T12: AUC from dose, 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


Total tried:       Correct:       Wrong:

The elimination rate constant of a drug is 0.103 h-1. If its volume of distribution in a patient is 61.05 L, then determine the AUC obtained if 167.28 mg of the drug is injected.


Click on the button below to see the answer and explanations


Notice: Undefined variable: input_value1 in /home/jshzub/web/rxcal.org/public_html/practice/practice/includes/body.php on line 220
lb equals 26.6 mg/L h kg

    We know that, `AUC = (dose)/(cl earance)`; but Cl is the product of elimination rate constant (K) and volume of distribution (V).

    ` ∴ AUC = X_{0}/ (K × V) `.

    Plugging the values in this equation we get:
    ` AUC = (167.28 \quad mg)/ (0.103 \quad h^{-1} × 61.05 \quad L ) = 26.6 \quad (mg)/L × h ` Ans


Notice: Undefined index: TOTALTRY in /home/jshzub/web/rxcal.org/public_html/practice/practice/includes/body.php on line 675