Exercise 8 | Acids and Bases

Ka >> Kw   ⇒   we can neglect the autoprotonation source of H₃O⁺

[H₃O⁺] depends only on the dissociation reaction: HA (aq) + H₂O (l) $\rightleftarrows$ H₃O⁺ (aq) + A^- (aq)

At = 0, [HA] = 0.050 M and [H₃O⁺] = [A^-] = 0 M

At the equilibrium, [HA] = 0.050 – X M and [H₃O⁺] = [A^-] = X M

Ka = $\frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}$ = 1.4 x 10^-3 M

⇒ at the equilibrium: $\frac{{\mathrm{X}}^2}{0.050 - \mathrm{X}}$ = 1.4 x 10^-3

⇒ X² + 1.4 x 10^-3 X - 7.0 x 10^-5 = 0

⇒ X = 7.7 x 10^-3  or X = -9.1 x 10^-3

⇒ The concentration cannot be negative ⇒ X = 7.7 x 10^-3 

⇒ at the equilibrium, [HA] = 4.2 x 10^-2 M and [H₃O⁺] = [A^-] = 7.7 x 10^-3 M

pH = -log [H₃O⁺] = 2.1