T15: K from Dose, AUC and V

50. Pharmacokinetics 50.1) PK Basic parameters 50.1.3) Area Under Curve (AUC) AUC basic concepts Normal 1

Total tried:       Correct:       Wrong:

Determine the elimination rate constant of a drug from the provided information. The volume of distribution is 36.58 L after administering 99.5 mg of the drug. The area under the plasma drug concentration over time (AUC) was found to be 9.13 mg × h/L.

Click on the button below to see the answer and explanations

lb equals 0.3 h-1 kg

    We know that, `AUC=(dose)/(K × V)`.

    `⇒ K=(dose)/(V × AUC)`

    Plugging the values in this equation we get:
    ` K = (99.5 \quad mg)/(36.58 \quad L × 9.13 (mg × h)/L)`

    ` ∴ K=0.298 \quad h^{-1}`. Ans.