Key Equations
\[pK_a = -\log K_a \qquad K_a = 10^{-pK_a}\]
\[pK_b = -\log K_b \qquad K_b = 10^{-pK_b}\]
\[pK_a + pK_b = pK_w = 14.00 \text{ (at 25}^\circ\text{C)}\]
💡 Key Point
Smaller pKa = stronger acid. Smaller pKb = stronger base. The "p" scale inverts the relationship.
Worked Example
Q: CH₃COOH has Ka = 1.76 × 10⁻⁵. Find pKa and pKb of acetate.
\(pK_a = -\log(1.76 \times 10^{-5}) = \textbf{4.75}\)
\(pK_b = 14.00 - 4.75 = \textbf{9.25}\)
📋 Exam Tip
Never compare a Ka directly with a Kb. Convert to the same format (e.g., both pKa) before deciding which species is stronger.