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tutorial 1 assignment

Tutorial 1
1. pK = 3.4
DH = ASA
Plasma pH = 7.3 Stomach pH = 1.5 = 10pH – pK = 10pH – pK = 107.3 – 3.4 = 101.5 – 3.4 = 7943.2823 = 0.01259

[Total Drug] = [D-] + [DH] [Total Drug] = [D-] + [DH] = 7943.2823 + 1 = 0.0126 + 1 = 7944.2823 = 1.0126

[Total Drug] in stomach relative to plasma = 7944.2823/1.0126 = 7845.43

Conclusion: High absorption of ASA from the stomach lumen. Absorption from stomach is 7845.43 times greater than in the plasma.

2. At pH 1.4 At pH 7.4 At pH 1.4 At pH 7.4
A1 (pK = 3.4) A1 (pK = 3.4) B1 (pK = 7.4) B1 (pK = 7.4) = 101.4 – 3.4 = 107.4 – 3.4 = 101.4 – 7.4 = 107.4 – 7.4 = 0.01 = 10000 = = 1
%D- = x100% %D- = x100% %DH+ = x100% %DH+ = x100% = 0.99% = 99.99% = 100.00% = 50.00% A2 (pK = 4.4) A2 (pK = 4.4) B2 (pK = 8.4) B2 (pK = 8.4) = 101.4 – 4.4 = 107.4 – 4.4 = 101.4 – 8.4 = 107.4 – 8.4 = 0.001 = 1000 = = 0.1
%D- = x100% %D- = x100% %DH+ = x100% %DH+ = x100% = 0.10% = 99.90% = 100.00% = 90.91%

A3 (pK = 5.4) A3 (pK = 5.4) B3 (pK = 9.4) B3 (pK = 9.4) = 101.4 – 5.4 = 107.4 – 5.4 = 101.4 – 9.4 = 107.4 – 9.4 = 0.0001 = 100 = = 0.01
%D- = x100% %D- = x100% %DH+ = x100% %DH+ = x100% = 0.01% = 99.01% = 100.00% = 99.01%

3. DH = sulfonylureas

At pH = 5 At pH =6.5
Metsulfuron-methyl (pK = 3.3) Metsulfuron-methyl (pK = 3.3) = 105 – 3.3 = 106.5 – 3.3 = 50.1187 = 1584.8932 %DH = x 100% %DH = x 100% = 1.96% = 0.063%

Chlorsulfuron (pK = 3.7) Chlorsulfuron (pK = 3.7) = 105 – 3.7 = 106.5 – 3.7 = 19.9526 = 630.9573 %DH = x 100% %DH = x 100% = 4.77% = 0.16%

Triasulfuron (pK = 4.6) Triasulfuron (pK = 4.6) = 105 – 4.6

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