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