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 0.241 h-1. Determine its volume of distribution in a patient who received 19.92 mg of the drug parenterally and the AUC was found to be 0.62 mg × h/L.
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lb equals 132.77 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 = (19.92 \quad mg)/(0.241 \quad h^{-1} × 0.62 (mg × h)/L)`
` ∴ V=132.77 \quad L`. Ans.