PKb question?

Question

The AIDS drug zalcitabine (also known as ddC) is a weak base with the structure shown below and a pKb of 9.8.

The molecular formular of zalcitabine is C10H12N2O4.

What percentage of the base is protonated in an aqueous zalcitabine solution containing 580 mg/L?

Any help would be appreciated. Thanks.

Answer 1

C10H12N2O4 has a molar mass of 195.2g/mol

0.580 g/L @ 195.2 g/mol = 0.00297moles / litre

C10H12N2O4 --> HC10H12N2O4 & OH-

@ pKb = 9.8, the Kb = 1.58 e-10

1.58 e-10 = [HC10H12N2O4] [OH-] / [C10H12N2O4]

1.58 e-10 = [x] [x] / [0.00297]

x= 6.86e-7

--------------------------------------...

percentage of the base is protonated:

6.86e-7 / [0.00297] times 100 = 0.023 %

your answer is 0.023 %

(but i will point out the pKb had only 1 sigfig, & your answer should be rounded off to 0.02%, unless the pKb was actually 9.80)