3.1.11 (HL)—Conjugate pair relationship

$Ka × Kb = Kw$

For a conjugate pair, calculate the missing constant by dividing Kw by the known Ka or Kb, keeping the pair direction consistent.

Match the constants to one conjugate pair: Ka(HA) × Kb(A⁻) = Kw. A stronger acid therefore has a weaker conjugate base at the same temperature; do not multiply constants belonging to unrelated species.

Worked conjugate-constant example at 298 K: methylamine has $\mathrm{p}K_b=3.34$ , so for its conjugate acid $\ce{CH3NH3+}$ , $\mathrm{p}K_a=\mathrm{p}K_w-\mathrm{p}K_b=14.00-3.34=10.66$ . Equivalently, $K_aK_b=K_w$ . This relationship applies only to a conjugate acid–base pair at the same temperature.

Representative question

[Maximum number: 1]

Calculate the $K_{b}$ of the conjugate base of ethanoic acid using sections 2 and 21 of the data booklet.

$K_a=10^{-4.76}=1.7\times10^{-5}$ $K_w=K_aK_b$ $K_b=\frac{1.0\times10^{-14}}{1.7\times10^{-5}} =5.8\times10^{-10}$

Accept $5.7\times10^{-10}$ to $5.9\times10^{-10}$ .

Retrieve the route: track proton transfer and conjugates, calculate pH and Kw, distinguish strength, balance neutralization, read titration curves, use Ka/Kb and hydrolysis, select indicators, and explain and calculate buffer behaviour.

Check donor versus acceptor, one-proton differences, logarithm direction, ion comparison, strength versus concentration, equivalence versus endpoint, pKa landmarks, conjugate equations and dilution ratios.

3.1.11 (HL)—Conjugate pair relationship

Conjugate Acid–Base Constants

Calculating a Conjugate Constant

Assessment in practice

Proton Transfer Reactions Summary

Question 1