Exercise 8 | The Properties of Acids and Bases
General Chemistry 3 - Exercise 8
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