Starter question.
Consider propanoic acid (CH3CH2COOH)
reacting with water…..
 1. Write a symbol equation for the
equilibrium that is established.
 2. Identify the conjugate base and acid on
the right hand side of the equation.
 3. Write a Kc expression for the equilibrium.

Starter answers





CH3CH2COOH (aq) + H20(l) ↔ H3O+(aq) + CH3CH2COO-(aq)
Conjugate base….CH3CH2COOConjugate acid H3O+
Ka = [H+(aq)][CH3CH2COO-(aq)]
[CH3CH2COOH (aq)]
The pH scale
The pH scale
Devised by a Danish chemist called Soren
Sorensen.
 It indicates how much acid or alkali is
present in a solution.
 The ‘p’ in pH stands for ‘potens’, which is
latin for power.

 pH
= -log[H+ (aq)]
Calculating the pH of a strong
acid.
Calculating the pH of a
strong acid
The pH of strong acids are relatively simple
to calculate.
 If we assume that the acid molecules fully
dissociate then [H+] = [HA]
 Therefore pH = -log [HA]


E.g.
1 mol HCl, pH = –log[1] = 0

E.g.
0.1 mol HCl, pH = -log[0.1] = 1
Calculating the pH of a weak
acid.
Calculating the pH of a weak
acid. An example.

What is the pH of 0.1moldm-3 ethanoic
acid, which has a Ka of….1.74X105moldm-3?)
You need to…
 1. Write out the equilibrium expression
(Ka)
 2. convert hydrogen concentration [H+]
to pH

Step 1 the Ka expression.
Equilibrium equation.
CH3COOH (aq) ↔ H+ (aq) + CH3COO-(aq)
So the general Ka expression will be……
Ka = [CH3COO-(aq)][H+(aq)]
[CH3COOH(aq)]
But we need to consider when full equilibrium has been
reached and take account of our initial concentration of acid.
Step 2 the equilibrium
position.
CH3COOH ↔
H+
+
CH3COO-
CH3COOH
H+
CH3COO-
Start Conc.
0.100
0
0
At equilibrium.
0.100- [CH3COO-]
[H+]
[CH3COO-]
CH3COOH
H+
CH3COO-
For every ethanoic acid molecule
that dissociates an ethanoate ion
and a H+ also forms.
Writing the new Ka
expression at equilibrium.





COO-(aq)]
[H+ (aq)]
Initial concentration
minus the dissociation
concentration of the
ions.
Ka = [CH3
eq
eq
0.100 – [CH3COO-(aq)]eq
As [CH3COO-] = [H+] at equilibrium we can write…
Ka =
[H+]2
0.100 – [H+]
Because the disassociation of the weak acid is so small
we assume 0.100 – [H+] = 0.100
Plugging the numbers in..
Substituting in the values gives….





1.74 X 10-5 = [H+]2
0.100
[H+]2= 0.100 X 1.74 X 10-5
[H+] = √(1.74 X 10-6)
[H+]= 1.319 X 10-3moldm-3
=
pH = 2.88
(using –log)
Comparing strong and
weak acids
Factors
Strong Acid
Weak Acid
[H+ (aq)] becomes 100 x
smaller
[H+ (aq)] becomes 10 x
smaller
pH increases by 2
pH increases by 1
Conductivity
Higher as more free ions
Lower as less free ions.
Reaction Rate
Reacts far quicker due to
greater dissociation
Reacts slower as relies on
H+ being removed from
equilibrium to completely
dissociate acid molecules.
Acid diluted by a factor of
100
Strength vs. Concentration

Concentration is a measure of the amount of
substance in a given volume of solution.

Strength is a measure of the extent to which
an acid can donate H+. Measured as pKa
values:

pKa = -log Ka
Calculating the pH of a strong
base.
Calculating pH of a strong
base.



To begin with we need to understand another
term, Kw, this is the ionic product of water.
For water, Ka = [H+(aq)] [OH-(aq)] / [H2O(l)]
As water is always present in excess we ignore
the term, [H2O(l)] to give:


Kw = [H+(aq)] [OH-(aq)]
At 298K Kw = 1x10-14 mol2 dm-6
Calculating pH of a strong
base.

Kw = [H+(aq)] [OH-(aq)]

At 298K Kw = 1x10-14 mol2 dm-6

Kw = 1x10-14 = [H+]2

Therefore [H+] = 1x10-7

=pH7 at 298K, pH will fall as temperature
increases.
Calculating pH of a strong
base.

What is the pH of a 0.1 mol solution of
sodium hydroxide?
Kw = 1 x 10-14 = [H+] x [OH-]
 Kw = 1 x 10-14 = [H+] x [0.1]
 [H+] = 1 x 10-14 / 0.1
 [H+] = 1 x 10-13
 pH = -log [1 x 10-13] = 13
