Maths and Chemistry for Biologists

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Maths and Chemistry for Biologists
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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

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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
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How do they work?
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
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Essential facts
Buffers are only effective in the pH range pKa ?
1 Buffers have their maximum buffering capacity
when pH pKa
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Titration curve for acetic acid
The pH changes rapidly at the beginning and end
but slowly in the middle this is the buffering
range
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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
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 -
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H-H contd
Take logs of both sides
so
Hence
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H-H contd
So for example if we make a buffer consisting of
0.075 M acetic acid (pKa 4.76) and 0.025 M
sodium acetate
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Example calculations
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
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Examples contd
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
This gives b 0.174 -1.74b and b 0.064 M
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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?
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Answer
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.
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Buffers are not perfect
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?
?pH 0.17
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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
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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

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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
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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
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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
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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

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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
so
and
From this we can calculate that
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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
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HisH goes up to 3.93 mM His goes down to
6.70 mM New pH is given by
Change in pH is 7.40 7.16 0.24
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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
H2CO3becomes 2.5 mM and HCO3- becomes 29 mM.
So
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