T08: Clearance from K, dose and concentration

50. Pharmacokinetics 50.1) PK Basic parameters 50.1.2) Clearance

50.1.2.1) Clearance basic concepts Normal 1


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157.78 mg of a drug was administered to a patient as an IV push, which resulted the plasma concentration of 4.58 mg/L. The elimination rate constant of the drug is 0.24 h-1. Determine the clearance of the drug in this patient.


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lb equals 8.268 L/h kg

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

    ` Cl = K × V `.

    In this case, we hav the dose of the drug administered (`X_{0}`) and the plasma concentration (`C_{0}`). We know that:
    `C_{0} = X_{0}/V ⇒ V = X_{0}/C_{0}`.

    Therefore, clearance (Cl) will be:
    `Cl = 0.24 \quad h^{-1} \quad × (157.78 \quad mg)/(4.58 \quad (mg)/L) = 8.27 \quad L/h`. Ans


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