Covers Lessons 7–12: Conjugate Pairs & Amphiprotic Substances, pH/pOH for Strong Acids & Bases, Ka/Kb & ICE Tables for Weak Acids/Bases, Enthalpy Comparison, Mastery, and Ka/pKa Comparisons.
Lesson Summaries — Quick Review
A conjugate acid–base pair differs by exactly one H⁺. In every Brønsted-Lowry reaction, two pairs are present. Amphiprotic substances (e.g. HCO₃⁻, HSO₄⁻, H₂O, amino acids) can act as either acid or base depending on conditions. Water autoionises: 2H₂O ⇌ H₃O⁺ + OH⁻ with Kw = [H₃O⁺][OH⁻] = 10⁻¹⁴ at 25°C.
pH = −log[H₃O⁺]. For strong acids: [H₃O⁺] = concentration (complete dissociation). For strong bases: find [OH⁻] first, then pOH = −log[OH⁻], then pH = 14 − pOH. Dilution: track the new concentration after mixing. pH + pOH = 14 at 25°C. For polyprotic strong acids, account for stoichiometry (H₂SO₄ → 2H⁺).
Weak acids use Ka = [H⁺][A⁻]/[HA]. Set up an ICE table, assume x << [HA]₀ for simplification (valid when Ka is very small), then solve for x = [H⁺]. pH = −log(x). Similarly Kb for weak bases. Key: never assume complete dissociation for weak acids/bases. If the 5% rule is violated, use the quadratic formula.
Strong acid + strong base gives the most exothermic ΔH ≈ −57 kJ mol⁻¹ (all energy from bond formation H⁺ + OH⁻ → H₂O). Weak acid + strong base or strong acid + weak base gives less exothermic values because energy is also consumed breaking the weak electrolyte's partial association. The difference in ΔH is the enthalpy of ionisation of the weak acid/base.
Consolidation of all pH calculation types: strong acid, strong base, weak acid (ICE), weak base (ICE), mixing problems, and dilution. Band 6 responses connect the mathematical answer to particle-level chemistry. Common errors: using [H⁺] = concentration for weak acids; forgetting that mixing changes volume; losing track of significant figures.
pKa = −log Ka. A lower pKa means a stronger acid (larger Ka). Use Ka/pKa to rank acid strengths and predict the direction of equilibrium in acid–base reactions. The stronger acid reacts with the stronger base to form the weaker conjugate pair. Ka and Kb are related: Ka × Kb = Kw = 10⁻¹⁴ at 25°C.
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Multiple Choice — 20 Questions (1 mark each)
Which of the following ions can act as BOTH an acid AND a base according to the Brønsted-Lowry model?
What is the pH of a 0.025 mol L⁻¹ Ca(OH)₂ solution at 25°C?
Propanoic acid (Ka = 1.34 × 10⁻⁵) has a concentration of 0.100 mol L⁻¹. Using the approximation method, what is its approximate pH?
At 25°C, the Ka of acetic acid (CH₃COOH) is 1.8 × 10⁻⁵. What is the Kb of the acetate ion (CH₃COO⁻)?
A student measures the enthalpy of neutralisation for CH₃COOH(aq) + NaOH(aq) and finds it is less exothermic than −57 kJ mol⁻¹. The most likely reason is:
Which pair is a true conjugate acid-base pair?
At 50?C, Kw is greater than 1.0 ? 10???, so neutral water has:
25.0 mL of 0.200 mol L?? HNO? is mixed with 35.0 mL of 0.100 mol L?? NaOH. What is the pH of the final solution?
For which weak acid solution is the square-root approximation LEAST likely to be valid without checking the 5% rule?
What is the approximate pH of 0.100 mol L?? NH?(aq) at 25?C? Kb = 1.8 ? 10??.
Which acid is strongest?
20.0 mL of 0.100 mol L?? CH?COOH is mixed with 10.0 mL of 0.100 mol L?? NaOH. Which method is now required to calculate pH?
Using the Module 6 strong-acid treatment of sulfuric acid, what is the pH of 0.0100 mol L?? H?SO? at 25?C?
Which salt solution is expected to be acidic?
If an acid has Ka = 3.2 ? 10??, its pKa is closest to:
Which neutralisation is expected to be closest to ?57 kJ mol???
Which solution has the LOWER pH?
If the pKa of HA is 4.76 at 25?C, the Kb of A? is closest to:
At 50?C, Kw = 5.5 ? 10???. The pH of neutral water is closest to:
Which statement is correct?
Short Answer — 3 Questions
5 marksA 0.20 mol L⁻¹ solution of ammonia (NH₃) has Kb = 1.8 × 10⁻⁵. Using an ICE table, calculate the pH of this solution. Show all working and state any assumptions you make.
4 marks50.0 mL of 0.10 mol L⁻¹ HCl is mixed with 50.0 mL of 0.10 mol L⁻¹ NaOH. Calculate the pH of the resulting solution. Then, recalculate the pH if only 30.0 mL of NaOH is added instead.
3 marksThree acids have the following Ka values: HF (Ka = 6.8 × 10⁻⁴), CH₃COOH (Ka = 1.8 × 10⁻⁵), HCN (Ka = 6.2 × 10⁻¹⁰). Rank these acids from strongest to weakest, calculate their pKa values, and predict which acid will have the lowest pH at equal concentrations.
HCO₃⁻ (hydrogen carbonate) is amphiprotic. As an acid: HCO₃⁻ → H⁺ + CO₃²⁻. As a base: HCO₃⁻ + H⁺ → H₂CO₃. Cl⁻, SO₄²⁻ and NO₃⁻ are conjugate bases of strong acids and do not act as Brønsted-Lowry acids.
Ca(OH)₂ is a strong base dissociating fully: Ca(OH)₂ → Ca²⁺ + 2OH⁻. [OH⁻] = 2 × 0.025 = 0.050 mol L⁻¹. pOH = −log(0.050) = 1.30. pH = 14 − 1.30 = 12.70.
ICE: [H⁺] = √(Ka × C) = √(1.34×10⁻⁵ × 0.100) = √(1.34×10⁻⁶) = 1.158×10⁻³ mol L⁻¹. pH = −log(1.158×10⁻³) ≈ 2.94. Check: x/C = 1.16% < 5% → approximation valid.
Ka × Kb = Kw = 1.0×10⁻¹⁴. Kb = 1.0×10⁻¹⁴ / 1.8×10⁻⁵ = 5.56×10⁻¹⁰ ≈ 5.6×10⁻¹⁰. Note: options C and D appear the same — both should show D as correct with the correct value.
Acetic acid is a weak acid: it does not fully dissociate. Some energy must be absorbed to drive its ionisation. This ionisation enthalpy is endothermic, so the net enthalpy of the overall neutralisation is less negative (less exothermic) than −57 kJ mol⁻¹.
True conjugate pairs differ by exactly one proton. HSO?? and SO??? differ by one H?, so they are a conjugate acid-base pair. H?SO? and SO??? differ by two protons, which is the classic one-proton-rule error.
Neutral means [H?] = [OH?], not ?pH = 7? under all conditions. When Kw increases with temperature, both [H?] and [OH?] increase equally, so neutral water can have pH below 7 while still remaining neutral.
n(HNO?) = 0.0250 ? 0.200 = 5.00 ? 10?? mol. n(NaOH) = 0.0350 ? 0.100 = 3.50 ? 10?? mol. Excess H? = 1.50 ? 10?? mol. V(total) = 0.0600 L. [H?] = 1.50 ? 10?? / 0.0600 = 0.0250 mol L??. pH = 1.60.
The approximation is least safe when Ka is relatively large and concentration is relatively low, because x is then a larger fraction of the initial concentration. Option D combines the largest Ka here with the smallest concentration, making the 5% rule most likely to fail.
For NH?: [OH?] = ?(Kb ? c) = ?(1.8 ? 10?? ? 0.100) = 1.34 ? 10?? mol L??. pOH = 2.87. pH = 14.00 ? 2.87 = 11.13.
Lower pKa means larger Ka and therefore a stronger acid. Of the three, HF has the lowest pKa (3.17), so it is strongest. HCN, with pKa 9.21, is the weakest.
Partial neutralisation of a weak acid by strong base produces both HA and A? in significant amounts. That is a buffer, so Henderson-Hasselbalch is the correct method, not a strong-acid excess calculation or a pure weak-acid ICE table.
Using the Module 6 treatment, H?SO? contributes 2H? per formula unit. [H?] = 2 ? 0.0100 = 0.0200 mol L??. pH = ?log(0.0200) = 1.70.
NH?? is the conjugate acid of the weak base NH?, so it hydrolyses to produce H? and makes solution acidic. NaNO? is neutral, while Na?CO? and CH?COONa are basic because their anions are conjugate bases of weak acids.
pKa = ?log(3.2 ? 10??) = 4.49. This is just the logarithmic form of Ka and should always be interpreted with lower pKa meaning stronger acid.
Only strong acid + strong base neutralisation is consistently close to ?57 kJ mol??, because no additional ionisation step is needed. Any weak acid or weak base pair gives a less exothermic value due to the endothermic ionisation enthalpy.
0.0010 mol L?? HCl gives [H?] = 0.0010 and pH = 3.00. For 1.0 mol L?? CH?COOH, [H?] ? ?(1.8 ? 10?? ? 1.0) = 4.24 ? 10??, so pH ? 2.37. The concentrated weak acid has the lower pH because concentration and strength both matter.
pKa = 4.76 implies Ka = 10????? = 1.74 ? 10??. Therefore Kb = Kw / Ka = 1.0 ? 10??? / 1.74 ? 10?? = 5.75 ? 10???. The closest answer is 5.8 ? 10???.
For neutral water, [H?] = [OH?] = ?(Kw) = ?(5.5 ? 10???) = 2.35 ? 10?? mol L??. pH = ?log(2.35 ? 10??) = 6.63. Neutral is therefore below pH 7 at this higher temperature.
The most important first step in Module 6 calculations is classification: is the final solution strong acid/base excess, pure weak acid/base, salt hydrolysis, or a buffer? Once that chemical identity is clear, the correct maths follows. Most high-frequency errors come from skipping this identification step.
Reaction: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
ICE table: I: 0.20, 0, 0 | C: −x, +x, +x | E: 0.20−x, x, x (1 mark)
Assumption: x << 0.20 (valid if Kb small). (1 mark)
Kb = x²/0.20 → x² = 1.8×10⁻⁵ × 0.20 = 3.6×10⁻⁶ → x = 1.897×10⁻³ mol L⁻¹. (1 mark)
Check: x/0.20 = 0.95% < 5% → valid. (1 mark)
[OH⁻] = 1.897×10⁻³ → pOH = 2.72 → pH = 14 − 2.72 = 11.28. (1 mark)
Scenario 1 (equal volumes, equal concentrations): n(HCl) = 0.050 × 0.10 = 0.005 mol; n(NaOH) = 0.005 mol. They exactly neutralise → salt solution only → pH = 7.00. (2 marks)
Scenario 2 (30 mL NaOH): n(NaOH) = 0.030 × 0.10 = 0.003 mol. Excess HCl = 0.005 − 0.003 = 0.002 mol. Total volume = 80 mL = 0.080 L. [H⁺] = 0.002/0.080 = 0.025 mol L⁻¹. pH = −log(0.025) = 1.60. (2 marks)
Ranking (strongest → weakest): HF > CH₃COOH > HCN. (1 mark — based on Ka values, larger Ka = stronger acid)
pKa values: HF: pKa = −log(6.8×10⁻⁴) = 3.17; CH₃COOH: pKa = 4.74; HCN: pKa = 9.21. (1 mark)
HF will have the lowest pH at equal concentrations because it has the largest Ka → greatest [H⁺] at equilibrium. (1 mark)
Tick when you've finished the quiz and reviewed the marking guide.