Question

I.) The pKa of aspirin is 3.5. The rate of absorption is based on the polarity of molecules charged and polar species are absorbed more slowly than neutral and hydrophobic species. The pH of the stomahc is 1.5, while that of the small intestines is 6. Calculate the percentage of charged aspirin at each pH. % @ pH 1.5 = _______ and %@pH 6 = _________

II.) Where is more aspirin absorbed?

Answer 1

We need t apply

Henderson Hasselbach equation

in which we know that:

pH = pKa + log([A-]/[HA])

is valid for acidic species

for aspirin, let it be "HA"

then

pKa = 3.5

in pH = 1.5

find % charged, that is "A-"

pH = pKa + log([A-]/[HA])

Assume

A- + HA = 1; so HA = 1-A-

1.5 = 3.5 + log(A-/(1-A-))

10^-(1.5-3.5) = (A-/(1-A-))

100*(1-A) = A

100 - 100A = A

100= 101A

A = 100/101 = 0.99009

so

HA = 1-A = 1-0.99009 = 0.00991

ratio

% A = A / (HA + A ) = 0.99009/(0.00991+0.99009) * 100% = 99.009% is charged as A-

then...

pH = 6

6= 3.5 + log(A-/(1-A-))

10^-(6-3.5) = (A-/(1-A-))

0.00316 = (A-/(1-A-))

0.00316 *(1-A) = A

0.00316 - 0.00316 A= A

0.00316 = (1+0.00316 )A

A = 0.00316 /(1+0.00316 ) = 0.003150

HA = 1-A = 1-0.003150 = 0.99685

now

%A = A/(HA+A) *100%

%A = 0.003150/ ( 0.99685+0.003150) * 100 = 0.315 %

more is absorbed in the pH = 6 media, i.e. small intestines