1. HCOOH (aq) + H2O (l) ⇋ HCOO- (aq) + H3O+ (aq)
Learning objective:
To observe the effect of weak and strong acids
To explain the phenomenon in terms of equilibrium
17/03/2017
Explain the difference between a strong and a weak acid
Identify the acid/base conjugate pairs in this reaction:
HCOOH (aq) + H2O (l) ⇋ HCOO- (aq) + H3O+ (aq)
acid 1 base 2 base acid 2
Define pH
pH = -log10[H3O+]

2. pH of a strong acid
A strong acid fully dissociates, HA + H2O  H3O+ + A-
so [H3O+] = [HA] (where HA is any acid)
pH 1 moldm-3 HCl = -log 10[HCl] = 0.00
Try finding the pH of:
0.1 moldm-3 HCl (aq) 1.00
10 moldm-3 HCl (aq)

3. More pH of a strong acid
Find pH of HCl at the following concentrations:
0.01 moldm-3
0.001 moldm-3
moldm-3
1.00x10-5 moldm-3
1.00x10-6 moldm-3
1.00x10-7 moldm-3
1.00x10-8 moldm-3
Have you noticed anything odd…
…or even impossible?

4. How can a neutral solution have a pH of 7?
Calculate the [H3O+] in a neutral solution
pH = -log10[H3O+]
so: log10[H3O+] = -pH
and: [H3O+] = 10-pH (use shift/2nd fn log)
[H3O+] = 10-7
= 1x10-7 moldm-3
Where has it come from?
Water dissociates (very slightly)
H2O + H2O ⇋ H3O+ + OH-
Write an expression for Kc for this reaction

5. Dissociation constant for water
H2O + H2O ⇋ H3O+ + OH-
Kc= [H3O+] [OH-] / [H2O]2
Calculate the concentration of water
1dm3 = 1000g
Amount H2O in 1dm3 = 1000/18=55.56moldm-3
i.e. about 1.8x10-7% dissociated
[H2O]2 is very large and effectively constant so we rearrange the equation:
Kc[H2O]2 = [H3O+] [OH-]
Kw = [H3O+] [OH-] (the dissociation constant of water)

6. Calculations with Kw Kw = [H3O+] [OH-]
Kw is constant (for a given temperature)
If we know Kw (given in questions) and [OH-] we can calculate [H3O+] and, hence pH
pH 0.10 moldm-3 NaOH (Kw at 25oC = 1.00x10-14 mol2dm-6)
assume the alkali fully dissociates so [OH-] = 0.10
So [H3O+] = Kw/ [OH-]
[H3O+] = 1.00x10-14/0.1
[H3O+] = 1.00x10-13
pH = -log10(1.00x10-13)
13.00

7. CH3COOH (aq) + H2O (l) ⇋ CH3COO- (aq) + H3O+ (aq)
Weak acids
Weak acids only partially dissociate:
CH3COOH (aq) + H2O (l) ⇋ CH3COO- (aq) + H3O+ (aq)
When a weak acid reacts with a base the H3O+ (aq) reacts:
H3O+ (aq) + OH-  2 H2O (l)
What will happen in the first reaction when the H3O+ reacts?
Use these ideas to explain the demonstration

8. Dissociation constant for a weak acid
CH3COOH (aq) + H2O (l) ⇋ CH3COO- (aq) + H3O+ (aq)
Write an expression for Kc for this reaction
Kc = [CH3COO- (aq)] [H3O+ (aq)]/ [CH3COOH (aq)] [H2O (l) ]
But [H2O (l)] is large and we can assume it is constant
Kc[H2O (l) ] = [CH3COO- (aq)] [H3O+ (aq)]/ [CH3COOH (aq)]
Ka = [CH3COO- (aq)] [H3O+ (aq)]/ [CH3COOH (aq)] (The acid dissociation constant)

9. Example
Calculate the pH of 0.1 moldm-3 HCl and CH3COOH (Ka = 1.7x10-5 moldm-3)
HCl is a strong acid so [H3O+] = [HCl]
pH = -log10[HCl]
pH = 1.00
CH3COOH is a weak acid
Ka = [CH3COO- (aq)] [H3O+ (aq)]/ [CH3COOH (aq)]
Assuming all H3O+ comes from the dissociation then [CH3COO- (aq)]=[H3O+ (aq)]
Ka = [H3O+ (aq)]2/ [CH3COOH (aq)]

10. Example Assume [CH3COOH (aq)]eqm = [CH3COOH (aq)]initial
Ka = [H3O+ (aq)]2/ [CH3COOH (aq)]
[H3O+ (aq)]2 = Ka[CH3COOH (aq)]
[H3O+ (aq)] = √Ka[CH3COOH (aq)]
Assume [CH3COOH (aq)]eqm = [CH3COOH (aq)]initial
[H3O+ (aq)] = √ 1.7x10-5x0.1
[H3O+ (aq)] = √ 1.7x10-6
[H3O+ (aq)] = moldm-3
pH = -log10( )
pH = 2.88