### T11: Mixed problems from T1 - T11.

50. Pharmacokinetics 50.1) PK Basic parameters 50.1.2) Clearance

50.1.2.1) Clearance basic concepts Normal 1

Total tried:       Correct:       Wrong:

The elimination half life of a drug is 8.35 h. It is administered as an IV bolus to a 158 Kg patient. If the volume of distribution of the drug is 0.74 L/Kg, then calculate the clearance of the drug in this patient's body. Click on the button below to see the answer and explanations

lb equals 9.704 L/h kg

We know that, clearance (Cl) is the product of elimination rate constant (K) and volume of distribution (V).

 Cl = K × V .

From the relationship of t_{½}=0.693/K, we can calculate K as 0.693/t_{½}=0.693/(8.35 h)= 0.083 \quad h^{-1}.

Also, the total volume of distribution will be the product of volume per Kg body weight and the weight of the patient. Thus,
 V = 0.74 \quad L/(Kg) × 158 \quad Kg =116.92 \quad L.

Plugging the values in the first equation we get:
 Cl = 0.083 \quad h^{-1} × 116.92 \quad L = 9.7 \quad L/h  Ans