When an acid and a base react they neutralise each other to form a salt and water. Acid-base titrations use this stoichiometry to find the concentration of an unknown solution with great precision.
11.1 Neutralisation and Salts
Definition
Neutralisation
A neutralisation reaction is one in which an acid and a base react to form a salt and water. Acid + Base → Salt + Water.
Definition
Salt
A salt is a neutral ionic compound formed from the cation of the base and the anion of the acid in a neutralisation reaction.
To predict the salt formed: take the cation from the base and the anion from the acid. For example, NaOH (cation Na⁺) + HCl (anion Cl⁻) → NaCl (sodium chloride). The water molecule comes from the H⁺ of the acid and the OH⁻ of the base.
Common neutralisation reactions
- HCl + NaOH → NaCl + H₂O (sodium chloride)
- H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O (sodium sulfate)
- HNO₃ + KOH → KNO₃ + H₂O (potassium nitrate)
- 2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O (calcium chloride)
Worked Example
Write the balanced equation for the reaction between H₂SO₄ and Ca(OH)₂. Name the salt formed.
Given
- Acid: H₂SO₄
- Base: Ca(OH)₂
Find
Balanced equation and name of salt
Solution
- 1Identify the salt: cation from base = Ca²⁺; anion from acid = SO₄²⁻ → CaSO₄
- 2Unbalanced: H₂SO₄ + Ca(OH)₂ → CaSO₄ + H₂O
- 3Balance: H₂SO₄ + Ca(OH)₂ → CaSO₄ + 2H₂O ✓
Exam Tip
A salt does NOT have to taste salty or be neutral in solution. NaCl is neutral but NH₄Cl is acidic and Na₂CO₃ is basic. The CAPS definition focuses on how the salt is formed, not its properties.
Strong vs Weak Acids and Bases
| Property | Strong | Weak |
|---|---|---|
| Ionisation | Completely ionised in water | Partially ionised in water |
| Equilibrium | Reaction goes to completion | Equilibrium mixture of ions and molecules |
| Conductivity | High (many ions) | Low (fewer ions) |
| Acid examples | HCl, H₂SO₄, HNO₃ | CH₃COOH, H₂CO₃, HF |
| Base examples | NaOH, KOH, Ca(OH)₂ | NH₃, NH₄OH |
Watch Out
Do not confuse STRONG with CONCENTRATED. A strong acid is one that fully ionises (regardless of concentration). A concentrated acid simply has a high number of moles per dm³. A concentrated weak acid is still a weak acid.
Practice Question
Write the balanced neutralisation equation between HNO₃ and Mg(OH)₂. Name the salt and state whether Mg(OH)₂ is a strong or weak base.
(5 marks)
11.2 Titrations
A titration is a technique used to find the concentration of an unknown solution by reacting it with a solution of known concentration. At the equivalence point, the moles of acid exactly neutralise the moles of base (adjusted for the stoichiometric ratio from the balanced equation).
Formula
Titration formula (1:1 reactions)
c₁ = concentration of acid (mol·dm⁻³), V₁ = volume of acid (dm³), c₂ = concentration of base (mol·dm⁻³), V₂ = volume of base (dm³)
SI unit: mol
Note
For reactions where the mole ratio is NOT 1:1, use: (c₁V₁)/n₁ = (c₂V₂)/n₂ where n₁ and n₂ are the stoichiometric coefficients from the balanced equation.
Worked Example
In a titration, 25,0 cm³ of NaOH solution reacts completely with 20,0 cm³ of 0,1 mol·dm⁻³ HCl. Calculate the concentration of the NaOH solution. (HCl + NaOH → NaCl + H₂O)
Given
- V(HCl) = 20,0 cm³ = 0,020 dm³
- c(HCl) = 0,1 mol·dm⁻³
- V(NaOH) = 25,0 cm³ = 0,025 dm³
- Mole ratio HCl : NaOH = 1 : 1
Find
c(NaOH) = ?
Solution
- 1c₁V₁ = c₂V₂ (since ratio is 1:1)
- 20,1 × 0,020 = c(NaOH) × 0,025
- 3c(NaOH) = (0,1 × 0,020) / 0,025
- 4c(NaOH) = 0,002 / 0,025 = 0,08 mol·dm⁻³
Worked Example
25,0 cm³ of H₂SO₄ reacts completely with 30,0 cm³ of 0,2 mol·dm⁻³ NaOH. Calculate c(H₂SO₄). (H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O)
Given
- V(H₂SO₄) = 25,0 cm³ = 0,025 dm³
- V(NaOH) = 30,0 cm³ = 0,030 dm³
- c(NaOH) = 0,2 mol·dm⁻³
- Mole ratio H₂SO₄ : NaOH = 1 : 2
Find
c(H₂SO₄) = ?
Solution
- 1n(NaOH) = c × V = 0,2 × 0,030 = 0,006 mol
- 2n(H₂SO₄) = n(NaOH) / 2 = 0,006 / 2 = 0,003 mol
- 3c(H₂SO₄) = n/V = 0,003 / 0,025 = 0,12 mol·dm⁻³
Watch Out
Always convert volumes from cm³ to dm³ before substituting into the formula: divide by 1000. (25,0 cm³ = 0,025 dm³). Using cm³ directly is the single most common titration error.
Practice Question
20,0 cm³ of 0,15 mol·dm⁻³ HCl is titrated against KOH solution. If 18,0 cm³ of KOH is required to reach the equivalence point, calculate c(KOH). (HCl + KOH → KCl + H₂O)
(4 marks)
11.3 Practical Titration and Indicators
In a practical titration, an indicator is added to signal the equivalence point by changing colour. The choice of indicator depends on whether the reaction produces a neutral, acidic, or basic salt solution.
Common indicators and their ranges
- Methyl orange: red in acid (pH < 3,1), orange at end point, yellow in base (pH > 4,4) — used for strong acid vs strong base
- Phenolphthalein: colourless in acid (pH < 8,3), pink in base (pH > 10) — used for weak acid vs strong base
- Universal indicator: full colour range pH 1–14
Concentrated vs Dilute — Strong vs Weak
| Property | Concentrated/Dilute | Strong/Weak |
|---|---|---|
| What it measures | Amount of acid per unit volume (c = n/V) | Degree of ionisation in water |
| Can be combined | Yes — can have concentrated weak acid | Yes — can have dilute strong acid |
| Example | Concentrated CH₃COOH = many moles of acid per dm³ | Weak acid = partial ionisation regardless of concentration |
Worked Example
Classify each of the following as strong/weak AND concentrated/dilute: (a) 0,001 mol·dm⁻³ HCl (b) 10 mol·dm⁻³ CH₃COOH.
Given
- HCl = strong acid
- CH₃COOH = weak acid
Find
Classification
Solution
- 1(a) HCl: strong (fully ionises). c = 0,001 mol·dm⁻³ → dilute. Answer: strong and dilute.
- 2(b) CH₃COOH: weak (partially ionises). c = 10 mol·dm⁻³ → concentrated. Answer: weak and concentrated.
Exam Tip
Exam tip: the question 'Is HF a strong or weak acid?' is asking about ionisation, NOT concentration. HF is a weak acid (partial ionisation) regardless of how concentrated the solution is.
Practice Question
A student says 'concentrated sulfuric acid must be a strong acid because it is concentrated.' Is the student correct? Explain.
(3 marks)