Exercise 6 | Acids and Bases

The equation for the reaction is:

HX (aq) + H₂O (l) $\rightleftarrows$ 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⁺]

K_a = $\frac{\left[{\mathrm{X}}^{-}\right] \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}{\left[\mathrm{HX}\right]}$ 

K_a = $\frac{{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}^2}{{\left[\mathrm{HX}\right]}_0 - \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}$ = 7.5 x 10^-6

pKa = - log K_a = 5.1