Equilibrium

Chemical equilibrium, Kc and Kp, Le Chatelier, ionic equilibrium, pH, buffer, solubility product.

Chemical equilibrium and Kc/Kp

Equilibrium constant, units, relation Kp = Kc(RT)^Δn.

No published notes for this topic yet.

Le Chatelier's principle

Effect of concentration, temperature, pressure on equilibrium.

No published notes for this topic yet.

pH, buffers and solubility product

Acid-base equilibria, Henderson-Hasselbalch, Ksp.

pH, Ka, Kb and the buffer equation
Notes

pH measures the activity (effectively concentration) of H⁺ ions in solution:

pH = −log[H⁺]

Pure water at 25°C has [H⁺] = 10⁻⁷ M, so pH = 7. Acidic: pH < 7. Basic: pH > 7.

Water self-ionization: [H⁺][OH⁻] = K_w = 10⁻¹⁴ (at 25°C). So pH + pOH = 14.

Strong acids/bases dissociate completely:

  • 0.01 M HCl → [H⁺] = 0.01 → pH = 2.
  • 0.001 M NaOH → [OH⁻] = 0.001 → pOH = 3 → pH = 11.

Weak acids partially dissociate:

HA ⇌ H⁺ + A⁻, K_a = [H⁺][A⁻] / [HA]

For a weak acid with initial concentration C and small dissociation (α ≪ 1):

[H⁺] ≈ √(K_a · C), pH = ½ (pK_a − log C).

Buffer solutions resist pH change. They contain a weak acid + its conjugate base (or weak base + its conjugate acid).

Henderson-Hasselbalch equation:

pH = pK_a + log([A⁻] / [HA])

When [A⁻] = [HA], pH = pK_a — the maximum buffer capacity. Buffers work effectively within ±1 pH unit of pK_a.

Worked example. Acetic acid (pK_a = 4.74) and sodium acetate buffer with [HA] = 0.1 M, [A⁻] = 0.1 M:

pH = 4.74 + log(0.1/0.1) = 4.74 + 0 = 4.74.

If you add 0.01 mol of HCl per litre to this buffer, [HA] becomes 0.11, [A⁻] becomes 0.09:

pH = 4.74 + log(0.09/0.11) = 4.74 − 0.087 = 4.65.

Adding the same HCl to pure water (pH 7) would give pH 2 — buffer changed by 0.09 units vs pure water by 5 units. That's the buffering action.