T08: Clearance from K, 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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325.32 mg of a drug was administered to a patient as an IV push, which resulted the plasma concentration of 2.06 mg/L. The elimination rate constant of the drug is 0.147 h-1. Determine the clearance of the drug in this patient.
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lb equals 23.214 L/h kg
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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}`.
Therefore, clearance (Cl) will be:
`Cl = 0.147 \quad h^{-1} \quad × (325.32 \quad mg)/(2.06 \quad (mg)/L) = 23.21 \quad L/h`. Ans