# Exercise 9 | The Properties of Acids and Bases

1) Calculate the pH of a 0.050 M aqueous solution of HClO at 25.0°C without using the method of successive approximations.

2) Calculate the pH of this solution using the method of successive approximations.

Data: Ka (HClO/ClO-) = 4.0 x 10-8 M

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

[H3O+] depends only on the dissociation reaction: HClO (aq) + H2O (l) $⇄$ 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 =  = 4.0 x 10-8 M

1) At the equilibrium:  = 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:  = 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