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The pH scale provides a convenient way to express the acidity or alkalinity of a solution. It was introduced by Soren Sorensen in 1909 and is defined as:
pH = -log₁₀[H⁺]
where [H⁺] is the concentration of hydrogen ions in mol dm⁻³. Because the logarithm is base 10, each unit change in pH represents a tenfold change in hydrogen ion concentration.
The reverse calculation is equally important:
[H⁺] = 10⁻ᵖᴴ
These two equations are the foundation of every pH calculation you will meet at A-Level. You must be able to move fluently between pH values and hydrogen ion concentrations in both directions.
Why logarithms? Hydrogen ion concentrations in aqueous solutions span an enormous range — from about 10 mol dm⁻³ for concentrated HCl down to 10⁻¹⁴ mol dm⁻³ for 1 mol dm⁻³ NaOH. The logarithmic pH scale compresses this 15-order-of-magnitude range into a manageable 0 to 14 scale.
Calculate the pH of a 0.050 mol dm⁻³ solution of HCl.
HCl is a strong acid, so it fully dissociates:
HCl → H⁺ + Cl⁻
Therefore [H⁺] = 0.050 mol dm⁻³.
pH = -log₁₀(0.050) pH = -(-1.30) pH = 1.30
This is straightforward because the strong acid fully dissociates — every molecule of HCl produces one H⁺ ion.
A solution has a pH of 3.40. Calculate the hydrogen ion concentration.
[H⁺] = 10⁻³·⁴⁰ [H⁺] = 3.98 × 10⁻⁴ mol dm⁻³
Always give your answer to the appropriate number of significant figures (usually 3 s.f. for Edexcel).
Calculate the pH of 0.0025 mol dm⁻³ HNO₃.
HNO₃ is a strong monoprotic acid: HNO₃ → H⁺ + NO₃⁻
[H⁺] = 0.0025 mol dm⁻³
pH = -log₁₀(0.0025) = -log₁₀(2.5 × 10⁻³) = 2.60
The pH scale typically runs from 0 to 14 for aqueous solutions at 25°C, though values outside this range are possible with very concentrated solutions:
| pH | [H⁺] / mol dm⁻³ | Character |
|---|---|---|
| 0 | 1.0 | Very strongly acidic |
| 1 | 0.10 | Strongly acidic |
| 3 | 1.0 × 10⁻³ | Weakly acidic |
| 7 | 1.0 × 10⁻⁷ | Neutral (at 25°C) |
| 11 | 1.0 × 10⁻¹¹ | Weakly alkaline |
| 14 | 1.0 × 10⁻¹⁴ | Strongly alkaline |
Notice that pH 7 is only neutral at 25°C. This is because neutrality means [H⁺] = [OH⁻], and the exact concentrations depend on the ionic product of water, which changes with temperature.
Can pH be negative? Yes. A 2.0 mol dm⁻³ solution of HCl has [H⁺] = 2.0 mol dm⁻³ and pH = -log₁₀(2.0) = -0.30. Negative pH values occur with concentrated strong acids, though they are rare in exam questions.
Water undergoes a very slight self-ionisation (autoprotolysis):
H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)
or more simply:
H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)
The equilibrium constant for this process is the ionic product of water:
Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 25°C
This relationship holds in ALL aqueous solutions, not just pure water. It is the key to calculating the pH of strong bases.
Why does Kw hold in all aqueous solutions? Because the autoionisation equilibrium exists in every aqueous solution, regardless of what else is dissolved. Adding acid increases [H⁺] and decreases [OH⁻] (the equilibrium shifts left), but the product [H⁺][OH⁻] remains constant at a given temperature.
Calculate the pH of a 0.020 mol dm⁻³ solution of NaOH at 25°C.
NaOH fully dissociates:
NaOH → Na⁺ + OH⁻
Therefore [OH⁻] = 0.020 mol dm⁻³.
Using Kw:
[H⁺] = Kw / [OH⁻] [H⁺] = (1.00 × 10⁻¹⁴) / 0.020 [H⁺] = 5.00 × 10⁻¹³ mol dm⁻³
pH = -log₁₀(5.00 × 10⁻¹³) pH = 12.30
This two-step process (find [OH⁻], then use Kw to find [H⁺], then calculate pH) is essential for all strong base pH problems.
Calculate the pH of 0.015 mol dm⁻³ Ba(OH)₂ at 25°C.
Ba(OH)₂ is a strong base that produces two OH⁻ ions per formula unit:
Ba(OH)₂ → Ba²⁺ + 2OH⁻
[OH⁻] = 2 × 0.015 = 0.030 mol dm⁻³
[H⁺] = Kw / [OH⁻] = (1.00 × 10⁻¹⁴) / 0.030 = 3.33 × 10⁻¹³ mol dm⁻³
pH = -log₁₀(3.33 × 10⁻¹³) = 12.48
Key point: Always check whether the base produces 1 or 2 moles of OH⁻ per mole of base. Ba(OH)₂, Ca(OH)₂, and Sr(OH)₂ all give 2 OH⁻.
The self-ionisation of water is endothermic (ΔH is positive). By Le Chatelier's principle, increasing the temperature shifts the equilibrium to the right, producing more H⁺ and OH⁻ ions. This means Kw increases with temperature.
| Temperature / °C | Kw / mol² dm⁻⁶ | pH of pure water |
|---|---|---|
| 10 | 0.29 × 10⁻¹⁴ | 7.27 |
| 25 | 1.00 × 10⁻¹⁴ | 7.00 |
| 37 | 2.40 × 10⁻¹⁴ | 6.81 |
| 50 | 5.50 × 10⁻¹⁴ | 6.63 |
| 100 | 51.3 × 10⁻¹⁴ | 6.14 |
At 50°C, Kw ≈ 5.5 × 10⁻¹⁴. At this temperature, pure water has:
[H⁺] = √(5.5 × 10⁻¹⁴) = 2.3 × 10⁻⁷ mol dm⁻³ pH = -log₁₀(2.3 × 10⁻⁷) = 6.63
The pH of pure water at 50°C is 6.63, not 7.00. But the water is still neutral because [H⁺] = [OH⁻]. Neutrality means equal concentrations, not pH 7.
Sulfuric acid is diprotic — each molecule can donate two protons:
H₂SO₄ → 2H⁺ + SO₄²⁻
For a 0.010 mol dm⁻³ solution of H₂SO₄ (assuming full dissociation of both protons, which is a reasonable approximation for dilute solutions):
[H⁺] = 2 × 0.010 = 0.020 mol dm⁻³ pH = -log₁₀(0.020) = 1.70
Always check whether the question specifies H₂SO₄ — if so, remember the factor of 2.
Note on H₂SO₄: The first dissociation is strong (H₂SO₄ → H⁺ + HSO₄⁻), but the second dissociation of HSO₄⁻ is actually a moderately strong acid (Ka₂ ≈ 0.012 mol dm⁻³). For Edexcel at dilute concentrations, you can assume both protons are fully donated.
| Formula | When to use |
|---|---|
| pH = -log₁₀[H⁺] | Always — the definition |
| [H⁺] = 10⁻ᵖᴴ | Finding [H⁺] from pH |
| Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ at 25°C | Linking [H⁺] and [OH⁻] |
| [H⁺] = Kw / [OH⁻] | pH of bases |
| For H₂SO₄: [H⁺] = 2c | Diprotic strong acid |
Edexcel 9CH0 specification, Topic 12: Acid-base equilibria, sub-strand 12.2 requires you to define pH = −log₁₀[H⁺(aq)]; calculate pH given [H⁺] and vice versa; define the ionic product of water Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K; calculate pH of strong acids and strong bases (refer to the official specification document for exact wording). The base-10 logarithm convention is universal: pH, pOH, pKa and pKw are all log₁₀-based. The temperature dependence of Kw is examined: water dissociation is endothermic, so heating shifts the equilibrium to the right, increases [H⁺] and [OH⁻] equally, and reduces pH below 7 — without making pure water acidic. The Edexcel data booklet provides Kw at 298 K but expects students to reason about its temperature variation.
Question (8 marks):
(a) Calculate the pH of 0.05 mol dm⁻³ hydrochloric acid. (2)
(b) Calculate the pH of 0.10 mol dm⁻³ aqueous sodium hydroxide at 298 K. Use Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶. (3)
(c) The pH of pure water at 50°C is 6.63. Explain why the water is not acidic. (3)
Solution with mark scheme:
(a) HCl is a strong, monoprotic acid, so [H⁺] = 0.05 mol dm⁻³.
pH = −log₁₀(0.05) = 1.30 (3 s.f.)
M1 — recognises that for a strong monoprotic acid [H⁺] = c. A1 — pH = 1.30 (or 1.3) to appropriate precision.
(b) NaOH is a strong base, so [OH⁻] = 0.10 mol dm⁻³.
[H⁺] = Kw / [OH⁻] = 1.0 × 10⁻¹⁴ / 0.10 = 1.0 × 10⁻¹³ mol dm⁻³.
pH = −log₁₀(1.0 × 10⁻¹³) = 13.00.
M1 — uses Kw to convert [OH⁻] to [H⁺]. M1 — correct numerical [H⁺]. A1 — pH = 13.00 (or 13.0).
(c) At 50°C, water still contains equal concentrations of H⁺ and OH⁻ (each = 10⁻⁶·⁶³ ≈ 2.3 × 10⁻⁷ mol dm⁻³). The water is neutral because [H⁺] = [OH⁻], even though [H⁺] is greater than 10⁻⁷ mol dm⁻³.
The shift arises because the autoionisation 2H₂O ⇌ H₃O⁺ + OH⁻ is endothermic; raising the temperature shifts the equilibrium to the right (Le Chatelier), increasing Kw and lowering pH. Acidic and neutral are not defined by pH = 7 but by the relative concentrations of H⁺ and OH⁻.
M1 — neutrality criterion is [H⁺] = [OH⁻], not pH = 7. M1 — links increase in Kw to endothermic autoionisation and Le Chatelier. A1 — explicit conclusion that pure water at 50°C is neutral but not pH 7.
Total: 8 marks (M5 A3, split as shown).
Question (6 marks): A student prepares a solution by dissolving 0.40 g of solid NaOH in distilled water and making the volume up to 250 cm³.
(a) Calculate the concentration of OH⁻(aq) in the solution. (M(NaOH) = 40.0 g mol⁻¹.) (2)
(b) Calculate the pH of the solution at 298 K. (3)
(c) State, with a reason, how the pH of the solution would change if the experiment were repeated at 60°C. (1)
Mark scheme decomposition by AO:
(a)
(b)
(c)
Total: 6 marks split AO1 = 2, AO2 = 4. Edexcel weights this topic toward AO2 because the calculations test confident manipulation of logarithms and Kw, not memorised facts.
Connects to:
Mathematics (logarithms): pH and pOH are only meaningful if students can confidently use the change-of-base relation log₁₀(ab) = log₁₀ a + log₁₀ b and the inverse 10⁻ˣ = antilog. Slips here — e.g. forgetting that pH = 1.30 means [H⁺] = 10⁻¹·³⁰ ≈ 5.0 × 10⁻² — are the most common cause of lost marks.
Topic 16 — Thermodynamics: Kw is an equilibrium constant, so its temperature dependence is governed by the van't Hoff equation. The endothermic autoionisation of water (ΔH ≈ +56 kJ mol⁻¹) explains quantitatively why Kw rises from 1.0 × 10⁻¹⁴ at 25°C to ≈ 5.5 × 10⁻¹⁴ at 50°C.
Topic 12.3 — Weak acid pH (Lesson 3): the calculation pH = −log₁₀[H⁺] is identical, but [H⁺] for a weak acid must come from Ka rather than from c directly. The same logarithm conventions apply.
Topic 12.4 — Buffers (Lesson 5): the Henderson-Hasselbalch equation is a rearrangement of the pH formula combined with the Ka expression. Confidence with logarithms here is a prerequisite.
Biology (pH homeostasis): physiological pH is 7.35–7.45; deviations of just 0.4 units cause acidosis or alkalosis. Quantitative pH calculation here connects to Lesson 5 buffer chemistry and to A-Level Biology topics on respiration.
pH-calculation questions on 9CH0 split AO marks as follows:
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