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Maths and Chemistry for
Biologists
Chemistry 3
Acids, Bases and pH
This section of the course covers –
• the nature of acids and bases
• how their strengths are defined
• the hydrogen ion concentrations of solutions of
acids and bases
• the pH scale
What are acids and bases?
Acids are molecules that can donate protons to
other substances. The substances that accept
the proton are called bases.
If we represent the acid by HA and the base by B
then the reaction is
HA + B  A- + BH+
The product of this reaction is a salt. For example
HCl + NH3  NH4+ClHere hydrochloric acid is reacting with ammonia to
produce ammonium chloride
Acids in water
When an acid is dissolved in water the water acts
as a base
HA + H2O 
A+
H3O+
acid base conjugate base conjugate acid
The ion A- is called the conjugate base of the acid
because it is capable of reacting with some other
acid to go back to HA
H3O+, which is called the hydroxonium ion, is the
conjugate acid of water because it can donate a
proton to some other base and go back to H2O
Bases in water
When a base is dissolved in water the water acts
as an acid
B + H2O 
BH+
+
OHbase acid
conjugate acid conjugate base
The reasons why BH+ is called the conjugate acid
of B and OH- is called the conjugate base of H2O
are just the same as in the previous slide
OH- is called the hydroxide ion and occurs in
substances such as NaOH which react with
acids to form salts plus water
NaOH + HCl  NaCl + H2O
Important facts
Water acts as an acid or a base depending on what
is dissolved in it
An acid/base reaction is always of the form
Acid + Base  Conjugate base + Conjugate acid
of the acid
of the base
Acids in biology
These are of two main types Carboxylic acids where the
-O-H is the acidic group. R means
any other chemical group.
(NB alcohols, R-OH, are not acidic)
Phosphates. Here one or both
of the Rs can be H so there can be
two or even three acid groups
O
R C
OH
RO
OH
P
RO
O
Bases in biology
One main type – amines
These can be primary (one R
group), secondary (two R
groups) or tertiary (three R
groups)
H
R N H
R
R N H
(NB amides are not basic)
O
R C NH2
R
R N R
Amino acids
A very important group
of biomolecules are the
amino acids. They have both
a carboxylic acid and an
amine group. So they are both
acids and bases. The structure
shown is glycine. In water the
molecule is ionised as shown
in the lower structure.
NH2
CH2
COOH
NH3
CH2
COO
Strengths of acids
Measured by the extent to which they dissociate in
water; that is, for the process below (now shown as
an equilibrium), what fraction of HA dissociates to AHA + H2O
A- + H3O+
For strong acids such as HCl, H2SO4 and HNO3 the
process is essentially complete
So in a 0.1 M solution of HCl [H3O+] = 0.1 M (note
that the [ ] indicate the concentration of H30+)
Strengths of weak acids
Weak acids dissociate only partially. For the process
HA + H2O
A- + H3O+
we can write an equilibrium expression
[A - ][H 3O ]
K
[HA][H 2O]
where K is the equilibrium constant and the
concentrations are those at equilibrium (for more
details about equilibria see section Chem 6)
contd
[A - ][H 3O ]
K
[HA][H 2O]
We can simplify this by noting that the concentration
of water is very high (55.6 M)* and effectively
constant so the it can be included in K to give Ka
(Ka = 55.6 x K). In addition it is usual to write H+ as
a shorthand for H3O+ so the equation becomes
[A - ][H  ]
Ka 
[HA]
Ka is called the acid dissociation constant
*(1,000 g of H2O is 1000/18 = 55.6 mol)
Meaning of Ka
The bigger is Ka then the stronger is the acid; that is,
the more H+ there will be in the solution
What sort of values does Ka have?
For CH3COOH, Ka = 1.74 x 10-5 M
For HCOOH, Ka = 1.78 x 10-4 M
So formic acid is about 10 x stronger than acetic acid
These numbers are inconvenient and there is a better
way of doing it
pKa
We DEFINE a quantity called pKa as
pKa = -log10 Ka
Then for acetic acid pKa = 4.76
and for formic acid pKa = 3.75
This gives us a scale of acid strength where the
numbers are small and positive and where the value
changes by 1 as the acid strength changes 10x
The SMALLER the pKa the STRONGER the acid
One for you to do
Phosphoric acid, H3PO4, has three dissociable
hydrogens with the Ka values shown. Calculate
the pKa values
H3PO4
Ka

H2PO4- 
HPO42- 
PO43-
7.08 x 10-3 M 6.31 x 10-8 M 4.68 x 10-13 M
Answer
pKa1 = 2.15,
pKa2 = 7.20,
pKa3 = 12.33
Note that the species are progressively
weaker acids. HPO42- is 1010 x weaker than
is H3PO4. In fact PO43- is quite a strong
base
What about bases?
We could write an equation
BH+ + OH-
B + H20
and then an equilibrium expression
[BH  ][OH - ]
K
[B][H 2O]
or
[BH  ][OH - ]
Kb 
[B]
and then define pKb = -log10 Kb
- but we usually don’t!
The pKa of the conjugate acid
Instead we talk about the pKa of the conjugate acid.
That is, we consider the process
BH+ + H2O
from which
B
+
H3O+
[B][H  ]
Ka 

[BH ]
e.g. for ammonia (NH3) we quote the pKa of the
conjugate acid (NH4+, pKa = 9.2) as a measure of
base strength
The hydrogen ion concentration
The hydrogen ion concentration (or more properly
[H3O+]) is of crucial importance in biology
For a strong acid it is equal to the concentration
0.1 M HCl, [H+] = 0.1 M, 0.1 M H2SO4, [H+] = 0.2 M
For weak acids it depends on both the
concentration and the pKa
0.1 M acetic acid, [H+] = 1.32 x 10-3 M or 1.32 mM
The pH
Again, these numbers are not convenient so we
DEFINE the quantity pH = -log10 [H+]
For [H+] = 1.32 x 10-3 M, pH = 2.88
NB the HIGER the [H+] the LOWER is the pH
[H+] = 10-4 M, pH = 4, [H+] = 10-3 M, pH = 3
Calculation of [H+] for weak acids
Suppose we make a 0.1 M solution of acetic acid
and that some of it dissociates to give x M of H+
CH3COO- + H+
CH3COOH
At equilibrium (0.1 – x) M
xM
-

xM
2
[CH3COO ][H ]
x
Ka 

[CH3COOH]
0.1 - x
We know that Ka = 1.74 x 10-5 M so we can
solve for x
Water is both an acid and a base
Water undergoes self-dissociation
2 H2O
Or, in shorthand,
H3O+ + OHH 2O
H+ + OH-
The ionic product of water Kw is DEFINED as
Kw = [H+][OH-] M2 and pKw = -log10 Kw
At 25 oC, Kw = 10-14 M2 and pKw = 14
The pH of neutral water
In neutral water [H+] = [OH-]
Since [H+][OH-] = 10-14 M2 then [H+] = 10-7 M
so the pH = 7
So a solution with a pH less than 7 is acidic and a
solution with a pH greater than 7 is basic
pH of solutions of strong bases
We can calculate these using the value of Kw
For example in 0.1 M NaOH, [OH-] = 0.1 M
But
[H+][OH-]
=
10-14 M2
-14
10
or [H ] 
[OH - ]

So [H+] = 10-14/0.1 = 10-13 and pH = 13
Relationship between pKa and pKb

[BH
][OH
]
For a base K b 
[B]
For its conjugate acid
So
[B][H  ]
Ka 

[BH ]
[B][H  ] [BH  ][OH - ]
Ka x Kb 
x

[B]
[BH ]
Hence Ka.Kb = [H+][OH-] = Kw or pKa + pKb = pKw
Ones for you to do
What is the pH of a solution with [H+] of a) 5 mM, b)
50 mM?
What is the [H+] of a solution of pH = 6?
What is the pH of a 1mM solution of NaOH?
Answers
5 mM is 5 x 10-3 M so pH = -log (5 x 10-3) = 2.30
50 mM is 5 x 10-2 M so pH = -log (5 x 10-2) = 1.30
[H+] = antilog (-pH) so if pH = 6.0,
[H+] = antilog -6 = 10-6 M or 1 M
[OH-] = 10-3 M so [H+] = 10-14/10-3 M = 10-11 M
pH = -log 10-11 = 11