A-Level Chemistry 9701 / Topic 25 / A2

Putting a number on acid strength.

Water itself ionises a tiny amount, and that fixes the link between [H⁺] and [OH⁻]. Strong acids and bases give their pH straight from the concentration; weak acids need the acid dissociation constant, K_a, and a square root.

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The key idea

Water ionises slightly: H₂O ⇌ H⁺ + OH⁻, and the ionic product of water K_w = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K. pH = −log[H⁺]; for a strong acid [H⁺] equals the acid concentration, and for a strong base [H⁺] = K_w ÷ [OH⁻], so 0.0100 mol dm⁻³ NaOH has pH 12.0. A weak acid HA is only partly ionised, described by the acid dissociation constant K_a = [H⁺][A⁻] ÷ [HA], and pK_a = −log K_a (smaller pKa, stronger acid). Assuming [H⁺] = [A⁻] and that very little HA ionises, [H⁺] = √(K_a × c): 0.100 mol dm⁻³ ethanoic acid (K_a 1.74 × 10⁻⁵ mol dm⁻³) has pH 2.88.

Fig. 1: the pH scale at 298 K, with a strong acid, a weak acid of the same concentration and a strong base

Section 01

Water, Kw and strong acids and bases.

Choose a strong acid, then a strong base, in the simulator and change the concentration. Watch how [H⁺] and [OH⁻] always multiply to the same value.

Solution	Key concentration	Working	pH
0.100 mol dm⁻³ HCl	[H⁺] = 0.100	−log 0.100	1.00
0.0100 mol dm⁻³ NaOH	[OH⁻] = 0.0100	[H⁺] = 1.00 × 10⁻¹⁴ ÷ 0.0100 = 1.00 × 10⁻¹²	12.0
0.0250 mol dm⁻³ NaOH	[OH⁻] = 0.0250	[H⁺] = 4.00 × 10⁻¹³	12.4
0.0500 mol dm⁻³ Ba(OH)₂	[OH⁻] = 0.100 (two OH⁻ per formula)	[H⁺] = 1.00 × 10⁻¹³	13.0

Neutral means [H⁺] = [OH⁻], not pH 7. Ionisation of water is endothermic, so K_w rises with temperature: at 323 K, K_w = 5.48 × 10⁻¹⁴ and pure water has pH 6.63 but is still neutral.
Tenfold dilution of a strong acid raises its pH by exactly 1 unit.

Stage 1 · Learn

Check what the sim just showed you

Four quick checks. Each correct answer earns XP and records this skill on your profile.

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What is the expression and value of K_w at 298 K?

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What is the pH of 0.0100 mol dm⁻³ sodium hydroxide at 298 K?

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Acid P has pK_a 3.75 and acid Q has pK_a 4.76. Which statement is correct?

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What is the pH of 0.100 mol dm⁻³ ethanoic acid (K_a = 1.74 × 10⁻⁵ mol dm⁻³)?

Section 02

Weak acids and Ka.

Switch the simulator to a weak acid and compare its pH with a strong acid of the same concentration. Most of the weak acid molecules stay un-ionised.

Step	0.100 mol dm⁻³ ethanoic acid, K_a = 1.74 × 10⁻⁵ mol dm⁻³
Equilibrium	CH₃COOH ⇌ CH₃COO⁻ + H⁺
Expression	K_a = [H⁺][CH₃COO⁻] ÷ [CH₃COOH]
Assumptions	[H⁺] = [CH₃COO⁻] (water's H⁺ ignored); [CH₃COOH] at equilibrium ≈ 0.100
[H⁺]	√(1.74 × 10⁻⁵ × 0.100) = 1.32 × 10⁻³ mol dm⁻³
pH	2.88 (compare 1.00 for 0.100 mol dm⁻³ HCl)

pKa: pK_a = −log(1.74 × 10⁻⁵) = 4.76. A smaller pKa means a larger K_a and a stronger acid.
Ka from pH: 0.200 mol dm⁻³ of a weak acid has pH 2.72, so [H⁺] = 10^−2.72 = 1.91 × 10⁻³ and K_a = (1.91 × 10⁻³)² ÷ 0.200 = 1.81 × 10⁻⁵ mol dm⁻³.
Dilution: tenfold dilution of a weak acid raises its pH by only 0.5 (2.88 to 3.38), because [H⁺] depends on √c.

Examiner trap

Do not set [H⁺] equal to the concentration of a weak acid; that is only true for strong acids. For a strong base, find [OH⁻] first, then use K_w; quoting −log[OH⁻] as the pH is the classic slip (that is pOH). Remember that Ba(OH)₂ releases two OH⁻ per formula unit, and that pure water is neutral at every temperature even when its pH is not 7.

Stage 2 · Exam

Exam-style questions

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What is the pH of 0.0250 mol dm⁻³ NaOH at 298 K?

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A 0.200 mol dm⁻³ solution of a weak monobasic acid has pH 2.72. What is K_a?

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0.100 mol dm⁻³ ethanoic acid (pH 2.88) is diluted to 0.0100 mol dm⁻³. What is the new pH?

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At 323 K, K_w = 5.48 × 10⁻¹⁴ mol² dm⁻⁶. Which describes pure water at 323 K?

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Six original Cambridge-style questions

Kw, pH of strong acids and bases, Ka and pKa, and the pH of weak acids from Ka and concentration.