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 0.269 h-1. Determine its volume of distribution in a patient who received 703.49 mg of the drug parenterally and the AUC was found to be 16.43 mg × h/L.


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lb equals 159.16 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 = (703.49 \quad mg)/(0.269 \quad h^{-1} × 16.43 (mg × h)/L)`

    ` ∴ V=159.16 \quad L`. Ans.


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