Chemistry • Year 12 • Module 6 • Lesson 9

pH of Weak Acids & Bases: Ka, Kb & ICE Tables

Synthesise ICE table calculations, evaluate data from real Australian chemistry contexts, and construct extended responses that reach an evidence-based judgement.

Master • Band 5–6 • Synthesis & Evaluation

1. ICE table mastery — citric acid in Australian citrus processing

Scenario

Citrus Australia reports that freshly squeezed lemon juice contains approximately 5.4 g of citric acid per 100 mL (M = 192 g/mol; treat as a monoprotic acid using Ka1 = 7.4 × 10−4). A food scientist at a Mildura citrus processing plant needs to calculate the pH of pure lemon juice, determine whether the simplifying assumption is valid, and compare citric acid with acetic acid (Ka = 1.8 × 10−5) at equal concentrations.

The scientist is also provided with pH-meter data from three 0.280 mol/L citric acid samples: pH readings of 1.78, 1.79 and 1.79. The food scientist notes a discrepancy between the pH-meter reading and a colleague's ICE table calculation that used the simplifying assumption.

0 5 10 15 20 25 % error in pH 0 0.010 0.020 0.030 0.040 0.050 Ka/c ratio 5% threshold Ka/c = 0.0025 CH₃COOH (0.280 mol/L) Citric acid (0.280 mol/L) HF (0.020 mol/L) Assumption valid region Use quadratic
Figure 1. Approximate % error in calculated pH when using the simplifying assumption vs the Ka/c ratio. Highlighted points show CH3COOH, citric acid and HF at stated concentrations. Calculated from ICE table exact vs approximate solutions.

In your response you must:

  1. Calculate the molar concentration of citric acid in lemon juice (5.4 g / 100 mL; M = 192 g/mol).
  2. Check the validity of the simplifying assumption using Ka/c and degree of ionisation.
  3. Use the ICE table to calculate pH (use the simplifying assumption only if valid; otherwise use the quadratic).
  4. Use the Figure 1 data point for citric acid to explain why the food scientist's colleague found a discrepancy when using the assumption for 0.280 mol/L solutions, but not for 0.10 mol/L solutions.
  5. Compare citric acid and acetic acid strength using Ka and pKa, giving one named Australian application for each.
8 marks

Write your extended response below. Show all ICE table working.

Stuck? Step 1: c = (5.4/192) / 0.100 = ? mol/L. Step 2: Ka/c = ? Compare to 0.0025. Step 3: Build ICE table, solve for x. Step 4: Read the Figure — how does the error change with Ka/c?

2. Ka, conjugate relationships and partial neutralisation — AWRI wine buffering

Scenario

The Australian Wine Research Institute (AWRI) investigates the buffering capacity of wine during malolactic fermentation, where acidity changes affect flavour. A winemaker adds 0.0400 mol of NaOH to 500 mL of wine containing 0.200 mol/L acetic acid (CH3COOH, Ka = 1.8 × 10−5). The winemaker also wants to know the pH of the acetate ion solution that forms if 100 mL of the partially neutralised wine is diluted to 250 mL and all remaining acetic acid is removed (theoretical).

Table 1. Ka, pKa and Kb for the acetic acid/acetate system at 25°C.
Species Ka pKa Kb (conjugate base) pKb
CH3COOH (acetic acid) 1.8 × 10−5 4.74 Kb(CH3COO−) = ? ?
CH3COO− (acetate ion) — — calculated from Kw/Ka —

In your response you must:

  1. Calculate n(CH3COOH) and n(NaOH) in the initial wine solution; determine which reagent limits the neutralisation.
  2. Use stoichiometry to find n(CH3COOH) remaining and n(CH3COO−) formed after neutralisation goes to completion.
  3. Apply Henderson-Hasselbalch to calculate the pH of the resulting mixture. State why the ICE table method is not appropriate here.
  4. Calculate Kb(CH3COO−) using Ka × Kb = Kw, then use an ICE table to find the pH of a 0.100 mol/L pure sodium acetate solution. Show the three-step [OH−] → pOH → pH pathway.
  5. Evaluate the claim: “Adding NaOH to acetic acid wine until the half-equivalence point gives a pH equal to pKa = 4.74.” State whether this is correct, and explain why using the Henderson-Hasselbalch equation.
8 marks

Write your extended response below. Show all working including ICE table, stoichiometry, and H-H calculation.

Stuck? Step 1: n = c × V. Step 2: HA + OH− → A− + H2O, 1:1 stoichiometry. Step 3: H-H needs n(A−) and n(HA). Step 4: Kb = Kw/Ka; ICE table with Kb gives [OH−]. Step 5: at half-equivalence, n(A−) = n(HA), so log ratio = 0.

3. Source critique — identify the scientific flaw

Read the following excerpt from an online chemistry study guide and answer the questions below. 6 marks

“For any weak acid HA with concentration c and acid dissociation constant Ka, the pH can always be calculated using the simplified formula pH = ½(pKa − log c). This means pH depends only on pKa and the concentration, and the simplifying assumption is always valid because weak acids by definition have small Ka values.”

— Online Study Guide, Year 12 Chemistry, accessed 2026.

3.1 Identify the specific scientific flaw in the excerpt’s claim about the simplifying assumption. 2 marks

3.2 Construct a counterexample using HF (Ka = 6.8 × 10−4, c = 0.010 mol/L): calculate the degree of ionisation using the simplified formula and show it exceeds 5%. 2 marks

3.3 Explain how an experimentalist could detect that the simplified formula was giving an incorrect answer without using the quadratic formula. Reference the pH meter data from Scenario 1 in Q1. 2 marks

Stuck? For 3.1: the claim says small Ka is sufficient. But is Ka alone the relevant quantity? For 3.2: x = √(Ka × c), then α = x/c × 100%. For 3.3: compare the calculated pH to the pH-meter reading.
Answers — Do not peek before attempting

Q1 — Marking criteria (8 marks)

  • 1 mark: Correct concentration: c = (5.4 / 192) / 0.100 = 0.281 mol/L (accept 0.28).
  • 1 mark: Assumption check: Ka/c = 7.4 × 10−4 / 0.281 = 2.63 × 10−3, just above 0.0025 threshold. α = √(2.63 × 10−3) × 100% = 5.1% > 5% — assumption is marginally invalid; the quadratic is required for full precision. (Accept: assumption marginally invalid; OR if student accepts <5% rounding, still award mark with correct reasoning.)
  • 2 marks: Correct ICE table and calculation. Quadratic: x = (−7.4 × 10−4 + √((7.4 × 10−4)2 + 4 × 7.4 × 10−4 × 0.281)) / 2 = (−7.4 × 10−4 + √(5.476 × 10−7 + 8.316 × 10−4)) / 2 ≈ (−7.4 × 10−4 + 9.12 × 10−3) / 2 = 4.19 × 10−3 mol/L. pH = −log(4.19 × 10−3) ≈ 2.38. Simplified: x = √(7.4 × 10−4 × 0.281) = 4.56 × 10−3; pH = 2.34 (this is the approximate value showing the discrepancy). Accept either value with working.
  • 2 marks: Figure 1 explanation: the Figure shows that for acetic acid (Ka/c very small), the error is <1% — assumption is fine. For citric acid at 0.280 mol/L, Ka/c = 2.64 × 10−3, placing the data point just past the 5% threshold, giving ~1–3% error in pH. At 0.10 mol/L citric acid, Ka/c = 7.4 × 10−3 — higher ratio — error is greater — so the discrepancy is more pronounced at lower concentrations. Student should reference the trend in the Figure correctly. 1 mark for Figure reference; 1 mark for correct trend explanation.
  • 2 marks: Correct comparison: Ka(citric) = 7.4 × 10−4 > Ka(acetic) = 1.8 × 10−5; pKa(citric) = 3.13 < pKa(acetic) = 4.74 → citric acid is the stronger weak acid. Named applications: acetic acid in Australian wine (AWRI); citric acid in Mildura/Queensland citrus processing, fruit preservatives, or cleaning products. 1 mark per correct pKa/Ka comparison with direction; 1 mark for named Australian applications of each.

Q2 — Marking criteria (8 marks)

  • 1 mark: n(CH3COOH) = 0.200 × 0.500 = 0.100 mol. n(NaOH) = 0.0400 mol. NaOH is the limiting reagent (0.0400 < 0.100).
  • 1 mark: n(CH3COOH) remaining = 0.100 − 0.0400 = 0.0600 mol. n(CH3COO−) formed = 0.0400 mol (1:1 stoichiometry).
  • 2 marks: H-H: pH = pKa + log(n(A−)/n(HA)) = 4.74 + log(0.0400/0.0600) = 4.74 + log(0.6667) = 4.74 − 0.176 = 4.56. The ICE table is inappropriate here because the solution already contains significant [CH3COO−] from neutralisation — the ICE table assumes [A−] = 0 initially, which is violated; this is a buffer system requiring Henderson-Hasselbalch. 1 mark pH; 1 mark explanation of ICE table limitation.
  • 2 marks: Kb(CH3COO−) = Kw/Ka = (1.0 × 10−14)/(1.8 × 10−5) = 5.56 × 10−10. ICE table for 0.100 mol/L CH3COO−: check Kb/c = 5.56 × 10−9 << 0.0025 → valid. [OH−] = √(5.56 × 10−10 × 0.100) = √(5.56 × 10−11) = 7.46 × 10−6 mol/L. pOH = −log(7.46 × 10−6) = 5.13. pH = 14 − 5.13 = 8.87. 1 mark for Kb calculation; 1 mark for correct 3-step pH pathway with answer.
  • 2 marks: Claim is correct. At the half-equivalence point, exactly half the acid has been neutralised: n(A−) = n(HA). Henderson-Hasselbalch: pH = pKa + log(n(A−)/n(HA)) = pKa + log(1) = pKa + 0 = 4.74. The log term is zero because the ratio equals 1. 1 mark for identifying claim as correct; 1 mark for H-H derivation showing log(1) = 0.

Q3 — Source critique (6 marks)

3.1 The flaw is that the guide claims the simplifying assumption is “always valid” for weak acids because they have small Ka. This is incorrect: validity depends on the ratio Ka/c, not Ka alone. A small Ka paired with a very small concentration c can still give Ka/c > 0.0025 and degree of ionisation > 5%. 1 mark for identifying that Ka/c (not Ka alone) is the criterion; 1 mark for explaining the role of concentration.

3.2 HF (Ka = 6.8 × 10−4, c = 0.010): x = √(6.8 × 10−4 × 0.010) = √(6.8 × 10−6) = 2.61 × 10−3. α = (2.61 × 10−3 / 0.010) × 100% = 26.1% > 5%. This demonstrates the assumption fails despite Ka being “small” (10−4 order). 1 mark for α calculation; 1 mark for “> 5%, assumption fails” conclusion.

3.3 An experimentalist can compare the pH calculated by the simplified formula with the pH meter reading. In Scenario 1, the pH meter gave 1.79 for 0.280 mol/L citric acid. The simplified formula gives pH = ½(pKa − log c) = ½(3.13 − log 0.280) = ½(3.13 + 0.553) = 1.84 — a difference of ~0.05 pH units. When the experimental pH and the calculated pH diverge by more than measurement uncertainty (≈0.01–0.02 pH units), this signals that the simplified formula is inaccurate. 1 mark for describing pH meter comparison method; 1 mark for correct calculation showing discrepancy using Scenario 1 data.

HSCScience • Chemistry • Year 12 • Module 6 • Lesson 9 • Worksheet 3 (Master) • Outcomes: CH11/12-1, CH11/12-2, CH11/12-3, CH11/12-5, CH12-13