Grade 11 · Chemistry · Lesson 3
Acids & Bases
Compare Arrhenius and Brønsted-Lowry theories, identify conjugate acid-base pairs, calculate pH, and understand the role of indicators and buffer systems.
Curriculum:

Arrhenius Theory (1884)

Svante Arrhenius proposed the first modern definitions of acids and bases based on their behaviour in water.

HCl(aq) → H⁺ + Cl⁻
H₂SO₄(aq) → 2H⁺ + SO₄²⁻
NaOH(aq) → Na⁺ + OH⁻
Ca(OH)₂(aq) → Ca²⁺ + 2OH⁻

Limitation: The Arrhenius model is only applicable to aqueous (water) solutions. It cannot explain why NH₃ — which contains no OH⁻ — acts as a base, since it doesn't dissolve to release OH⁻ directly.

Brønsted-Lowry Theory (1923)

Johannes Brønsted and Thomas Lowry independently proposed a broader, more general theory:

This model does not require water. It explains NH₃ as a base because NH₃ accepts a proton (H⁺) from water, forming NH₄⁺ and leaving OH⁻:

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
Key insight: The Brønsted-Lowry model explains ALL Arrhenius acids and bases, but also covers reactions in non-aqueous solvents and gases reacting. For the NSC exam, Brønsted-Lowry is the required framework.

Conjugate Acid-Base Pairs

When an acid donates H⁺, the species that remains is its conjugate base — it differs from the acid by exactly one proton (H⁺). Every acid-base reaction involves two conjugate pairs simultaneously.

HA + H₂O ⇌ H₃O⁺ + A⁻
Conjugate pairs: HA/A⁻ and H₂O/H₃O⁺
AcidBaseConjugate pair 1Conjugate pair 2
HClH₂OHCl / Cl⁻H₂O / H₃O⁺
CH₃COOHH₂OCH₃COOH / CH₃COO⁻H₂O / H₃O⁺
NH₄⁺H₂ONH₄⁺ / NH₃H₂O / H₃O⁺
H₂ONH₃H₂O / OH⁻NH₃ / NH₄⁺

Amphoteric Substances

An amphoteric substance can act as EITHER an acid OR a base depending on what it reacts with.

H₂O is the classic amphoteric substance: it acts as a base with HCl (accepts H⁺ to form H₃O⁺) and as an acid with NH₃ (donates H⁺ to form OH⁻).

Strong vs Weak Acids and Bases

The strength of an acid or base refers to the degree of ionisation in water — not the concentration.

The pH Scale

pH is a logarithmic measure of the hydrogen ion concentration in solution at 25°C:

pH = −log[H₃O⁺]    pOH = −log[OH⁻]    pH + pOH = 14

The ion product of water: Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C. This means pH + pOH = 14 always holds at 25°C.

pH Calculations:

Neutralisation

When an acid and a base react, they form a salt and water in an exothermic reaction:

Acid + Base → Salt + Water
Net ionic: H₃O⁺ + OH⁻ → 2H₂O

The reaction is exothermic — heat is released to the surroundings.

Hydrolysis of Salts

The pH of a salt solution depends on the relative strengths of the acid and base that formed it:

SaltFormed frompH of solution
NaClStrong acid + strong base~7 (neutral)
CH₃COONaWeak acid + strong base>7 (basic)
NH₄ClStrong acid + weak base<7 (acidic)
CH₃COONH₄Weak acid + weak basedepends on Ka vs Kb

Explanation: CH₃COONa — the acetate ion (CH₃COO⁻) is the conjugate base of a weak acid; it accepts H⁺ from water, producing OH⁻ → basic solution. NH₄Cl — the ammonium ion (NH₄⁺) is the conjugate acid of a weak base; it donates H⁺ to water, producing H₃O⁺ → acidic solution.

Acid-Base Indicators

Indicators are weak acids (HIn) whose ionised form (In⁻) has a different colour from the unionised form:

HIn ⇌ H⁺ + In⁻
(colour 1)      (colour 2)
IndicatorAcid colourpH rangeBase colourBest used for
Methyl orangeRed3.1–4.4YellowStrong acid / weak base
LitmusRed4.5–8.3BlueOnly qualitative
Bromothymol blueYellow6.0–7.6BlueStrong acid / strong base
PhenolphthaleinColourless8.2–10.0Pink/magentaWeak acid / strong base
IEB Extension: Ka, Kb, Buffers & Polyprotic Acids

Ka and pKa: pKa = −log(Ka). A smaller pKa means a stronger weak acid. This is analogous to pH but for acid strength.

Ka × Kb = Kw: For any conjugate acid-base pair at 25°C, Ka × Kb = 1.0 × 10⁻¹⁴. This allows you to calculate Kb from Ka (and vice versa) for conjugate pairs.

Buffer solutions: A buffer is a mixture of a weak acid (HA) and its conjugate base (A⁻, typically added as a sodium salt such as CH₃COONa) in similar concentrations. Buffers resist pH change when small amounts of acid or base are added:

  • Add H⁺: H⁺ + A⁻ → HA (base component reacts) — pH barely changes
  • Add OH⁻: OH⁻ + HA → A⁻ + H₂O (acid component reacts) — pH barely changes
Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA])

When [A⁻] = [HA], pH = pKa. The buffer is most effective within ±1 pH unit of pKa.

Polyprotic acids donate H⁺ in steps, with Ka₁ >> Ka₂ >> Ka₃:

  • H₂SO₄: Ka₁ very large (effectively strong), Ka₂ = 1.2 × 10⁻²
  • H₃PO₄: Ka₁ = 7.5 × 10⁻³, Ka₂ = 6.2 × 10⁻⁸, Ka₃ = 4.8 × 10⁻¹³

In practice, only the first ionisation contributes significantly to [H₃O⁺] in dilute solution.

Simulation Mode

pH Scale & Calculation

Solution Type
Concentration (mol·L⁻¹)
0.100
Ka / Kb
1.8×10⁻⁵
Calculated Values
pH
pOH
[H₃O⁺]
mol·L⁻¹
Select a solution type and concentration above.
0/6
NSC Questions complete
IEB Extension Questions
These questions are written in the style of NSC exam questions. Show all working where calculations are required. Answers are not provided here — use your notes and discuss with your teacher.
Question 1
For the following reaction: HSO₄⁻ + H₂O ⇌ SO₄²⁻ + H₃O⁺

(a) Identify the Brønsted-Lowry acid and base in the forward reaction.
(b) Write down both conjugate acid-base pairs.
(c) Is HSO₄⁻ amphoteric? Justify your answer.

[5 marks]
Question 2
Calculate the pH of the following solutions at 25°C:

(a) 0.050 mol·L⁻¹ HNO₃ (strong acid)
(b) 2.00 × 10⁻³ mol·L⁻¹ NaOH (strong base)

Show all working.
[4 marks]
Question 3
Calculate the pH of 0.250 mol·L⁻¹ formic acid (HCOOH), given Ka = 1.77 × 10⁻⁴. Use the approximation [H₃O⁺] ≈ √(Ka × c). Show full working.

[4 marks]
Question 4
Three salt solutions — NaCl, Na₂CO₃, and NH₄NO₃ — are tested with universal indicator. One is acidic, one is neutral, one is basic. Match each salt to its expected pH category (acidic / neutral / basic) and explain your reasoning based on the relative strengths of the acids and bases that formed each salt.

[6 marks]
Question 5
A student wants to detect the endpoint of an acid-base titration in which a strong acid is slowly added to a solution of sodium carbonate (Na₂CO₃, a salt of a weak acid). Explain:

(a) Why the equivalence point will be at a pH below 7.
(b) Which of methyl orange or phenolphthalein would be the better indicator choice, and why.

[4 marks]
Question 6 — Titration Data Table
25.0 cm³ of HCl(aq) of unknown concentration is titrated with 0.100 mol·L⁻¹ NaOH(aq), added from a burette. The pH of the mixture is recorded after each addition:

Volume NaOH added (cm³)0510151820222530
pH1.31.51.82.53.57.010.511.812.3
(a) Using the data table, identify the volume of NaOH at the equivalence point. Explain how you can tell from the pattern of pH values (without needing to sketch the curve).
(b) Calculate the number of moles of NaOH used at the equivalence point.
(c) Using the mole ratio in the reaction HCl + NaOH → NaCl + H₂O, calculate the concentration of the original HCl solution.
(d) Explain why the pH changes only gradually between 0 and 15 cm³, but very sharply between 18 and 22 cm³.

[7 marks]