Problem+Set+4

Here is the ProblemSet #4 file. The on-line due day is 07-02-2010. Please download it, answer the questions assigned to your LG (LG-WX, LG from Worcester; LG-MX, LG from Manchester), and upload your answer file back accordingly (below). If you have any problems to upload your file back, please send your file directly to me at jonathan.sheng@mcphs.edu and l will do it for you. Each question will be answered by two LGs independently.

Question A. LG-W21:

pH 2 %I= 9.999x10^-5% pH 5.5 %I= 0.315% pH 8 %I= 50% As the pH gets more basic we see more and more of the drug become ionized. LG-W22:

Calculate the % ionization of amobarbital at pHs 2.0, 5.5, and 8.0. What trend is seen?

Amobarbital is an acid with a pKa of 7.8 For acid % ionization = 100/(1+10pKa – pH)

At pH 2, % ionized ≈ 0% At pH 5.5, % ionized ≈ 0.5% At pH 8.0, % ionized ≈ 61%

The trend that we see here is that an acidic drug in solution with a pH less than the pKa will be predominantly in the unionized state (acid form) as we see at pHs of 2 and 5.5. The more acidic the pH is (and the further it is from the pKa), the higher the percentage of the unionized form of the drug. At a pH equal to the pKa there will be a 50/50 mixture of ionized/unionized form of the drug. Finally, as the pH gets higher than the pKa, the ionized form of the drug (the conjugate base) becomes the predominate form (as we see at a pH of 8).

Question B. LG-W23: Phenylpropanolamine// : % ionization at pH 2.0= // % ionization = 100/(1+10(pH-pKa))=//100/(1+10(2.0-9.4))=// // 100.000 //// % // Phenylpropanolamine// : % ionization at pH 5.5= // % ionization = 100/(1+10(pH-pKa))= //100/(1+10(5.5-9.4)) =// // 99.987 //// % // Phenylpropanolamine// : % ionization at pH 8.0= // % ionization = 100/(1+10(pH-pKa))= //100/(1+10(8.0-9.4)) =// // 96.171 //// % // //Phenylpropanolamine has a higher pKa than amobarbital, so it stays more ionized than amobarbital// at //pH 5.5 and pH 8.0. Both phenylpropanolamine and amobarbital are too far from their pKa at pH 2.0, so they are both 100% ionized at this pH.//

LG-W24:

pKa of propanolamine is 9.4

% ionization = (10^(9.4-**2.0**))/(10^(9.4-2.0)+1) x 100% = 99.99%

% ionization = (10^(9.4-**5.5**))/(10^(9.4-5.5)+1) x 100% = 99.99%

% ionization = (10^(9.4-**8.0**))/(10^(9.4-8.0)+1) x 100% = 71.53%

The trend here is that a basic drug in solution with a pH less than the pKa will be predominantly in the ionized state. The more acidic the pH is (and the further it is from the pKa), the higher the percentage of the ionized form of the drug. At a pH equal to the pKa there will be a 50/50 mixture of ionized/unionized form of the drug. Theoretically, as the pH gets higher than the pKa, the unionized form of the drug would become the predominate form, but none of the pHs used in this problem are above the pKa value.

Question C. LG-W25:

LG-W26: pka = 5.4 Stomach (pH 2): % ionization = 99.95 % Duodenum (pH 6.5): % ionization = 7.36 % Ileum (pH 7.5): % ionization = 0.79%

Question D LG-W27: Structures in ionized form.

LG-W28: