁴) into a manageable positive number, usually between 0 and 14. There is nothing magical about the numerical range; very concentrated acids genuinely have pH below 0 (e.g. 5 mol dm⁻³ HCl has pH ≈ −0.7), and very concentrated alkalis have pH above 14. A-Level questions almost always stay within 0–14, but the scale itself is unbounded.
The inverse relationship is:
[H⁺] = 10⁻^pH
Because the scale is logarithmic, each unit change in pH corresponds to a tenfold change in [H⁺]. A solution of pH 3 has [H⁺] ten times higher than a solution of pH 4, and one hundred times higher than pH 5. This is the single most important conceptual point about pH: differences in pH that seem small numerically (e.g. blood pH falling from 7.40 to 7.20) correspond to large multiplicative changes in [H⁺] (here, a 60% increase) that are physiologically devastating.
The pH of pure water at 25 °C is 7.00. Solutions with pH < 7 at 25 °C are acidic ([H⁺] > 10⁻⁷ mol dm⁻³); solutions with pH > 7 at 25 °C are alkaline ([H⁺] < 10⁻⁷ mol dm⁻³); pH = 7 at 25 °C is neutral. The qualifier "at 25 °C" is essential — at other temperatures, the pH of neutral water is not 7 (see the Kw section below).
A parallel scale, pOH = −log₁₀[OH⁻], is occasionally useful. From Kw it follows that pH + pOH = pKw = 14.00 at 25 °C. Most A-Level work uses pH directly and computes [H⁺] via the Kw bridge when only [OH⁻] is given, but pOH is a legitimate shortcut.
Worked Example 1 — pH from [H⁺]. Calculate the pH of a solution with [H⁺] = 3.20 × 10⁻⁴ mol dm⁻³.
pH = −log₁₀(3.20 × 10⁻⁴) = 3.49 (3 s.f.).
A calculator gives 3.4949…; rounded to 2 d.p. (the conventional precision for pH), that is 3.49.
Worked Example 2 — [H⁺] from pH. A solution has pH 5.62. Calculate [H⁺].
[H⁺] = 10⁻⁵·⁶² = 2.40 × 10⁻⁶ mol dm⁻³.
Use the 10ˣ key on the calculator with x = −5.62. Quote to 3 s.f.
A strong acid is one that fully dissociates in aqueous solution. The dissociation equilibrium lies so far to the right that we treat it as complete. The standard A-Level strong acids are HCl, HBr, HI, HNO₃ and (in the first ionisation) H₂SO₄. Other species occasionally encountered as fully dissociated include HClO₄ and HClO₃.
For a strong monoprotic acid:
HA(aq) → H⁺(aq) + A⁻(aq)
Every mole of acid releases exactly one mole of H⁺, so:
[H⁺] = [HA] (assuming the acid is the dominant H⁺ source).
The pH is then a direct logarithm of the acid concentration.
Worked Example 3 — Strong monoprotic acid. Calculate the pH of 0.10 mol dm⁻³ HCl at 25 °C.
HCl is a strong monoprotic acid → [H⁺] = 0.10 mol dm⁻³.
pH = −log₁₀(0.10) = 1.00.
Worked Example 4 — Strong monoprotic acid. Calculate the pH of 0.050 mol dm⁻³ HNO₃.
[H⁺] = 0.050 mol dm⁻³; pH = −log₁₀(0.050) = 1.30.
Worked Example 5 — Strong monoprotic acid, different concentration. Calculate the pH of 0.010 mol dm⁻³ HCl.
[H⁺] = 0.010; pH = −log₁₀(0.010) = 2.00.
Notice the relationship between examples 3 and 5: a tenfold dilution of HCl (from 0.10 to 0.010 mol dm⁻³) raises the pH by exactly one unit (from 1.00 to 2.00). This is a general feature of strong acids in the "dilute but not very dilute" range (10⁻¹ to 10⁻⁵ mol dm⁻³), and is exploited routinely in pH-management calculations.
Sulfuric acid, H₂SO₄, is a strong acid in its first ionisation but only moderately strong in its second:
At A-Level we approximate H₂SO₄ as fully diprotic — both protons fully released — unless the question explicitly directs otherwise. This is exact to within ~5% for dilute H₂SO₄ and is the conventional treatment.
For a fully dissociated diprotic acid:
[H⁺] = 2 × [H₂A]
Worked Example 6 — Strong diprotic acid. Calculate the pH of 0.025 mol dm⁻³ H₂SO₄.
[H⁺] = 2 × 0.025 = 0.050 mol dm⁻³.
pH = −log₁₀(0.050) = 1.30.
Worked Example 7 — Strong diprotic acid, low concentration. Calculate the pH of 5.0 × 10⁻⁴ mol dm⁻³ H₂SO₄.
[H⁺] = 2 × 5.0 × 10⁻⁴ = 1.0 × 10⁻³ mol dm⁻³.
pH = −log₁₀(1.0 × 10⁻³) = 3.00.
Exam Tip: A common error is to compute pH from [H₂SO₄] directly, giving pH 1.60 for 0.025 mol dm⁻³ acid. The factor of 2 is worth a mark on its own. Read the question carefully — if asked for the pH of H₂SO₄ assuming full dissociation, multiply by 2.
Pure water itself partially autoionises:
H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)
This is a genuine equilibrium with a (very small) equilibrium constant. Because [H₂O] is essentially constant (~55.5 mol dm⁻³ in pure water and dilute aqueous solutions), it is absorbed into the constant to give the ionic product of water:
Kw = [H⁺][OH⁻]
At 298 K (25 °C), Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶.
This relationship is not restricted to pure water — Kw holds in any aqueous solution at the stated temperature. It is the master equation linking [H⁺] and [OH⁻]: knowing one determines the other.
In pure water at 25 °C, autoionisation produces equal amounts of H⁺ and OH⁻:
[H⁺] = [OH⁻] = √Kw = √(1.00 × 10⁻¹⁴) = 1.00 × 10⁻⁷ mol dm⁻³
→ pH = −log₁₀(1.00 × 10⁻⁷) = 7.00 at 25 °C.
The autoionisation of water is endothermic — breaking the O–H bond to liberate H⁺ requires energy (ΔH° ≈ +57 kJ mol⁻¹ for the autoionisation). By Le Chatelier's principle, increasing the temperature shifts the equilibrium to the right, increasing both [H⁺] and [OH⁻] and therefore increasing Kw.
| Temperature (°C) | Kw (mol² dm⁻⁶) | [H⁺] in pure water (mol dm⁻³) | pH of pure water |
|---|---|---|---|
| 0 | 1.14 × 10⁻¹⁵ | 3.38 × 10⁻⁸ | 7.47 |
| 25 | 1.00 × 10⁻¹⁴ | 1.00 × 10⁻⁷ | 7.00 |
| 50 | 5.48 × 10⁻¹⁴ | 2.34 × 10⁻⁷ | 6.63 |
| 100 | 5.13 × 10⁻¹³ | 7.16 × 10⁻⁷ | 6.14 |
Two points students must internalise:
Worked Example 8 — pH of pure water at 50 °C. Given Kw = 5.48 × 10⁻¹⁴ mol² dm⁻⁶ at 50 °C, calculate the pH of pure water.
[H⁺] = [OH⁻] = √Kw = √(5.48 × 10⁻¹⁴) = 2.34 × 10⁻⁷ mol dm⁻³.
pH = −log₁₀(2.34 × 10⁻⁷) = 6.63.
Comment: water is still neutral ([H⁺] = [OH⁻]), but its pH has fallen below 7.
Strong bases are species that fully dissociate to release OH⁻. The standard A-Level strong bases are the Group 1 hydroxides (NaOH, KOH) and the Group 2 hydroxides where soluble (Ba(OH)₂; Ca(OH)₂ is only sparingly soluble, often quoted as a strong base for the small amount that does dissolve).
For a strong monoprotic base (one OH⁻ per formula unit):
NaOH → Na⁺ + OH⁻
[OH⁻] = [NaOH].
To find the pH we cannot compute −log[OH⁻] directly — pH is defined in terms of [H⁺]. We must go through Kw:
[H⁺] = Kw / [OH⁻]
Then pH = −log₁₀([H⁺]).
Worked Example 9 — Strong monoprotic base. Calculate the pH of 0.10 mol dm⁻³ NaOH at 25 °C.
[OH⁻] = 0.10 mol dm⁻³.
[H⁺] = Kw / [OH⁻] = 1.00 × 10⁻¹⁴ / 0.10 = 1.00 × 10⁻¹³ mol dm⁻³.
pH = −log₁₀(1.00 × 10⁻¹³) = 13.00.
Worked Example 10 — Strong monoprotic base, low concentration. Calculate the pH of 0.0040 mol dm⁻³ KOH.
[OH⁻] = 4.0 × 10⁻³; [H⁺] = 1.00 × 10⁻¹⁴ / 4.0 × 10⁻³ = 2.5 × 10⁻¹² mol dm⁻³.
pH = −log₁₀(2.5 × 10⁻¹²) = 11.60.
Barium hydroxide, Ba(OH)₂, fully dissociates to give two OH⁻ per formula unit:
Ba(OH)₂ → Ba²⁺ + 2 OH⁻
So [OH⁻] = 2 × [Ba(OH)₂].
Worked Example 11 — Diprotic strong base. Calculate the pH of 0.025 mol dm⁻³ Ba(OH)₂ at 25 °C.
[OH⁻] = 2 × 0.025 = 0.050 mol dm⁻³.
[H⁺] = Kw / [OH⁻] = 1.00 × 10⁻¹⁴ / 0.050 = 2.00 × 10⁻¹³ mol dm⁻³.
pH = −log₁₀(2.00 × 10⁻¹³) = 12.70.
Exam Tip: For any strong base, the workflow is identical: (1) write the dissociation equation; (2) compute [OH⁻], multiplying by the stoichiometric factor (1 for NaOH/KOH, 2 for Ba(OH)₂); (3) use Kw to get [H⁺]; (4) take −log₁₀ for pH. Going through [H⁺] explicitly is required at A-Level — answers that compute pH = 14 + log₁₀[OH⁻] (i.e. 14 − pOH) without showing the Kw step typically lose a working mark.
When strong acid and strong base are mixed, they neutralise stoichiometrically until one is exhausted. The pH of the resulting solution depends on which species is in excess.
Worked Example 12 — Excess acid. 25.0 cm³ of 0.20 mol dm⁻³ HCl is mixed with 20.0 cm³ of 0.10 mol dm⁻³ NaOH. Calculate the pH of the resulting mixture.
Moles HCl = 25.0 × 10⁻³ × 0.20 = 5.00 × 10⁻³ mol.
Moles NaOH = 20.0 × 10⁻³ × 0.10 = 2.00 × 10⁻³ mol.
Reaction: H⁺ + OH⁻ → H₂O. Moles H⁺ in excess = 5.00 × 10⁻³ − 2.00 × 10⁻³ = 3.00 × 10⁻³ mol.
Total volume = 25.0 + 20.0 = 45.0 cm³ = 0.0450 dm³.
[H⁺] = 3.00 × 10⁻³ / 0.0450 = 0.0667 mol dm⁻³.
pH = −log₁₀(0.0667) = 1.18.
Worked Example 13 — Excess base. 30.0 cm³ of 0.20 mol dm⁻³ NaOH is mixed with 20.0 cm³ of 0.10 mol dm⁻³ HCl. Calculate the pH.
Moles NaOH = 0.0300 × 0.20 = 6.00 × 10⁻³; moles HCl = 0.0200 × 0.10 = 2.00 × 10⁻³.
Excess OH⁻ = 4.00 × 10⁻³ mol; total volume = 0.0500 dm³.
[OH⁻] = 4.00 × 10⁻³ / 0.0500 = 0.0800 mol dm⁻³.
[H⁺] = 1.00 × 10⁻¹⁴ / 0.0800 = 1.25 × 10⁻¹³; pH = 12.90.
For a strong monoprotic acid in the "normal" concentration range (10⁻¹ to ~10⁻⁵ mol dm⁻³), diluting tenfold increases pH by exactly one unit. This follows trivially from the log relationship: if [H⁺] falls by a factor of 10, log₁₀[H⁺] falls by 1, so pH rises by 1.
Worked Example 14 — Dilution. 10.0 cm³ of 0.10 mol dm⁻³ HCl is diluted to 100 cm³. Calculate the pH of the diluted solution.
Moles HCl = 0.0100 × 0.10 = 1.00 × 10⁻³ mol (conserved under dilution).
New [H⁺] = 1.00 × 10⁻³ / 0.100 = 0.0100 mol dm⁻³.
pH = −log₁₀(0.0100) = 2.00 (one unit above the original pH 1.00).
The simple "pH = −log[acid]" relationship breaks down at very low strong-acid concentrations (below about 10⁻⁶ mol dm⁻³). At these concentrations, the H⁺ produced by water autoionisation (10⁻⁷ mol dm⁻³ at 25 °C) becomes comparable to or larger than the H⁺ from the acid, and the contribution from water can no longer be ignored.
For example, "10⁻⁸ mol dm⁻³ HCl" does not have pH 8 — water autoionisation supplies more H⁺ than the acid itself. The correct treatment uses a charge-balance/mass-balance approach:
Solving the resulting quadratic gives [H⁺] ≈ 1.05 × 10⁻⁷ mol dm⁻³, so pH ≈ 6.98 — barely below pure water. As the strong-acid concentration falls below 10⁻⁶, the pH asymptotically approaches 7 from below (never exceeding 7 — adding any amount of acid can only lower pH, never raise it).
A-Level questions almost always stay safely above 10⁻⁶ mol dm⁻³ so this regime can be ignored; but A* candidates should recognise that pH = 8 for "10⁻⁸ M HCl" is nonsense and be able to explain why.
Subscribe to continue reading
Get full access to this lesson and all 8 lessons in this course.