A physician-validated, board-style question from the Active Transport QBank. Try it, then check the reasoning for every option.
A 25-year-old woman with an extensive psychiatric history is suspected of having metabolic acidosis after ingesting a large amount of aspirin in a suicide attempt. Labs are drawn and the values from the ABG are found to be: PCO2: 25, and HCO3: 15, but the pH value is smeared on the print-out and illegible. The medical student is given the task of calculating the pH using the pCO2 and HCO3 concentrations. He recalls from his first-year physiology course that the pKa of relevance for the bicarbonate buffering system is approximately 6.1. Which of the following is the correct formula the student should use, using the given values from the incomplete ABG?
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A
6.1 + log[15/(0.03*25)]Correct. This is the correct Henderson-Hasselbalch equation: pH = 6.1 + log[HCO3 / (0.03 x PCO2)] = 6.1 + log[15/(0.03 x 25)].
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B
10^6.1 + 15/0.03*25Incorrect. This formula incorrectly uses 10^6.1 and lacks the logarithm - it does not represent the Henderson-Hasselbalch equation.
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C
6.1 + log[0.03/15*25)Incorrect. The numerator and denominator are flipped (and 25 is in the wrong position); this does not match Henderson-Hasselbalch.
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D
6.1 + log [25/(15*0.03)]Incorrect. The numerator and denominator are inverted - the bicarbonate should be in the numerator, not the denominator.
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E
log[6.1 + 15/(0.03*25)]Incorrect. Placing the entire pH expression inside a single logarithm misapplies the Henderson-Hasselbalch equation, where 6.1 (pKa) is added to the log of the bicarbonate/PCO2 ratio.
↑ Tap an answer to reveal the reasoning
Answer: A. The Henderson-Hasselbalch equation describes the relationship between pH and the bicarbonate buffer system: pH = pKa + log([HCO3-]/[0.03 x PCO2]). The pKa for the bicarbonate buffer is 6.1, and the factor 0.03 converts PCO2 (in mmHg) to dissolved CO2 concentration in mEq/L.
For this aspirin-overdose patient with PCO2 = 25 mmHg and HCO3 = 15 mEq/L, plugging into the formula: pH = 6.1 + log[15/(0.03 x 25)] = 6.1 + log[15/0.75] = 6.1 + log(20) = 6.1 + 1.30 = 7.40. Despite the low bicarbonate (metabolic acidosis), respiratory compensation has produced a near-normal pH - a typical picture for salicylate overdose, which causes BOTH a primary metabolic acidosis (from accumulated organic acids and uncoupled oxidative phosphorylation) AND a primary respiratory alkalosis (from direct stimulation of the medullary respiratory center).
Aspirin overdose classically produces a mixed acid-base disorder with high anion gap metabolic acidosis plus primary respiratory alkalosis - a distinctive pattern. Treatment includes activated charcoal (if recent ingestion), urinary alkalinization with sodium bicarbonate (traps ionized salicylate in urine), and hemodialysis for severe toxicity.
**Why each option:**
**A.** This is the correct Henderson-Hasselbalch equation: pH = 6.1 + log[HCO3 / (0.03 x PCO2)] = 6.1 + log[15/(0.03 x 25)].
**B.** This formula incorrectly uses 10^6.1 and lacks the logarithm - it does not represent the Henderson-Hasselbalch equation.
**C.** The numerator and denominator are flipped (and 25 is in the wrong position); this does not match Henderson-Hasselbalch.
**D.** The numerator and denominator are inverted - the bicarbonate should be in the numerator, not the denominator.