# Exercise 8 | The Properties of Acids and Bases

Calculate the pH of a 0.050 M aqueous solution of a weak acid with a Ka = 1.4 x 10-3 M at 25.0°C.

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

[H3O+] depends only on the dissociation reaction: HA (aq) + H2O (l) $⇄$ 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 =  = 1.4 x 10-3 M

⇒ at the equilibrium:  = 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