Exercise 7 | Acids and Bases

1) Dissociation reaction:

C₆H₅COOH (aq) + H₂O (l) $\rightleftarrows$ H₃O⁺ (aq) + C₆H₅COO^- (aq)

2) Ka = 6.3 x 10^-5 M and Kw = 1.0 x 10^-14 M ²

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

[H₃O⁺] depends only on the dissociation reaction

3) A t = 0, [C₆H₅COOH] = 0.250 M and [H₃O⁺] = [C₆H₅COO^-] = 0 M

At the equilibrium, [C₆H₅COOH] = 0.250 – X M and [H₃O⁺] = [C₆H₅COO^-] = X M

Ka = $\frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \left[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}\right]}{\left[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}\right]}$ = 6.3 x 10^-5 M

⇒ at the equilibrium: $\frac{{\mathrm{X}}^2}{0.250 - \mathrm{X}}$ = 6.3 x 10^-5

⇒ X² + 6.3 x 10^-5 X - 1.6 x 10^-5 = 0

⇒ X = 3.9 x 10^-3  or X = -4.0 x 10^-3

⇒ The concentration cannot be negative ⇒ X = 3.9 x 10^-3

⇒ at the equilibrium, [C₆H₅COOH] = 2.46 x 10^-1 M and [H3O⁺] = [C₆H₅COO^-] = 3.9 x 10^-3 M

4) pH = -log [H₃O⁺] = 2.4