### 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 0.253 h-1. If its volume of distribution in a patient is 51.84 L, then determine the quantity of drug to be administered as an IV push which will produce an AUC of 8.14 mg × h/L. Click on the button below to see the answer and explanations

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lb equals 106.79 mg kg

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} = 8.14 \quad (mg × h)/L × 0.253 \quad h^{-1} × 51.84 \quad L

= 106.79 \quad mg Ans

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