1. Maths and Chemistry for Biologists

2. 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

3. 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

4. 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 CH 3 COO - provided by process 2) Add OH - - then OH - + H +  H 2 O the H + being provided by process 1) shifting to the right

5. Essential facts Buffers are only effective in the pH range pK a  1 Buffers have their maximum buffering capacity when pH = pK a

6. Titration curve for acetic acid The pH changes rapidly at the beginning and end but slowly in the middle – this is the buffering range

7. 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 -

8. H-H contd Take logs of both sides so Hence

9. H-H contd So for example if we make a buffer consisting of 0.075 M acetic acid (pK a = 4.76) and 0.025 M sodium acetate

10. 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

11. 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

12. 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?

13. 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.

14. 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

15. 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 M 2 Hence [H+] = 10 -14 /10 -2 = 10 -12 M pH = 12 and  pH = 5

16. Things to remember about buffers Strong buffers are better than weak ones at resisting pH change Buffers work best at pH = pK a Buffers only work well one pH unit either side of the pK a, i.e. in the pH range pK a  1

17. 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 CO 2 CO 2 + H 2 O H 2 CO 3 HCO 3 - + H + (lungs) (blood) pK a = 6.1 At pH 7.4 the carbonic acid/bicarbonate reaction lies far to the right ([HCO 3 - ]  30 mM, [H 2 CO 3 ]  1.5 mM) Add H + to the blood – combines with HCO 3 - to form H 2 CO 3 which breaks down to CO 2 and H 2 O (catalysed by carbonic anhydrase) and CO 2 is breathed out

18. Buffering by proteins Carbonate/bicarbonate system not a very effective as a buffer because the pH is too far away from the pK a 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 pK a 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

19. A difficult problem The concentration of albumin in blood serum is about 4 g /100 ml and the pH is 7.4 The M r of serum albumin is 66,500 and each molecule of the protein contains 16 histidines with a pK a 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

20. 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 M r = 66,500  [albumin] = = 6.02 x 10 -4 M or 0.602 mM [histidine] = 16 x [albumin] = 9.63 mM

21. 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 pK a is that of the conjugate acid (HisH + ) and the neutral molecule (His) is the equivalent of the salt soand From this we can calculate that

22. 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

23. [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

24. What about if the buffer had been the carbonic acid/bicarbonate system? At pH 7.4 [HCO 3 - ]  30 mM, [H 2 CO 3 ]  1.5 mM) If we add 1 mM H + then [H 2 CO 3 ]becomes 2.5 mM and [HCO 3 - ] 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