An ABG result shows pH of 7.35, PaCO2 of 30 mm Hg, and HCO3 of 18 mEq/L. Which of the following is ...

(Answer to Q. 1)  ANS: A
In cases in which compensation has occurred, if the pH is on the acidic side of 7.40 (7.35 to 7.39), the component that would cause an acidosis (either increased PaCO2 or decreased plasma HCO3) is generally the primary cause of the original acid-base imbalance.

(Answer to Q. 2)  ANS: A
This completely compensated respiratory acidosis is characterized by the same originally ob-served high PaCO2, a pH that is now in the 7.35 to 7.39 range, and a plasma HCO3 that is greater than it was before complete compensation took place.

(Answer to Q. 3)  ANS: C
For example, if the pH was lower than 7.35 (denoting an acidosis) and the PaCO2 was higher than 45 mm Hg, according to the H-H equation, the high PaCO2 would indeed lower the pH (i.e., produce an acidosis). Therefore, the respiratory system is at least in part, if not entirely, re-sponsible for the acidosis.

(Answer to Q. 4)  ANS: B
Alkalemia is defined as a blood pH greater than 7.45. Acidemia is defined as a blood pH less than 7.35.

(Answer to Q. 5)  ANS: A
The lungs normally compensate quickly for metabolic acid-base defects because ventilation can change the PaCO2 within seconds. The kidneys require more time to retain or excrete significant amounts of HCO3, and thus compensate for respiratory defects at a much slower pace.

(Answer to Q. 6)  ANS: C
The kidneys compensate by retaining HCO3, returning the plasma HCO3/dissolved CO2 ratio to almost 20:1. The conversion of PCO2 to mEq is done by multiplying by 0.03.
Thus 60  0.03 = 1.8. 36 to 1.8 is equal to a 20 to 1 ratio, thus the pH should be 7.40.

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