### 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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121.64 mg of a drug was administered to a patient as an IV push, which resulted the plasma concentration of 2.08 mg/L. The plasma half life of the drug is 28.88 h. Determine the clearance of the drug in this patient. Click on the button below to see the answer and explanations

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lb equals 1.404 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/(28.88 h)= 0.024 \quad h^{-1}.

Therefore, clearance (Cl) will be:
Cl = 0.024 \quad h^{-1} \quad × (121.64 \quad mg)/(2.08 \quad (mg)/L) = 1.4 \quad L/h. Ans

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