### 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 0.276 h-1. Determine its volume of distribution in a patient who received 177.39 mg of the drug parenterally and the AUC was found to be 13.22 mg × h/L.

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lb equals 48.6 L kg

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 = (177.39 \quad mg)/(0.276 \quad h^{-1} × 13.22 (mg × h)/L)

 ∴ V=48.6 \quad L. Ans.

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