# 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 = $\frac{\left[{\mathrm{H}}_{3}{\mathrm{O}}^{+}\right]}{{\left[\mathrm{HX}\right]}_{0}}$ x 100

⇒ [H3O+] = % dissociation x $\frac{{\left[\mathrm{HX}\right]}_{0}}{100}$ = 8.6 x 10-4 M

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

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

Ka =

Ka =  = 7.5 x 10-6

pKa = - log Ka = 5.1