Chemistry • Year 12 • Module 6 • Lesson 9
HSC Exam Practice
pH of Weak Acids & Bases: Ka, Kb & ICE Tables
Kw = 1.0 × 10−14 at 25°C. Use pH + pOH = 14.00 throughout. Provide full working for all calculation questions.
Short answer
1.Short answer — definitions and principles
Define the acid dissociation constant Ka for a weak acid HA. Include the equilibrium expression.
Explain why the simplifying assumption x << c is valid when Ka/c < 0.0025 but not when Ka/c = 0.04.
Identify the conjugate base of HCN. Outline how Kb for this conjugate base can be calculated from Ka(HCN) and Kw.
Describe the three-step procedure for calculating the pH of a weak base solution from its Kb and initial concentration. Include the name of each step.
Distinguish between the ICE table method and the Henderson-Hasselbalch equation, identifying the condition under which each is the appropriate method to calculate pH after partial neutralisation of a weak acid.
Data response
2.Data response — pH vs pKa for Australian weak acids
The graph below shows the calculated pH of three Australian-context weak acids at concentrations from 0.001 to 0.500 mol/L, all at 25°C. Acetic acid (Ka = 1.8 × 10−5, wine/AWRI), hydrocyanic acid (Ka = 6.2 × 10−10, cyanide leach), and hydrofluoric acid (Ka = 6.8 × 10−4, aluminium processing).
(a) Describe the trend in pH across the three acids at the same concentration of 0.100 mol/L, identifying which acid gives the lowest pH and explaining why in terms of Ka. 2 marks
(b) Estimate the pH of 0.010 mol/L HCN from the graph. Compare this to the pH of 0.010 mol/L HCl (a strong acid at the same concentration). Explain the difference. 3 marks
(c) Account for why the HF curve (red) would diverge from the simplified formula at low concentrations (< 0.010 mol/L). Refer to the Ka/c ratio. 2 marks
3.Data response — Ka from pH measurement
A chemist at an AWRI laboratory measures the pH of a 0.0500 mol/L solution of a weak acid HA found in a wine sample. The pH meter reads 3.56.
(a) Calculate [H+] in the solution. 1 mark
(b) Calculate Ka of HA using the exact reverse-ICE expression Ka = [H+]2 / (c − [H+]). 2 marks
(c) Calculate the degree of ionisation (α) and determine whether the simplifying assumption would have been valid. 2 marks
Extended response
4.Multi-step calculation
A process chemist at Newcrest Mining works with a cyanide leach solution containing 0.0800 mol/L NaCN dissolved in water. HCN has Ka = 6.2 × 10−10.
(a) Write the equilibrium equation for the reaction of CN− with water. Calculate Kb(CN−) using Ka × Kb = Kw. 2 marks
(b) Set up a full ICE table for the equilibrium in (a). Check whether the simplifying assumption is valid (Kb/c). Calculate [OH−]. 3 marks
(c) Calculate pOH, then pH of the 0.0800 mol/L NaCN solution. State whether the solution is acidic, basic or neutral and explain why NaCN does not give a neutral solution despite being formed from a neutralisation reaction. 3 marks
5.Extended response — no scaffold
Analyse the following experimental scenario and evaluate the student’s conclusion.
“A student dissolves 60.0 mL of 0.150 mol/L ammonia (NH3, Kb = 1.8 × 10−5) in 40.0 mL of 0.150 mol/L hydrochloric acid (HCl). The student calculates the pH of the resulting solution using an ICE table with Kb(NH3) = 1.8 × 10−5, obtaining pH = 11.0. The student concludes that because NH3 is in excess, the solution is basic.”
Evaluate the student’s method and conclusion. In your response, determine the actual pH of the resulting solution and identify all errors in the student’s approach.
Chemistry • Year 12 • Module 6 • Lesson 9
Answer Key & Marking Guidelines
Section 1 • Short answer • 2 marks • Band 3
Sample response. The acid dissociation constant Ka is the equilibrium constant for the partial ionisation of a weak acid HA in water: Ka = [H+][A−] / [HA], where all concentrations are equilibrium values. A larger Ka indicates a stronger weak acid (greater degree of ionisation).
Marking notes. 1 mark for correct expression Ka = [H+][A−] / [HA]; 1 mark for identifying these as equilibrium concentrations OR correctly stating Ka measures degree of ionisation / acid strength.
Section 1 • Short answer • 3 marks • Band 4
Sample response. The simplifying assumption x << c is valid when the degree of ionisation is small (<5%). Since x = √(Ka × c), the ratio x/c = √(Ka/c). This is <5% (or 0.05) when Ka/c < 0.0025. When Ka/c = 0.04, √(0.04) = 0.20 = 20% degree of ionisation — a substantial fraction of HA has ionised, so the (c − x) denominator cannot be approximated as c. The error in [H+] would be ~20%, making the simplified pH unacceptably inaccurate.
Marking notes. 1 mark for linking validity to degree of ionisation (<5%) or Ka/c ratio criterion; 1 mark for calculating or stating that Ka/c = 0.04 gives α = 20%; 1 mark for explaining why a 20% degree of ionisation makes c − x ≠ c and the approximation fails.
Section 1 • Short answer • 2 marks • Band 3
Sample response. The conjugate base of HCN is CN− (cyanide ion), formed when HCN donates a proton. Kb(CN−) = Kw / Ka(HCN) = (1.0 × 10−14) / (6.2 × 10−10) = 1.61 × 10−5.
Marking notes. 1 mark for identifying CN− as conjugate base; 1 mark for correctly applying Kb = Kw/Ka (numerical answer not required).
Section 1 • Short answer • 3 marks • Band 3–4
Sample response. Step 1 (ICE table): Set up I, C, E rows for the equilibrium B + H2O ⇌ BH+ + OH− and solve Kb = x2/(c − x) to find [OH−] = x (using the simplifying assumption if valid). Step 2 (pOH): Calculate pOH = −log[OH−]. Step 3 (pH): Calculate pH = 14 − pOH.
Marking notes. 1 mark per step correctly described (3 marks total). The step names (ICE table, pOH, pH) must be present or implied for each mark.
Section 1 • Short answer • 3 marks • Band 4
Sample response. The ICE table method applies when only the pure weak acid (or pure weak base) is present in solution, with [A−] ≈ 0 at the start — for example, calculating the pH of 0.10 mol/L acetic acid before any NaOH is added. The Henderson-Hasselbalch equation (pH = pKa + log([A−]/[HA])) applies when both the weak acid and its conjugate base are present simultaneously — specifically after partial neutralisation, when [A−] is significant. Using the ICE table in a buffer situation (after partial neutralisation) is a conceptual error because the assumption [A−] = 0 is violated.
Marking notes. 1 mark for correct ICE table condition ([A−] = 0 / pure weak acid); 1 mark for correct H-H condition (both HA and A− present / buffer / partial neutralisation); 1 mark for explicitly stating why using ICE table in the buffer situation is incorrect.
Section 2 • Data response • 7 marks • Band 4–5
(a) 2 marks. At 0.100 mol/L: HF gives the lowest pH (approx 2.1) — it is the strongest of the three because it has the largest Ka (6.8 × 10−4) and therefore ionises most. CH3COOH gives intermediate pH (~2.9, Ka = 1.8 × 10−5) and HCN the highest pH (~5.1, Ka = 6.2 × 10−10, weakest). 1 mark for correct order (HF lowest, HCN highest); 1 mark for Ka-based explanation.
(b) 3 marks. Reading the graph at c = 0.010: HCN pH ≈ 5.6 (accept 5.4–5.7). For 0.010 mol/L HCl (strong acid), [H+] = 0.010 mol/L exactly → pH = 2.00. The difference (~3.6 pH units) arises because HCN is a very weak acid (Ka = 6.2 × 10−10) and ionises only partially; only a tiny fraction of HCN molecules donate protons, so [H+] << 0.010 mol/L. 1 mark for reading HCN pH correctly; 1 mark for pH(HCl) = 2.00; 1 mark for explanation (partial ionisation of weak acid vs complete ionisation of strong acid).
(c) 2 marks. At low concentrations, Ka/c increases for HF. E.g. at c = 0.001 mol/L: Ka/c = 6.8 × 10−4 / 0.001 = 0.68, so α = √(0.68) = 82% — far above the 5% threshold. The simplified formula pH ≈ ½(pKa − log c) assumes (c − x) ≈ c, which is completely invalid at 82% ionisation. The actual [H+] is much lower than the simplified formula predicts, so the real curve would sit higher (less acidic) than the plotted line at very low concentrations. 1 mark for Ka/c ratio argument at low c; 1 mark for stating the real pH would be higher (less acidic) and explaining why (c − x ≠ c).
Section 2 • Data response • 5 marks • Band 4–5
(a) 1 mark. [H+] = 10−3.56 = 2.754 × 10−4 mol/L (accept 2.75–2.76 × 10−4).
(b) 2 marks. Ka = [H+]2 / (c − [H+]) = (2.754 × 10−4)2 / (0.0500 − 2.754 × 10−4) = 7.584 × 10−8 / 0.04975 = 1.52 × 10−6. 1 mark for correct setup; 1 mark for correct answer.
(c) 2 marks. α = (2.754 × 10−4 / 0.0500) × 100% = 0.55% < 5% → the simplifying assumption was valid. Ka/c = 1.52 × 10−6 / 0.0500 = 3.04 × 10−5 << 0.0025 — confirms validity. 1 mark for α calculation; 1 mark for conclusion (valid).
Section 3 • Multi-step calculation • 8 marks • Band 4–5
(a) 2 marks. Equation: CN−(aq) + H2O(l) ⇌ HCN(aq) + OH−(aq). Kb(CN−) = Kw/Ka(HCN) = (1.0 × 10−14) / (6.2 × 10−10) = 1.613 × 10−5. 1 mark equation; 1 mark Kb.
(b) 3 marks. ICE table (CN− / HCN / OH−): I = 0.0800 / 0 / 0; C = −x / +x / +x; E = (0.0800 − x) / x / x. Check: Kb/c = 1.613 × 10−5 / 0.0800 = 2.02 × 10−4 < 0.0025 → assumption valid. [OH−] = x = √(Kb × c) = √(1.613 × 10−5 × 0.0800) = √(1.290 × 10−6) = 1.136 × 10−3 mol/L. Verify: α = 1.136 × 10−3/0.0800 = 1.42% < 5% ✓. 1 mark ICE table; 1 mark assumption check; 1 mark [OH−].
(c) 3 marks. pOH = −log(1.136 × 10−3) = 2.945. pH = 14.000 − 2.945 = 11.06. The solution is basic. NaCN does not give a neutral solution because CN− is the conjugate base of the weak acid HCN (Ka = 6.2 × 10−10). The term “neutralisation” refers to the reaction process, not the resulting pH. When HCN and NaOH react, the CN− formed is a moderately strong base (Kb = 1.61 × 10−5) that hydrolyses water to produce OH−, making the solution basic. Only the salt of a strong acid + strong base gives pH = 7. 1 mark pOH and pH; 1 mark “basic” conclusion; 1 mark explanation (CN− is conjugate base of weak acid HCN, hydrolyses water to give OH−).
Section 3 • Extended response • 6 marks • Band 5–6
Sample response. The student’s method contains two fundamental errors, and the conclusion requires qualification. Error 1 (stoichiometry): The student ignores the reaction between NH3 and HCl. n(NH3) = 0.150 × 0.0600 = 9.00 × 10−3 mol; n(HCl) = 0.150 × 0.0400 = 6.00 × 10−3 mol. HCl reacts completely with NH3: NH3 + HCl → NH4+ + Cl−. After reaction: n(NH3) remaining = 9.00 × 10−3 − 6.00 × 10−3 = 3.00 × 10−3 mol; n(NH4+) formed = 6.00 × 10−3 mol. Total volume = 100.0 mL = 0.1000 L. The solution contains both NH3 and NH4+ — it is a buffer, not a pure NH3 solution. Error 2 (method): The student applies the Kb ICE table to a solution that already contains significant NH4+. This violates the ICE table assumption [BH+] = 0 initially. The correct method is Henderson-Hasselbalch: pKa(NH4+) = 14 − pKb(NH3) = 14 − 4.74 = 9.26. pH = pKa + log(n(NH3)/n(NH4+)) = 9.26 + log(3.00 × 10−3 / 6.00 × 10−3) = 9.26 + log(0.5) = 9.26 − 0.301 = 8.96. Conclusion evaluation: The student’s conclusion that the solution is basic is correct (pH 8.96 > 7) because NH3 is in excess. However, the pH of 11.0 is grossly incorrect: the actual pH is 8.96, approximately 2 pH units lower. The large error arises from ignoring both the neutralisation reaction and the common ion suppression effect of NH4+ on NH3 ionisation.
Marking notes: 1 mark — calculates n(NH3) and n(HCl) correctly; 1 mark — identifies neutralisation reaction occurs, giving n(NH3)rem = 3.00 × 10−3 mol and n(NH4+) = 6.00 × 10−3 mol; 1 mark — identifies the solution as a buffer and states ICE table is inappropriate (NH4+ ≠ 0); 1 mark — correctly applies Henderson-Hasselbalch using pKa(NH4+) = 9.26; 1 mark — correct pH = 8.96 (accept 8.9–9.0); 1 mark — evaluates the student’s conclusion: correct direction (basic) but incorrect pH value, identifying ignoring neutralisation and common ion effect as causes of error.