Key Equations
HL Extension
Logarithmic pKa, pKb and Conjugate Law
\[\text{p}K_{\text{a}} = -\log_{10} K_{\text{a}} \Longleftrightarrow K_{\text{a}} = 10^{-\text{p}K_{\text{a}}}\]
Smaller \(\text{p}K_{\text{a}}\) = larger \(K_{\text{a}}\) = stronger acid.
\[\text{p}K_{\text{b}} = -\log_{10} K_{\text{b}} \Longleftrightarrow K_{\text{b}} = 10^{-\text{p}K_{\text{b}}}\]
Smaller \(\text{p}K_{\text{b}}\) = larger \(K_{\text{b}}\) = stronger base.
Worked Example
Worked Example
Calculating pKa and Conjugate pKb
Problem: Ethanoic acid has \(K_{\text{a}} = 1.76 \times 10^{-5}\). Calculate its \(\text{p}K_{\text{a}}\) and the \(\text{p}K_{\text{b}}\) of the ethanoate ion (\(\text{CH}_3\text{COO}^-\)).
\[\text{p}K_{\text{a}} = -\log_{10}(1.76 \times 10^{-5}) = 4.75\] \[\text{p}K_{\text{b}} = 14.00 - 4.75 = 9.25\]
Examiner Trap
Direct Comparison Pitfall
Never compare \(K_{\text{a}}\) directly with \(K_{\text{b}}\) to determine whether an acid is stronger than a base. Convert both to \(\text{p}K\) scales or compare relative to water's dissociation constants.
AQA GCSE & IB Chemistry
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