Chemistry • Year 12 • Module 6 • Lesson 9

pH of Weak Acids & Bases: K_a, K_b & ICE Tables

Lock in the core vocabulary, recall the ICE table structure, check the simplifying assumption, and connect K_a × K_b = K_w before you attempt any calculations.

Build • Band 3–4 • Vocab & Structure

1. Term–definition match

Write the matching term from this list into the right-hand column. Terms: acid dissociation constant (K_a), base dissociation constant (K_b), ICE table, small-x (simplifying) assumption, degree of ionisation (α), pK_a, conjugate pair, K_w. 8 marks

#	Definition	Matching term
1.1	The equilibrium constant for the partial ionisation of a weak acid: K = [H⁺][A⁻] / [HA].
1.2	The equilibrium constant for the partial ionisation of a weak base: K = [BH⁺][OH⁻] / [B].
1.3	A three-row table (Initial, Change, Equilibrium) used to track concentrations from start to equilibrium for a weak acid or base.
1.4	The approximation c − x ≈ c, valid when K_a/c < 0.0025 (degree of ionisation < 5%).
1.5	The percentage of weak acid molecules that actually ionise in solution: ([H⁺]/c) × 100%.
1.6	Equal to −log₁₀(K_a); lower value means stronger acid.
1.7	A specific acid-base pair related by one proton transfer, e.g. CH₃COOH and CH₃COO⁻.
1.8	The product K_a × K_b for a conjugate pair at 25°C; equals 1.0 × 10⁻¹⁴.

Stuck? Revisit the Key Terms panel and formula strip at the top of Lesson 9.

2. Build and label an ICE table for acetic acid

Acetic acid (CH₃COOH, K_a = 1.8 × 10⁻⁵) partially ionises in water according to the equation below. Complete the ICE table by filling in every blank cell. Use x for the unknown equilibrium change. 9 marks (1 per blank)

CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq)

Row	[CH₃COOH] (mol/L)	[H⁺] (mol/L)	[CH₃COO⁻] (mol/L)
Initial (I)	c
Change (C)			+x
Equilibrium (E)		x

2.1 Write the K_a expression using the Equilibrium row values from your table.

2.2 State the simplifying assumption and the simplified expression for [H⁺] when it is valid.

Stuck? Revisit Card 1 (ICE table structure) in Lesson 9.

3. True or false — with correction

Circle T or F. If false, write the corrected statement. 10 marks (1 T/F + 1 correction each)

3.1 For a strong acid at concentration c, [H⁺] = c because ionisation is complete, so no ICE table is needed. T / F

3.2 In the Initial row of an ICE table for a pure weak acid, [H⁺] should be set to 10⁻⁷ mol/L (from water autoionisation). T / F

3.3 The simplifying assumption x << c is valid whenever K_a has a negative exponent (i.e. K_a < 1). T / F

3.4 For a weak base B, the three-step route to pH is: find [OH⁻] from the ICE table → pOH = −log[OH⁻] → pH = 14 − pOH. T / F

3.5 K_a × K_b = K_w applies to any acid and any base dissolved in the same solution at 25°C. T / F

Stuck? Revisit Cards 1, 2 and 3 in Lesson 9 for ICE table rules, the assumption check, and conjugate pair relationships.

4. Interpret the K_a / K_b data table

The table below lists real K_a and K_b values for five species relevant to Australian chemistry contexts. Use it to answer questions 4.1–4.5. 8 marks

Weak acid / base	Formula	K_a or K_b (25°C)	pK_a or pK_b	Australian context
Acetic (ethanoic) acid	CH₃COOH	K_a = 1.8 × 10⁻⁵	4.74	Wine fermentation (AWRI research)
Hydrofluoric acid	HF	K_a = 6.8 × 10⁻⁴	3.17	Aluminium ore processing
Hydrocyanic acid	HCN	K_a = 6.2 × 10⁻¹⁰	9.21	Cyanide leach, Newcrest Mining
Ammonia	NH₃	K_b = 1.8 × 10⁻⁵	4.74	Fertiliser production (Haber process)
Methylamine	CH₃NH₂	K_b = 4.4 × 10⁻⁴	3.36	Pharmaceutical synthesis

4.1 Rank the three weak acids (CH₃COOH, HF, HCN) from strongest to weakest, giving a reason based on the K_a values. 2 marks

4.2 Calculate K_b for the cyanide ion (CN⁻), the conjugate base of HCN. Show your working. 2 marks

4.3 Notice that K_a(CH₃COOH) = K_b(NH₃) = 1.8 × 10⁻⁵. Does this mean acetic acid and ammonia are a conjugate pair? Explain. 2 marks

4.4 HCN is used in cyanide leach processes at Newcrest Mining operations. The leach solution pH is carefully controlled. Would an increase in pH (more basic conditions) favour more or less HCN dissociation? Explain using the equilibrium expression. 2 marks

Stuck? For 4.2 use K_b = K_w / K_a. For 4.3 think about what “conjugate pair” actually means.

5. Cloze passage — complete the procedure

Fill each blank with one word or expression from the word bank below. Each word is used once only. 8 marks

Word bank: ICE table • x²/(c − x) • simplifying assumption • 5% • quadratic • pOH • 14 • degree of ionisation

To find the pH of a weak acid solution, first draw a complete (i) _______________________ with three rows labelled I, C and E. Substituting the Equilibrium row values into the K_a expression gives K_a = (ii) _______________________. Before solving, check whether the (iii) _______________________ is valid by computing K_a/c. If this ratio is less than 0.0025, the approximation is valid and x = √(K_a × c). Always verify afterwards: calculate the (iv) _______________________ as (x/c) × 100% and confirm it is less than (v) _______________________. If the check fails, solve the full (vi) _______________________ equation instead. For a weak base, the ICE table gives [OH⁻] = x; then calculate (vii) _______________________ = −log[OH⁻], and finally pH = (viii) _______________________ − pOH.

Stuck? Revisit Cards 1, 2 and 3 in Lesson 9 and the formula strip at the top of the lesson.

Answers — Do not peek before attempting

Q1 — Term–definition matches

1.1 acid dissociation constant (K_a). 1.2 base dissociation constant (K_b). 1.3 ICE table. 1.4 small-x (simplifying) assumption. 1.5 degree of ionisation (α). 1.6 pK_a. 1.7 conjugate pair. 1.8 K_w.

Q2 — ICE table for acetic acid

Complete table:

I row: [CH₃COOH] = c; [H⁺] = 0; [CH₃COO⁻] = 0.
C row: [CH₃COOH] = −x; [H⁺] = +x; [CH₃COO⁻] = +x.
E row: [CH₃COOH] = c − x; [H⁺] = x; [CH₃COO⁻] = x.

2.1 K_a = x² / (c − x).

2.2 Simplifying assumption: when x << c, c − x ≈ c, so K_a ≈ x²/c, giving [H⁺] = x = √(K_a × c). Valid when K_a/c < 0.0025 (degree of ionisation < 5%).

Q3 — True / false with correction

3.1 True. Strong acids fully dissociate; [H⁺] = c directly.

3.2 False. Correction: In the Initial row, [H⁺] = 0. The water autoionisation contribution (10⁻⁷ mol/L) is negligible compared to any weak acid at practical concentrations.

3.3 False. Correction: The assumption is valid when K_a/c < 0.0025 — both K_a and concentration c matter. A small K_a with a very small c can still give a high degree of ionisation.

3.4 True.

3.5 False. Correction: K_a × K_b = K_w applies only to a specific conjugate acid–base pair (HA and its conjugate base A⁻). It does not apply to an arbitrary acid and base dissolved together.

Q4 — K_a / K_b data table questions

4.1 Strongest to weakest: HF (K_a = 6.8 × 10⁻⁴) > CH₃COOH (K_a = 1.8 × 10⁻⁵) > HCN (K_a = 6.2 × 10⁻¹⁰). Higher K_a = greater degree of ionisation = stronger weak acid. 1 mark for correct order; 1 mark for reason linked to K_a magnitude.

4.2 K_b(CN⁻) = K_w / K_a(HCN) = (1.0 × 10⁻¹⁴) / (6.2 × 10⁻¹⁰) = 1.61 × 10⁻⁵. 1 mark working; 1 mark answer.

4.3 No. Acetic acid and ammonia are not a conjugate pair. A conjugate pair is related by the gain or loss of exactly one proton from the same species (e.g. CH₃COOH / CH₃COO⁻). CH₃COOH and NH₃ are two completely different molecular species; the fact that their K values happen to be equal is coincidental, not a conjugate relationship. 1 mark for “no”; 1 mark for explaining conjugate pair definition.

4.4 Increasing pH (adding OH⁻) removes H⁺ from the right-hand side of HCN ⇌ H⁺ + CN⁻. By Le Chatelier’s principle, the equilibrium shifts right, favouring more HCN dissociation. So more basic conditions lead to more dissociation of HCN. 1 mark equilibrium shift; 1 mark correct direction conclusion.

Q5 — Cloze passage

(i) ICE table • (ii) x²/(c − x) • (iii) simplifying assumption • (iv) degree of ionisation • (v) 5% • (vi) quadratic • (vii) pOH • (viii) 14.