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

Here you will learn outcomes 14.1 to 14.14, the opening half of Topic 14 (Acid-base Equilibria), as set out by the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. That means proton transfer, pH, strong and weak acids, Ka, Kw, pKa and pKw, and the analysis of pH data. Topic 14 is assessed in Unit 4, the International A2 (IA2) unit titled Rates, Equilibria and Further Organic Chemistry. Expect the Unit 2, Topic 9B equilibrium ideas to be extended here. Titration curves, indicators, buffers and Core Practical 11 (outcomes 14.15 onwards) have their own titration curves and buffers guide.

Other pages for this topic: the condensed pH, Ka and Kw revision notes and the pH, Ka and Kw practice questions. For the Unit 2 groundwork on equilibrium, see introduction to kinetics and equilibria. You can also take a free diagnostic, print the course checklist or browse every unit from the course hub.

Outcomes 14.1 to 14.14 in one table

Outcome	Skill you need
14.1	Define Brønsted–Lowry acids and bases; see acid-base reactions as proton transfer
14.2	Identify conjugate acid-base pairs
14.3, 14.4, 14.5	Define pH; convert [H⁺] to pH and pH to [H⁺]
14.6	Tell strong and weak acids apart by degree of dissociation
14.7	Calculate the pH of a strong acid
14.8, 14.9	Write Ka; calculate the pH of a weak acid from Ka or pKa
14.10, 14.11, 14.12	Define Kw, pKa and pKw; calculate the pH of a strong base
14.13	Analyse pH readings of equimolar solutions, and of diluted strong and weak acids
14.14	Calculate Ka from the pH of a solution made from a known mass of acid

14.1 and 14.2 Brønsted–Lowry acids, bases and conjugate pairs

In Brønsted–Lowry theory, an acid donates protons (H⁺ ions) and a base accepts them. So every acid-base reaction is a proton transfer: one species loses H⁺ and another gains it.

When an acid gives away its proton, what is left can take a proton back, so it is a base. The two species form a conjugate acid-base pair. They differ by exactly one H⁺.

HA + B ⇌ A⁻ + BH⁺

Here HA and A⁻ are one pair (acid 1, base 1), and BH⁺ and B are the other (acid 2, base 2).

Water can act either way. With an acid it accepts a proton to form H₃O⁺; with a base it donates one to form OH⁻. In equations, H⁺(aq) is shorthand for H₃O⁺(aq).

Worked example 1: labelling pairs

In a mixture of concentrated nitric and sulfuric acids:

HNO₃ + H₂SO₄ ⇌ H₂NO₃⁺ + HSO₄⁻

• H₂SO₄ loses H⁺ to become HSO₄⁻, so H₂SO₄ is the acid and HSO₄⁻ is its conjugate base.
• HNO₃ gains H⁺ to become H₂NO₃⁺, so here HNO₃ is the base and H₂NO₃⁺ is its conjugate acid.

The point: “acid” and “base” describe what a species does in a particular reaction. Nitric acid acts as a base when it meets a stronger acid.

A second case, with ethylamine in water:

C₂H₅NH₂ + H₂O ⇌ C₂H₅NH₃⁺ + OH⁻

The pairs are C₂H₅NH₃⁺/C₂H₅NH₂ and H₂O/OH⁻. Write each pair with the acid first, and check the two members differ by one H⁺ only.

14.3 to 14.5 Defining and using pH

pH = −log₁₀[H⁺], with [H⁺] measured in mol dm⁻³. Rearranged, [H⁺] = 10⁻ᵖᴴ.

Because the scale is logarithmic, a change of one pH unit means a tenfold change in [H⁺]. A lower pH means a higher [H⁺].

Worked example 2: both directions

(a) A sample of rainwater has [H⁺] = 2.9 × 10⁻⁴ mol dm⁻³. pH = −log₁₀(2.9 × 10⁻⁴) = 3.54

(b) A soil extract has pH 5.36. [H⁺] = 10⁻⁵·³⁶ = 4.37 × 10⁻⁶ mol dm⁻³

Give pH to two decimal places unless told otherwise, and give [H⁺] to three significant figures.

14.6 and 14.7 Strong acids and their pH

A strong acid is (almost) fully dissociated in water: HCl(aq) → H⁺(aq) + Cl⁻(aq). A weak acid is only partly dissociated, so most of it stays as HA molecules at equilibrium: CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq).

Strength is about the degree of dissociation. Concentration is about how much acid is in each dm³. So you can have a strong acid that is dilute, or a weak acid that is concentrated.

For a strong monoprotic acid, [H⁺] = concentration of the acid.

Worked example 3: strong acid pH

(a) Nitric acid, 0.0360 mol dm⁻³. [H⁺] = 0.0360 mol dm⁻³, so pH = −log₁₀(0.0360) = 1.44.

(b) 25.0 cm³ of 0.200 mol dm⁻³ hydrochloric acid is made up to 250 cm³ with water. The new concentration is 0.200 × 25.0/250 = 0.0200 mol dm⁻³, so pH = −log₁₀(0.0200) = 1.70.

14.8 and 14.9 Ka and the pH of a weak acid

For HA(aq) ⇌ H⁺(aq) + A⁻(aq), the acid dissociation constant is

Ka = [H⁺][A⁻] / [HA]

All concentrations are at equilibrium; Ka has units of mol dm⁻³. A larger Ka means a stronger (more dissociated) acid.

pKa = −log₁₀Ka, so Ka = 10⁻ᵖᴷᵃ. A smaller pKa means a stronger acid.

To find the pH of a weak acid, make two assumptions:

1. [H⁺] = [A⁻]: all the H⁺ comes from the acid, and the tiny contribution from water’s own ionisation is ignored.
2. [HA] at equilibrium ≈ the starting concentration: so little of the acid dissociates that its concentration hardly falls.

Then Ka = [H⁺]² / [HA], which gives [H⁺] = √(Ka × [HA]). Under 14.9 the specification notes that quadratic equations will not be expected, so this approximation is the method to use.

Worked example 4: pH of pentanoic acid

Pentanoic acid has pKa = 4.84 (use this value). Find the pH of a 0.0620 mol dm⁻³ solution.

1. Ka = 10⁻⁴·⁸⁴ = 1.45 × 10⁻⁵ mol dm⁻³
2. [H⁺]² = Ka × [HA] = 1.45 × 10⁻⁵ × 0.0620
3. [H⁺] = 9.47 × 10⁻⁴ mol dm⁻³
4. pH = −log₁₀(9.47 × 10⁻⁴) = 3.02

Check assumption 2: 9.47 × 10⁻⁴ / 0.0620 × 100 ≈ 1.5%, so only about 1.5% of the acid has dissociated. The approximation is sound. This percentage is also a direct measure of the degree of dissociation from 14.6.

14.10 to 14.12 Kw, pKw and strong bases

Water self-ionises to a very small extent. The equilibrium is H₂O(l) ⇌ H⁺(aq) + OH⁻(aq), and for it the ionic product of water is

Kw = [H⁺][OH⁻]

Water’s concentration is so large that it is effectively constant, so it is not included. At 298 K, Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶.

pKw = −log₁₀Kw, so pKw = 14.00 at 298 K. Taking logs of Kw gives a handy link: pH + pOH = pKw.

Like any equilibrium constant, Kw changes with temperature (Topic 13). For a value at another temperature, use the one you are given. In pure water [H⁺] = [OH⁻] at every temperature, so pure water is always neutral, even when its pH is not 7.00.

A strong base such as NaOH is fully dissociated, so [OH⁻] equals its concentration (twice its concentration for a hydroxide such as Ba(OH)₂). Then [H⁺] = Kw/[OH⁻].

Worked example 5: barium hydroxide

Find the pH of 0.00650 mol dm⁻³ Ba(OH)₂ at 298 K.

1. Ba(OH)₂ → Ba²⁺ + 2OH⁻, so [OH⁻] = 2 × 0.00650 = 0.0130 mol dm⁻³
2. [H⁺] = Kw/[OH⁻] = 1.00 × 10⁻¹⁴ / 0.0130 = 7.69 × 10⁻¹³ mol dm⁻³
3. pH = −log₁₀(7.69 × 10⁻¹³) = 12.11

Using pKw: pOH = −log₁₀(0.0130) = 1.89, and pH = 14.00 − 1.89 = 12.11. Both routes agree.

14.13 Analysing pH data

(i) Equimolar solutions

The readings below are values you would expect for 0.0700 mol dm⁻³ solutions at 298 K.

Solution	Type	pH
Hydrochloric acid	strong acid	1.15
Ethanoic acid	weak acid	2.96
Ammonium chloride	salt of strong acid + weak base	5.20
Sodium chloride	salt of strong acid + strong base	7.00
Sodium ethanoate	salt of weak acid + strong base	8.80
Ammonia	weak base	11.05
Sodium hydroxide	strong base	12.85

How to read the pattern:

• At the same concentration, the strong acid has the lower pH because it is fully dissociated. The weak acid releases far fewer H⁺ ions.
• The same logic applies to bases: NaOH gives a higher pH than NH₃.
• Salts are not all neutral. NH₄⁺ is a weak acid (it can donate H⁺ to water), so ammonium chloride is slightly acidic. CH₃COO⁻ is a weak base (it accepts H⁺ from water, releasing OH⁻), so sodium ethanoate is slightly alkaline. Na⁺ and Cl⁻ do neither, so sodium chloride is neutral.

(ii) Dilution by 10, 100 and 1000 times

Take 0.400 mol dm⁻³ hydrochloric acid and 0.400 mol dm⁻³ ethanoic acid (pKa 4.76) and dilute each.

Dilution factor	HCl pH	Ethanoic acid pH
none	0.40	2.58
10	1.40	3.08
100	2.40	3.59
1000	3.40	4.12

• Strong acid: [H⁺] falls tenfold with each tenfold dilution, so pH rises by 1.00 each time.
• Weak acid: [H⁺] = √(Ka × c), so a tenfold fall in c lowers [H⁺] by only √10. The pH rises by about 0.5 each time. As the acid gets more dilute, a larger fraction dissociates, so the gap between the strong and weak acids narrows.
• At the highest dilution the rise for ethanoic acid is a little more than 0.5, because the fraction dissociated is no longer small and assumption 2 starts to fail.

14.14 Ka from a known mass of acid and a pH reading

Worked example 6: phenylethanoic acid

A 1.088 g sample of phenylethanoic acid, C₆H₅CH₂COOH, is dissolved and the volume brought to 250 cm³ in a volumetric flask. A calibrated pH meter reads 2.90. Calculate Ka.

1. Mr(C₆H₅CH₂COOH) = 136.0, so n = 1.088 / 136.0 = 8.00 × 10⁻³ mol
2. [HA] = 8.00 × 10⁻³ / 0.250 = 0.0320 mol dm⁻³
3. [H⁺] = 10⁻²·⁹⁰ = 1.26 × 10⁻³ mol dm⁻³, and [A⁻] = [H⁺]
4. Ka = (1.26 × 10⁻³)² / 0.0320 = 4.95 × 10⁻⁵ mol dm⁻³ (using the unrounded [H⁺])

The same two assumptions apply. Here [H⁺] is about 4% of [HA], so treating [HA] as 0.0320 is reasonable. You could then quote pKa = −log₁₀(4.95 × 10⁻⁵) = 4.31.

Common errors

• Writing a conjugate “pair” that differs by more than one H⁺, such as H₂SO₄ and SO₄²⁻.
• Forgetting the minus sign in pH = −log₁₀[H⁺], giving a negative pH for a dilute acid.
• Using the strong-acid method for a weak acid, or the square-root method for a strong acid.
• Forgetting to double [OH⁻] for a hydroxide with two OH⁻ ions per formula.
• Leaving [H₂O] in the Kw expression, or giving Kw the wrong units.
• Using 1.00 × 10⁻¹⁴ for Kw when the question gives a value at another temperature.
• In 14.14 working, using the mass of acid or the volume in cm³ directly instead of the concentration in mol dm⁻³.
• Stating the weak-acid assumptions vaguely; say which concentration is assumed equal to which.

Next steps

Try the pH, Ka and Kw practice set next, and keep the condensed notes for the last weeks before the Unit 4 exam. The free diagnostics will show which topics need more work.

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.