### T11: Clearance from plasma half-life, dose and concentration

50. Pharmacokinetics 50.1) PK Basic parameters 50.1.2) Clearance

50.1.2.1) Clearance basic concepts Normal 1

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191.27 mg of a drug was administered to a patient as an IV push, which resulted the plasma concentration of 3.45 mg/L. The plasma half life of the drug is 3.77 h. Determine the clearance of the drug in this patient. Click on the button below to see the answer and explanations

lb equals 10.201 L/h kg

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

 Cl = K × V .

In this case, we hav the dose of the drug administered (X_{0}) and the plasma concentration (C_{0}). We know that:
C_{0} = X_{0}/V ⇒ V = X_{0}/C_{0}.

From the relationship of t_{½}=0.693/K, we can calculate K as 0.693/t_{½}=0.693/(3.77 h)= 0.184 \quad h^{-1}.

Therefore, clearance (Cl) will be:
Cl = 0.184 \quad h^{-1} \quad × (191.27 \quad mg)/(3.45 \quad (mg)/L) = 10.2 \quad L/h. Ans