Lidocaine pKa Calculation — MFDS Part 1 MCQ

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HardPharmacologyLidocaine pKa CalculationMFDS Part 1

For this calculation, treat lidocaine as a monoprotic weak base and use a pKa of 7.90. At a tissue pH of 7.40, what percentage of the total lidocaine is present as the non-ionised base (B), to the nearest whole percentage?

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Correct answer: D — Approximately 24%

Explanation lettering: E = shown as A · A = shown as D · D = shown as E

The correct answer is A, approximately 24%. For a weak base, the Henderson–Hasselbalch relationship gives [BH+]/[B] = 10^(pKa−pH). Here this ratio is 10^0.5 = 3.16, meaning there are 3.16 parts ionised drug for each part of non-ionised drug. The non-ionised fraction is therefore 1/(1+3.16) = 0.240, or 24%. Option B is the complementary ionised fraction, approximately 76%. Option C incorrectly treats the 3.16:1 ratio as though the non-ionised fraction were simply 1/3.16. Options D and E arise from confusing the ratio with a percentage or misplacing the decimal point. The free-base form is the species that predominantly crosses lipophilic nerve membranes.

Reference: Pope RLE, Brown AM. A primer on tissue pH and local anesthetic potency. Advances in Physiology Education. 2020;44(3):305–308. https://pubmed.ncbi.nlm.nih.gov/32484400/