Question

14. When taken orally the weak acid sulfisomidine (pka 7.47) exists as a 0.073M solution in the upper intestinal tra sulfisomidine in the neutral form in the solution of the stomach to the nearest tenth ofa percent. aquesous ct where the pH is 5.83. Calculate the percent of

Answer 1

Let HA represent sulfisomidine in the neutral form

Total concentration of sulfisomidine = [HA] + [A-] = 0.073 M

[A-] = 0.073 - [HA]

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

= -log Ka + log([A-]/[HA])

= -log Ka + log((0.073 - [HA])/[HA])

5.83 = 7.47 + log((0.073 - [HA])/[HA])

log((0.073 - [HA])/[HA]) = -1.64

(0.073 - [HA])/[HA] = 10^(-1.64)

= 0.02291

1.02290 [HA] = 0.073

[HA] =  0.073 / 1.02290

[HA] = 0.071365 M

Percent of sulfisomidine in the neutral form

= [HA] / Total concentration of sulfisomidine x 100%

= 0.071365 / 0.073 x 100%

= 97.76%