Note that water slightly dissociates in equilibrium with hydrogen and
hydroxide ions:
H2O  H+ + OHSo, the equilibrium constant is:
Kw = [H+][OH-] (product constant of water)
Experimentally, it has been determined in a neutral solution at 250C:
[H+] = [OH-] = 10-7
(a very small amount!)
So,
Kw = (10-7)(10-7) = 10-14
This shows how
far to the left this
equilibrium is
Temperature effects on Kw
T (°C)
Kw (mol2
dm-6)
pH
Kw = (10-7)(10-7) = 10-14
0
0.114 x 10-14
7.47
10
0.293 x 10-14
7.27
This value applies to 250C
20
0.681 x 10-14
7.08
25
1.008 x 10-14
7.00
30
1.471 x 10-14
6.92
40
2.916 x 10-14
6.77
50
5.476 x 10-14
6.63
100
51.3 x 10-14
6.14
In the previous slide we saw:
Kw changes with
temperature variations and
so does the pH!
Note: this does not mean that water is more acidic at higher temperatures
and more alkaline at lower temperatures. Water is still neutral. It simply
means that neutral is a different pH value at different temperatures.
Temperature effects on Kw
1. Given the values in the table,
what are the [H+] and [OH-] at 00C
and 1000C?
2. What do the values of Kw tell you
about whether the ionisation of
water is endothermic or
exothermic? Explain.
Answers:
1. At 0C : 3.38 x10-7; at 100C: 7.16x10-7
(square root of Kw)
2. An increase of temperature increases
the value of Kw. This suggests the
products are favoured. According to
LC Principle, this means the forward
reaction is endothermic (absorbs heat)
T (°C)
Kw (mol2
dm-6)
pH
0
0.114 x 10-14
7.47
10
0.293 x 10-14
7.27
20
0.681 x 10-14
7.08
25
1.008 x 10-14
7.00
30
1.471 x 10-14
6.92
40
2.916 x 10-14
6.77
50
5.476 x 10-14
6.63
100
51.3 x 10-14
6.14
Since
Kw = (10-7)(10-7) = 10-14
We can use this constant value to calculate [H+] or [OH-]
Calculating [H+]
Calculating [OH-]
An alkali with a hydroxide ion
concentration of 0.01M (10-2)
An acid with a hydrogen ion
concentration of 0.001M (10-3)
Kw = (10-2)[H+] = 10-14
[H+] = 10-12
Kw = (10-3)[OH-] = 10-14
[OH-] = 10-11
pH
We have seen previously that
pH = -log[H30+]
pOH
This is similar to pH
pOH = -log[OH-]
The ‘p’ is a mathematical operation that means ‘-log’
pKw
Since Kw = [H+][OH-]
pKw = pH + pOH = 14
Calculating pH example 1
An alkali with a hydroxide ion
concentration of 0.01M (10-2)
Calculating pH example 2
A substance has been found to have
a hydroxide ion concentration of 10-11
Kw = (10-2)[H+] = 10-14
[H+] = 10-12
pH = 12
[OH-] = 10-11
pOH = 11
pH = 14-11 = 3
Recall the inverse of the pH
calculation:
Example
pH = -log[H30+]
The pH of a solution is found
to be 4.6
The inverse:
Find[H+]
[H30+] = 10-pH
or
[H+] = 10-pH
[H+] = 10-pH
On your calculator, this may be
‘2nd’ LOG or ‘10x’
= 10-4.6
= 2.5x10-5 mol dm-3
Problem 1
Calculate the pH of solutions with the following H3O+ concentrations in mol dm-3:
a)10-8
b)6.8 x 10-3 c)0.035
Problem 2
Calculate the pH of solutions with the following OH- concentrations in mol dm-3:
a)10-2
b)7.6 x 10-3 c)0.055
Problem 3
Calculate the H3O+ concentrations in solutions with the following pH values:
a)0.00
b)4.3
c)2.35
d)13.7
Problem 4
What is the pH of a 3.5M solution of H2SO4? (assume complete ionisation)
Problem 5
What is the pH of a 0.001M solution of Ba(OH)2? (Hint: use balanced equation)
Unlike strong acids, weak acids only partially dissociate:
HA + H2O  H3O+ + AThe same is true for weak bases:
B + H2O  BH+ + OHDissociation Constants
Because of these equilibria, we
have known values for the ratio
between products and reactants.
Formulae
+
Ka =
Kb =
𝐵𝐻 [𝑂𝐻 ]
[𝐵]
+
These are known as dissociation
constants:
Ka – acid dissociation constant
Kb – base dissociation constant
−
𝐻3𝑂 [𝐴 ]
[𝐻𝐴]
−
Notice water is not in these calculations. Its
concentration has been incorporated into
the dissociation constant.
CH3COOH(aq) ↔ H+(aq) + CH3COO-(aq)
Ka expression for acetic acid (ethanoic acid) above is: (constant at
set temp)
Ka
=
[H+] [CH3COO-] mol dm3
[CH3COOH]
Ka can be used to find the pKa (like pH)
pKa
=
- log Ka
The larger the value of Ka the stronger the acid
(more it dissociates – breaks apart)
What about pKa?
CH3COOH(aq) ↔ H+(aq) + CH3COO-(aq)
Ka
=
[H+] [CH3COO-] mol dm3
[CH3COOH]
Using the equilibrium constant expression (above) we are able to
calculate [H+] and thus the pH of the acid
Initial conc
Final conc
CH3COOH(aq) ↔ H+(aq) + CH3COO-(aq)
x
0
0
x-y
y
y
𝑦 [𝑦]
𝑦2
𝑦2
𝐾𝑎 =
=
≈
[𝑥 − 𝑦]
𝑥 − 𝑦 (𝑦 ≈ 0)
[𝑥]
Assumption made based on relative
concentrations of ‘x’ vs. ‘y’ to simplify calculations
NH3(aq) + H2O(l) ↔ NH4+(aq) + OH-(aq)
Kb
=
[NH4+] [OH-]
[NH3]
This works the same as in the acid example
Initial conc
Final conc
NH3(aq)
x
x-y
↔
NH4+(aq) + OH-(aq)
0
0
y
y
𝑦 [𝑦]
𝑦2
𝐾𝑏 =
≈
[𝑥 − 𝑦]
[𝑥]
Consider the following generic acid + water reaction:
HA + H2O  A- + H3O+
ACID
Conjugate BASE
Now, if the conjugate base reacts with water in this reaction:
A- + H2O  HA + OH-
𝐾𝑎 =
𝐴
−
+
[𝐻3𝑂 ]
[𝐻𝐴]
So, Ka x Kb
𝐴
−
𝐻𝐴 [𝑂𝐻 ]
𝐾𝑏 =
−
[𝐴 ]
+
−
+
−
[𝐻3𝑂 ]
𝐻𝐴 [𝑂𝐻 ]
𝑋
−
[𝐻𝐴]
[𝐴 ]
−
= [𝐻3𝑂 ][𝑂𝐻 ] = Kw
Ka x Kb = Kw
From the previous slide,
Ka x Kb = Kw
If we take the ‘p’ of each of these values (-log) we find,
pKa + pKb = pKw = 14
+

Ka =
−
𝐻3𝑂 [𝐴 ]
[𝐻𝐴]
+
(acid dissociation constant; larger the number, stronger the acid)
−
𝐵𝐻 [𝑂𝐻 ]
(base dissociation constant;
[𝐵]

Kb =

pKa = - log Ka (smaller the value, stronger the acid) (pKb is the same)

Ka x Kb = Kw= 10-14

pKa + pKb = pKw = 14 = pH + pOH
larger the number stronger the base)
Calculate the H3O+ concentrations, and the pH value for the following
weak acids:
a)
HCO2H
conc. = 0.0100 mol dm-3
Ka = 2.04 x 10-4
b)
CH3CO2H
conc. = 0.1 mol dm-3
Ka = 1.77 x 10-5
c)
HCN
conc. = 1.00 mol dm-3
Ka = 3.96 x 10-10
Calculate Ka for the following acids:
d)
HF
conc. = 0.200 mol dm-3
e)
HClO2
conc. = 0.0200 mol dm-3
pH = 2.23
pH = 1.85
Calculate the OH- and H3O+ concentrations, and the pH value for the
following weak bases:
f)
NH3
conc. = 0.0100 mol dm-3
Kb = 1.80 x 10-5
g)
C5H5N
conc. = 0.1 mol dm-3
Kb = 1.40 x 10-9
h)
CH3NH2
conc. = 1.00 mol dm-3
Kb = 4.38 x 10-4

A buffer is a solution that resists changes
in pH when small amounts of acid or
base are added to it

There are 2 types:
› Acidic
› Alkaline
An acidic buffer has a pH less than 7
It is often made from equal molar
concentrations of a weak acid and it’s
conjugate base
Example: ethanoic acid and ethanoate ion
CH3COOH 
(weak acid)
CH3COO(conj. base)
+
H+
Take a 0.1M solution of ethanoic acid:
At equil:
CH3COOH

(0.09583M)
What’s the pH of this solution?
CH3COO(0.00417M)
+
H+
(0.00417M)
-log [H+] = 2.38
Now, say you add HCl to this equilibrium. The acetate ion is the only
species available to reduce the added H+ and these are very
low in concentration, so the pH will drop dramatically.
What can you do to reduce this effect?
Add acetate ions (sodium
acetate)
CH3COOH

CH3COO
-
+
H
+
If we buffer the solution by adding 1M sodium acetate, what will happen?
This will have the following effect:
1. Increase the amount of acetate ions, shifting the equilibrium to the left
2. Increase the pH to 4.76
+
+
3. When adding H ions, the extra acetate ions will react to reduce the [H ]
moving the equilibrium to the left – producing more weak acid.
4. The solution resists changes to pH, meaning it is buffered
Add OH
-
CH3COOH

-
CH3COO
+
H
+
Adding base results in more OH- being added to the
equilibrium. This extra amount of ions is removed by:
-  CH COO- + H O
CH
COOH
+
OH

3
3
2
 hydroxide ions are removed by reaction with undissociated
ethanoic acid, shifting equilibrium right making a weak
base (CH COO-)
3
pH
vol. of 0.1M
acetic acid (ml)
vol. of 0.1M
sodium acetate (ml)
3
982.3
17.7
4
847.0
153.0
5
357.0
653.0
6
52.2
947.8
Note: you can also get various pH buffers by changing the acid/base pair.
An alkaline buffer has a pH greater than 7
It is often made from a equal molar
concentrations of a weak base and it’s
conjugate acid
Example: ammonia and ammonium ion
+
NH3 + H2O   NH4 + OH
(weak base)
(conj. acid)
-
+
-
NH3 + H2O   NH4 + OH
(weak base)
(conj. acid)
Due to the weak nature of ammonia, this equilibrium will be well
to the left. We can create a buffered solution by adding
ammonium chloride or a strong acid.
What will this do to the equilibrium?
Addition of ammonium ions (adding ammonium chloride) will shift the equilibrium
even further to the left. The pH of this solution would be 9.25
You can also add ~half the moles of a strong acid (e.g. HCl) which will react
with ammonia to form more ammonium ions (NH3 + H+  NH4+)
+
-
NH3 + H2O   NH4 + OH
What will happen if you add acid to this solution?
Reaction of the acid with the weak base


NH3 + H+   NH4+
removal by reaction with ammonia to produce more
ammonium ion – a weak acid)
NH3 + H2O  
+
-
NH4 + OH
What will happen to the equilibrium if you add base?
Adding base effectively adds OH-. This means:
The ammonium ion reacts with the OH- to shift the equilibrium to
the left, consuming most of the OH- ions.
NH4+ + OH-   NH3 + H2O
Buffer solutions resist changes in pH when
acids and alkalis are added
 Buffers generally contain:

› Sufficient concentrations of a weak acid and it’s
conjugate base OR weak base and it’s
conjugate acid

The pH of buffer solutions depend on the
concentrations and type of conjugate
acid/base pairs that are used.
Go here 
http://www.pearsonhotlinks.co.uk/978043599440
2.aspx for good animations of this (chapter 8)
Consider the dissociation of a weak acid
HA(aq) + H2O(l)  A-(aq) + H3O+
The acid dissociation constant expression is
𝐾𝑎 =
𝐴
−
Commercially available buffer
solutions are used to calibrate pH
meters
+
[𝐻3𝑂 ]
[𝐻𝐴]
+
This can be rearranged to find [𝐻3𝑂 ]
𝐻3𝑂
+
= 𝐾𝑎
[𝐻𝐴]
−
𝐴
HA(aq) + H2O(l)  A-(aq) + H3O+
Assumptions:
1. As a large quantity of conjugate base (A-) has been added, the equilibrium
shifts far left, so that equilibrium concentration of the acid is approximately
equal to the initial concentration of the weak acid :
[HA]eq ≈ [HA]i = [acid]
2. The equilibrium concentration of the conjugate base ion (A- ) is
approximately equal to the concentration of the salt that was added to the
equilibrium.
[A-]eq ≈ [A-]i = [salt]
Recall from a previous slide that
𝐻3𝑂
+
= 𝐾𝑎
[𝐻𝐴]
−
𝐴
So, considering our 2 assumptions:
[HA]eq ≈ [HA]I = [acid]
[A-]eq ≈ [A-]i = [salt]
𝐻3𝑂
When [acid] = [salt]
pH= pKa
+
[𝑎𝑐𝑖𝑑]
= 𝐾𝑎 x
𝑠𝑎𝑙𝑡
𝑝𝐻 = 𝑝𝐾𝑎 − 𝑙𝑜𝑔
[𝑎𝑐𝑖𝑑]
𝑠𝑎𝑙𝑡
Alternatively, you
may solve for [H+]
first and then solve
for pH using
pH=-log[H+]
Similarly for a base equilibrium
B(aq) + H2O(l)  HB+(aq) + OH-(aq)
𝑂𝐻
When [base] = [salt]
pOH= pKb
−
= 𝐾𝑏
[𝐵]
+
𝐻𝐵
= 𝐾𝑏
𝑝𝑂𝐻 = 𝑝𝐾𝑏 − 𝑙𝑜𝑔
[𝑏𝑎𝑠𝑒]
𝑠𝑎𝑙𝑡
[𝑏𝑎𝑠𝑒]
𝑠𝑎𝑙𝑡
Alternatively, you
may solve for [OH-]
first and then solve
for pOH using
pOH=-log[OH-]
An aqueous solution of 0.1M ammonia and 0.1M ammonium chloride
has a pH of 9.3. The reaction is: NH3 + H2O  NH4+ + OHa) Calculate the Kb for ammonia
b) If a pH of 9.0 is needed, what should be added? Explain
c) Calculate the new concentration of the substance added in b)
Answers:
a) 𝐾𝑏 =
−
+
𝑂𝐻 [𝑁𝐻4 ]
[𝑁𝐻3]
=
10−4.7 [0.1]
[0.1]
= 10-4.7 = 2.00 x 10-5
b) Ammonium chloride should be added as this will shift the equil to the
left(towards ammonia), reducing the [OH-] and decreasing the pH
c) pH = 9.0 means pOH = 5, so [OH-] = 10-5
+
𝐾𝑏[𝑁𝐻3]
(2𝑥10−5 )(0.1)
-3
−
[𝑁𝐻4 ] = [𝑂𝐻
=
=
0.2
mol
dm
−5
[10 ]
]
Notice the additional volume has a
negligible effect on [NH3] and has
been ignored
If you wanted to make a buffer solution of pH=4.46 using ethanoic acid and
sodium ethanoate, what concentrations could you use? pKa (ethanoic acid) =
4.76. The reaction is: CH3COOH+ H2O  CH3COO- + H3O+
Answer:
[𝑎𝑐𝑖𝑑]
[𝑠𝑎𝑙𝑡]
[𝑎𝑐𝑖𝑑]
log
[𝑠𝑎𝑙𝑡]
pH = pKa – log
4.5 = 4.76 –
log
[𝑎𝑐𝑖𝑑]
[𝑠𝑎𝑙𝑡]
[𝑎𝑐𝑖𝑑]
=
[𝑠𝑎𝑙𝑡]
[𝑎𝑐𝑖𝑑]
[𝑠𝑎𝑙𝑡]
= 4.76 – 4.46 = 0.30
100.30
= 2.0
So, we need a solution with twice as
concentrated acid as ethanoate salt
Note: it should be expected that the acid
concentration will be higher than the salt.
At equal concentrations, the pH = pKa = 4.76
At pH 4.46 (a lower pH), we need to add acid
to shift equil right, increasing [H3O+], decreasing
pH
An aqueous solution of 0.025M ethanoic acid and 0.050M sodium
ethanoate is prepared.
a) Calculate the pH of this solution given Ka = 1.74x10-5 mol dm-3
b) If 1.0cm-3 of 1.0M NaOH is added to 250cm3 of buffer, what will
happen to the pH?
Answers:
a) 𝑝𝐻 = 𝑝𝐾𝑎 − 𝑙𝑜𝑔
b)
[𝑎𝑐𝑖𝑑]
=
𝑠𝑎𝑙𝑡
CH3COOH + OH-

= 4.76+0.30=4.96
CH3COO- + H2O
-
So,
[𝑎𝑐𝑖𝑑]
pH = pKa – log [𝑠𝑎𝑙𝑡]
+0.001
-
pH = 4.76 – log [0.054]
0
0.0135
-
0
0.054
-
ni (mol)
0.00625
nc(mol)
-0.001
-0.001
ne(mol)
0.00525
[ ] (mol
0.021
dm-3)
[.025]
.050
-log (1.74x10-4.7) – 𝑙𝑜𝑔
0.001
0.0125
[0.021]
pH = 4.76 + 0.41
pH = 5.17
Notice the additional volume has again
been ignored as it is assumed to have
little effect on the overall pH
Titration is an analytical technique that is used to determine the end
point of a reaction (often Acid + Base) using an indicator or pH meter.
Acid-Base Titration (basic steps):
1. Unknown: Accurately measure
a volume of base of unknown
concentration using a pipette
and place into a flask
2. Indicator: Add an appropriate
indicator which will change
colour at the end point of the
reaction
3. Titrate: Carefully add an acid
of known concentration until
the end point is reached as
indicated by the colour
change
Pipette
Titration
Indicator addition
End point
Equivalence Point
End point
The equivalence point in an
acid-base reaction is the
point where neutralisation
occurs.
The end point is where an
indicator solution just
changes colour
permanently.
The molar ratios have been
reached
Indicators change colours at specific pH ranges and are chosen to
be as close to the equivalence point as possible. More later…
Investigating the titration between:
1M HCl and 1M NaOH
41
KeMsoft06
HCl
10 ml NaOH
Start with 10ml of alkali
and slowly add acid
measuring the pH
42
KeMsoft06
pH During a Titration
15
14
13
12
HCl
11
10
pH
9
8
7
6
5
4
3
2
1
0
0
2
4
6
8
10
Volume of acid added / ml
10 ml NaOH
+ almost 10
ml HCl
43
KeMsoft06
12
14
16
18
20
pH During a Titration
15
14
13
12
HCl
11
10
pH
9
8
7
6
5
4
3
2
1
0
0
2
4
6
8
10
12
14
16
Volume of acid added / ml
10 ml NaOH + 10 ml HCl
Now we have equivalent amounts of strong acid and
strong base – notice the pH changes dramatically
44
KeMsoft06
18
20
pH During a Titration
15
14
13
12
HCl
11
10
pH
9
8
7
6
5
4
3
2
1
0
0
2
4
6
8
10
12
14
16
Volume of acid added / ml
10 ml NaOH
Now as we continue to add acid past pH = 7, the pH
drops quickly at first and then more slowly
45
KeMsoft06
18
20
pH During a Titration
15
14
1NaOH
NaOH ++HCl
1HCl
= 
NaClNaCl
+ H2O
+ H2O
1M
1M
13
12
11
10ml
10
10ml
pH
9
8
Equivalence point at pH7
7
6
Solutions mixed in the right
proportions according to the
equation.
5
4
3
2
1
0
0
5
10
15
Volume of acid added / ml
46
KeMsoft06
20
25
Running acid into alkali
Running alkali into acid
This nearly horizontal section
of the graph is called the
buffer region. Here the pH
stays relatively constant due
to a buffering between the
acid and the salt
pOH = pKb
pH = pKa
This point is called half-equivalence where half of the acid
has been neutralised. Here, there are equivalent amounts
The same applies to a base being neutralised by an acid
of acid and salt, so 𝑝𝐻 = 𝑝𝐾𝑎 − 𝑙𝑜𝑔
pH = pKa
47
𝑎𝑐𝑖𝑑
𝑠𝑎𝑙𝑡
= pKa − 0 or
Investigating the titration between:
1M HCl and 1M NH3
48
KeMsoft06
Weak base so initial pH
value is less than 14
pH starts to fall quickly as
acid is added
acid
pH falls less quickly as
buffer soln formed (excess
NH3 and NH4Cl present)
1NH3 + 1HCl  NH4Cl
1M
1M
25ml
25ml
Soln at equivalence
point is slightly acidic
because ammonium
ion is slightly acidic
NH4+ + H2O  NH3 + H3O+
alkali
49
KeMsoft06
After equivalence point the soln
contains NH3 and NH4Cl – a buffer
soln and so resists large increase
in pH so graph flattens out.
alkali
Very low pH indicative of a strong
acid solution
acid
50
KeMsoft06
< pH 7
Running acid into alkali
Running alkali into acid
51
KeMsoft06
Investigating the titration between:
1M CH3COOH and 1M NaOH
52
KeMsoft06
1CH3COOH + 1NaOH  CH3COONa
Very high pH – indicative of
a strong alkali solution
acid
1M
Soln at equivalence
point is slightly
alkaline because
ethanoate ion is
slightly alkaline
CH3COO-+ H2O  CH3COOH + OH-
alkali
53
KeMsoft06
25ml
1M
25ml
pH falls less quickly as
buffer soln formed (excess
CH3COOH and CH3COOpresent)
Excess of alkali present –
graph same as when adding
strong alkali to strong acid
alkali
acid
pH rises less quickly as
buffer soln formed (excess
CH3COOH and CH3COOpresent)
pH starts to rise quickly as
alkali is added
54
KeMsoft06
Running acid into alkali
Running alkali into acid
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Investigating the titration between:
1M CH3COOH and 1M NH3
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1CH3COOH + 1NH3 = CH3COO- + 1NH4+
About as weak as each other acid
alkali
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No steep section – small addition of acid
causes a large change in pH, so…
acid
Very difficult to do a
titration between a weak
acid and a weak base.
alkali
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


The pH of a salt depends upon the
relative strength of the ions that make
up the salt.
Very few salts are neutral
Salts completely dissociate into their
ions when sufficiently dilute
NaCl(s)  Na+(aq) + Cl-(aq)

It is possible for these ions to interact
with water to produce H+ or OH- ions
which results in acidic or alkaline
solutions. These are known as
hydrolysis reactions
Remember: the stronger the acid/base, the weaker it’s conjugate
pH = 7



Neutral anions are formed from
strong acids
Neutral cations are formed from
strong bases
NaCl is a neutral salt because
the ions that are formed derive
from a strong acid and base
› HCl + NaOH  NaCl + H2O

Na+ and Cl- are weak
conjugates, so there is no
tendency for these ions to
undergo hydrolysis reactions.
pH > 7
Weak acids form conjugate bases that can
react with water to form hydroxide ions
› H2CO3 + 2NaOH  Na2CO3 + H2O
› Na2CO3  2Na+ + CO32› CO32- + H2O  HCO3- + OH-
In the above reactions:
 NaOH is a strong base so the weak
conjugate, Na+ will not react with water
to form hydrogen ions
 Carbonic acid is weak, so the carbonate
ion will react with water to a small extent
to form OH- ions
Basic anions are formed by weak acids
pH < 7
HCl + NH3  NH4+ + Cl
A donatable hydrogen must be
available on the cation.
NH4+ + H2O  NH3 + H3O+
The ammonium ion comes from the
weak base, ammonia, so a hydrolysis
reaction can occur to a small extent
producing hydronium ions.

When combining weak acids and bases,
the pH of the salt will depend on their
relative strengths.
›
›
›
If Ka > Kb – acidic
If Kb > Ka – basic
If Ka = Kb - neutral
Example:
CH3COOH + NH3  NH4+ + CH3COO
Both of the salts formed in this reaction
will react with water, but the effects
cancel resulting in a nearly neutral
solution
Ammonium Acetate
Small ions with multiple charges can hydrolyse
water. The high charge density pulls electrons
towards the metal ion causing a proton to be
released, decreasing the pH
H
H
H
H
H
O
O
O
H
Fe3+
H
O
H
O
O
H
H
H
H
H
H

An O-H
bond may
be broken
releasing a
proton
H
H
2+
H
O
O
O
H
Fe3+
H
O
H
O
O
H
+ H+
H
H
Other metal ions that are able to
hydrolyse water include Be2+ and Al3+
Type of
reaction
Example
Reaction
Salt
produced
Ion that
hydrolyses
water
Nature of
final solution
Strong acid HCl + NaOH
Strong Base
NaCl
Neither ion Neutral
pH = 7
Weak acid CH3COOH +
Strong Base NaOH
NaCH3COO
Anion
Basic
pH>7
Strong Acid HCl + NH3
Weak Base
NH4Cl
Cation
Acidic
pH<7
Weak Acid
Weak Base
CH3COOH + NH3 NH4CH3COO
Anion &
Cation
Depends on
Ka/Kb
Metal
complex
[Al(H20)6]3+
[Al(H20)5OH]2+
+ H+
Metal
complex
with
multiple
charge
Acidic
pH<7
NA
Indicators can change colours
depending upon the pH of the
solution they are in.
Indicators are themselves weak
acids or weak bases and as they
lose or gain H+, they form
substances that have distinct
colours.
In general, an indicator molecule (In)
Hin(aq)  H+(aq) + In-(aq)
Let’s look at some examples…
Add base (hydroxide)
“HLit” is litmus – a weak
acid – is red in solution
Base
When hydroxide (base) is
added to this weak base,
“Lit-” is formed and is blue
When hydrogen ions (acid)
is added, we shift back to
the red “HLit” .
Add acid (hydrogen ion)
Acid
So,
• Acid turns blue litmus red
• Alkali turns red litmus blue
Methyl orange
is yellow below
pH 3.7
Methyl orange is red
above pH 3.7
Notice the transfer of a proton
between the different
coloured compounds
Here is a common acid-base indicator
that changes colour at pH 9.3
What colour is the acid?
What colour is the base?
Here is a common acid-base indicator
that changes colour at pH 9.3
What colour is the acid? colourless
What colour is the base? pink
Recall the generic equation for an indicator
Hin(aq)  H+(aq) + In-(aq)
We can write an Equilibrium expression for this weak acid
+
−
𝐻 [𝐼𝑛 ]
𝐾𝑎 =
[𝐻𝑖𝑛]
−
When the equilibrium is balanced, [𝐼𝑛 ]= [𝐻𝑖𝑛] so,
+
−
𝐻 [𝐼𝑛 ]
𝐾𝑎 =
[𝐻𝑖𝑛]
This means
Ka = [H+] or
pKa = pH
So the pKa of the
indicator, tells us where
the colour change will
occur
Choosing an Indicator
3.7
9.3
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Strong Acid - Strong Base
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Strong Acid - Weak Base
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Weak Acid - Strong Base
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Weak Acid – Weak Base
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