Edexcel International A Level Chemistry: Acid-base equilibria: pH, Ka and Kw (YCH11) – Revision Notes

Use these notes for fast recall of outcomes 14.1 to 14.14, from Topic 14 (Acid-base Equilibria). The source document is the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. Everything here is Unit 4 (Rates, Equilibria and Further Organic Chemistry) content, so it is International A2 (IA2), and it extends Unit 2, Topic 9B. For full explanations and worked examples, use the pH, Ka and Kw study guide.

When the self-test below feels easy, move on to the practice questions. Track your progress with the printable checklist, try a free diagnostic, or return to the course hub.

Definitions (14.1, 14.3, 14.6, 14.8, 14.10, 14.12)

Term	Meaning
Brønsted–Lowry acid	Proton (H⁺) donor
Brønsted–Lowry base	Proton (H⁺) acceptor
Acid-base reaction	A transfer of a proton from acid to base
Conjugate pair	Two species that differ by one H⁺
pH	−log₁₀[H⁺], with [H⁺] in mol dm⁻³
Strong acid	Fully (or almost fully) dissociated in water
Weak acid	Only partly dissociated in water
Ka	[H⁺][A⁻]/[HA] at equilibrium, for HA ⇌ H⁺ + A⁻
pKa	−log₁₀Ka
Kw	[H⁺][OH⁻] in water or any aqueous solution
pKw	−log₁₀Kw

Keep “strong” and “concentrated” apart. Strength is the degree of dissociation. Concentration is the amount of acid per dm³.

Formulas

Use	Formula
pH from [H⁺]	pH = −log₁₀[H⁺]
[H⁺] from pH	[H⁺] = 10⁻ᵖᴴ
Strong monoprotic acid	[H⁺] = c
Weak acid	[H⁺] = √(Ka × c)
Ka from pKa	Ka = 10⁻ᵖᴷᵃ
Ionic product	Kw = [H⁺][OH⁻]; its value at 298 K is 1.00 × 10⁻¹⁴ mol² dm⁻⁶
Strong base	[H⁺] = Kw/[OH⁻]
Log form	pH + pOH = pKw (14.00 at 298 K)
Ka from data	Ka = [H⁺]²/c

Units: Ka in mol dm⁻³; Kw in mol² dm⁻⁶; pH, pKa and pKw have no units.

Conjugate pairs in steps (14.1, 14.2)

1. Find which species loses H⁺ (the acid) and which gains it (the base).
2. Pair each with the species it turns into.
3. Check each pair differs by exactly one H⁺.

Example: HSO₄⁻ + HCO₃⁻ ⇌ SO₄²⁻ + H₂CO₃. Pairs: HSO₄⁻/SO₄²⁻ and H₂CO₃/HCO₃⁻. Here HCO₃⁻ is the base.

Some species, such as H₂O, HCO₃⁻ and HSO₄⁻, can act as either an acid or a base.

Method: pH of a strong acid (14.7)

1. A strong monoprotic acid (HCl, HNO₃, HBr) is fully dissociated, so [H⁺] equals the acid concentration.
2. If the acid has been diluted, find the new concentration first: c₂ = c₁ × V₁/V₂.
3. pH = −log₁₀[H⁺].

Worked reminder: 0.0475 mol dm⁻³ hydrobromic acid. [H⁺] = 0.0475; pH = 1.32.

Degree of dissociation (14.6)

The degree of dissociation is the fraction (or percentage) of acid molecules that have split into ions. For a strong acid it is close to 100%. For a weak acid, find it as [H⁺]/c × 100. For the pKa 4.13 acid in the reminder below, 2.86 × 10⁻³/0.110 × 100 is about 2.6%, so the weak-acid assumptions are safe. If this percentage climbs above a few per cent, the square-root method starts to overestimate [H⁺].

Method: pH of a weak acid (14.9)

1. If given pKa, convert: Ka = 10⁻ᵖᴷᵃ.
2. Write Ka = [H⁺]²/[HA].
3. Rearrange: [H⁺] = √(Ka × [HA]).
4. pH = −log₁₀[H⁺].

The two assumptions:

• [H⁺] = [A⁻], because H⁺ from the ionisation of water is negligible.
• [HA] at equilibrium equals the starting concentration, because the degree of dissociation is small.

You will not be asked to solve a quadratic (the specification says so), so always use these assumptions.

Worked reminder: an acid with pKa 4.13 at 0.110 mol dm⁻³. Ka = 7.41 × 10⁻⁵; [H⁺] = √(7.41 × 10⁻⁵ × 0.110) = 2.86 × 10⁻³; pH = 2.54.

Method: pH of a strong base (14.11)

1. Find [OH⁻]. For NaOH or KOH it equals the concentration. For Ba(OH)₂ or Sr(OH)₂ double it.
2. Either [H⁺] = Kw/[OH⁻] and pH = −log₁₀[H⁺], or pOH = −log₁₀[OH⁻] and pH = pKw − pOH.

Worked reminder: 0.0215 mol dm⁻³ KOH. [H⁺] = 1.00 × 10⁻¹⁴/0.0215 = 4.65 × 10⁻¹³; pH = 12.33.

Kw and temperature

The ionisation of water is an equilibrium, so Kw depends on temperature. Use 1.00 × 10⁻¹⁴ mol² dm⁻⁶ only at 298 K, and use the value given for any other temperature. Whatever the temperature, the H⁺ and OH⁻ in pure water come from the same molecules in equal numbers, so pure water counts as neutral at any pH it reaches.

Method: Ka from mass and pH (14.14)

1. Moles of acid = mass / Mr.
2. Concentration = moles / volume in dm³.
3. [H⁺] = 10⁻ᵖᴴ.
4. Ka = [H⁺]² / concentration.
5. If asked, pKa = −log₁₀Ka.

Worked reminder: 0.390 g of ethanoic acid (Mr 60.0) is made up to 200 cm³, and the pH is 3.12.

• n = 0.390/60.0 = 6.50 × 10⁻³ mol
• c = 6.50 × 10⁻³/0.200 = 0.0325 mol dm⁻³
• [H⁺] = 10⁻³·¹² = 7.59 × 10⁻⁴ mol dm⁻³
• Ka = (7.59 × 10⁻⁴)²/0.0325 = 1.77 × 10⁻⁵ mol dm⁻³

Check: [H⁺] is about 2.3% of c, so the assumption that [HA] stays at 0.0325 mol dm⁻³ holds.

Reading pH data (14.13)

Equimolar solutions (same concentration):

• Strong acid gives the lowest pH; weak acid a higher one.
• Strong base gives the highest pH; weak base a lower one.
• Salt of a strong acid and a strong base (NaCl, KNO₃): pH about 7.
• Salt of a strong acid and a weak base (NH₄Cl): slightly acidic, since NH₄⁺ donates H⁺ to water.
• Salt of a weak acid and a strong base (CH₃COONa): slightly alkaline, since CH₃COO⁻ accepts H⁺ from water and leaves OH⁻.

Dilution by 10, 100 and 1000 times:

Acid	pH change per tenfold dilution	Reason
Strong	+1	[H⁺] = c, so [H⁺] falls tenfold
Weak	about +0.5	[H⁺] = √(Ka × c), so [H⁺] falls by √10

At very high dilution the weak acid’s rise creeps above 0.5, because a larger fraction dissociates and the second assumption weakens.

Must-know distinctions

• Strong vs concentrated: degree of dissociation versus amount per dm³.
• Ka vs pKa: larger Ka means stronger acid; smaller pKa means stronger acid.
• Kw vs pKw: Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ and pKw = 14.00, both at 298 K only.
• Neutral vs pH 7: a solution is neutral when its H⁺ and OH⁻ concentrations match; that happens at pH 7.00 only when the temperature is 298 K.
• H⁺ vs H₃O⁺: the same species in aqueous equations; H₃O⁺ shows water acting as a base.

Quick self-test

1. Give the conjugate base of HSO₄⁻ and the conjugate acid of HSO₄⁻.
2. Calculate the pH of a solution with [H⁺] = 5.2 × 10⁻⁴ mol dm⁻³.
3. Calculate [H⁺] in a solution of pH 3.65.
4. Calculate the pH of 0.0018 mol dm⁻³ nitric acid.
5. Calculate the pH of 0.0230 mol dm⁻³ sodium hydroxide at 298 K.
6. A weak acid has Ka = 7.4 × 10⁻⁵ mol dm⁻³. Calculate pKa.
7. A weak monoprotic acid at 0.0850 mol dm⁻³ gives a pH reading of 2.96. Calculate Ka.
8. By how much does the pH of a strong acid change when it is diluted 100 times?
9. By roughly how much does the pH of a weak acid change when it is diluted 100 times?
10. Predict whether aqueous sodium ethanoate is acidic, neutral or alkaline, and give the reason.
11. State the two assumptions used to calculate the pH of a weak acid.

Answers

1. Conjugate base: SO₄²⁻. Conjugate acid: H₂SO₄.
2. pH = −log₁₀(5.2 × 10⁻⁴) = 3.28.
3. [H⁺] = 10⁻³·⁶⁵ = 2.24 × 10⁻⁴ mol dm⁻³.
4. Strong acid, so [H⁺] = 0.0018; pH = 2.74.
5. [H⁺] = 1.00 × 10⁻¹⁴/0.0230 = 4.35 × 10⁻¹³ mol dm⁻³; pH = 12.36.
6. pKa = −log₁₀(7.4 × 10⁻⁵) = 4.13.
7. [H⁺] = 10⁻²·⁹⁶ = 1.10 × 10⁻³; Ka = (1.10 × 10⁻³)²/0.0850 = 1.41 × 10⁻⁵ mol dm⁻³ (using the unrounded [H⁺]).
8. It rises by 2 units.
9. It rises by about 1 unit (0.5 per tenfold dilution).
10. Alkaline: the ethanoate ion is a weak base and accepts H⁺ from water, leaving OH⁻.
11. [H⁺] = [A⁻] (ionisation of water ignored), and [HA] at equilibrium equals the starting concentration (dissociation small).

Where marks are usually lost

• Naming a pair such as H₂SO₄/SO₄²⁻ as conjugate; the members must differ by one H⁺.
• Using a one-way arrow when showing a weak acid dissociating; the ⇌ sign is what shows partial dissociation.
• Writing Ka as [H⁺]²/[HA] in a definition; the general expression uses [H⁺][A⁻]/[HA].
• Rounding [H⁺] early, then quoting a pH that is out by 0.01 or more.
• Forgetting to double [OH⁻] for Ba(OH)₂, Sr(OH)₂ or Ca(OH)₂.
• Using pH = 14 − pOH at a temperature where pKw is not 14.00.
• Giving Kw the units of Ka, or giving pH or pKa units.
• Using cm³ instead of dm³ when finding the concentration of an acid from its mass.
• Saying “a weak acid is dilute” instead of “a weak acid is only partly dissociated”.
• Calling a solution with pH below 7 acidic when the question is about pure water at a higher temperature.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited. Pages in this set teach Topic 14: Acid-base Equilibria (Unit 4), outcomes 14.1 to 14.14.