📘 IB Understanding
Ka and Kb are equilibrium constants for weak acid and weak base dissociation. For any conjugate pair: Ka × Kb = Kw.
The Expressions
\(\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\)
\[K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\]\(\text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^-\)
\[K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}\]The Conjugate Relationship
\[K_a \times K_b = K_w = 1.00 \times 10^{-14} \text{ (at 25}^\circ\text{C)}\]
💡 The See-Saw
The stronger an acid (larger Ka), the weaker its conjugate base (smaller Kb), and vice versa. This is a mathematical consequence of Ka × Kb = Kw.
Worked Example
Q: Ka = 1.0 × 10⁻² for a weak acid. Find Kb for its conjugate base.
\(K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-2}} = \textbf{1.0} \times \textbf{10}^{\textbf{-12}}\)
📋 Exam Tip
Never include [H₂O] in Ka or Kb expressions. Water is the solvent and its concentration is constant.