Acids and bases are at the heart of chemistry in everyday life — from stomach acid to antacids, household cleaners to battery acid. In this chapter you will learn the Brønsted-Lowry model of acid-base behaviour, master the language of conjugate pairs, and use titration calculations to find the concentration of an unknown solution.
13.1 Brønsted-Lowry Acids and Bases
The Brønsted-Lowry model defines acids and bases in terms of proton (H⁺) transfer. This is more powerful than the older Arrhenius model because it works in any solvent, not just water. The key question to ask in any reaction is: which species gives away an H⁺, and which species accepts it?
Definition
Acids (Brønsted-Lowry)
A Brønsted-Lowry acid is a substance that gives away protons (hydrogen cations H⁺), and is therefore called a proton donor.
Definition
Bases (Brønsted-Lowry)
A Brønsted-Lowry base is a substance that takes up protons (hydrogen cations H⁺), and is therefore called a proton acceptor.
To identify the acid and base in a reaction, look for the H⁺ transfer. The species that LOSES an H⁺ (has one fewer H after the reaction) is the ACID. The species that GAINS an H⁺ (has one more H after the reaction) is the BASE. For example, in HCl + H₂O → H₃O⁺ + Cl⁻: HCl loses an H⁺ (acid) and H₂O gains an H⁺ (base).
Definition
Conjugate acid-base pair
A conjugate acid-base pair consists of two species that differ by exactly one proton (H⁺). The acid donates H⁺ to form the conjugate base; the base accepts H⁺ to form the conjugate acid.
When an acid donates an H⁺, the species left behind is its conjugate base (it can now accept an H⁺ to re-form the acid). When a base accepts an H⁺, the species formed is its conjugate acid. Every Brønsted-Lowry reaction therefore involves TWO conjugate pairs. In HCl + H₂O ⇌ H₃O⁺ + Cl⁻ the pairs are: (HCl / Cl⁻) and (H₂O / H₃O⁺).
Exam Tip
EXAM TIP — To write a conjugate base: remove one H⁺ from the acid and reduce the charge by 1. To write a conjugate acid: add one H⁺ to the base and increase the charge by 1. E.g. conjugate base of H₂SO₄ is HSO₄⁻; conjugate acid of NH₃ is NH₄⁺.
Watch Out
COMMON MISTAKE — Students often confuse conjugate pairs with the two sides of a reaction. The conjugate pair must differ by EXACTLY one H⁺ and the charge must change by exactly 1. HCl and Cl⁻ are a conjugate pair; HCl and NaCl are NOT.
Worked Example
Identify the Brønsted-Lowry acid, base, conjugate acid, and conjugate base in the reaction: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
Given
- Reaction: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
Find
Acid, base, conjugate acid, conjugate base, and the two conjugate pairs.
Solution
- 1Step 1: Find the H⁺ transfer. H₂O loses an H⁺ → it is the ACID. NH₃ gains an H⁺ → it is the BASE.
- 2Step 2: H₂O (acid) loses H⁺ to form OH⁻ — so OH⁻ is the conjugate base of H₂O. Conjugate pair 1: H₂O / OH⁻.
- 3Step 3: NH₃ (base) gains H⁺ to form NH₄⁺ — so NH₄⁺ is the conjugate acid of NH₃. Conjugate pair 2: NH₄⁺ / NH₃.
Practice Question
Identify the two conjugate acid-base pairs in the reaction: H₂SO₄ + H₂O → H₃O⁺ + HSO₄⁻
(4 marks)
13.2 Strong vs Weak Acids and Bases; Concentrated vs Dilute
These are two completely independent descriptions of an acid or base, and mixing them up is one of the most common errors in Grade 11 chemistry. 'Strong/weak' describes HOW COMPLETELY the acid or base dissociates (breaks apart) in solution. 'Concentrated/dilute' describes HOW MUCH acid or base is dissolved per litre of solution. You can have any combination of the four.
Strong/Weak vs Concentrated/Dilute
| Property | Strong / Weak | Concentrated / Dilute |
|---|---|---|
| What it describes | Degree of dissociation — what fraction of molecules ionise | Amount of solute dissolved per unit volume (mol·dm⁻³) |
| Strong example | HCl: almost 100% → H⁺ + Cl⁻ | Not relevant — depends on moles added |
| Weak example | CH₃COOH: only ~1% ionises at room temp | Not relevant — depends on moles added |
| Concentrated example | Not relevant — depends on ionisation | High mol·dm⁻³, e.g. 6 mol·dm⁻³ HCl |
| Dilute example | Not relevant — depends on ionisation | Low mol·dm⁻³, e.g. 0.01 mol·dm⁻³ HCl |
| Can they combine? | Yes — concentrated weak acid is possible | Yes — dilute strong acid is possible |
Strong acids (HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄) dissociate completely (→, one-way arrow). Weak acids (CH₃COOH, H₂CO₃, HF) dissociate only partially (⇌, reversible arrow — an equilibrium is set up). The same principle applies to bases: NaOH and KOH are strong bases; NH₃ is a weak base.
Exam Tip
EXAM TIP — The six strong acids can be remembered as: HCl, HBr, HI (the hydrohalic acids except HF), plus HNO₃, H₂SO₄, and HClO₄. All other acids in your syllabus are weak. HF is weak despite being a hydrogen halide acid.
Watch Out
COMMON MISTAKE — Do NOT say 'a strong acid has a lower pH than a weak acid' without specifying equal concentration. A very dilute strong acid can have a higher pH than a concentrated weak acid. Always specify the conditions.
Worked Example
Classify each of the following as strong/weak AND concentrated/dilute: (a) 0.001 mol·dm⁻³ HCl (b) 5 mol·dm⁻³ CH₃COOH
Given
- HCl: strong acid (complete dissociation), concentration = 0.001 mol·dm⁻³
- CH₃COOH: weak acid (partial dissociation), concentration = 5 mol·dm⁻³
Find
Strength classification and concentration classification for each.
Solution
- 1Step 1: HCl is one of the six strong acids — STRONG. 0.001 mol·dm⁻³ is very low — DILUTE.
- 2Step 2: CH₃COOH (acetic acid) partially ionises — WEAK. 5 mol·dm⁻³ is very high — CONCENTRATED.
Practice Question
Explain the difference between a strong acid and a concentrated acid. Give an example of each.
(4 marks)
Practice Question
Two solutions both have a concentration of 0.1 mol·dm⁻³: Solution A is HCl and Solution B is CH₃COOH. Which solution has a lower pH? Explain.
(4 marks)
13.3 Neutralisation Reactions and Titration Calculations
A neutralisation reaction occurs when an acid reacts with a base to form a salt and water. The H⁺ ions from the acid combine with the OH⁻ ions from the base to form water. In a titration, we use a standard solution of known concentration to determine the unknown concentration of another solution. The key relationship at the equivalence point is derived from the mole ratio in the balanced equation.
Formula
Titration equation (1:1 mole ratio)
na = moles of acid used; Va = volume of acid (dm³); nb = moles of base used; Vb = volume of base (dm³). For 1:1 reactions, na = nb = 1, so c(acid)×V(acid) = c(base)×V(base).
SI unit: mol
When the mole ratio is not 1:1 (e.g. H₂SO₄ + 2NaOH), you must adjust. The general form is: [c(acid) × V(acid)] / na = [c(base) × V(base)] / nb, where na and nb are the stoichiometric coefficients of acid and base respectively from the balanced equation. Always write the balanced equation first.
Exam Tip
EXAM TIP — Convert ALL volumes to dm³ (divide cm³ by 1000) before substituting into the titration formula. The formula n = cV requires volume in dm³, not cm³.
Worked Example
25.0 cm³ of HCl solution requires 20.0 cm³ of 0.1 mol·dm⁻³ NaOH for complete neutralisation. Calculate the concentration of the HCl solution.
Given
- V(HCl) = 25.0 cm³ = 0.0250 dm³
- V(NaOH) = 20.0 cm³ = 0.0200 dm³
- c(NaOH) = 0.1 mol·dm⁻³
- Balanced equation: HCl + NaOH → NaCl + H₂O (1:1 mole ratio)
Find
c(HCl)
Solution
- 1Step 1: Calculate moles of NaOH used. n(NaOH) = c × V = 0.1 × 0.0200 = 0.002 mol
- 2Step 2: Use mole ratio. From HCl + NaOH → NaCl + H₂O, the ratio is 1:1, so n(HCl) = n(NaOH) = 0.002 mol
- 3Step 3: Calculate c(HCl). c = n/V = 0.002 / 0.0250 = 0.08 mol·dm⁻³
Worked Example
What volume of 0.2 mol·dm⁻³ H₂SO₄ is needed to neutralise 30.0 cm³ of 0.3 mol·dm⁻³ NaOH? (H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O)
Given
- c(NaOH) = 0.3 mol·dm⁻³
- V(NaOH) = 30.0 cm³ = 0.0300 dm³
- c(H₂SO₄) = 0.2 mol·dm⁻³
- Mole ratio: H₂SO₄ : NaOH = 1 : 2
Find
V(H₂SO₄)
Solution
- 1Step 1: n(NaOH) = 0.3 × 0.0300 = 0.009 mol
- 2Step 2: From mole ratio 1:2, n(H₂SO₄) = 0.009 / 2 = 0.0045 mol
- 3Step 3: V(H₂SO₄) = n / c = 0.0045 / 0.2 = 0.0225 dm³ = 22.5 cm³
Watch Out
COMMON MISTAKE — Using the formula naVa = nbVb directly with a 1:2 ratio without adjusting gives the wrong answer. Always write the balanced equation first and use the mole ratio explicitly. The 'safe' method is always: calculate n of known → apply ratio → calculate V of unknown.
Practice Question
15.0 cm³ of 0.20 mol·dm⁻³ NaOH neutralises 25.0 cm³ of HNO₃. Calculate the concentration of the HNO₃ solution. (HNO₃ + NaOH → NaNO₃ + H₂O)
(5 marks)
Practice Question
Write a balanced neutralisation equation for sulfuric acid reacting with potassium hydroxide. Identify the salt formed.
(3 marks)
Steps for any titration calculation
- Write the balanced chemical equation for the neutralisation reaction.
- Identify the known concentration and volume (standard solution).
- Convert all volumes from cm³ to dm³ (divide by 1000).
- Calculate the moles of the known solution: n = c × V.
- Use the mole ratio from the balanced equation to find moles of the unknown.
- Calculate the unknown concentration: c = n / V.