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 0.239 h-1 is administered as an IV bolus to a patient. The weight of the patient is 125 Kg. If the volume of distribution of the drug is 1.1 L/Kg, then calculate the clearance of the drug in this patient's body.


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lb equals 32.863 L/h kg

    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 = 1.1 \quad L/(Kg) × 125 \quad Kg `.

    Plugging the values in the first equation we get:
    ` Cl = 0.239 \quad h^{-1} × 1.1 \quad L/(Kg) × 125 \quad Kg = 32.86 \quad L/h ` Ans


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