Question

explain in detail, please. Thanks

A patient diagnosed with atrial fibrillation is administered an oral dose of warfarin of 10 mg; the drug is an anticoagulant and is a weak acid with a pka of approximately 5.0 (five). What is the amount of this drug that could be absorbed from the stomach at pH 3.0 (three)? (a) (b) (c) (d) (e) 9.95 mg 9.9 mg 500 micrograms 5 mg 0.5 mg

Answer 1

The stomach absorbs mostly nonpolar molecules. In this case, this is the protonated weak acid. To calculate the fraction of the acid that will be protonated, we can use the Henderson- Hasselbach equation, where we will call the protonated acid HA and the deprotonated acid A^-:

$pH=pKa+log\left ( \frac{[A^{-}]}{[HA]} \right )$

The ratio of concentration is then given by:

$\frac{[A^{-}]}{[HA]}=10^{pH-pKa}=10^{3.0-5.0}=0.01$

This means that the concentration of protonated acid will be 100 tomes higher than that of the protonated acid, which means that apporximately 99% will be in the absorbable form. This is: 9.9 mg