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Maths and Chemistry for
Biologists
Chemistry 4
Buffers
This section of the course covers –
• buffer solutions and how they work
• the Henderson-Hasselbalch equation and how
to use it to make buffers
• the ability of buffer solutions to resist changes in
pH
• rules for making effective buffers
• buffering of the blood
What are buffers?
Buffers are solutions that resist changes in pH on
addition of acid or base
They consist of either
a weak acid and a salt of that acid
or a weak base and a salt of that base
For example
a solution of acetic acid and sodium acetate
How do they work?
Consider acetic acid/sodium acetate as an example
In solution we have the following two processes:
1)
CH3COOH
CH3COO
+ H
This equilibrium lies far to the left
2)
CH3COONa
CH3COO
+ Na
This equilibrium lies far to the right
Add H+ - process 1) shifts to the left with
CH3COO- provided by process 2)
Add OH- - then OH- + H+  H2O the H+ being
provided by process 1) shifting to the right
Essential facts
Buffers are only effective in the pH range pKa  1
Buffers have their maximum buffering capacity
when pH = pKa
Titration curve for acetic acid
The pH changes rapidly at the beginning and end
but slowly in the middle – this is the buffering range
The Henderson-Hasselbalch equation
Suppose that we have an acid HA concentration a M
and its sodium salt NaA concentration b M
HA
H+ + A- (low degree of dissociation)
NaA  Na+ + A- (complete dissociation)
For the acid dissociation
[A - ][H  ]
Ka 
HA
But [A-] will be almost equal to the concentration of
salt (b M) and HA will be almost equal to the
concentration of acid (a M) so -
H-H contd

b[H ]
Ka 
a
Take logs of both sides

b[H ]
b

log K a  log
 log[ H ]  log
a
a
so
[salt]
- log [H ]  - log K a  log
[acid]
Hence

[salt]
pH  pK a  log
[acid]
H-H contd
[salt]
pH  pK a  log
[acid]
So for example if we make a buffer consisting of
0.075 M acetic acid (pKa = 4.76) and 0.025 M
sodium acetate
0.025
pH  4.76  log
 4.28
0.075
Example calculations
[salt]
pH  pK a  log
[acid]
pH of 0.100 M acetic acid plus 0.075 M NaOH
Here the [salt] is equal to the concentration of
NaOH added (0.075 M) because it will react
completely with acetic acid to make sodium
acetate and the [acid] is the amount of acetic acid
left (0.100 – 0.075 M). So
0.075
pH  4.76  log
 5.24
0.100 - 0.075
[salt]
Examples contd pH  pK a  log
[acid]
What concentration of NaOH must be added to
0.100 M acetic acid to give a pH of 5.0?
Let the concentration of NaOH be b M
b
5.00  4.76  log
0.100 - b
b
b
 0.24 and
 1.74
Hence log
0.100 - b
0.100 - b
This gives b = 0.174 -1.74b and b= 0.064 M
One for you to do
What is the pH if 750 ml of 0.10 M formic acid, pKa
3.76, is added to 250 ml of 0.10 M NaOH to give a
final volume of 1 L of buffer?
Answer
[salt]
pH  pK a  log
[acid]
[salt] is equal to the concentration of NaOH added
i.e. 0.025 M (250 ml added to a final volume of 1L
so there is a dilution of 1 in 4)
[acid] is equal to that left after partial neutralisation
by the NaOH i.e. 0.075 – 0.025 = 0.05 M.
0.025
pH  3.76  log
 3.46
0.050
Buffers are not perfect
Consider a buffer made from 0.10 M acetic acid plus
0.05 M NaOH
0.05
pH  4.76  log
 4.76
0.10 - 0.05
Increase NaOH concentration by 0.01 M. What is
the new pH?
0.06
pH  4.76  log
 4.93
0.10 - 0.06
pH = + 0.17
But they are pretty good!
Take water at pH 7.0 and add NaOH to 0.01M
[OH-] = 0.01 and [OH-][H+] = 10-14 M2
Hence [H+] = 10-14/10-2 = 10-12 M
pH = 12 and pH
=5
Things to remember about buffers
Strong buffers are better than weak ones at
resisting pH change
Buffers work best at pH = pKa
Buffers only work well one pH unit either side of the
pKa, i.e. in the pH range pKa  1
Buffering the blood
Vitally important to keep the pH of the blood constant
at around 7.4. Blood has a way of getting rid of acid
CO2
CO2 + H2O
(lungs)
(blood)
H2CO3
HCO3- + H+
pKa = 6.1
At pH 7.4 the carbonic acid/bicarbonate reaction lies
far to the right ([HCO3-]  30 mM, [H2CO3]  1.5 mM)
Add H+ to the blood – combines with HCO3- to form
H2CO3 which breaks down to CO2 and H2O
(catalysed by carbonic anhydrase) and CO2 is
breathed out
Buffering by proteins
Carbonate/bicarbonate system not a very effective as a
buffer because the pH is too far away from the pKa
Another important buffer in blood is protein
Blood proteins contain a high concentration of the
amino acid histidine the side chain of which has a pKa
of about 6.8
These systems co-operate in resisting pH change and
the carbonate/bicarbonate system reverses the small
changes of pH that do occur
A difficult problem
The concentration of albumin in blood serum is
about 4 g /100 ml and the pH is 7.4
The Mr of serum albumin is 66,500 and each
molecule of the protein contains 16 histidines with
a pKa of 6.8
Calculate the change in pH if 1 mmol of HCl is
added to 1 L of serum assuming that the albumin
histidine residues are the only buffer present
Calculate the change in pH that would occur if the
carbonate/bicarbonate system was the only buffer
Answer
First calculate the concentration of albumin in the
serum and hence the concentration of histidines
4 g/100 ml is 400 g/L. The Mr = 66,500
40
[albumin] =
= 6.02 x 10-4 M or 0.602 mM
66,500
[histidine] = 16 x [albumin] = 9.63 mM
Now calculate the concentrations of neutral and
protonated histidines at pH 7.4 using the
Henderson-Hasselbalch equation.
In this case, because histidine is a base the pKa is
that of the conjugate acid (HisH+) and the neutral
molecule (His) is the equivalent of the salt
[His]
[His]
so pH  pK a  log
and 7.4  6. 8  log

[HisH ]
[HisH  ]
From this we can calculate that
[His]
 3.98

[HisH ]
or [His]  3.98 x [HisH  ]
But we know that the total concentration of both
forms of histidine, [His] + [HisH+] = 9.63 mM
So 3.98 x [HisH+] + [HisH+] = 9.63 mM
[HisH+] =
9.63 = 1.93 mM
4.98
and [His] = 9.63 – 1.93 = 7.70 mM
Now calculate what happens when 1 mmol of H+
is added remembering that the volume is 1 L
[HisH+] goes up to 3.93 mM
[His] goes down to 6.70 mM
New pH is given by
6.70
pH  6.8  log
 7.16
2.93
Change in pH is 7.40 – 7.16 = 0.24
What about if the buffer had been the carbonic
acid/bicarbonate system?
At pH 7.4 [HCO3-]  30 mM, [H2CO3]  1.5 mM)
If we add 1 mM H+ then [H2CO3]becomes 2.5 mM
and [HCO3-] becomes 29 mM. So
29
pH  6.1  log
 7.16
2.5
Hence the change in pH is again 0.24 units. This
means that protein and the carbonate/bicarbonate
system make about equal contributions to the
buffering of the blood