Acids, Bases and pH

This lesson covers Brønsted-Lowry theory, strong and weak acids, Ka and pKa, pH calculations, the ionic product of water Kw, and pH of strong bases.

Brønsted-Lowry Theory

Key Definition: A Brønsted-Lowry acid is a proton (H⁺) donor. A Brønsted-Lowry base is a proton (H⁺) acceptor.

In the reaction: HCl + H₂O → H₃O⁺ + Cl⁻

HCl donates a proton (acid), H₂O accepts a proton (base). H₃O⁺ and Cl⁻ are the conjugate acid and conjugate base respectively.

Conjugate acid-base pairs differ by one proton (H⁺).

pH = −log₁₀[H⁺]

Conversely: [H⁺] = 10^(−pH)

pH is a logarithmic scale — each unit decrease in pH represents a tenfold increase in [H⁺].
pH 7 is neutral (at 25°C), pH < 7 is acidic, pH > 7 is alkaline.

Strong acids fully dissociate in aqueous solution.

HCl → H⁺ + Cl⁻ (complete dissociation)

For a strong monoprotic acid: [H⁺] = [acid]

Example 1: Calculate the pH of 0.050 mol dm⁻³ HCl.

[H⁺] = 0.050 mol dm⁻³ pH = −log₁₀(0.050) = 1.30

For a strong diprotic acid like H₂SO₄: [H⁺] = 2 × [acid]

Example 2: Calculate the pH of 0.010 mol dm⁻³ H₂SO₄.

[H⁺] = 2 × 0.010 = 0.020 mol dm⁻³ pH = −log₁₀(0.020) = 1.70

Weak acids partially dissociate in aqueous solution.

CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq)

The acid dissociation constant Ka is:

Ka = [CH₃COO⁻][H⁺] / [CH₃COOH]

For a weak acid HA: Ka = [H⁺]² / [HA] (assuming [H⁺] = [A⁻] and that dissociation is small enough that [HA]equilibrium ≈ [HA]initial)

Therefore: [H⁺] = √(Ka × [HA])

pKa = −log₁₀(Ka) and Ka = 10^(−pKa)

A smaller Ka (larger pKa) means a weaker acid.

Example 3: Calculate the pH of 0.10 mol dm⁻³ ethanoic acid (Ka = 1.74 × 10⁻⁵ mol dm⁻³).

[H⁺] = √(Ka × [HA]) = √(1.74 × 10⁻⁵ × 0.10) = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³

pH = −log₁₀(1.32 × 10⁻³) = 2.88

Check: [H⁺] = 1.32 × 10⁻³ is much less than 0.10, so the assumption that [HA] ≈ 0.10 is valid.

Water undergoes autoionisation: H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)

Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 25°C

This applies to all aqueous solutions at 25°C, not just pure water.