S4 Chemistry · Unit 16

Acids and Bases

Arrhenius · Brønsted-Lowry · Lewis theories · pH and pOH · Strong vs weak · Ka and Kb · Neutralisation · Buffers · Indicators

16.1 Acid-Base Theories 16.2 Strong & Weak Acids/Bases 16.3 pH and Calculations 16.4 Ka, Kb and pKa 16.5 Neutralisation 16.6 Salt Hydrolysis 16.7 Buffers 16.8 Indicators & Titrations Exercises Quiz Unit Test
16.1

Acid-Base Theories

Arrhenius Theory (1884)

The earliest theory, applicable only to aqueous solutions:

Arrhenius Acid A substance that dissociates in water to produce H⁺ ions (protons). e.g. HCl → H⁺ + Cl⁻; H₂SO₄ → 2H⁺ + SO₄²⁻
Arrhenius Base A substance that dissociates in water to produce OH⁻ ions. e.g. NaOH → Na⁺ + OH⁻; Ca(OH)₂ → Ca²⁺ + 2OH⁻

Limitations: Cannot explain why NH₃ is a base (no OH⁻ in its formula), or acid-base reactions in non-aqueous solvents.

Brønsted-Lowry Theory (1923)

A broader theory based on proton (H⁺) transfer:

Brønsted-Lowry Acid A proton (H⁺) donor — loses a proton to another species.
Brønsted-Lowry Base A proton (H⁺) acceptor — gains a proton from an acid.

In every Brønsted-Lowry acid-base reaction, a proton is transferred. There is always a conjugate acid-base pair:

HCl + H₂O → H₃O⁺ + Cl⁻ acid₁ + base₂ → acid₂ + base₁ (HCl/Cl⁻ is conjugate pair 1; H₃O⁺/H₂O is conjugate pair 2) NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ base₁ + acid₂ → acid₁ + base₂ (NH₄⁺/NH₃ is conjugate pair 1; H₂O/OH⁻ is conjugate pair 2)
💡
Conjugate acid-base pairs differ by ONE proton (H⁺) HCl and Cl⁻ are a conjugate pair (differ by H⁺). NH₄⁺ and NH₃ are a conjugate pair. H₂O can act as either acid (donates H⁺ → OH⁻) or base (accepts H⁺ → H₃O⁺) — water is amphiprotic.

Lewis Theory (1923)

The most general theory — based on electron pair donation:

Lewis Acid An electron pair acceptor — accepts a lone pair of electrons to form a coordinate bond.
Lewis Base An electron pair donor — donates a lone pair to a Lewis acid.
BF₃ + NH₃ → F₃B←NH₃ (BF₃ = Lewis acid, has empty p orbital; NH₃ = Lewis base) AlCl₃ + Cl⁻ → AlCl₄⁻ (AlCl₃ = Lewis acid; Cl⁻ = Lewis base) Fe³⁺ + 6H₂O → [Fe(H₂O)₆]³⁺ (Fe³⁺ = Lewis acid; H₂O = Lewis base) Cu²⁺ + 4NH₃ → [Cu(NH₃)₄]²⁺ (Cu²⁺ = Lewis acid; NH₃ = Lewis base)

Lewis theory explains all Brønsted-Lowry reactions (H⁺ is a Lewis acid — accepts lone pair from a base) and extends to reactions without H⁺ transfer, including complex ion formation and many organic reactions.

TheoryAcidBaseScope
ArrheniusProduces H⁺ in waterProduces OH⁻ in waterAqueous only
Brønsted-LowryH⁺ donorH⁺ acceptorAqueous + non-aqueous; explains NH₃ as base
Lewise⁻ pair acceptore⁻ pair donorMost general; includes complex ions, BF₃ reactions
16.2

Strong and Weak Acids and Bases

Strong Acids

A strong acid is one that is completely (fully) dissociated in aqueous solution — all molecules ionise:

HCl(aq) → H⁺(aq) + Cl⁻(aq) 100% ionised HNO₃(aq) → H⁺(aq) + NO₃⁻(aq) 100% ionised H��SO₄(aq) → 2H⁺(aq) + SO₄²⁻(aq) 100% first ionisation (2nd: Ka₂ = 0.012)

The six common strong acids: HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄. Strong acids give pH very close to −log[acid].

Weak Acids

A weak acid is partially dissociated in aqueous solution — only a small fraction of molecules ionise. The dissociation is an equilibrium:

CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq) Ka = 1.8×10⁻⁵ mol dm⁻³ HCN(aq) ⇌ H⁺(aq) + CN⁻(aq) Ka = 4.9×10⁻¹⁰ mol dm⁻³ H��CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq) Ka₁ = 4.3×10⁻⁷ mol dm⁻³ H��PO₄(aq) ⇌ H⁺(aq) + H₂PO₄⁻(aq) Ka₁ = 7.5×10⁻³ mol dm⁻³

Examples of weak acids: CH₃COOH (ethanoic/acetic), HCN, H₂CO₃, H₂SO₃, H₃PO₄, HF, organic carboxylic acids. Most are molecular organic acids.

Strong and Weak Bases

Strong bases: completely ionised in water — Group 1 hydroxides (NaOH, KOH, LiOH) and Ca(OH)₂ (sparingly soluble but fully ionised when dissolved).

NaOH(aq) → Na⁺(aq) + OH⁻(aq) 100% ionised

Weak bases: partially ionised — accept proton from water in a reversible equilibrium:

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) Kb = 1.8×10⁻⁵ mol dm⁻³ CH₃NH₂(aq) + H₂O ⇌ CH₃NH₃⁺(aq) + OH⁻(aq) (methylamine, slightly stronger base than NH₃)
🧠
Strong vs weak: strength ≠ concentration! A "concentrated weak acid" (e.g. concentrated ethanoic acid) is NOT the same as a strong acid. Strength refers to the degree of ionisation; concentration refers to amount of acid per unit volume. 0.1 mol dm⁻³ HCl is a dilute strong acid. 10 mol dm⁻³ CH₃COOH is a concentrated weak acid. Dilute HCl has a much lower pH than concentrated ethanoic acid.
16.3

pH, pOH, and Calculations

The pH Scale

pH = −log₁₀[H⁺] (or −log₁₀[H₃O⁺]) [H⁺] = 10^(−pH) pOH = −log₁₀[OH⁻] pH + pOH = 14 (at 25°C) The ionic product of water: Kw = [H⁺][OH⁻] = 1.0×10⁻¹⁴ mol² dm⁻⁶ at 25°C Therefore: pKw = 14 = pH + pOH
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ACIDIC NEUTRAL BASIC gastric acid: ~1 | vinegar: 3 | rain: 5.6 | pure water: 7 | blood: 7.4 | seawater: 8.1 | bleach: 12
pH scale 0–14: acidic (0–7), neutral (7), basic (7–14) at 25°C

pH Calculations

Strong Acid

HCl fully ionises → [H⁺] = [HCl] 0.01 mol dm⁻³ HCl: [H⁺] = 0.01 = 10⁻² → pH = 2 H��SO₄ (dilute, both protons): [H⁺] = 2×[H₂SO₄] 0.05 mol dm⁻³ H₂SO₄: [H⁺] = 0.10 → pH = 1

Strong Base

NaOH fully ionises → [OH⁻] = [NaOH] 0.01 mol dm⁻³ NaOH: [OH⁻] = 0.01 → pOH = 2 → pH = 14 − 2 = 12

Weak Acid (using Ka)

For HA ⇌ H⁺ + A⁻: Ka = [H⁺][A⁻] / [HA] Assume [H⁺] = [A⁻] = x and [HA] ≈ c (initial conc., valid when Ka << c) Then: Ka = x² / c → x = √(Ka × c) → [H⁺] = √(Ka × c) Example: 0.10 mol dm⁻³ CH₃COOH, Ka = 1.8×10⁻⁵ [H⁺] = √(1.8×10⁻⁵ × 0.10) = √(1.8×10⁻⁶) = 1.34×10⁻³ mol dm⁻³ pH = −log(1.34×10⁻³) = 2.87
Worked Example 16.1 — pH Calculations

Q1: Calculate pH of 0.050 mol dm⁻³ HNO₃ (strong acid).

1

HNO₃ fully ionises: [H⁺] = 0.050 mol dm⁻³. pH = −log(0.050) = −log(5.0×10⁻²) = 2 − log(5) = 2 − 0.699 = 1.30

Q2: Calculate pH of 0.20 mol dm⁻³ NaOH.

2

[OH⁻] = 0.20 mol dm⁻³. pOH = −log(0.20) = 0.699. pH = 14 − 0.699 = 13.30

Q3: Calculate pH of 0.050 mol dm⁻³ HF, Ka = 6.3×10⁻⁴.

3

[H⁺] = √(Ka × c) = √(6.3×10⁻⁴ × 0.050) = √(3.15×10⁻⁵) = 5.61×10⁻³. pH = −log(5.61×10⁻³) = 2.25

Q4: A solution has pH = 9.0. Calculate [H⁺] and [OH⁻].

4

[H⁺] = 10⁻⁹·⁰ = 1.0×10⁻⁹ mol dm⁻³. [OH⁻] = 10⁻(14-9) = 10⁻⁵ = 1.0×10⁻⁵ mol dm⁻³.

16.4

Acid Dissociation Constant Ka, Base Kb, and pKa

Ka and pKa

For the equilibrium: HA(aq) ⇌ H⁺(aq) + A⁻(aq)

Ka = [H⁺][A⁻] / [HA] (units: mol dm⁻³) pKa = −log₁₀(Ka) Stronger acid → larger Ka → smaller pKa
AcidFormulaKa (mol dm⁻³)pKaStrength
Hydrochloric acidHCl~10⁷ (essentially ∞)~−7Strong
Sulfuric acid (1st)H₂SO₄~10³~−3Strong
Phosphoric acidH₃PO₄7.5×10⁻³2.12Weak
Hydrofluoric acidHF6.3×10⁻⁴3.20Weak
Ethanoic acidCH₃COOH1.8×10⁻⁵4.75Weak
Carbonic acidH₂CO₃4.3×10⁻⁷6.37Very weak
Hydrogen cyanideHCN4.9×10⁻¹⁰9.31Very weak
Water (as acid)H₂OKw/[H₂O] ≈ 10⁻¹⁶~16Extremely weak

Kb and the Relationship Ka × Kb = Kw

For conjugate acid-base pairs:

Ka(HA) × Kb(A⁻) = Kw = 1.0×10⁻¹⁴ (at 25°C) pKa + pKb = 14 Example: CH₃COOH and CH₃COO⁻ (conjugate pair) Ka(CH₃COOH) = 1.8×10⁻⁵ Kb(CH₃COO⁻) = Kw/Ka = 10⁻¹⁴/1.8×10⁻⁵ = 5.6×10⁻¹⁰ Stronger acid → weaker conjugate base (smaller Kb) Weaker acid → stronge
🧠
Neutral ≠ pH 7 at all temperatures "Neutral" means [H⁺] = [OH⁻]. At 25°C, this gives pH 7. At higher temperatures, Kw increases → [H⁺]=[OH⁻] at a higher value → pH < 7 for neutral solution. A solution at 50°C with pH 6.8 may still be neutral if [H⁺] = [OH⁻] at that temperature.
Worked Example 16.1 — pH Calculations

(a) Calculate the pH of 0.050 mol dm⁻³ HCl.

1

HCl is a strong acid — fully dissociates. [H⁺] = 0.050 mol dm⁻³.

2

pH = −log(0.050) = −log(5.0×10⁻²) = −(log 5.0 + log 10⁻²) = −(0.699 − 2) = 1.30

(b) Calculate [H⁺] and [OH⁻] for a solution of pH 3.5 at 25°C.

1

[H⁺] = 10⁻³·⁵ = 3.16 × 10⁻⁴ mol dm⁻³

2

[OH⁻] = Kw/[H⁺] = 10⁻¹⁴ / 3.16×10⁻⁴ = 3.16 × 10⁻¹¹ mol dm⁻³

(c) Calculate the pH of 0.10 mol dm⁻³ NaOH.

1

[OH⁻] = 0.10 mol dm⁻³ (strong base, full dissociation).

2

pOH = −log(0.10) = 1.00. pH = 14 − pOH = 14 − 1.00 = 13.00

Or: [H⁺] = Kw/[OH⁻] = 10⁻¹⁴/0.10 = 10⁻¹³ → pH = 13

16.4

Ka, Kb, pKa and the Acid Dissociation Constant

The Acid Dissociation Constant Ka

For a weak acid HA ⇌ H⁺ + A⁻:

Ka = [H⁺][A⁻] / [HA] (mol dm⁻³) pKa = −log₁₀(Ka) Larger Ka → stronger weak acid → smaller pKa → lower pH Smaller Ka → weaker acid → larger pKa → higher pH
AcidKa (mol dm⁻³)pKaStrength
HClO₄ (perchloric)~10⁷−7Very strong
HCl, HNO₃, H₂SO₄~10³ to 10⁶Very negativeStrong
H₃PO₄ (1st)7.5×10⁻³2.1Moderate weak
HF6.3×10⁻⁴3.2Weak
CH₃COOH (ethanoic)1.7×10⁻⁵4.76Weak
H₂CO₃ (carbonic, 1st)4.3×10⁻⁷6.4Very weak
NH₄⁺5.6×10⁻¹⁰9.25Very weak
H₂O (as acid)~10⁻¹⁶~16Extremely weak

Calculating pH of a Weak Acid from Ka

For weak acid HA at concentration c, with degree of dissociation α << 1 (weak acid approximation):

Ka = [H⁺]² / [HA]initial (if α << 1, i.e. [HA] barely changes) → [H⁺] = √(Ka × c) → pH = ½(pKa − log c) or pH = ½(pKa + pC) where pC = −log c
Worked Example 16.2 — Weak Acid pH

Calculate the pH of 0.10 mol dm⁻³ CH₃COOH (Ka = 1.7 × 10⁻⁵ mol dm⁻³).

1

CH₃COOH ⇌ H⁺ + CH₃COO⁻. Ka = [H⁺]² / [CH₃COOH]initial (weak acid approximation)

2

[H⁺]² = Ka × c = 1.7×10⁻⁵ × 0.10 = 1.7×10⁻⁶

3

[H⁺] = √(1.7×10⁻⁶) = 1.30×10⁻³ mol dm⁻³

4

pH = −log(1.30×10⁻³) = 2.89

5

Check: degree of dissociation α = [H⁺]/c = 1.30×10⁻³/0.10 = 1.3% — valid since α << 1 ✓

Relationship: Ka and Kb for Conjugate Pairs

For a conjugate acid-base pair (HA and A⁻):

Ka × Kb = Kw = 1.0 × 10⁻¹⁴ (at 25°C) pKa + pKb = pKw = 14 Example: CH₃COOH (Ka = 1.7×10⁻⁵) ↔ CH₃COO⁻ (Kb = Kw/Ka = 10⁻¹⁴/1.7×10⁻⁵ = 5.9×10⁻¹⁰) Strong acid → very weak conjugate base (Kb ≈ 0) Weak acid → stronger conjugate base
16.5

Buffer Solutions

Definition — Buffer Solution A solution that resists significant changes in pH when small amounts of acid or alkali are added, or when the solution is diluted. It typically consists of a weak acid and its conjugate base (or a weak base and its conjugate acid) in comparable amounts.

Acidic Buffer — Weak Acid + Conjugate Base Salt

Example: CH₃COOH / CH₃COONa (ethanoic acid / sodium ethanoate). pH ≈ 4–6.

How it works:

Buffer pH (Henderson-Hasselbalch equation):

pH = pKa + log([A⁻] / [HA]) = pKa + log([salt] / [acid]) At equal concentrations of acid and salt: pH = pKa (simplest case)
💡
A buffer is most effective when [HA] ≈ [A⁻] Equal amounts of weak acid and its conjugate base → pH = pKa. Buffer capacity is greatest at this point. Buffer is generally effective within ±1 pH unit of pKa.

Basic Buffer — Weak Base + Conjugate Acid Salt

Example: NH₃ / NH₄Cl (ammonia / ammonium chloride). pH ≈ 8–10.

When H⁺ added: NH₃ + H⁺ → NH₄⁺ (NH₃ mops up H⁺) When OH⁻ added: NH₄⁺ + OH⁻ → NH₃ + H₂O (NH₄⁺ mops up OH⁻) pH = 14 − pKb − log([NH₄⁺]/[NH₃]) or use pKa of NH₄⁺ (9.25) in Henderson-Hasselbalch
Worked Example 16.3 — Buffer pH Calculation

Calculate the pH of a buffer made from 0.20 mol dm⁻³ CH₃COOH and 0.40 mol dm⁻³ CH₃COONa. (Ka = 1.7×10⁻⁵)

1

pKa = −log(1.7×10⁻⁵) = 4.77

2

pH = pKa + log([A⁻]/[HA]) = 4.77 + log(0.40/0.20) = 4.77 + log(2) = 4.77 + 0.30 = 5.07

Because [A⁻] > [HA], pH > pKa (as expected — more base than acid).

Biological Importance of Buffers

16.6

Neutralisation and Salt Solutions

Neutralisation Reactions

Strong acid + Strong base → Neutral salt + water (pH 7) HCl + NaOH → NaCl + H₂O (ionic: H⁺ + OH⁻ → H₂O) Weak acid + Strong base → Alkaline salt + water (pH > 7) CH₃COOH + NaOH → CH₃COONa + H₂O (CH₃COO⁻ is the conjugate base of weak acid — slight alkaline hydrolysis) Strong acid + Weak base → Acidic salt + water (pH < 7) HCl + NH₃ → NH₄Cl (NH₄⁺ is conjugate acid of weak base NH₃ — slight acidic hydrolysis) Weak acid + Weak base → pH depends on relative Ka, Kb

Salt hydrolysis — why some salt solutions are not neutral:

Salt typeExamplepHReason
Strong acid + Strong baseNaCl, KNO₃7Neither Na⁺ nor Cl⁻ hydrolyses
Weak acid + Strong baseCH₃COONa, Na₂CO₃>7 (alkaline)CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻ (hydrolysis)
Strong acid + Weak baseNH₄Cl, FeCl₃<7 (acidic)NH₄⁺ ⇌ NH₃ + H⁺ (hydrolysis)
Weak acid + Weak baseCH₃COONH₄~7 (depends)Both hydrolyse — nearly cancel
16.7

Indicators and Acid-Base Titrations

Acid-Base Indicators

An indicator is a weak acid (HIn) where the acid and conjugate base forms have different colours. The indicator changes colour near its pKa:

HIn ⇌ H⁺ + In⁻ (colour 1) (colour 2) Colour change occurs at pH ≈ pKa(indicator) ± 1 range
IndicatorpKa (HIn)pH rangeAcid colourBase colour
Methyl orange3.53.1 – 4.4RedYellow
Litmus6.55.0 – 8.0RedBlue
Bromothymol blue7.16.0 – 7.6YellowBlue
Phenolphthalein9.68.3 – 10.0ColourlessPink/magenta

Titration Curves and Choosing the Right Indicator

The choice of indicator depends on the pH at the equivalence point:

Titration Curves: Strong/Weak Acid vs Strong Base 0 4 7 10 14 pH Volume NaOH added → Strong/Strong Weak/Strong pH 7 pH ~9
Titration curves: equivalence point pH 7 (strong/strong) vs pH ~9 (weak acid/strong base)
Titration typeEquivalence pHSuitable indicatorUnsuitable
Strong acid vs strong base~7Litmus, bromothymol blue, methyl orange, phenolphthalein (all work — large pH jump)Any works
Weak acid vs strong base~8–9 (alkaline)Phenolphthalein (range 8.3–10)Methyl orange (changes too early)
Strong acid vs weak base~5 (acidic)Methyl orange (range 3.1–4.4)Phenolphthalein (changes too late)
Weak acid vs weak base~7 (variable)No suitable indicator — gradual pH change at equivalenceMost indicators
16.8

Lewis Acid-Base Theory

Lewis Theory (1923) Lewis acid: an electron pair acceptor.
Lewis base: an electron pair donor.

Lewis Theory — The Broadest Definition

Lewis theory is the most general acid-base theory. It includes all Brønsted-Lowry reactions but also covers many reactions where no proton transfer occurs:

BF₃ + :NH₃ → F₃B←NH₃ (BF₃ = Lewis acid; NH₃ = Lewis base; dative bond formed) Al³⁺ + 6H₂O → [Al(H₂O)₆]³⁺ (Al³⁺ = Lewis acid; H₂O = Lewis base) Cu²⁺ + 4NH₃ → [Cu(NH₃)₄]²⁺ (Cu²⁺ = Lewis acid; NH₃ = Lewis base — complex ion) Ag⁺ + 2NH₃ → [Ag(NH₃)₂]⁺ (Ag⁺ = Lewis acid; NH₃ = Lewis base)

Lewis acids typically: electron-deficient atoms (BF₃, AlCl₃), transition metal cations (Fe³⁺, Cu²⁺), compounds with empty low-energy orbitals.

Lewis bases typically: molecules/ions with lone pairs (NH₃, H₂O, F⁻, CN⁻, Cl⁻, OH⁻, CO).

TheoryAcidBaseScope
ArrheniusH⁺ producer in waterOH⁻ producer in waterNarrowest — aqueous only
Brønsted-LowryH⁺ donorH⁺ acceptorIntermediate — any solvent
LewisElectron pair acceptorElectron pair donorBroadest — includes non-proton reactions

No videos added yet for this unit.

✏️

Exercises

🧪

Multiple Choice Quiz — 25 Questions

Unit 16: Acids and Bases

25 Questions
Q1

According to Brønsted-Lowry theory, a base is:

Brønsted-Lowry base = proton (H⁺) acceptor. This is broader than Arrhenius (OH⁻ producer) because it includes substances like NH₃ which accept H⁺ without having OH⁻. Lewis base = electron pair donor (even broader). All Arrhenius bases are Brønsted-Lowry bases, but not vice versa.
Q2

What is the conjugate base of H₂SO₄?

Conjugate base = acid − H⁺. H₂SO₄ − H⁺ = HSO₄⁻. The conjugate base of H₂SO₄ is HSO₄⁻ (hydrogen sulfate ion). H₂SO₄ → H⁺ + HSO₄⁻ (1st dissociation). SO₄²⁻ is the conjugate base of HSO₄⁻ (2nd dissociation). The conjugate base is formed by removing one H⁺ from the acid.
Q3

The pH of 0.010 mol dm⁻³ HNO₃ is:

HNO₃ is a strong acid (fully dissociates): [H⁺] = 0.010 = 10⁻² mol dm⁻³. pH = −log(10⁻²) = 2. Quick rule: −log(10⁻ⁿ) = n. 0.010 = 10⁻² → pH = 2. (0.10 mol dm⁻³ → pH 1; 0.001 → pH 3).
Q4

Which of the following is a weak acid?

CH₃COOH (ethanoic acid) is a weak acid — only partially dissociates: Ka = 1.7×10⁻⁵. Strong acids (essentially 100% dissociation): HCl, HNO₃, H₂SO₄, HBr, HI, HClO₄. Common weak acids: CH₃COOH, HF, H₂CO₃, H₃PO₄, HNO₂, H₂SO₃.
Q5

At 25°C, the ionic product of water Kw = 10⁻¹⁴. This means in pure water:

Kw = [H⁺][OH⁻] = 10⁻¹⁴. In pure water [H⁺] = [OH⁻] = √(10⁻¹⁴) = 10⁻⁷ mol dm⁻³. pH = −log(10⁻⁷) = 7. "Neutral" at 25°C = pH 7. Above 25°C, Kw increases → [H⁺] increases → pH < 7 for neutral — but still neutral because [H⁺]=[OH⁻].
Q6

Ka for CH₃COOH = 1.7×10⁻⁵. The pH of 0.10 mol dm⁻³ CH₃COOH is approximately:

[H⁺] = √(Ka × c) = √(1.7×10⁻⁵ × 0.10) = √(1.7×10⁻⁶) = 1.30×10⁻³. pH = −log(1.30×10⁻³) = 2.89. Note: 4.76 = pKa (the pH when [acid]=[conjugate base] in a buffer, not for pure acid solution). pH 2.89 confirms weak acid: much higher than HCl at same concentration (pH 1.0).
Q7

A buffer solution is best described as:

A buffer RESISTS (partially opposes) pH change — it does not prevent all change. Buffer contains: (1) weak acid (to mop up added OH⁻) and (2) conjugate base (to mop up added H⁺). Examples: CH₃COOH / CH₃COO⁻ (acidic buffer, pH ≈ 4–6); NH₃ / NH₄⁺ (basic buffer, pH ≈ 8–10). Buffer does NOT mean pH = 7 — buffers can maintain any pH near their pKa.
Q8

The Henderson-Hasselbalch equation gives the pH of a buffer as:

Henderson-Hasselbalch: pH = pKa + log([conjugate base]/[weak acid]) = pKa + log([A⁻]/[HA]). When [A⁻] = [HA]: log(1) = 0 → pH = pKa. This is the half-equivalence point on a titration curve. When [A⁻] > [HA]: pH > pKa. When [A⁻] < [HA]: pH < pKa. Buffer is most effective within ±1 pH unit of pKa.
Q9

The pH of blood must be maintained at 7.35–7.45. The main buffer system in blood is:

Blood buffer: CO₂(aq) + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻. Lungs remove CO₂ (shifts left → reduces [H⁺] → raises pH when too acidic). Kidneys regulate [HCO₃⁻] (excrete or retain HCO₃⁻). This coupled system maintains blood pH within the narrow 7.35–7.45 range. pH <7.0 = acidosis; pH >7.8 = alkalosis — both fatal within minutes.
Q10

What is the pH of 0.050 mol dm⁻³ NaOH at 25°C?

NaOH is a strong base: [OH⁻] = 0.050 mol dm⁻³. pOH = −log(0.050) = 1.30. pH = 14 − pOH = 14 − 1.30 = 12.70. Alternatively: [H⁺] = Kw/[OH⁻] = 10⁻¹⁴/0.050 = 2.0×10⁻¹³. pH = −log(2.0×10⁻¹³) = 12.70.
Q11

For a titration of weak acid with strong base (e.g. CH₃COOH with NaOH), the most suitable indicator is:

Weak acid + strong base: equivalence point pH ≈ 8–9 (alkaline — conjugate base CH₃COO⁻ hydrolyses). The steep pH jump occurs between pH ~7 and ~11. Phenolphthalein (8.3–10) falls within this steep jump → sharp colour change → suitable. Methyl orange (3.1–4.4) changes in the buffer region (before equivalence, gradual slope) → poor endpoint. Litmus (5–8) might just work but has a wide gradual range — phenolphthalein is preferred.
Q12

A solution of Na₂CO₃ in water is alkaline because:

Na₂CO₃ → 2Na⁺ + CO₃²⁻. CO₃²⁻ is the conjugate base of HCO₃⁻ (itself derived from weak acid H₂CO₃). CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻ (hydrolysis — CO₃²⁻ accepts H⁺ from water → OH⁻ produced → alkaline). Na⁺ does not hydrolyse. Na₂CO₃ is a common base used in washing soda, drain cleaner, glass making (pH ≈ 11).
Q13

The pKa of CH₃COOH is 4.76. A buffer of equal concentrations of CH₃COOH and CH₃COONa has pH:

Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]). Equal concentrations: [A⁻] = [HA] → log(1) = 0 → pH = pKa = 4.76. This is the most important point on a titration curve: the half-equivalence point where [acid] = [conjugate base] → pH = pKa → used to determine Ka from titration data.
Q14

Lewis acid BF₃ reacts with Lewis base NH₃ to form the adduct F₃B←NH₃. The new bond formed is:

Lewis acid-base reaction: NH₃ (lone pair on N) donates to empty 2p orbital on B in BF₃ → forms a dative (coordinate) covalent bond where BOTH electrons come from the Lewis base (NH₃). Result: B now has 4 bonds (sp³ hybridised), tetrahedral; N has 4 bonds (positively charged, as in NH₄⁺). This is identical to how metal complexes form ([Cu(NH₃)₄]²⁺): metal = Lewis acid; NH₃ = Lewis base; dative bonds throughout.
Q15

The Ka × Kb relationship for a conjugate acid-base pair at 25°C is:

For any conjugate acid-base pair (HA/A⁻): Ka(HA) × Kb(A⁻) = Kw = 10⁻¹⁴ at 25°C. Also: pKa + pKb = pKw = 14. This means: strong acid (large Ka) → very weak conjugate base (tiny Kb). Weak acid (small Ka) → stronger conjugate base (larger Kb). Example: CH₃COOH Ka = 1.7×10⁻⁵ → CH₃COO⁻ Kb = 10⁻¹⁴/1.7×10⁻⁵ = 5.9×10⁻¹⁰.
Q16

HCO₃⁻ (hydrogen carbonate ion) is described as amphiprotic because:

HCO₃⁻ as acid (donates H⁺): HCO₃⁻ ⇌ H⁺ + CO₃²⁻ (Ka₂ of H₂CO₃ = 4.7×10⁻¹¹). HCO₃⁻ as base (accepts H⁺): HCO₃⁻ + H⁺ ⇌ H₂CO₃ (or equivalently, + H₃O⁺ ⇌ H₂CO₃ + H₂O). Amphiprotic = can be both acid AND base in B-L theory. Other amphiprotic species: H₂O, HSO₄⁻, HS⁻, H₂PO₄⁻, HPO₄²⁻. Amphoteric = can act as acid/base — applies especially to oxides and hydroxides.
Q17

A buffer is made from equal volumes of 0.1 mol dm⁻³ NH₃ and 0.1 mol dm⁻³ NH₄Cl mixed together. The pH of this buffer (pKa NH₄⁺ = 9.25) is:

Use NH₄⁺ as the "acid" with pKa = 9.25. Henderson-Hasselbalch: pH = pKa(NH₄⁺) + log([NH₃]/[NH₄⁺]). Equal volumes of equal concentrations → [NH₃] = [NH₄⁺] (after mixing, volumes double but concentrations halved equally — ratio unchanged). log(1) = 0. pH = 9.25. This is the half-equivalence point of the NH₃/HCl titration curve. Buffer range: ~8.25–10.25.
Q18

A solution of NH₄Cl (ammonium chloride) is:

NH₄Cl = salt of weak base (NH₃) + strong acid (HCl). NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺ — NH₄⁺ partially donates its proton to water → H₃O⁺ released → solution acidic (pH ~5 for 0.1 mol dm⁻³). Cl⁻ is conjugate base of strong acid HCl → too weak as a base to hydrolyse water → Cl⁻ does nothing. Rule: salt of weak base + strong acid → acidic solution.
Q19

Which of the following pairs can act as a buffer?

A buffer requires a weak acid + its conjugate base (or weak base + conjugate acid) in comparable amounts. CH₃COOH (weak acid) + CH₃COONa (sodium salt provides CH₃COO⁻, the conjugate base) → buffer. HCl + NaCl: HCl is strong acid; NaCl has no buffering capacity. NaOH + NaCl: NaOH is strong base; NaCl neutral. HCl + NaOH equal amounts → complete neutralisation → NaCl + H₂O (no buffer — equivalence point).
Q20

The pH of 0.040 mol dm⁻³ H₂SO₄ (assume complete dissociation of both protons) is:

H₂SO₄ → 2H⁺ + SO₄²⁻ (diprotic strong acid, assuming both protons fully dissociate). [H⁺] = 2 × 0.040 = 0.080 mol dm⁻³. pH = −log(0.080) = −log(8.0×10⁻²) = −(log 8.0 + log 10⁻²) = −(0.903 − 2) = 1.10. Note: in reality the 2nd ionisation of H₂SO₄ is not quite complete at this concentration, but at A-level both protons are assumed to fully dissociate.
Q21

A strong acid and a weak acid at the same concentration have different pH values. At 0.1 mol dm⁻³, HCl has pH 1 but CH₃COOH has pH ~2.9. This is because:

HCl (strong): [H⁺] = 0.1 mol dm⁻³ (100% dissociation). pH = 1. CH₃COOH (weak): Ka = 1.7×10⁻⁵; only ~1.3% dissociates at 0.1 mol dm⁻³. [H⁺] = √(1.7×10⁻⁵ × 0.1) = 1.3×10⁻³ mol dm⁻³. pH = 2.89. At the same concentration, the weak acid has fewer H⁺ ions → higher pH. The difference in pH reflects the degree of ionisation (extent of reaction).
Q22

Phenolphthalein is colourless below pH 8.3 and pink above pH 10.0. It is suitable for titration of:

Phenolphthalein changes colour in pH range 8.3–10.0. Weak acid + strong base: equivalence point ~pH 8–9 → falls within the steep part of the curve → phenolphthalein gives a sharp endpoint. Strong acid + weak base: equivalence at pH ~5 → phenolphthalein changes after equivalence → poor endpoint. Strong acid + strong base: equivalence at pH 7 → phenolphthalein might work (jump from 4–10 includes 8.3–10) but a lower-range indicator is sharper.
Q23

Which species acts as a Lewis acid in the formation of [Fe(H₂O)₆]³⁺?

Lewis acid = electron pair acceptor. Fe³⁺ has empty d orbitals → accepts lone pairs from H₂O (Lewis base, electron pair donor via O lone pairs). 6 H₂O molecules each donate a lone pair to Fe³⁺ → 6 dative bonds → [Fe(H₂O)₆]³⁺ complex ion. All transition metal cations act as Lewis acids in forming complex ions. The ligands (H₂O, NH₃, CN⁻, etc.) are all Lewis bases.
Q24

The pKa of HF is 3.2. Which is more acidic: HF or CH₃COOH (pKa 4.76)?

Smaller pKa = larger Ka = stronger acid. pKa 3.2 (HF) < pKa 4.76 (CH₃COOH) → HF is more acidic. Ka(HF) = 10⁻³·² = 6.3×10⁻⁴; Ka(CH₃COOH) = 10⁻⁴·⁷⁶ = 1.7×10⁻⁵. HF dissociates more than ethanoic acid at the same concentration → lower pH → more acidic. However, both are weak acids (Ka << 1).
Q25

If [H⁺] in a solution is doubled, the pH:

pH = −log[H⁺]. If [H⁺] doubles: new pH = −log(2[H⁺]) = −log 2 − log[H⁺] = original pH − log(2) = original pH − 0.301. pH decreases by 0.30 (since doubling [H⁺] makes solution more acidic → lower pH). Example: pH 3.0 → [H⁺] = 10⁻³; double → [H⁺] = 2×10⁻³ → pH = −log(2×10⁻³) = 2.70. Change = −0.30 pH units.
📝

Unit Test — 50 Marks

Section A — Short Answer

30 marks
Q1 [5 marks]

Compare the Arrhenius, Brønsted-Lowry, and Lewis definitions of acids and bases. Give one example of a reaction that illustrates the Lewis definition but cannot be explained by the Arrhenius definition. [5]

Arrhenius: acid = H⁺ producer in water; base = OH⁻ producer. Limitation: only aqueous; cannot explain NH₃ as base (no OH⁻). [1] Brønsted-Lowry: acid = H⁺ donor; base = H⁺ acceptor. Broader — works in any solvent; explains NH₃ as base (accepts H⁺). [1] Lewis: acid = electron pair acceptor; base = electron pair donor. Broadest — includes reactions with no H⁺ transfer. [1] Lewis reaction not in Arrhenius: BF₃ + NH₃ → F₃B←NH₃. No H⁺ transfer — BF₃ accepts NH₃'s lone pair → dative bond. Neither BF₃ nor NH₃ produces H⁺ or OH⁻ in water. Lewis acid = BF₃; Lewis base = NH₃. [1] Also acceptable: AlCl₃ + Cl⁻ → [AlCl₄]⁻; or [Fe(H₂O)₆]³⁺ formation. [1 mark for clear explanation]
Q2 [5 marks]

Calculate the pH of: (a) 0.020 mol dm⁻³ HCl [1] (b) 0.50 mol dm⁻³ NaOH [1] (c) 0.15 mol dm⁻³ CH₃COOH (Ka = 1.7×10⁻⁵) [2] (d) a mixture of 25.0 cm³ of 0.10 mol dm⁻³ HCl and 25.0 cm³ of 0.10 mol dm⁻³ NaOH [1]. [5]

(a) HCl strong acid: [H⁺] = 0.020. pH = −log(0.020) = −log(2×10⁻²) = 1.70. [1] (b) NaOH strong base: [OH⁻] = 0.50. pOH = −log(0.5) = 0.30. pH = 14 − 0.30 = 13.70. [1] (c) CH₃COOH weak acid: [H⁺] = √(Ka × c) = √(1.7×10⁻⁵ × 0.15) = √(2.55×10⁻⁶) = 1.597×10⁻³. pH = −log(1.597×10⁻³) = 2.80. [2] (d) Equal volumes and equal concentrations of strong acid + strong base → exact equivalence → complete neutralisation → NaCl solution → pH = 7. [1]
Q3 [5 marks]

Describe how a buffer solution of pH 5.00 could be prepared using ethanoic acid (pKa = 4.76). Include the concentrations/ratio needed and explain how the buffer resists pH change when small amounts of acid and alkali are added. [5]

Preparation: use CH₃COOH and CH₃COONa (sodium ethanoate). Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]) → 5.00 = 4.76 + log([CH₃COO⁻]/[CH₃COOH]) → log ratio = 0.24 → ratio = 10⁰·²⁴ = 1.74. [1.5] So [CH₃COO⁻]/[CH₃COOH] = 1.74 : 1. Example: dissolve 0.174 mol CH₃COONa and 0.100 mol CH₃COOH in 1 dm³ water. [0.5] How it resists: when H⁺ added: CH₃COO⁻ + H⁺ → CH₃COOH (conjugate base mops up H⁺) → small decrease in [A⁻], small increase in [HA] → ratio barely changes → pH barely changes. [1.5] When OH⁻ added: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O (weak acid neutralises OH⁻) → small increase in [A⁻], small decrease in [HA] → ratio barely changes → pH barely changes. [1.5]
Q4 [5 marks]

Explain why salt solutions of the following are acidic, alkaline, or neutral, writing a hydrolysis equation where relevant: (a) KNO₃ (b) Na₂CO₃ (c) FeCl₃ (d) (NH₄)₂SO₄ (e) CH₃COONH₄. [5]

(a) KNO₃: neutral pH 7. K⁺ (strong base conjugate cation) + NO₃⁻ (strong acid conjugate anion) — neither hydrolyses. [1] (b) Na₂CO₃: alkaline. CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻ — CO₃²⁻ is conjugate base of weak acid HCO₃⁻; hydrolyses → OH⁻ → alkaline (pH ~11). Na⁺ inert. [1] (c) FeCl₃: acidic. [Fe(H₂O)₆]³⁺ + H₂O ⇌ [Fe(H₂O)₅OH]²⁺ + H₃O⁺ — Fe³⁺ has very high charge density; polarises coordinated H₂O → H⁺ released → acidic (pH ~3). Cl⁻ inert. [1] (d) (NH₄)₂SO₄: acidic. NH₄⁺ ⇌ NH₃ + H⁺ — NH₄⁺ is conjugate acid of weak base NH₃ → hydrolyses to give H⁺ → acidic (pH ~5). SO₄²⁻ inert (from strong acid H₂SO₄). [1] (e) CH₃COONH₄: nearly neutral (~7). Both CH₃COO⁻ (hydrolyses → OH⁻) and NH₄⁺ (hydrolyses → H⁺) — they nearly cancel. Exact pH depends on relative Ka(NH₄⁺) vs Kb(CH₃COO⁻). [1]
Q5 [5 marks]

Sketch and annotate the titration curves for: (a) 25 cm³ of 0.10 mol dm⁻³ NaOH added to 25 cm³ of 0.10 mol dm⁻³ HCl and (b) 0.10 mol dm⁻³ NaOH added to 0.10 mol dm⁻³ CH₃COOH. Label the equivalence point pH, buffer region, and the appropriate indicator for each. [5]

(a) Strong acid vs strong base: starts pH ~1 (HCl side). Gradual rise. Very steep jump from pH ~3 to pH ~11 at equivalence (25 cm³ NaOH added). Equivalence pH = 7. Levels off at pH ~13 (excess NaOH). Suitable indicator: any broad-range indicator (methyl orange, litmus, phenolphthalein all work — steep jump covers all their ranges). [2.5] (b) Weak acid vs strong base: starts pH ~2.9 (CH₃COOH). Buffer region (pH ~4–6, gradual slope) between start and half-equivalence point. Half-equivalence point at 12.5 cm³ NaOH: pH = pKa = 4.76 (buffer region midpoint). Steep jump from pH ~7 to pH ~11 at 25 cm³ (equivalence). Equivalence pH ~8–9 (CH₃COO⁻ slightly alkaline). Levels off at pH ~13. Indicator: phenolphthalein (8.3–10) — changes within the steep jump at equivalence. Methyl orange unsuitable (changes in buffer region, before equivalence). [2.5]
Q6 [5 marks]

The Kw of water at 37°C (body temperature) is approximately 2.4 × 10⁻¹⁴. (a) Calculate the pH of pure water at 37°C. (b) Is this solution acidic, neutral, or alkaline? Explain. (c) What does this tell us about the pH of "neutrality" at different temperatures? [5]

(a) Kw = [H⁺][OH⁻] = 2.4×10⁻¹⁴ at 37°C. Pure water: [H⁺] = [OH⁻] = √(2.4×10⁻¹⁴) = 1.55×10⁻⁷ mol dm⁻³. pH = −log(1.55×10⁻⁷) = 6.81. [2] (b) The solution is NEUTRAL — even though pH = 6.81 (less than 7). Neutral means [H⁺] = [OH⁻], which is true here. The solution is neither acidic (excess H⁺) nor alkaline (excess OH⁻). [2] (c) The pH of neutrality is NOT always 7 — it equals −½log(Kw), which depends on temperature. At 25°C: pKw = 14, so neutral pH = 7. At higher T, Kw increases (water ionisation is endothermic — higher T favours more ions) → [H⁺] increases → neutral pH < 7. The practical implication: pH of "neutral" blood at 37°C is 6.81, but blood is slightly alkaline at 7.4 (relative to the neutral point of 6.81). [1]

Section B — Extended Answer

20 marks
Q7 [10 marks]

(a) Explain, with Ka values and equations, the difference between a strong acid (HCl) and a weak acid (CH₃COOH) at the same concentration. Describe how you would distinguish them experimentally using three different observations. [5]
(b) Describe the manufacture of a buffer of pH 7.4 (blood pH) using dihydrogen phosphate ions. The pKa of H₂PO₄⁻ is 7.2. Calculate the ratio [HPO₄²⁻]/[H₂PO₄⁻] needed. Why is the phosphate buffer important inside cells? [5]

(a) HCl (strong): HCl → H⁺ + Cl⁻ (complete, no equilibrium). Ka → ∞ (or very large). At 0.1 mol dm⁻³: [H⁺] = 0.1 → pH = 1.0. CH₃COOH (weak): CH₃COOH ⇌ H⁺ + CH₃COO⁻ (partial). Ka = 1.7×10⁻⁵ (small). At 0.1 mol dm⁻³: [H⁺] = 1.3×10⁻³ → pH = 2.9. [2] Experimental distinctions: (1) pH meter/indicator: HCl pH 1; CH₃COOH pH ~3 (at same concentration). (2) Electrical conductivity: HCl much higher conductivity (more ions in solution → more current). (3) Rate of reaction with Mg: HCl reacts faster (more H⁺ available → higher initial rate of H₂ evolution); both eventually produce same total H₂ if enough Mg used. (4) Reaction with Na₂CO₃: both produce CO₂ at same rate initially? — Actually HCl reacts faster (more H⁺). (5) Titration curve: HCl shows much steeper initial pH; CH₃COOH shows buffer region. [3] (b) Buffer: H₂PO₄⁻ (weak acid, pKa 7.2) + HPO₄²⁻ (conjugate base). Henderson-Hasselbalch: pH = pKa + log([HPO₄²⁻]/[H₂PO₄⁻]) → 7.4 = 7.2 + log(ratio) → log(ratio) = 0.2 → ratio = 10⁰·² = 1.58. [2] So [HPO₄²⁻]/[H₂PO₄⁻] = 1.58 : 1. Importance: pKa 7.2 is close to intracellular pH (~7.0–7.2) → phosphate buffer is most effective near intracellular pH → effectively maintains the pH around which most cellular enzymes function optimally. Phosphate is abundant in cells (from ATP, DNA backbone). [3]
Q8 [10 marks]

(a) Explain the concept of conjugate acid-base pairs in Brønsted-Lowry theory. State the relationship Ka × Kb = Kw and use it to explain why a strong acid has a very weak conjugate base. Calculate Kb for the ethanoate ion (CH₃COO⁻) given Ka(CH₃COOH) = 1.7 × 10⁻⁵. [5]
(b) Explain why a titration of weak acid with weak base (e.g. CH₃COOH with NH₃) gives a poor endpoint and cannot use a standard indicator. Sketch the shape of this titration curve and compare it with a strong acid-strong base curve. [5]

(a) Conjugate pairs: when acid HA donates H⁺, it forms A⁻ (conjugate base). When base B accepts H⁺, it forms BH⁺ (conjugate acid). For any pair: Ka(HA) × Kb(A⁻) = Kw = 10⁻¹⁴ at 25°C. [1] Why strong acid → very weak conjugate base: HCl is strong acid (Ka → very large). Kb(Cl⁻) = Kw/Ka = 10⁻¹⁴ / (very large) = essentially zero. Cl⁻ has almost no tendency to accept H⁺ from water → very weak base. Conversely, weak acid CH₃COOH (small Ka) → its conjugate base CH₃COO⁻ has a non-negligible Kb → can accept some H⁺ → appreciable base → hydrolyses water slightly → alkaline salt solution. [2] Kb(CH₃COO⁻) = Kw / Ka(CH₃COOH) = 10⁻¹⁴ / 1.7×10⁻⁵ = 5.9×10⁻¹⁰ mol dm⁻³. pKb = 9.23. (Check: pKa + pKb = 4.77 + 9.23 = 14 ✓) [2] (b) Weak acid vs weak base titration: both species have their own buffering effects — before, at, and after equivalence, the pH changes gradually. The equivalence point is NOT sharply defined — the curve barely has a steep section. No common indicator has a range that falls within a steep pH jump, because there is none. Any indicator would change colour gradually over a wide volume range → no precise endpoint. [2] Titration curve shape: starts at pH ~3 (CH₃COOH, weak acid). Rises gradually through the buffer region (~pH 4–6). Near equivalence: pH changes very slowly and steadily — there is NO steep jump at all (compare strong/strong which has a jump of ~8 pH units in 0.1 cm³). After equivalence: gradual rise to ~pH 11. Compared to strong/strong curve: strong/strong has a dramatic near-vertical jump of 6–8 pH units centred at pH 7 around equivalence. The weak/weak curve has a slow S-shape with no steep section. The pH at equivalence is approximately 7 for weak acid/weak base when Ka ≈ Kb, but there is no indicator that can reliably detect this as a sharp endpoint. Potentiometric titration (pH meter and chart recorder) is needed instead. [3]
← Unit 15: Equilibrium S4 Course Home Unit 17: Redox →

💳 Pay to Download

Unit notes — instant PDF download

Price
300 RWF
One-time · Download immediately after payment
Phone Number
Rwanda — MTN MoMo or Airtel Money
Network
MTN MoMo
Airtel Money
Other
Processing Payment...
Check your phone — enter your PIN to confirm.
✅
Payment Confirmed!
Generating your PDF...
❌
Payment Not Completed
WhatsApp
⚙️
Payment service is being set up.
Online payment is not yet available for this unit.
Please contact your teacher to get access.
16.5

Neutralisation Reactions

Types of Neutralisation

Neutralisation is the reaction between an acid and a base to produce a salt and water. The net ionic equation for any strong acid + strong base neutralisation is always:

H⁺(aq) + OH⁻(aq) → H₂O(l) ΔH = −57.1 kJ mol⁻¹ (enthalpy of neutralisation)
Strong acid + strong base: HCl + NaOH → NaCl + H₂O Strong acid + weak base: HCl + NH₃ → NH₄Cl (NH₄Cl is the salt — acidic solution) Weak acid + strong base: CH₃COOH + NaOH → CH₃COONa + H₂O (CH₃COONa — basic solution) Weak acid + weak base: CH₃COOH + NH₃ → CH₃COONH₄ (depends on relative Ka and Kb) Acid + carbonate: H��SO₄ + Na₂CO₃ → Na₂SO₄ + H₂O + CO₂↑ 2HCl + CaCO₃ → CaCl₂ + H₂O + CO₂↑ Acid + metal oxide: H��SO₄ + CuO → CuSO₄ + H₂O Acid + metal: Zn + H₂SO₄(dil.) → ZnSO₄ + H₂↑

Enthalpy of Neutralisation

The standard enthalpy of neutralisation (ΔHₙ) is the enthalpy change when 1 mol of water is formed in a neutralisation reaction.

Acid + BaseΔHₙ (kJ/mol)Reason for difference
HCl + NaOH−57.1Strong + strong → net ionic equation only: H⁺ + OH⁻ → H₂O
HNO₃ + KOH−57.3Strong + strong — same net reaction
CH₃COOH + NaOH−55.2Weak acid — energy needed to ionise CH₃COOH; less exothermic
HCl + NH₃−52.2Weak base — energy needed to ionise NH₃; less exothermic
CH₃COOH + NH₃−50.4Both weak — most energy used in ionisation; least exothermic

All strong acid + strong base reactions have ΔHₙ ≈ −57 kJ/mol because the net reaction is always H⁺ + OH⁻ → H₂O. Weak acids/bases are less exothermic because some energy is used to ionise the weak electrolyte (endothermic partial ionisation step subtracts from the total heat).

16.6

Salt Hydrolysis — pH of Salt Solutions

Why Are Some Salt Solutions Not Neutral?

When a salt dissolves in water, the ions may react with water (hydrolyse), producing H⁺ or OH⁻ → the solution is not neutral. The pH depends on the parent acid and base:

Salt typeExamplepH in waterReason
Strong acid + Strong baseNaCl, KNO₃, Na₂SO₄7 (neutral)Na⁺ and Cl⁻ neither hydrolyse — spectator ions
Strong acid + Weak baseNH₄Cl, NH₄NO₃<7 (acidic)NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺ — ammonium hydrolyses
Weak acid + Strong baseCH₃COONa, Na₂CO₃>7 (basic)CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻ — acetate hydrolyses
Weak acid + Weak baseCH₃COONH₄≈7 (depends on Ka vs Kb)Both ions hydrolyse; pH depends on relative strength
16.7

Buffer Solutions

Definition — Buffer Solution A solution that resists changes in pH when small amounts of acid or base are added. It maintains a nearly constant pH. Buffers contain a weak acid and its conjugate base (or a weak base and its conjugate acid) in comparable concentrations.

How a Buffer Works — Acidic Buffer

An acidic buffer contains a weak acid (HA) and its conjugate base (A⁻), typically as its sodium salt NaA:

Example: CH₃COOH / CH₃COO⁻Na⁺ buffer (pKa = 4.75; effective range pH 3.75–5.75) When acid (H⁺) is added: CH₃COO⁻ + H⁺ → CH₃COOH (base component neutralises added H⁺ → pH barely changes) When base (OH⁻) is added: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O (acid component neutralises added OH⁻ → pH barely changes)

The buffer works because there is a large reservoir of both the weak acid (to neutralise OH⁻) and its conjugate base (to neutralise H⁺).

Henderson-Hasselbalch Equation

pH = pKa + log([A⁻]/[HA]) where [A⁻] = concentration of conjugate base (salt) [HA] = concentration of weak acid Special case: when [A⁻] = [HA] → log(1) = 0 → pH = pKa (half-equivalence point)

The buffer has maximum capacity (resists pH change best) when [A⁻] = [HA], i.e. when pH = pKa. Buffers work effectively in the range pKa ± 1.

Worked Example 16.2 — Buffer pH Calculation

A buffer contains 0.20 mol dm⁻³ CH₃COOH and 0.30 mol dm⁻³ CH₃COONa. Ka = 1.8×10⁻⁵. Calculate the pH.

1

pKa = −log(1.8×10⁻⁵) = 4.74

2

pH = pKa + log([A⁻]/[HA]) = 4.74 + log(0.30/0.20) = 4.74 + log(1.5) = 4.74 + 0.176 = 4.92

3

Adding 0.01 mol HCl to 1 dm³ of buffer: H⁺ + CH₃COO⁻ → CH₃COOH. New [CH₃COO⁻] = 0.29; [CH₃COOH] = 0.21. pH = 4.74 + log(0.29/0.21) = 4.74 + 0.141 = 4.88. Change = only 0.04 pH units (vs ~2 pH units in unbuffered water).

Biological Importance of Buffers

16.8

Indicators and Acid-Base Titrations

How Acid-Base Indicators Work

An indicator is a weak acid (HIn) where the undissociated form (HIn) and the conjugate base (In⁻) have different colours:

HIn(aq) ⇌ H⁺(aq) + In⁻(aq) colour A colour B In acidic solution: equilibrium shifts left → colour A dominates In basic solution: equilibrium shifts right → colour B dominates Transition occurs around pH ≈ pKa(HIn) ± 1
IndicatorpKapH rangeAcid colourBase colour
Methyl orange3.53.1–4.4RedYellow
Methyl red5.04.4–6.2RedYellow
Litmus6.55.0–8.0RedBlue
Phenolphthalein9.38.3–10.0ColourlessPink/red
Universal indicator—1–14Red → orange → yellow → green → blue → purpleUsed for rough pH estimation

Choosing the Right Indicator for a Titration

At the equivalence point, the pH changes very sharply (the vertical portion of the titration curve). The indicator must change colour within the sharp pH jump at the equivalence point:

Titration typeEquivalence point pHSuitable indicator(s)
Strong acid + Strong base~7Any — methyl orange, litmus, or phenolphthalein (all change in the steep region)
Strong acid + Weak base (e.g. HCl + NH₃)<7 (~5.2)Methyl orange or methyl red (change at pH 3–5); phenolphthalein unsuitable
Weak acid + Strong base (e.g. CH₃COOH + NaOH)>7 (~8.7)Phenolphthalein (changes at pH 8.3–10); methyl orange unsuitable
Weak acid + Weak base~7 (variable)No single indicator is suitable — gradual pH change, no sharp jump
⚠️
Weak acid + weak base: no sharp endpoint For CH₃COOH + NH₃, there is no sharp pH jump at the equivalence point — the pH changes gradually over a wide range. No common indicator shows a clear sharp colour change — this titration must be done using a pH meter or conductimetric method.

Titration Curves

Strong acid vs Strong base: Start: pH ~1 (acid). Rises slowly, then STEEP jump at equivalence point (~7). End: pH ~13. Steep region: pH 4–10 (huge jump). All indicators work. Strong acid vs Weak base (e.g. HCl + NH₃): Start: pH ~1. Steep jump at equivalence ~5.2. End: ~12. Use: methyl orange or methyl red only. Weak acid vs Strong base (e.g. CH₃COOH + NaOH): Start: pH ~3. Buffer region around pKa (4.75). Half-equivalence point: pH = pKa = 4.75. Equivalence ~8.7. End: ~13. Use: phenolphthalein only. Weak acid vs Weak base: No steep jump → no simple indicator reliable.
✏️

Exercises

🧪

Multiple Choice Quiz — 25 Questions

Unit 16: Acids and Bases

25 Questions
Q1

According to Brønsted-Lowry theory, an acid is:

Brønsted-Lowry: acid = H⁺ donor; base = H⁺ acceptor. This definition works in aqueous AND non-aqueous solvents and explains NH₃ as a base (it accepts H⁺ even though it has no OH⁻). Option A = Arrhenius base. Option C = Lewis acid. Option D = Arrhenius acid (limited).
Q2

In the reaction: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, water acts as:

H₂O donates H⁺ to NH₃ → OH⁻ remains. Therefore H₂O is the Brønsted-Lowry acid (H⁺ donor) in this reaction. NH₃ is the B-L base (H⁺ acceptor → NH₄⁺). Conjugate pairs: H₂O/OH⁻ (differ by H⁺) and NH₄⁺/NH₃. Water is amphiprotic — can act as acid or base.
Q3

Which of the following is a Lewis acid but NOT a Brønsted-Lowry acid?

BF₃: has no hydrogen atom → cannot donate H⁺ → NOT a B-L acid. But B has an empty p orbital → accepts lone pair from bases (e.g. NH₃ → F₃B–NH₃) → Lewis acid. HCl, CH₃COOH, H₂SO₄: all have ionisable H → both B-L and Lewis acids (H⁺ itself acts as Lewis acid accepting lone pair of base).
Q4

What is the pH of 0.01 mol dm⁻³ HCl?

HCl → fully ionised: [H⁺] = 0.01 = 10⁻². pH = −log(10⁻²) = 2. Simple rule: for strong acids, pH = −log[acid]. For HCl concentrations: 1 M → pH 0; 0.1 M → pH 1; 0.01 M → pH 2; 0.001 M → pH 3.
Q5

A strong acid differs from a weak acid of the same concentration because:

Strength refers to degree of ionisation. Strong acid (HCl): 100% ionised — [H⁺] = [acid]. Weak acid (CH₃COOH): 1–3% ionised — [H⁺] << [acid]. At same concentration (e.g. 0.1 mol dm⁻³): HCl gives pH 1; CH₃COOH gives pH ~2.87. The key is the extent of dissociation, not concentration (which is the amount per unit volume).
Q6

For a weak acid HA with Ka = 4.0×10⁻⁵ and [HA] = 0.20 mol dm⁻³, [H⁺] is approximately:

[H⁺] = √(Ka × c) = √(4.0×10⁻⁵ × 0.20) = √(8.0×10⁻⁶) = 2.83×10⁻³ ≈ 2.8×10⁻³ mol dm⁻³. Check assumption: 2.8×10⁻³/0.20 = 1.4% << 5% ✓. This formula [H⁺] = √(Ka×c) is derived from Ka = x²/(c−x) ≈ x²/c when degree of ionisation is small.
Q7

The conjugate base of H₂PO₄⁻ is:

Conjugate base = species formed after the acid donates H⁺. H₂PO₄⁻ loses H⁺ → HPO₄²⁻. So HPO₄²⁻ is the conjugate base of H₂PO₄⁻. Note: H₃PO₄ would be the conjugate acid (H₂PO₄⁻ gains H⁺ → H₃PO₄). H₂PO₄⁻ is amphiprotic — can act as both acid and base.
Q8

What is the pH of 0.050 mol dm⁻³ NaOH?

[OH⁻] = 0.050 mol dm⁻³. pOH = −log(0.050) = −log(5×10⁻²) = 2 − log5 = 2 − 0.699 = 1.30. pH = 14 − 1.30 = 12.70. Quick check: [H⁺] = Kw/[OH⁻] = 10⁻¹⁴/0.05 = 2×10⁻¹³. pH = −log(2×10⁻¹³) = 13 − log2 = 13 − 0.30 = 12.70 ✓
Q9

A buffer solution resists changes in pH because:

Buffer mechanism: (1) Added H⁺ is neutralised by the conjugate base A⁻: A⁻ + H⁺ → HA (barely changes ratio). (2) Added OH⁻ is neutralised by the weak acid HA: HA + OH⁻ → A⁻ + H₂O. Both components are present in significant amounts ("reservoir"), so the ratio [A⁻]/[HA] changes only slightly → pH changes only slightly (Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA])).
Q10

The Henderson-Hasselbalch equation gives buffer pH = pKa + log([A⁻]/[HA]). At the half-equivalence point of a weak acid titration:

At the half-equivalence point: exactly half the acid has been neutralised → [A⁻] = [HA]. Substituting: pH = pKa + log([A⁻]/[HA]) = pKa + log(1) = pKa + 0 = pKa. This is an important result — used to determine Ka experimentally (find the half-equivalence volume on the titration curve, read off pH → that pH = pKa → Ka = 10⁻pKa).
Q11

An aqueous solution of NH₄Cl is:

NH₄Cl is the salt of a strong acid (HCl) + weak base (NH₃). NH₄⁺ hydrolyses: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺ → produces H₃O⁺ → acidic. Cl⁻ (from strong acid) does not hydrolyse. Result: acidic solution (pH ~5 for typical concentration). Remember: salt of strong acid + weak base → acidic; salt of weak acid + strong base → basic; salt of strong acid + strong base → neutral.
Q12

Kw = [H⁺][OH⁻] = 1.0×10⁻¹⁴ at 25°C. In pure water at 25°C, [H⁺] = [OH⁻] = ?

In pure water: [H⁺] = [OH⁻] = x. Kw = x² = 10⁻¹⁴ → x = 10⁻⁷ mol dm⁻³. pH = −log(10⁻⁷) = 7.00 (neutral). At temperatures other than 25°C, Kw changes (increases with T) → pH of neutral water is NOT 7 at other temperatures (e.g. at 37°C body temperature, Kw ≈ 2.4×10⁻¹⁴, neutral pH ≈ 6.81).
Q13

Which indicator is most suitable for titrating ethanoic acid against sodium hydroxide?

CH₃COOH + NaOH: weak acid + strong base. Equivalence point pH ≈ 8.7 (basic — CH₃COO⁻ is basic). Steep pH jump occurs from ~7 to ~11. Phenolphthalein (changes 8.3–10) lies within this steep region → sharp colour change at equivalence point. Methyl orange (3.1–4.4) changes in the buffer region before equivalence → no sharp endpoint. Litmus has too broad a range and no sharp colour change. Universal indicator gives approximate pH only.
Q14

The enthalpy of neutralisation for strong acid + strong base is approximately −57 kJ/mol. For CH₃COOH + NaOH it is −55 kJ/mol. Why is it less exothermic for the weak acid?

Net equation for H⁺ + OH⁻ → H₂O always releases the same energy (−57.1 kJ). For strong acid: HCl is already fully ionised, so all energy is from H⁺ + OH⁻ → H₂O. For weak acid: CH₃COOH must first be ionised (endothermic step: CH₃COOH → H⁺ + CH₃COO⁻ absorbs energy). The net enthalpy = −57.1 + (endothermic ionisation energy) = less exothermic. ΔHₙ(weak acid) < ΔHₙ(strong acid) because of the additional endothermic ionisation step.
Q15

Which of the following salts gives a basic aqueous solution?

Na₂CO₃ is the salt of a weak acid (H₂CO₃) + strong base (NaOH). CO₃²⁻ hydrolyses: CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻ → basic (pH ~11.6 for 0.1 mol dm⁻³). NH₄Cl: strong acid + weak base → acidic. KNO₃: strong acid + strong base → neutral. FeCl₃: strong acid + weak Lewis acid Fe³⁺ → acidic (Fe³⁺ hydrolyses to give H⁺).
Q16

For conjugate acid-base pair: Ka(acid) × Kb(conjugate base) = Kw. If Ka(HF) = 6.3×10⁻⁴, what is Kb(F⁻)?

Ka × Kb = Kw = 1.0×10⁻¹⁴. Kb(F⁻) = Kw/Ka(HF) = 1.0×10⁻¹⁴ / 6.3×10⁻⁴ = 1.59×10⁻¹¹ ≈ 1.6×10⁻¹¹. F⁻ is a weak base (Kb < 10⁻¹⁰ is very weak). This shows: stronger acid (large Ka) → weaker conjugate base (small Kb). HF is relatively strong for a weak acid → F⁻ is correspondingly weak as a base.
Q17

A buffer with pH = 5.00 is made from a weak acid with pKa = 4.74. The ratio [A⁻]/[HA] must be:

Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]). 5.00 = 4.74 + log([A⁻]/[HA]). log([A⁻]/[HA]) = 0.26. [A⁻]/[HA] = 10^0.26 = 1.82. So the salt/acid ratio = 1.82:1. This means more conjugate base than acid — the buffer is above the pKa, as expected (pH > pKa → [A⁻] > [HA]).
Q18

Water is described as "amphiprotic." This means:

Amphiprotic = can both donate AND accept protons (H⁺). H₂O as acid: H₂O + NH₃ → OH⁻ + NH₄⁺ (donates H⁺). H₂O as base: H₂O + HCl → H₃O⁺ + Cl⁻ (accepts H⁺). Also: H₂SO₄ is amphiprotic in some solvents; H₂PO₄⁻ is amphiprotic. Amphoteric refers to acid-base character with respect to full solutions (e.g. Al₂O₃ reacts with both acids and bases), while amphiprotic specifically refers to proton transfer.
Q19

A solution has [OH⁻] = 2.0×10⁻³ mol dm⁻³. Its pH is:

pOH = −log(2.0×10⁻³) = −log(2) − log(10⁻³) = 3 − 0.301 = 2.70. pH = 14 − pOH = 14 − 2.70 = 11.30. Alternatively: [H⁺] = Kw/[OH⁻] = 10⁻¹⁴ / 2.0×10⁻³ = 5.0×10⁻¹² mol dm⁻³. pH = −log(5.0×10⁻¹²) = 12 − log5 = 12 − 0.699 = 11.30 ✓
Q20

The degree of ionisation of a weak acid increases when:

For HA ⇌ H⁺ + A⁻, degree of ionisation α = √(Ka/c). When concentration c decreases (dilution), α increases. Physically: dilution shifts equilibrium right (products → lower concentration → Le Châtelier shifts right to restore). At infinite dilution, all weak acids are fully ionised. Adding A⁻ (common ion) shifts LEFT → less ionisation (decrease α). Increasing concentration → α decreases. Temperature: depends on ΔH of ionisation (usually endothermic → higher T → more ionised).
Q21

For the reaction: BF₃ + F⁻ → BF₄⁻, BF₃ acts as a:

BF₃ accepts the lone pair from F⁻ → forms a dative (coordinate) bond → BF₄⁻. BF₃ is the Lewis acid (e⁻ pair acceptor). F⁻ is the Lewis base (e⁻ pair donor). No proton transfer occurs — cannot be classified by Brønsted-Lowry. B in BF₃ has only 6 valence electrons (empty p orbital) → electron deficient → Lewis acid. This reaction is used in organic chemistry (BF₃ as a Lewis acid catalyst for Friedel-Crafts reactions).
Q22

The titration of a weak acid vs weak base has no sharp equivalence point. This is because:

For CH₃COOH + NH₃: the buffer on the acid side and the buffer on the base side both resist pH change. The titration curve has a very gradual slope throughout — no steep vertical section. Without a steep jump, no indicator changes sharply at the equivalence point. pH meter or conductimetric titration must be used instead. This is fundamentally different from strong acid + strong base (large steep jump from pH 4 to pH 10).
Q23

pKa of CH₃COOH = 4.75. A buffer made with equal concentrations of CH₃COOH and CH₃COO⁻ has pH:

pH = pKa + log([A⁻]/[HA]) = 4.75 + log(1) = 4.75 + 0 = 4.75. When [A⁻] = [HA], pH = pKa. This is a key result used in buffer preparation and in finding Ka from titration curves. The buffer has maximum capacity at this pH and is effective in the range pKa ± 1 = 3.75–5.75 for ethanoic acid.
Q24

Ka for HCN = 4.9×10⁻¹⁰. The pKa is approximately:

pKa = −log(Ka) = −log(4.9×10⁻¹⁰) = 10 − log(4.9) = 10 − 0.69 = 9.31 ≈ 9.3. HCN is a very weak acid (pKa = 9.3 → much weaker than CH₃COOH with pKa 4.75). High pKa = small Ka = weak acid. Rule: pKa = −log Ka; Ka = 10^(−pKa). You can estimate: Ka = 4.9×10⁻¹⁰ → exponent is −10, coefficient ~5 → pKa ≈ 9.3.
Q25

The blood buffer maintains pH 7.35–7.45. If blood pH falls below 7.35, the condition is called:

Blood pH 7.35–7.45 is maintained by H₂CO₃/HCO₃⁻ (carbonic acid/bicarbonate) buffer system. pH <7.35 = acidosis (blood too acidic) → caused by CO₂ retention (respiratory acidosis) or metabolic acid production. pH >7.45 = alkalosis (blood too basic) → caused by hyperventilation (blowing off CO₂) or metabolic alkalosis. Even small deviations from 7.4 can be life-threatening because enzymes have narrow pH optima.
📝

Unit Test — 50 Marks

Section A — Short Answer

30 marks
Q1 [5 marks]

Compare the Arrhenius, Brønsted-Lowry, and Lewis theories of acids and bases. For each theory: give the definition of an acid and a base, state a limitation or advantage, and give one example reaction. [5]

Arrhenius: Acid produces H⁺; Base produces OH⁻. Limitation: aqueous only; cannot explain NH₃ as base. Example: HCl → H⁺ + Cl⁻; NaOH → Na⁺ + OH⁻. [1] Brønsted-Lowry: Acid = H⁺ donor; Base = H⁺ acceptor. Advantage: works in any solvent; explains NH₃ (accepts H⁺ from water). Example: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ (NH₃ = base, H₂O = acid). [1.5] Lewis: Acid = e⁻ pair acceptor; Base = e⁻ pair donor. Most general — includes reactions without H⁺ transfer. Advantage: explains complex ion formation, BF₃ reactions. Example: BF₃ + NH₃ → F₃B–NH₃ (BF₃ = Lewis acid, NH₃ = Lewis base). [1.5] All B-L acids are Lewis acids (H⁺ is a Lewis acid — accepts lone pair from base). All Arrhenius reactions are B-L reactions. Lewis includes all others. [1]
Q2 [5 marks]

Calculate the pH of the following solutions at 25°C: (a) 0.050 mol dm⁻³ H₂SO₄ (assume both protons ionise) (b) 1.5×10⁻³ mol dm⁻³ NaOH (c) 0.20 mol dm⁻³ ethanoic acid (Ka = 1.8×10⁻⁵ mol dm⁻³) (d) a solution with [H⁺] = 3.2×10⁻⁸ mol dm⁻³. [5]

(a) H₂SO₄: [H⁺] = 2 × 0.050 = 0.10 mol dm⁻³. pH = −log(0.10) = 1.00. [1] (b) NaOH: [OH⁻] = 1.5×10⁻³. pOH = −log(1.5×10⁻³) = 3 − log1.5 = 3 − 0.176 = 2.82. pH = 14 − 2.82 = 11.18. [1] (c) CH₃COOH: [H⁺] = √(Ka × c) = √(1.8×10⁻⁵ × 0.20) = √(3.6×10⁻⁶) = 1.897×10⁻³. pH = −log(1.897×10⁻³) = 3 − log(1.897) = 3 − 0.278 = 2.72. [1.5] (d) pH = −log(3.2×10⁻⁸) = 8 − log(3.2) = 8 − 0.505 = 7.50. Solution is slightly basic. [1.5]
Q3 [5 marks]

Explain what is meant by a buffer solution. Describe how an ethanoic acid/sodium ethanoate buffer works when: (a) 0.002 mol HCl is added to 1 dm³ of buffer (b) 0.002 mol NaOH is added to 1 dm³ of buffer. Write equations for each case. A buffer contains 0.10 mol dm⁻³ CH₃COOH and 0.15 mol dm⁻³ CH₃COONa (Ka = 1.8×10⁻⁵). Calculate its pH. [5]

Buffer: a solution that resists changes in pH when small amounts of acid or base are added. Contains a weak acid and its conjugate base in comparable concentrations. [1] (a) HCl added: H⁺ + CH₃COO⁻ → CH₃COOH. The conjugate base (CH₃COO⁻) neutralises the added H⁺. [A⁻] decreases slightly; [HA] increases slightly. pH barely changes (small change in log ratio). [1] (b) NaOH added: OH⁻ + CH₃COOH → CH₃COO⁻ + H₂O. The weak acid neutralises the added OH⁻. [HA] decreases slightly; [A⁻] increases slightly. pH barely changes. [1] pH calculation: pKa = −log(1.8×10⁻⁵) = 4.74. pH = pKa + log([A⁻]/[HA]) = 4.74 + log(0.15/0.10) = 4.74 + log(1.5) = 4.74 + 0.176 = 4.92. [2]
Q4 [5 marks]

Explain how acid-base indicators work. Why is phenolphthalein the correct indicator for titrating NaOH against ethanoic acid, while methyl orange is not? Sketch the shape of the titration curve for this reaction and mark the half-equivalence point and the equivalence point. [5]

Indicator is a weak acid HIn where HIn (colour 1) ⇌ H⁺ + In⁻ (colour 2). In acid: excess H⁺ shifts left → colour 1. In base: OH⁻ removes H⁺ → shifts right → colour 2. Transition occurs near pH = pKa(HIn). [2] Weak acid + strong base equivalence point ≈ pH 8.7. Steep pH jump: ~7 to ~11. Phenolphthalein (pKa ≈ 9.3, range 8.3–10) lies within this steep region → sharp colour change (colourless → pink) at equivalence. Methyl orange (range 3.1–4.4) changes colour in the buffer region (before equivalence) where pH changes gradually → no clear endpoint. [2] Titration curve sketch: starts at low pH (~3), rises slowly (buffer region), steep jump ~8 to ~11 at equivalence, levels off at ~13. Half-equivalence point: at half the volume of base added, pH = pKa(CH₃COOH) = 4.75. Equivalence: steep jump mid-point ~8.7. [1]
Q5 [5 marks]

Predict the pH (acidic, neutral, or basic) of aqueous solutions of: (a) NaCl (b) Na₂SO₄ (c) NH₄Cl (d) Na₂CO₃ (e) AlCl₃. Write the hydrolysis equation for any ion that reacts with water. [5]

(a) NaCl: strong acid + strong base → neutral (pH 7). Na⁺ and Cl⁻ do not hydrolyse. [0.5] (b) Na₂SO₄: H₂SO₄ (strong) + NaOH (strong) → neutral (pH 7). SO₄²⁻ is conjugate base of strong acid H₂SO₄ → does not hydrolyse. [0.5] (c) NH₄Cl: strong acid (HCl) + weak base (NH₃) → acidic (pH < 7 ≈ 5). NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺ (ammonium ion hydrolyses). [1.5] (d) Na₂CO₃: weak acid (H₂CO₃) + strong base (NaOH) → basic (pH > 7 ≈ 11.6). CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻ (carbonate hydrolyses). [1.5] (e) AlCl₃: strong acid (HCl) + weak Lewis acid (Al³⁺ — high charge density) → acidic. Al³⁺ + 6H₂O → [Al(H₂O)₆]³⁺ → [Al(H₂O)₅OH]²⁺ + H⁺ → acidic (pH ≈ 3 for 0.1 mol dm⁻³). [1]
Q6 [5 marks]

Ka(ethanoic acid) = 1.8×10⁻⁵. (a) Calculate pKa. (b) Calculate the pH of 0.050 mol dm⁻³ ethanoic acid. (c) What is Ka for a dibasic weak acid that is exactly 2.0% ionised at 0.10 mol dm⁻³? (d) State the relationship Ka × Kb = Kw and use it to find Kb for CH₃COO⁻. [5]

(a) pKa = −log(1.8×10⁻⁵) = 5 − log(1.8) = 5 − 0.255 = 4.74. [1] (b) [H⁺] = √(Ka×c) = √(1.8×10⁻⁵ × 0.050) = √(9.0×10⁻⁷) = 9.49×10⁻⁴ mol dm⁻³. pH = −log(9.49×10⁻⁴) = 4 − log(9.49) = 4 − 0.977 = 3.02. [1] (c) 2.0% ionised at c = 0.10: degree α = 0.020. [H⁺] = α × c = 0.020 × 0.10 = 2.0×10⁻³ mol dm⁻³. Ka = [H⁺][A⁻]/[HA] = (2.0×10⁻³)² / (0.10 − 2.0×10⁻³) ≈ (2.0×10⁻³)² / 0.098 = 4.0×10⁻⁶ / 0.098 = 4.1×10⁻⁵ mol dm⁻³. [1.5] (d) Ka(CH₃COOH) × Kb(CH₃COO⁻) = Kw = 1.0×10⁻¹⁴. Kb = 10⁻¹⁴ / 1.8×10⁻⁵ = 5.6×10⁻¹⁰ mol dm⁻³. pKb = 9.26. This confirms: stronger acid (large Ka) → weaker conjugate base (small Kb); pKa + pKb = 14. [1.5]

Section B — Extended Answer

20 marks
Q7 [10 marks]

(a) Describe and compare the titration curves for: (i) strong acid + strong base (ii) weak acid + strong base (iii) strong acid + weak base. For each: give the approximate pH at the start, the equivalence point, and the end; explain the shape of the curve; state which indicator(s) are suitable. [6]
(b) Explain the biological importance of buffer solutions. Describe the blood buffering system in detail, including the equilibrium involved, how excess acid is handled, and how excess base is handled. What medical conditions arise when the blood buffer fails? [4]

(a)(i) Strong acid (HCl) + strong base (NaOH): Start pH ~1. Equivalence ~7. End ~13. Curve: slow rise, then very steep jump (~pH 4 to 10 in just a few drops). Salt NaCl formed (neutral). Indicators: any works — methyl orange, litmus, phenolphthalein all within steep region. [2] (ii) Weak acid (CH₃COOH) + strong base (NaOH): Start pH ~3. Buffer region during titration (gradual rise). Half-equivalence at pH = pKa ≈ 4.75. Equivalence ~8.7 (basic — CH₃COO⁻ hydrolyses). End ~13. Steep jump: ~7 to 11. Indicator: phenolphthalein only (range 8.3–10 within steep). Methyl orange unsuitable (changes in buffer region). [2] (iii) Strong acid (HCl) + weak base (NH₃): Start pH ~1 (HCl in burette) or ~11 (NH₃ in flask). Equivalence ~5.2 (acidic — NH₄Cl formed, NH₄⁺ hydrolyses). Steep jump: ~3 to 7. Indicator: methyl orange or methyl red only (range 3–5 within steep). Phenolphthalein changes before equivalence (in NH₃ buffer region) → unsuitable. [2] (b) Blood pH 7.35–7.45. Major buffer: H₂CO₃/HCO₃⁻. Equilibrium: CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻ (respiratory and metabolic control). [1] Excess acid (H⁺): HCO₃⁻ + H⁺ → H₂CO₃ → CO₂ + H₂O. CO₂ exhaled by faster breathing → H⁺ removed. Kidneys also excrete H⁺ (slower). [1] Excess base (OH⁻): H₂CO₃ + OH⁻ → HCO₃⁻ + H₂O. Also: CO₂ absorbed more slowly from lungs. [1] Medical conditions: Acidosis (pH <7.35): respiratory (CO₂ retention — slow breathing, lung disease) or metabolic (diabetes ketoacidosis, kidney failure). Alkalosis (pH >7.45): respiratory (hyperventilation, blowing off CO₂) or metabolic (vomiting, excess antacids). Both can cause confusion, seizures, cardiac arrhythmias, or death if severe. [1]
Q8 [10 marks]

(a) Discuss the factors that determine the strength of an oxyacid (oxoacid). Use HClO, HClO₂, HClO₃, HClO₄ as examples. Extend to explain why H₂SO₄ is a strong acid and H₂SO₃ is weak, and why H₃PO₄ is weaker than HNO₃. [5]
(b) Distinguish between monoprotic, diprotic, and polyprotic acids. For H₃PO₄ (triprotic), write the three ionisation equations with their Ka values (Ka₁ = 7.5×10⁻³; Ka₂ = 6.2×10⁻⁸; Ka₃ = 4.8×10⁻¹³). Explain why Ka₁ >> Ka₂ >> Ka₃. State which form predominates in blood (pH 7.4). [5]

(a) Oxyacid strength: more O atoms bonded to central atom → more electron withdrawal from O–H bond → weaker O–H bond → easier to donate H⁺ → stronger acid. Also: higher OS of central atom → more electron-withdrawing. [1] HClO (+1): weakest (pKa 7.5) — only 1 O atom; H–O–Cl — little withdrawal. HClO₂ (+3): pKa ~2. HClO₃ (+5): pKa ~−1 (strong). HClO₄ (+7): pKa ~−10 (strongest common acid) — 4 O atoms withdraw electron density from O–H bond maximally. [1.5] H₂SO₄ vs H₂SO₃: H₂SO₄ has 4 O atoms (S in +6), very strong first ionisation. H₂SO₃ has 3 O atoms (S in +4), weaker acid. More O atoms → stronger acid. [1] HNO₃ vs H₃PO₄: HNO₃ has 3 O atoms (N in +5) — strong acid. H₃PO₄ has 4 O atoms but P is less electronegative than N (P EN = 2.1, N EN = 3.0) → P withdraws electrons less effectively from O–H → H₃PO₄ weaker. Also: in HNO₃, NO₃⁻ conjugate base is highly stabilised by resonance (3 equivalent structures) → better conjugate base stability → stronger acid. [1.5] (b) Monoprotic: one ionisable H⁺ (HCl, HNO₃, CH₃COOH). Diprotic: two H⁺ (H₂SO₄, H₂CO₃). Polyprotic: three or more H⁺ (H₃PO₄, H₃C₆H₅O₇ citric acid). [1] H₃PO₄ ionisations: H₃PO₄ ⇌ H⁺ + H₂PO₄⁻ (Ka₁ = 7.5×10⁻³); H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻ (Ka₂ = 6.2×10⁻⁸); HPO₄²⁻ ⇌ H⁺ + PO₄³⁻ (Ka₃ = 4.8×10⁻¹³). [1.5] Why Ka₁ >> Ka₂ >> Ka₃: each successive ionisation removes H⁺ from an ion with increasing negative charge. The negative charge makes it progressively harder to remove a positively-charged H⁺ (electrostatic repulsion). Also: the second and third ionisations give increasingly negatively charged species (HPO₄²⁻, PO₄³⁻) which stabilise the remaining H⁺ better. [1.5] In blood pH 7.4: compare pKa₁ = 2.12, pKa₂ = 7.20, pKa₃ = 12.32. At pH 7.4, [HPO₄²⁻]/[H₂PO₄⁻] = 10^(7.4−7.2) = 10^0.2 = 1.58. So HPO₄²⁻ : H₂PO₄⁻ ≈ 1.6:1 → HPO₄²⁻ predominates slightly. (PO₄³⁻ negligible at pH 7.4; H₃PO₄ essentially absent.) [1]
← Unit 15: Equilibrium S4 Course Home Unit 17: Redox →