maths and chemistry for biologists. chemistry 4 buffers this section of the course covers – buffer...
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Maths and Chemistry for Biologists
Chemistry 4Buffers
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?
1) CH3COOH CH3COO + H
Consider acetic acid/sodium acetate as an example
In solution we have the following two processes:
This equilibrium lies far to the left
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
2) CH3COONa CH3COO + Na
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 equationSuppose 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
HA
]][H[A K
-
a
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
a]b[H
Ka
Take logs of both sides
ab
log]Hlog[ a
]b[H log K log a
[acid][salt]
log K log- ][H log- a so
[acid][salt]
log pK pH a Hence
H-H contd
[acid][salt]
log pK pH a
So for example if we make a buffer consisting of 0.075 M acetic acid (pKa = 4.76) and 0.025 M sodium acetate
28.4 0.075
0.025 log 4.76 pH
Example calculations
[acid][salt]
log pK pH a
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
24.5 0.075-0.100
0.075 log 4.76 pH
Examples contd [acid][salt]
log pK pH a
b - 0.100b
log 4.76 5.00
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
Hence 1.74 b - 0.100
b and 24.0
b - 0.100b
log
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
[acid][salt]
log pK pH a
[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.
3.46 050.0
0.025log 3.76 pH
Buffers are not perfect
4.76 0.05 -0.10
0.05log 4.76 pH
Consider a buffer made from 0.10 M acetic acid plus 0.05 M NaOH
Increase NaOH concentration by 0.01 M. What is the new pH?
4.93 0.06-0.10
0.06log 4.76 pH
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 H2CO3 HCO3- + H+
(lungs) (blood) 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
[albumin] = = 6.02 x 10-4 M or 0.602 mM
[histidine] = 16 x [albumin] = 9.63 mM
500,6640
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
][HisH
[His] log pK pH a
][HisH
[His] log 8 6. 7.4 so and
From this we can calculate that
][HisH 3.98 [His]or 98.3 ][HisH
[His] x
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+] = = 1.93 mM
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
98.463.9
[HisH+] goes up to 3.93 mM
[His] goes down to 6.70 mM
New pH is given by
7.16 2.936.70
log 6.8 pH
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
7.16 2.529
log 6.1 pH
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