### 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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472.83 mg of a drug was administered to a patient as an IV push, which resulted the plasma concentration of 4.14 mg/L. The elimination rate constant of the drug is 0.291 h-1. Determine the AUC. Click on the button below to see the answer and explanations

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lb equals 14.23 mg × h/L kg

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 = (4.14 \quad (mg)/L)/(0.291 \quad h^{-1})=14.23 \quad (mg × h)/L. Ans.

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