T13: Dose from AUC, K and V
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 0.217 h-1. If its volume of distribution in a patient is 68.37 L, then determine the quantity of drug to be administered as an IV push which will produce an AUC of 1.98 mg × h/L.
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lb equals 29.4 mg kg
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We know that, `AUC = (dose)/(cl earance)`
`⇒ dose=AUC × cl earance`.
However, Cl is the product of elimination rate constant (K) and volume of distribution (V).
` ∴ X_{0} = AUC × K × V`.
Plugging the values in this equation we get:
` X_{0} = 1.98 \quad (mg × h)/L × 0.217 \quad h^{-1} × 68.37 \quad L`
`= 29.4 \quad mg` Ans