T14: V from Dose, AUC and K

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 0.324 h-1. Determine its volume of distribution in a patient who received 56.24 mg of the drug parenterally and the AUC was found to be 2.25 mg × h/L.


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lb equals 77.04 L kg

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

    However, Cl is the product of elimination rate constant (K) and volume of distribution (V).

    `⇒ AUC=(dose)/(K × V)`.

    `⇒ V=(dose)/(K × AUC)`

    Plugging the values in this equation we get:
    ` V = (56.24 \quad mg)/(0.324 \quad h^{-1} × 2.25 (mg × h)/L)`

    ` ∴ V=77.04 \quad L`. Ans.