Henderson-Hasselbalch Equation — FRCA Primary MCQ
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Correct answer: C — 10:1
Explanation lettering: B = shown as A · A = shown as B · E = shown as C · C = shown as E
The correct answer is E, 10:1. A weak base exists as unionised B and protonated, ionised BH+. Using pH = pKa + log10([B]/[BH+]), the difference pKa − pH is 1.0. Therefore [BH+]/[B] = 10^1 = 10, giving 10 parts ionised drug to 1 part unionised drug. Option C reverses the requested ratio. A 100-fold ratio, as in A or B, would require a two-unit difference between pKa and pH. Option D would apply when pH equals pKa, at which point ionised and unionised forms are present in equal concentrations.
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://pubmed.ncbi.nlm.nih.gov/19503706/