T06: Clearance from K and V
50. Pharmacokinetics 50.1) PK Basic parameters 50.1.2) Clearance
50.1.2.1) Clearance basic concepts Normal 1
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A drug with the elimination rate constant of 0.043 h-1 is administered as an IV bolus to a patient. The weight of the patient is 159 Kg. If the volume of distribution of the drug is 0.49 L/Kg, then calculate the clearance of the drug in this patient's body.
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lb equals 3.35 L/h kg
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We know that, clearance (Cl) is the product of elimination rate constant (K) and volume of distribution (V).
` Cl = K × V `.
However, the total volume of distribution will be the product of volume per Kg body weight and the weight of the patient. Thus,
` V = 0.49 \quad L/(Kg) × 159 \quad Kg `.
Plugging the values in the first equation we get:
` Cl = 0.043 \quad h^{-1} × 0.49 \quad L/(Kg) × 159 \quad Kg = 3.35 \quad L/h ` Ans