Propofol pKa — FRCA Primary MCQ
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Correct answer: E — 99.975%
The correct answer is E. For a weak acid, the Henderson–Hasselbalch equation gives ionised:unionised = 10^(pH−pKa). Thus, at pH 7.4, this ratio is 10^−3.6, or approximately 0.000251:1. The unionised fraction is therefore 1/(1+0.000251) = 0.999749, equivalent to approximately 99.975%. Option A is approximately the ionised percentage, not the unionised percentage. Option D would apply when pH equals pKa. Option B may be selected by confusing ionisation with propofol's approximately 98% plasma-protein binding; these are distinct properties. The large pKa–pH difference means that nearly all propofol remains protonated and unionised.
Reference: Roche VF. Improving pharmacy students' understanding and long-term retention of acid-base chemistry. American Journal of Pharmaceutical Education. 2007;71(6):122. Figure 3: calculation of ionised/unionised ratios using the Henderson–Hasselbalch equation. https://www.medicines.org.uk/emc/product/11295/smpc