T13: Dose from AUC, K and V

50. Pharmacokinetics 50.1) PK Basic parameters 50.1.3) Area Under Curve (AUC)

50.1.3.1) AUC basic concepts Normal 1


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The elimination rate constant of a drug is 0.224 h-1. If its volume of distribution in a patient is 141.9 L, then determine the quantity of drug to be administered as an IV push which will produce an AUC of 20.63 mg × h/L.


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lb equals 655.58 mg kg

    We know that, `AUC = (dose)/(cl earance)`

    `⇒ dose=AUC × cl earance`.
    However, Cl is the product of elimination rate constant (K) and volume of distribution (V).

    ` ∴ X_{0} = AUC × K × V`.

    Plugging the values in this equation we get:
    ` X_{0} = 20.63 \quad (mg × h)/L × 0.224 \quad h^{-1} × 141.9 \quad L`

    `= 655.58 \quad mg` Ans