Question

In an aqueous solution of a certain acid with pKa = 7.33 the pH is 4.59. Calculate the percent of the acid that is dissociated in this solution. Round your answer to 2 significant digits.

Answer 1

pKa = -log Ka

7.33 = -log Ka

Ka = 4.677*10^-8

Let the concentration of HA be c

use:

pH = -log [H+]

4.59 = -log [H+]

[H+] = 2.57*10^-5 M

HA dissociates as:

HA -----> H+ + A-

c 0 0

c-x x x

x = [H+] = 2.57*10^-5 M

Ka = [H+][A-]/[HA]

Ka = x*x/(c-x)

4.677*10^-8 = 2.57*10^-5*2.57*10^-5/(c-2.57*10^-5)

c-2.57*10^-5 = 1.413*10^-2

c=1.415*10^-2 M

% dissociation = x*100/c

= (2.57*10^-5)*100 / (1.415*10^-2)

= 0.182 %

Answer: 0.18 %