Four hours after mechanical ventilation is initiated for a patient with a suspected drug overdose, the patient's arterial blood gas (ABG) results include a pH of 7.50, Pa0 2 of 80 mm Hg. PaCO2 of 29 mmHg, and HCO3 of 23 mEq/L (23 mmo/L). The nurse will anticipate the need to:
decrease the respiratory rate.
Leave the ventilator at the current settings.
increase the tidal volume (VT).
Increase the F102
The Correct Answer is A
The ABG results show a pH of 7.50, PaCO2 of 29 mmHg, and HCO3 of 23 mEq/L, indicating respiratory alkalosis. In respiratory alkalosis, there is a decrease in PaCO2 (hypocapnia), which can be caused by excessive ventilation.
To address the respiratory alkalosis, the nurse should decrease the respiratory rate. This would help reduce the amount of ventilation provided and allow the patient to retain more carbon dioxide (CO2), thereby increasing the PaCO2 levels and restoring acid-base balance.
B. Leaving the ventilator at the current settings in (option B) is incorrect because it may exacerbate respiratory alkalosis as it would maintain the same level of ventilation.
C. Increasing the tidal volume (VT) in (option C) is incorrect because it would not address the respiratory alkalosis. Tidal volume refers to the volume of air delivered with each breath, while the issue in this case is excessive ventilation leading to hypocapnia.
D. Increasing the FiO2 (fraction of inspired oxygen) in (option D) is incorrect because it is not indicated based on the given ABG results. The oxygenation (PaO2) level is within normal limits (80 mmHg), suggesting adequate oxygenation.
It is important to consult with the healthcare provider or respiratory therapist for further guidance on adjusting the ventilator settings based on the patient's condition and response to therapy.
Nursing Test Bank
Naxlex Comprehensive Predictor Exams
Related Questions
Correct Answer is B
Explanation
Lactate is a by-product of anaerobic metabolism that accumulates when there is insufficient oxygen supply to meet cellular metabolic demands. In the context of severe tissue hypoxia, such as in septic shock, the body may resort to anaerobic metabolism, leading to increased lactate production and elevated lactate levels in the blood.
Elevated lactate levels, typically above 4.0 mmol/L, are indicative of tissue hypoxia and inadequate oxygenation at the cellular level. Higher lactate levels, such as 9.0 mmol/L, suggest more severe tissue hypoxia and increased anaerobic metabolism.
A. Partial thromboplastin time (PTT) 64 seconds in (option A) is incorrect because: PTT is a laboratory test that evaluates the intrinsic pathway of the coagulation cascade. While coagulation abnormalities may occur in septic shock, PTT alone does not specifically indicate severe tissue hypoxia.
C. Potassium 2.8 mEq/L (2.8 mmol/L) (option C) is incorrect because Low potassium levels (hypokalemia) can be a concern in septic shock, but it does not directly indicate severe tissue hypoxia.
D. PaCO2 58 mm Hg in (option D) is incorrect because: PaCO2 refers to the partial pressure of carbon dioxide in arterial blood and is a measure of the respiratory status. While an elevated PaCO2 can be a sign of respiratory acidosis, it is not specific to severe tissue hypoxia.
Therefore, in a critically ill patient with septic shock, an elevated lactate level, such as 9.0 mmol/L, indicates severe tissue hypoxia and inadequate oxygenation at the cellular level
Correct Answer is ["9000"]
Explanation
4 mL × body weight in kg × percentage of burn = total fluid volume for the first 24 hours Then, divide the total fluid volume by 2 to determine the fluid volume for the first 8 hours.
In this case, the patient weighs 90 kg and has a 50% burn injury.
4 mL × 90 kg × 50% = 18,000 mL (total fluid volume for 24 hours)
18,000 mL / 2 = 9,000 mL (fluid volume for the first 8 hours)
Therefore, the nurse would infuse the intravenous fluid resuscitation at a rate of 9,000 mL over the first 8 hours.
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