Bronsted-Lowry Acids, Kw, Weak Acid Ka, pH Titration Curves & Buffer Solution Calculations

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Brønsted-Lowry Acids/Bases & The pH Scale

A Brønsted-Lowry acid is a proton (H⁺) donor; a Brønsted-Lowry base is a proton (H⁺) acceptor. Every acid-base reaction involves two conjugate acid-base pairs differing by H⁺: HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq). pH = -log₁₀[H⁺] and [H⁺] = 10^(-pH). For a strong monoprotic acid (HCl, HNO₃), [H⁺] = [Acid]; for a strong diprotic acid (H₂SO₄), [H⁺] ≈ 2 × [H₂SO₄].

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Ionic Product of Water (Kw), Strong Bases & Neutral Water at T > 298 K

Water undergoes endothermic auto-ionisation: H₂O(l) ⇌ H⁺(aq) + OH⁻(aq) (ΔH = +57 kJ mol⁻¹), where Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K. For a strong base (NaOH, KOH, Ba(OH)₂ where [OH⁻] = 2[Ba(OH)₂]), find [OH⁻], then [H⁺] = Kw / [OH⁻], then pH = -log₁₀[H⁺]. In pure water at any temperature, [H⁺] = [OH⁻], so [H⁺] = √Kw and the water is always neutral; as temperature increases, equilibrium shifts right (endothermic), Kw increases, [H⁺] increases, and the pH of pure neutral water decreases below 7.0.

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Weak Acids, Ka, pKa & The Pure Weak Acid Approximations

Weak acids (e.g. ethanoic acid, CH₃COOH) partially dissociate: HA(aq) ⇌ H⁺(aq) + A⁻(aq), with Ka = [H⁺][A⁻] / [HA] (units: mol dm⁻³) and pKa = -log₁₀(Ka) (a smaller pKa or larger Ka means a stronger weak acid). For a pure weak acid in water, two approximations are made: 1. [H⁺] ≈ [A⁻] (neglecting H⁺ from water dissociation), and 2. [HA]_eqm ≈ [HA]_initial (since dissociation is tiny), simplifying to: Ka ≈ [H⁺]² / [HA], or [H⁺] = √(Ka × [HA]).

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pH Titration Curves, Half-Neutralisation (pH = pKa) & Indicators

Plotting pH against volume of alkali added shows a vertical equivalence section: 1. Strong Acid + Strong Base: vertical pH 3 to 11 (equivalence at pH 7; methyl orange [3.1-4.4] and phenolphthalein [8.3-10.0] both work). 2. Weak Acid + Strong Base: starts ~pH 3, buffer region, vertical pH 7 to 11 (equivalence > pH 7; Only phenolphthalein works). 3. Strong Acid + Weak Base (NH₃): vertical pH 3 to 7 (equivalence < pH 7; Only methyl orange works). 4. Weak Acid + Weak Base: no vertical section (no indicator works; use pH meter). At the half-neutralisation volume of a weak acid with strong base, [HA] = [A⁻], so Ka = [H⁺] and pH = pKa.

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5

Buffer Solutions: Mechanism of Action (Acidic & Basic Buffers)

A buffer solution resists changes in pH when small amounts of acid or alkali are added (or on dilution). An acidic buffer (pH < 7) contains a large reservoir of weak acid (HA, e.g. CH₃COOH) and its conjugate base salt (A⁻, e.g. CH₃COONa), prepared either by mixing HA with solid/aqueous NaA OR by partially neutralising excess weak acid HA with NaOH. 1. On adding H⁺ (acid): the large reservoir of conjugate base ions A⁻ reacts with H⁺ (A⁻ + H⁺ → HA), shifting HA ⇌ H⁺ + A⁻ to the left to remove added H⁺. 2. On adding OH⁻ (alkali): OH⁻ reacts with H⁺ (H⁺ + OH⁻ → H₂O), and the large reservoir of HA dissociates (shifting equilibrium right: HA + OH⁻ → A⁻ + H₂O) to replace the H⁺.

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6

Buffer pH Calculations (never use [H⁺]² in a Buffer)

In a buffer solution, the salt (e.g. CH₃COONa) is 100% dissociated so [A⁻] ≠ [H⁺]! Never use Ka = [H⁺]² / [HA] for a buffer. Instead, rearrange the Ka expression directly: [H⁺] = Ka × ([HA] / [A⁻]) = Ka × (moles of HA / moles of A⁻) (since both are in the same total volume V, volumes cancel), then pH = -log₁₀[H⁺]. If a buffer is made by reacting excess weak acid (n_HA initial) with strong base (n_OH⁻) or adding acid/alkali to an existing buffer, use an ice mole subtraction first: n(HA)_after = n(HA)_initial - n(OH⁻) and n(A⁻)_after = n(A⁻)_initial + n(OH⁻).

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1. Q1

   Calculate the pH at 298 K (Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶) of: (a) 0.150 mol dm⁻³ barium hydroxide, Ba(OH)₂(aq), and (b) 0.200 mol dm⁻³ propanoic acid, CH₃CH₂COOH(aq) (Ka = 1.35 × 10⁻⁵ mol dm⁻³). Give your answers to 2 decimal places. (5 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. (a) Barium hydroxide is a strong dibasic alkali (Ba(OH)₂ → Ba²⁺ + 2OH⁻), so [OH⁻] = 2 × 0.150 = 0.300 mol dm⁻³ [1 mark]. 2. Calculate [H⁺] using Kw: [H⁺] = Kw ÷ [OH⁻] = (1.00 × 10⁻¹⁴) ÷ 0.300 = 3.333 × 10⁻¹⁴ mol dm⁻³, so pH = -log₁₀(3.333 × 10⁻¹⁴) = 13.48 [1 mark]. 3. (b) Propanoic acid is a pure weak monoprotic acid, so Ka ≈ [H⁺]² ÷ [CH₃CH₂COOH] [1 mark]. 4. Calculate [H⁺]: [H⁺] = √(Ka × [HA]) = √(1.35 × 10⁻⁵ × 0.200) = √(2.70 × 10⁻⁶) = 1.643 × 10⁻³ mol dm⁻³ [1 mark]. 5. Calculate pH = -log₁₀(1.643 × 10⁻³) = 2.78 [1 mark].

2. Q2

   A buffer solution is prepared by mixing 50.0 cm³ of 0.400 mol dm⁻³ ethanoic acid (CH₃COOH, Ka = 1.74 × 10⁻⁵ mol dm⁻³) with 25.0 cm³ of 0.300 mol dm⁻³ sodium hydroxide (NaOH). Calculate the pH of the resulting buffer solution to 2 decimal places. (5 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Calculate initial moles of weak acid and strong base: n(CH₃COOH)_initial = (50.0 ÷ 1000) × 0.400 = 0.0200 mol; n(NaOH) = (25.0 ÷ 1000) × 0.300 = 0.00750 mol [1 mark]. 2. NaOH reacts completely with CH₃COOH in a 1:1 ratio (CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O), so moles of unreacted acid remaining: n(CH₃COOH) = 0.0200 - 0.00750 = 0.0125 mol [1 mark]. 3. Moles of conjugate base salt formed: n(CH₃COO⁻) = 0.00750 mol [1 mark]. 4. Substitute into the buffer expression (total volume 75.0 cm³ cancels): [H⁺] = Ka × [n(CH₃COOH) ÷ n(CH₃COO⁻)] = (1.74 × 10⁻⁵) × (0.0125 ÷ 0.00750) = 2.90 × 10⁻⁵ mol dm⁻³ [1 mark]. 5. Calculate pH = -log₁₀(2.90 × 10⁻⁵) = 4.54 [1 mark].

3. Q3

   To the buffer solution in the previous question (containing 0.0125 mol of CH₃COOH and 0.00750 mol of CH₃COO⁻), a student adds 2.00 cm³ of 1.00 mol dm⁻³ hydrochloric acid (HCl). Calculate the new pH of the buffer solution and explain how the buffer minimised the change in pH. (4 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Moles of H⁺ added from HCl = (2.00 ÷ 1000) × 1.00 = 0.00200 mol [1 mark]. 2. The added H⁺ ions react with the large reservoir of ethanoate ions (CH₃COO⁻ + H⁺ → CH₃COOH), shifting the equilibrium to the left to remove almost all the added H⁺ [1 mark]. 3. New moles after reaction: n(CH₃COOH) increases to 0.0125 + 0.00200 = 0.0145 mol; n(CH₃COO⁻) decreases to 0.00750 - 0.00200 = 0.00550 mol [1 mark]. 4. New [H⁺] = (1.74 × 10⁻⁵) × (0.0145 ÷ 0.00550) = 4.587 × 10⁻⁵ mol dm⁻³, giving new pH = -log₁₀(4.587 × 10⁻⁵) = 4.34 (only a 0.20 unit drop) [1 mark].

4. Q4

   25.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid (HCl) is titrated with 40.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide (NaOH) at 298 K (Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶). Calculate the initial pH before any NaOH is added, the volume of NaOH at the equivalence point, and the final pH after all 40.0 cm³ of NaOH has been added. (5 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Initial pH of 0.100 mol dm⁻³ HCl (strong monoprotic acid): [H⁺] = 0.100 mol dm⁻³, so pH = -log₁₀(0.100) = 1.00 [1 mark]. 2. Since HCl + NaOH → NaCl + H₂O (1:1 ratio) and both solutions are 0.100 mol dm⁻³, the equivalence volume is 25.0 cm³ of NaOH [1 mark]. 3. After 40.0 cm³ of NaOH is added, calculate moles of excess OH⁻: n(HCl) = 0.00250 mol, n(NaOH) = 0.00400 mol → excess n(OH⁻) = 0.00400 - 0.00250 = 0.00150 mol [1 mark]. 4. Calculate total volume V_total = 25.0 + 40.0 = 65.0 cm³ (0.0650 dm³), and find [OH⁻] = 0.00150 ÷ 0.0650 = 0.02308 mol dm⁻³ [1 mark]. 5. Calculate [H⁺] = (1.00 × 10⁻¹⁴) ÷ 0.02308 = 4.333 × 10⁻¹³ mol dm⁻³, giving final pH = -log₁₀(4.333 × 10⁻¹³) = 12.36 [1 mark].

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