Lipoamide is a basic drug that has a pK_b of 4.66. The volume of distribution in

Question

Lipoamide is a basic drug that has a pK_b of 4.66. The volume of distribution in the tissues is 3.09 L and the volume of the plasma is 3.45 L. It was found to have a fraction of unbound of 5.06 times 10^-2 and a half-life of 4.70 hrs. The largest dose that can be safely given is 150 mg and its rate of absorption for an oral dose is 0.152 hr^-1. The molar mass of lipoamide is 419.1 g/mol. Consider the drug lipoamide described above: Determine the pKa of lipoamide. Calculate the concentration of ionized lipoamide in the plasma in m if the steady state concentration of un-ionized drug is found to be 45.1 mug/L and the patient has a blood pH of 7.47. Calculate the total mass of lipoamide and the mu moles of ionized lipoamide in the plasma Calculate the total volume of distribution in an average healthy adult male one hour after the administration of the maximum allowed dose if after an hour the plasma concentration was found to be 3.82 mug/mL when a blood sample was drawn from a vein in the arm. Calculate the tissue to blood equilibrium constant, and the rate distribution constant (k_T) the tissues if the average the tissues if the average blood fellow the issues is 82.5 L/hr and the average concentration in the tissues after an hour was found to be 1.93 mug/mL

Explanation / Answer

a) Since pKa + pKb = 14

Hence, pKa of lipoamide is 14-pKb = 14 - 4.66 = 9.34

b) Unionized drug (45.1 ug/L) ionized drug (?)

Percent ionized formula 100/(1+ 10)x(pH-pKa) where x = -1 if acid drug or 1 if basic drug. Since our drug is basic so x=1, pH of blood = 7.47, pKa =9.34

So put the values in formula and % of drug which is ionized is calculated which is 98.7 % and 1.3% of drug remains un ionized at steady state.

1.3% of total drug is unionized = 45.1 ug/L

13/1000 x = 45.1

x =3469.23ug/L

Ionized = 3469.23 - 45.1 = 3424.13 ug/L.

c) Total mass of lipoamide = 3469.23ug/L

umoles of ionized lipoamide = 3424.13/419.1 = 8.17 umoles/L

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