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 0.03 h-1. Determine its volume of distribution in a patient who received 113.66 mg of the drug parenterally and the AUC was found to be 94.67 mg × h/L.
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lb equals 40.02 L kg
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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 = (113.66 \quad mg)/(0.03 \quad h^{-1} × 94.67 (mg × h)/L)`
` ∴ V=40.02 \quad L`. Ans.