T19: AUC from K, dose and concentration
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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128.44 mg of a drug was administered to a patient as an IV push, which resulted the plasma concentration of 2.78 mg/L. The elimination rate constant of the drug is 0.094 h-1. Determine the AUC.
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lb equals 29.57 mg × h/L kg
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We know that, AUC may be calculated from dose (`X_{0}`) and clearance (`Cl`) OR initial concentration (`C_{0}`) and the elimination rate constant (`K`).
` AUC = (X_{0})/(Cl) = (C_{0})/K`.
By plugging the values of initial concentration and K, we get,
` AUC = (C_{0})/K = (2.78 \quad (mg)/L)/(0.094 \quad h^{-1})=29.57 \quad (mg × h)/L`. Ans.