Ka, Kb, and the Conjugate Product Law
Weak acid dissociation: \(\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\) → \(K_{\text{a}} = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\)
Weak base ionisation: \(\text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^-\) → \(K_{\text{b}} = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}\)
For any conjugate acid-base pair in water:
\[K_{\text{a}} \times K_{\text{b}} = K_{\text{w}} = 1.00 \times 10^{-14} \quad (\text{at } 298\text{ K})\]The Expressions
The Conjugate Relationship
The Conjugate Strength Inverse Law
Because \(K_{\text{a}} \times K_{\text{b}} = 10^{-14}\), an inverse relationship exists between the strength of an acid and its conjugate base:
- The stronger the acid (larger \(K_{\text{a}}\)), the weaker its conjugate base (smaller \(K_{\text{b}}\)).
- The conjugate base of a strong acid (\(\text{Cl}^-\)) is an extremely weak base (essentially non-basic).
Worked Example
Calculating Kb from Ka
Problem: Given \(K_{\text{a}}(\text{HF}) = 6.8 \times 10^{-4}\text{ at } 298\text{ K}\), calculate \(K_{\text{b}}\) for the fluoride ion (\(\text{F}^-\)).
\[K_{\text{b}} = \frac{K_{\text{w}}}{K_{\text{a}}} = \frac{1.00 \times 10^{-14}}{6.8 \times 10^{-4}} = 1.47 \times 10^{-11}\text{ mol dm}^{-3}\]Omission of Solvent [H₂O]
Never include water \([\text{H}_2\text{O}]\) in the equilibrium expression for \(K_{\text{a}}\) or \(K_{\text{b}}\). Because water is the solvent, its concentration is constant (\(\approx 55.5\text{ M}\)) and is incorporated into the equilibrium constant.
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