Exercise 9 | Acids and Bases

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

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

At = 0, [HClO] = 0.050 M and [H₃O⁺] = [ClO^-] = 0 M

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

Ka = $\frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \left[{\mathrm{ClO}}^{-}\right]}{\left[\mathrm{HClO}\right]}$ = 4.0 x 10^-8 M

1) At the equilibrium: $\frac{{\mathrm{X}}^2}{0.050 - \mathrm{X}}$ = 4.0 x 10^-8

⇒ X² + 4.0 x 10^-8 X - 2.0 x 10^-9 = 0

⇒ X = 4.5 x 10^-5  or X = -4.5 x 10^-5

⇒ The concentration cannot be negative ⇒ X = 4.5 x 10^-5 

⇒ At the equilibrium, [HClO] = 5.0 x 10^-2 M and [H₃O⁺] = [ClO^-] = 4.5 x 10^-5 M

pH = -log [H₃O⁺] = 4.3

2) At the equilibrium: $\frac{{\mathrm{X}}^2}{0.050 - \mathrm{X}}$ = 4.0 x 10^-8

Method of successive approximations: X << 0.050 ⇒ 0.050 – X ~ 0.050

$\frac{{\mathrm{X}}^2}{0.050}$ = 4.0 x 10^-8

⇒ X = 4.5 x 10^-5

Verification of the approximation: X = 4.5 x 10^-5 << 0.050 ⇒ the approximation is correct

At the equilibrium, [HClO] = 5.0 x 10^-2 M and [H₃O⁺] = [ClO^-] = 4.5 x 10^-5 M

pH = -log [H₃O⁺] = 4.3