Lab Values & ABGs Practice Test #5: ABG Compensation

Answer: B. Partially compensated respiratory acidosis

Rationale: Use typical adult ranges (pH 7.35 to 7.45, PaCO2 35 to 45 mm Hg, HCO3 22 to 26 mEq/L; check your lab's values). Step 1: the pH of 7.33 is low, so this is acidosis. Step 2: the PaCO2 of 55 is high, which is acid, so it matches the pH and the problem is respiratory. Step 3: the HCO3 of 28 is high, which is base. The kidneys are holding on to bicarbonate to push the pH back up, so compensation has started. Because the pH is still outside the normal range, the compensation is only partial. Option A is wrong because uncompensated means the HCO3 would still be normal. Option C is wrong because full compensation would bring the pH back inside 7.35 to 7.45. Option D is wrong because the pH is acidic, so the primary problem cannot be an alkalosis.

Answer: D. Fully compensated respiratory acidosis

Rationale: The pH of 7.36 is inside the normal range, but the PaCO2 (58) and HCO3 (32) are both abnormal, so this is not a normal gas (Option A). When the pH is normal, look at which side of 7.40 it sits on: 7.36 is on the acid side, so the primary problem is an acidosis. The high PaCO2 is acid and matches, so the primary problem is respiratory. The high HCO3 is the kidneys' long-term response, retaining bicarbonate over days. A normal pH with both values abnormal means full compensation. This pattern is common in clients who retain carbon dioxide over the long term. Option B is wrong because the pH leans acidic, not alkaline. Option C is wrong because the HCO3 is clearly raised and the pH is normal, so compensation has occurred. Knowing a client's baseline gas matters: these values may be this client's usual, but the same numbers in a previously healthy client would need urgent attention.

Answer: A. Rapid, deep respirations

Rationale: The pH of 7.29 is low (acidosis) and the HCO3 of 14 is low, which matches, so this is a metabolic acidosis, here from lactic acid produced during poor tissue perfusion. The low PaCO2 of 30 shows the lungs are compensating: the client breathes faster and deeper to blow off carbon dioxide (an acid) and raise the pH. In severe metabolic acidosis this deep, labored pattern is called Kussmaul breathing. The pH is still low, so the compensation is partial. Option B would retain carbon dioxide and worsen the acidosis. Option C reflects poor kidney perfusion from shock, not acid-base compensation. Option D is linked to low ionized calcium, which is more typical of alkalosis. The nurse treats the cause (perfusion and infection) per orders; faster breathing alone cannot fix lactic acidosis.

Answer: C. Partially compensated metabolic alkalosis

Rationale: The pH of 7.49 is high, so this is alkalosis. The HCO3 of 36 is high (base), which matches the pH, so the primary problem is metabolic. Losing stomach acid through vomiting leaves extra bicarbonate in the blood. The PaCO2 of 49 is also high: the lungs are slowing breathing to hold on to carbon dioxide (acid) and pull the pH down. The pH is still above 7.45, so the compensation is partial. Option A is wrong because the PaCO2 is no longer normal. Option B is wrong because the pH is alkaline, so the primary problem cannot be acidosis. Option D is wrong because the pH has not returned to the normal range. Respiratory compensation for metabolic alkalosis is limited, because slower breathing also lowers oxygen levels. Treatment targets the cause: fluid, chloride and potassium replacement as prescribed, and control of the vomiting.

Answer: C. "The kidneys need hours to days to retain extra bicarbonate, so compensation has not had time to develop."

Rationale: This is an acute, uncompensated respiratory acidosis: low pH, high PaCO2 and a normal HCO3. The lungs can change the PaCO2 within minutes, so they compensate quickly for metabolic problems. The kidneys are much slower. They need hours to begin retaining bicarbonate and several days (often about 3 to 5) to reach their full effect. That is why a sudden respiratory problem usually shows a normal HCO3 at first. Option A is wrong because the lungs are the source of this problem, and the bicarbonate would rise, not fall, to compensate. Option B is wrong because the matching value here is the high PaCO2, which points to a respiratory cause. Option D describes no real threshold. In myasthenic crisis a rising PaCO2 signals respiratory failure, so the nurse monitors respiratory function per facility protocol and notifies the provider promptly.

Answer: A. Fully compensated respiratory alkalosis, an expected change in pregnancy

Rationale: The pH of 7.43 is normal but on the alkaline side of 7.40. The low PaCO2 of 31 is the value that matches an alkalosis, so the primary process is respiratory alkalosis. The low HCO3 of 20 shows the kidneys excreting bicarbonate to compensate, and the normal pH means the compensation is full. In pregnancy, progesterone increases breathing, so a lower PaCO2 with a lower bicarbonate is an expected change, and pregnancy reference ranges differ from standard adult ranges. Option B is wrong because the pH is normal and leans alkaline. Option C is wrong because the PaCO2 is low, not high. Option D is wrong because the HCO3 is low, not high. Because a lower PaCO2 is expected in pregnancy, a 'normal' adult PaCO2 of 40 in a pregnant client with breathing difficulty may signal tiring and needs prompt evaluation. Always interpret results with the clinical picture.

Answer: D. Fully compensated metabolic acidosis

Rationale: Both the PaCO2 and HCO3 are low, and the pH is within normal range, so the question is which problem came first. The pH of 7.36 is on the acid side of 7.40, so the primary problem is an acidosis. The low HCO3 is the value that causes acidosis, so the primary problem is metabolic. Damaged kidneys cannot excrete enough acid or make enough new bicarbonate. The low PaCO2 is the lungs compensating by breathing off carbon dioxide. Option A has the same two low values, but a primary respiratory alkalosis would usually leave the pH on the alkaline side of 7.40. Option B is wrong because the pH is back within the normal range. Option C is wrong because the HCO3 is abnormal and the pH is not alkalotic. In practice, the lungs often cannot fully normalize the pH in metabolic acidosis, so always interpret the gas alongside the client's history and other labs.

Answer: B. This is a combined respiratory and metabolic acidosis, with no compensation present.

Rationale: The pH of 7.10 is severely low. The high PaCO2 (60) is acid, and the low HCO3 (18) is also acid. When both values push the pH in the same direction, two problems are happening at once: a mixed (combined) acidosis. Here, inadequate breathing during the arrest retained carbon dioxide, and poor perfusion produced lactic acid. Compensation always moves the second value in the direction that pushes the pH back toward normal, so a compensated respiratory acidosis would show a high HCO3, not a low one. That makes Options A and D wrong. Option C is wrong because the pH is far from normal and the PaCO2 is high rather than low. Mixed acidosis drops the pH quickly and is dangerous; the team's priorities are effective ventilation and circulation, then treating the cause.

Answer: A. Notify the provider promptly and anticipate ventilatory support, such as noninvasive ventilation

Rationale: This client's baseline is a fully compensated respiratory acidosis: a high PaCO2 balanced by a high HCO3, with a normal pH. The new gas shows the PaCO2 rising sharply (56 to 72) while the HCO3 has barely changed, because the kidneys cannot respond that fast. The pH has dropped to 7.28, so this is an acute respiratory acidosis on top of a chronic one, and the increasing drowsiness suggests rising carbon dioxide. The nurse should notify the provider promptly; ventilatory support such as noninvasive ventilation is often needed. Option B is wrong because the pH is no longer normal, so this is not the client's baseline. Option C is wrong because oxygen does not remove carbon dioxide, and high-flow oxygen in some clients who retain carbon dioxide can worsen it; titrate oxygen to the prescribed saturation target. Option D delays care for a client whose level of consciousness is falling.

Answer: D. "If compensation is strong enough, the pH will move past 7.40 to the opposite side."

Rationale: The body does not overcompensate. Compensation moves the pH back toward normal, but it does not push the pH past 7.40 to the opposite side, which is why the side of 7.40 the pH sits on points to the primary problem when the pH is normal. This statement is false, so it shows the student needs more teaching. Option A is true: the respiratory system responds quickly by changing the rate and depth of breathing. Option B is true: in a compensated disorder the PaCO2 and HCO3 move in the same direction, because one causes the imbalance and the other offsets it. If they move in opposite directions (for example, high PaCO2 and low HCO3), suspect a mixed disorder. Option C is true: full compensation means the pH is back within the normal range even though the PaCO2 and HCO3 are still abnormal.