Lab case 425 interpretation

Question 1:

PH = 7.51, that is mild alkalaemia

PCO2 = 28, that is less than 40. So we have respiratory alkalosis.

Compensation for acute respiratory alkalosis: HCO3 drops by 2 mmol/L for every 10 mmHg decrease in pCO2 from 40mmHg. Accordingly, HCO3 should drop by (12 x 0.2) = 2.4, SO expected HCO3 is 21.6 (that is close to 23) so we have pure respiratory alkalosis.

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