Exercise 6 | The Properties of Acids and Bases

A 0.10 M solution of a weak acid HX is 0.86 % ionized.

What is the pKa of this solution?

The equation for the reaction is:

HX (aq) + H2O (l)  X(aq) + H3O(aq)


% dissociation = H3O+HX0 x 100

⇒ [H3O+] = % dissociation x HX0100 = 8.6 x 10-4 M


From the stoichiometry of the reaction: [X-] = [H3O+]

And [HX] = [HX]0 - [H3O+]

Ka = X- H3O+HX 

Ka = H3O+2HX0 - H3O+ = 7.5 x 10-6

pKa = - log Ka = 5.1