p H of Weak Acids and Weak Bases

p. H of Weak Acids and Weak Bases

Weak Acids that only dissociate partly in aqueous solution HCOOH - methanoic acid, CH 3 COOH – ethanoic acid, HCN - hydrogen cyanide

Solutions of weak acids CH 3 COOH + H 2 O = CH 3 COO- + H 3 O+ dynamic equilibrium established Kc = [CH 3 COO-][H 3 O+] [CH 3 COOH][H 2 O] Ethanoic acid is a weak acid so only a tiny proportion is dissociated - lets say 2. 1% [H 2 O] is essentially constant since it is present in large excess [55. 5]

Solutions of weak acids �Because the water is in large excess we can leave it out of the equation Ka = [CH 3 COO-][H 3 O+] [CH 3 COOH] �Ka is called the acid dissociation constant of a weak acid �Ka is often called ionisation constant

* Mentioned earlier Solutions of weak acids In a sample of 0. 1 M ethanoic acid [H 2 O] = 55. 5 mole L-1 [H 3 O+] = 2. 1%* of 0. 1 = 0. 0021 M [H 2 O] : [H 3 O+] = 55. 5 : 0. 0021 water is 26428 times more concentrated than the hydroxonium ions it is so big it would swamp the other numbers in the Kc equation - so left out

Solutions of weak acids �When working out Ka the concentration of the acid is taken to be the original concentration before any of it dissociates �because so little of it dissociates that there is no significant difference in the concentration before and after dissociation.

p. H of Weak Acids • p. H = -log 10 (Ka * [HA]) • derivation Ka = [CH 3 COO-][H 3 O+] [CH 3 COOH] but [CH 3 COO-] = [H 3 O+] let them both = x [CH 3 COO-][H 3 O+] = x 2

p. H of Weak Acids CH 3 COOH is a weak acid so call it [HA] Ka = x 2 / [HA] x 2 = Ka * [HA] x = (Ka * [HA]) But x = [H 3 O+] so [H 3 O+] = (Ka * [HA]) p. H = -log 10[H 3 O+] = -log 10 (Ka * [HA])

• p. H = -log 10 (Ka * [HA]) • derivation Ka = [CH 3 COO-][H 3 O+] [CH 3 COOH] but [CH 3 COO-] = [H 3 O+] let them both = x • • [CH 3 COO-][H 3 O+] = x 2 CH 3 COOH is a weak acid so call it [HA] Ka = x 2 / [HA] x 2 = Ka * [HA] x = (Ka * [HA]) But x = [H 3 O+] so [H 3 O+] = (Ka * [HA]) p. H = -log 10[H 3 O+] = -log 10 (Ka * [HA])

p. H of Weak Bases �Weak base MOH �M stands for the metal ion �Kb = [M+][OH-] [MOH] �p. OH = -log 10 (Kb * [MOH] ) �p. H = 14 – p. OH �p. H = 14 – (-log 10 (Kb * [MOH]) )

Basicity of Acids The number of H atoms in the acid that can be replaced by a metal Monobasic – 1 H that can be replaced Monobasic: HNO 3, HCl, CH 3 COOH, NH 2 SO 3 H Dibasic : H 2 SO 4 Tribasic : H 3 PO 4 Equivalent for bases; mono, di and

Problems 1. Calculate the p. H of 0. 5 M solution of ethanoic acid. Ka= 1. 8 * 10 -4 [Answer 2. 02] 2. Calculate the approximate p. H of a vinegar solution that contains 9 g of ethanoic acid in 100 cm 3 given that Ka = 1. 8 * 10 -5 [Answer 2. 78] 3. A bottle of vinegar is labelled 3% (W/V) acetic [ethanoic] acid. The dissociation constant is 1. 8 * 10 -4. calculate the p. H. [Answer 2. 02] 4. Calculate the p. H of a 0. 003 M solution of methanoic acid (HCOOH). Ka value = 1. 8 * 10 -4 [Answer 3. 133]

5. Calculate the p. H of 0. 5 M solution of ammonium hydroxide, a weak base. Kb= 1. 8 * 10 -5 [Answer xxx] 6. Calculate the p. H of a 10% solution of the weak base NH 4 OH given that Kb= 1. 8 * 10 -5 [Answer xxx]