T06: Clearance from K and V

50. Pharmacokinetics 50.1) PK Basic parameters 50.1.2) Clearance Clearance basic concepts Normal 1

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A drug with the elimination rate constant of 0.078 h-1 is administered as an IV bolus to a patient. The weight of the patient is 119 Kg. If the volume of distribution of the drug is 0.98 L/Kg, then calculate the clearance of the drug in this patient's body.

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lb equals 9.096 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 = 0.98 \quad L/(Kg) × 119 \quad Kg `.

    Plugging the values in the first equation we get:
    ` Cl = 0.078 \quad h^{-1} × 0.98 \quad L/(Kg) × 119 \quad Kg = 9.1 \quad L/h ` Ans