The problems are on the acids pH buffers worksheet. Try them on your own first, then compare with the answers below.
Quick answer: A Brønsted–Lowry acid donates a proton and a base accepts one. These worked answers correct the 20 problems of the Chemistry 11-12 worksheet on acids pH buffers, from conjugate pairs in ammonia to a very dilute acid: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the equation, the substitution and the result with its unit. Check your logarithms with a calculator and compare the method as well as the final value, because small rounding differences in the last digit are acceptable.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB R3.1
Problem 1 Conjugate pairs in ammonia
1. A Brønsted–Lowry base is a proton (H⁺) acceptor.2. NH₃ accepts H⁺ from water, so NH₃ is the base and H₂O is the acid.
3. Pair 1: NH₃ / NH₄⁺ (base / conjugate acid). Pair 2: H₂O / OH⁻ (acid / conjugate base).
Problem 2 pH of dilute hydrochloric acid
1. HCl → H⁺ + Cl⁻ (complete, strong acid).2. [H⁺] = 0.0250 mol/dm³.
3. pH = −log(0.0250) = 1.60
Problem 3 Hydrogen ion concentration from pH
1. [H⁺] = 10⁻⁴·⁶⁰ = 2.5 × 10⁻⁵ mol/dm³ (2.51 × 10⁻⁵).2. Neutral water: [H⁺] = 1.0 × 10⁻⁷. Ratio = 2.51 × 10⁻⁵ ÷ 1.0 × 10⁻⁷ = about 250 times larger (10²·⁴ = 251).
Problem 4 pOH and pH of sodium hydroxide
1. NaOH is a strong base, so [OH⁻] = 0.0040 mol/dm³.2. pOH = −log(0.0040) = 2.40.
3. pH = 14 − 2.40 = 11.60
Problem 5 Comparing acid strength by pKa
1. pKa = −log(6.3 × 10⁻⁵) = 4.20.2. Chloroethanoic acid is stronger: its pKa is lower, so its Ka is larger and it ionises more.
3. Ka = 10⁻²·⁸⁵ = 1.4 × 10⁻³ mol/dm³.
Problem 6 pH of methanoic acid
1. Ka = [H⁺][HCOO⁻] / [HCOOH].2. [H⁺] = √(1.6 × 10⁻⁴ × 0.150) = √(2.4 × 10⁻⁵) = 4.9 × 10⁻³ mol/dm³.
3. pH = −log(4.9 × 10⁻³) = 2.31.
4. Fraction ionised = 4.9 × 10⁻³ ÷ 0.150 = 3.3 %, below 5 %, so [HA] ≈ c is acceptable.
Problem 7 Percentage ionisation of ethanoic acid
1. [H⁺] = √(1.8 × 10⁻⁵ × 0.0500) = √(9.0 × 10⁻⁷) = 9.5 × 10⁻⁴ mol/dm³.2. % ionised = 9.49 × 10⁻⁴ ÷ 0.0500 × 100 = 1.9 %.
3. It rises (to about 6 %). Dilution shifts CH₃COOH ⇌ CH₃COO⁻ + H⁺ towards the side with more particles, which is the ions.
Problem 8 Diluting nitric acid
1. n(H⁺) = 0.0100 dm³ × 1.00 mol/dm³ = 0.0100 mol; c = 0.0100 ÷ 0.500 = 0.0200 mol/dm³.2. pH = −log(0.0200) = 1.70.
3. Water added = 500 − 10.0 = 490 cm³.
Problem 9 Excess acid after partial neutralisation
1. n(H⁺) = 0.0300 × 0.120 = 3.60 × 10⁻³ mol; n(OH⁻) = 0.0200 × 0.100 = 2.00 × 10⁻³ mol.2. Excess H⁺ = 3.60 × 10⁻³ − 2.00 × 10⁻³ = 1.60 × 10⁻³ mol.
3. Total volume 50.0 cm³: [H⁺] = 1.60 × 10⁻³ ÷ 0.0500 = 0.0320 mol/dm³; pH = 1.49.
Problem 10 Ka from a measured pH
1. [H⁺] = 10⁻²·⁸⁵ = 1.41 × 10⁻³ mol/dm³.2. [A⁻] = [H⁺]; Ka = (1.41 × 10⁻³)² ÷ 0.200 = 1.0 × 10⁻⁵ mol/dm³.
3. pKa = 5.00.
Problem 11 A lactic acid buffer
1. pH = pKa + log([A⁻]/[HA]).2. pH = 3.85 + log(0.15 ÷ 0.30) = 3.85 − 0.30 = 3.55.
Problem 12 How a buffer neutralises added acid
1. CH₃COO⁻ + H⁺ → CH₃COOH.2. One A⁻ is used up for each H⁺: HA rises from 4 to 5 and A⁻ falls from 3 to 2.
3. The added ion is removed by the conjugate partner, so the ratio [A⁻]/[HA] changes but both species stay plentiful; log(ratio) moves very little, so the pH hardly changes.
Problem 13 Ionic product of water at 50 °C
1. [H⁺] = [OH⁻] = √(5.5 × 10⁻¹⁴) = 2.3 × 10⁻⁷ mol/dm³.2. pH = −log(2.35 × 10⁻⁷) = 6.63.
3. Neutral, because [H⁺] = [OH⁻]. Neutral pH is 7 only at 25 °C; ionisation of water is endothermic, so Kw grows with temperature.
Problem 14 pH of aqueous ammonia
1. [OH⁻] = √(1.8 × 10⁻⁵ × 0.100) = 1.34 × 10⁻³ mol/dm³.2. pOH = −log(1.34 × 10⁻³) = 2.87.
3. pH = 14.00 − 2.87 = 11.13
Problem 15 Reading pKa from a titration curve
1. The steep rise is centred at 24.0 cm³.2. At 12.0 cm³ the pH is 4.87 (half-equivalence, so pH = pKa).
3. pKa 4.87 matches propanoic acid; Ka = 10⁻⁴·⁸⁷ = 1.35 × 10⁻⁵ mol/dm³.
Problem 16 Equivalence point and indicator
1. Equivalence volume 25.0 cm³; n = 2.50 × 10⁻³ mol in 50.0 cm³: c = 0.0500 mol/dm³.2. Kb = 1.0 × 10⁻¹⁴ ÷ 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰. [OH⁻] = √(5.6 × 10⁻¹⁰ × 0.0500) = 5.3 × 10⁻⁶ mol/dm³; pOH = 5.28; pH = 8.72.
3. Phenolphthalein: its range (8.2–10.0) contains the equivalence pH and lies on the steep part of the curve. Methyl orange changes in the buffer region, far too early.
Problem 17 Preparing an ammonia buffer
1. 9.40 = 9.25 + log(ratio), so log(ratio) = 0.15 and ratio = 10⁰·¹⁵ = 1.41.2. n(NH₄⁺) = 0.100 × 0.200 = 0.0200 mol; n(NH₃) = 1.41 × 0.0200 = 0.0283 mol.
3. V = 0.0283 mol ÷ 1.00 mol/dm³ = 0.0283 dm³ = 28.3 cm³, then make up to 250 cm³ with water.
Problem 18 Adding acid to a buffer and to water
1. Equal concentrations, so pH = pKa = 4.74.2. Start: 0.0500 mol each. HCl converts A⁻ to HA: HA = 0.0600 mol, A⁻ = 0.0400 mol. pH = 4.74 + log(0.0400 ÷ 0.0600) = 4.74 − 0.18 = 4.56 (4.564).
3. [H⁺] = 0.0100 ÷ 0.500 = 0.0200 mol/dm³; pH = 1.70.
4. The buffer changes by about 0.18 units; the water changes by about 5.3 units.
Problem 19 Mixing weak acid and alkali
1. n(HA) = 0.0400 × 0.150 = 6.00 × 10⁻³ mol; n(OH⁻) = 0.0100 × 0.200 = 2.00 × 10⁻³ mol.2. HA + OH⁻ → A⁻ + H₂O: HA = 4.00 × 10⁻³ mol, A⁻ = 2.00 × 10⁻³ mol.
3. pH = 4.20 + log(2.00 ÷ 4.00) = 4.20 − 0.30 = 3.90.
4. V = 6.00 × 10⁻³ ÷ 0.200 = 0.0300 dm³ = 30.0 cm³.
Problem 20 A very dilute acid
1. pH 8.00 would be alkaline, but adding an acid to water cannot make it alkaline: water’s own ionisation (10⁻⁷ mol/dm³) is far bigger than the acid’s contribution and must be included.2. x = 1.0 × 10⁻⁸ + 1.0 × 10⁻¹⁴/x, so x² − 1.0 × 10⁻⁸x − 1.0 × 10⁻¹⁴ = 0. x = (1.0 × 10⁻⁸ + √(4.01 × 10⁻¹⁴)) ÷ 2 = 1.05 × 10⁻⁷ mol/dm³; pH = 6.98.
