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Physiology · Acid-base: Henderson-Hasselbalch
Sample itemAn arterial sample shows bicarbonate 12 mmol/L and PCO2 40 mmHg. Using pH = 6.1 + log [HCO3-] / (0.03 x PCO2), what is the pH?
Original practice item, not a recalled exam question. Nothing here is recorded.
The answer and the reasoning
C. 7.10
Dissolved carbon dioxide is 0.03 x 40 = 1.2 mmol/L. The ratio 12/1.2 equals 10, and log 10 is 1. So pH = 6.1 + 1 = 7.10. A normal ratio is 24/1.2 = 20, whose log is 1.3, giving 7.4.
Option by option
A. 7.25
The calculation does not give this value; log 10 equals exactly 1.
B. 6.90
6.90 would need a ratio of about 6.3, which means bicarbonate of about 7.6 at this PCO2.
C. 7.10 (correct)
Correct. 6.1 + log (12/1.2) = 7.10.
D. 7.40
A pH of 7.40 needs a bicarbonate of 24 at this PCO2.
Source note: Original NextPrep sample item — pending expert review. This item is original practice material awaiting independent expert review, so confirm important facts in a standard textbook.
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