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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The elimination rate constant of a drug is 0.138 h-1. If its volume of distribution in a patient is 99 L, then determine the clearance of the drug.


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lb equals 13.662 L/h kg

    We know that, clearance (Cl) is the product of elimination rate constant (K) and volume of distribution (V).

    ` Cl = K × V `.

    Plugging the values in this question we get:
    ` Cl = 0.138 \quad h^{-1} × 99 \quad L = 13.66 \quad L/h ` Ans


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