1. pH of Weak Acids and Weak Bases

2. Weak Acids Acids that only dissociate partly in aqueous solution
HCOOH - methanoic acid,
CH3COOH – ethanoic acid,
HCN - hydrogen cyanide

3. Solutions of weak acids
CH3COOH + H2O = CH3COO H3O+
dynamic equilibrium established
Kc = [CH3COO-][H3O+] [CH3COOH][H2O]
Ethanoic acid is a weak acid so only a tiny proportion is dissociated - lets say 2.1%
[H2O] is essentially constant since it is present in large excess [55.5]

4. Solutions of weak acids
Because the water is in large excess we can leave it out of the equation
Ka = [CH3COO-][H3O+]
[CH3COOH]
Ka is called the acid dissociation constant of a weak acid
Ka is often called ionisation constant

5. Solutions of weak acids
* Mentioned earlier
Solutions of weak acids
In a sample of 0.1 M ethanoic acid
[H2O] = 55.5 mole L-1
[H3O+] = 2.1%* of 0.1 = M
[H2O] : [H3O+] = 55.5 :
water is times more concentrated than the hydroxonium ions
it is so big it would swamp the other numbers in the Kc equation - so left out

6. 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.

7. pH of Weak Acids pH = -log10(Ka * [HA]) derivation
Ka = [CH3COO-][H3O+]
[CH3COOH] but [CH3COO-] = [H3O+] let them both = x [CH3COO-][H3O+] = x2

8. pH of Weak Acids CH3COOH is a weak acid so call it [HA] Ka = x2 / [HA]
x2 = Ka * [HA]
x =  (Ka * [HA])
But x = [H3O+] so
[H3O+] =  (Ka * [HA])
pH = -log10[H3O+] = -log10 (Ka * [HA])

9. pH = -log10(Ka * [HA])
derivation
Ka = [CH3COO-][H3O+]
[CH3COOH]
but [CH3COO-] = [H3O+]
let them both = x
[CH3COO-][H3O+] = x2
CH3COOH is a weak acid so call it [HA]
Ka = x2 / [HA]
x2 = Ka * [HA]
x =  (Ka * [HA])
But x = [H3O+] so
[H3O+] =  (Ka * [HA])
pH = -log10[H3O+] = -log10 (Ka * [HA])

10. pH of Weak Bases Weak base MOH M stands for the metal ion
Kb = [M+][OH-]
[MOH]
pOH = -log10  (Kb * [MOH] )
pH = 14 – pOH
pH = 14 – (-log10  (Kb * [MOH]) )

11. 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: HNO3, HCl, CH3COOH, NH2SO3H
Dibasic : H2SO4
Tribasic : H3PO4
Equivalent for bases; mono, di and triprotic

12. Problems
Calculate the pH of 0.5 M solution of ethanoic acid. Ka= 1.8 * 10-4 [Answer 2.02]
Calculate the approximate pH of a vinegar solution that contains 9 g of ethanoic acid in 100 cm3 given that Ka = 1.8 * 10-5 [Answer 2.78]
A bottle of vinegar is labelled 3% (W/V) acetic [ethanoic] acid. The dissociation constant is 1.8 * calculate the pH. [Answer 2.02]
Calculate the pH of a M solution of methanoic acid (HCOOH). Ka value = 1.8 * [Answer 3.133]

13. Calculate the pH of 0. 5 M solution of ammonium hydroxide, a weak base
Calculate the pH of 0.5 M solution of ammonium hydroxide, a weak base. Kb= 1.8 * 10-5 [Answer xxx]
Calculate the pH of a 10% solution of the weak base NH4OH given that Kb= 1.8 * 10-5 [Answer xxx]