### 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 0.269 h-1. Determine its volume of distribution in a patient who received 703.49 mg of the drug parenterally and the AUC was found to be 16.43 mg × h/L. Click on the button below to see the answer and explanations

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lb equals 159.16 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 = (703.49 \quad mg)/(0.269 \quad h^{-1} × 16.43 (mg × h)/L)

 ∴ V=159.16 \quad L. Ans.

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