1. pH and buffers

2. Definition of pH
pH= - log[H+]
if [H+] = 1mM = 0.001M = 10-3 M,
pH=3

3. Dissociation of H2O, ion product of water (Kw)
H2O is also an acid,
H2O ↔ H+ + OH-
Ka= [H+] [OH-]/ [H2O]
[H2O]=55.5 M (1000 g/l and Mw 18)
Ka=
[H+] [OH-] = Ka [H2O] = Kw (ion product of water)

4. Calculating concentrations of H+ and OH-
[H2O] is high, can be said to be constant throughout ionization, so Kw is a constant: x 55.5 = 1 x [H+][OH-]= If pH=7 then [H+]=10-7 and [OH-] = 10-7

5. HA H+ + A− pKa = -log(Ka) Dissociation of an acid Ka = [H+][A−]/[HA]
Acid dissociation constant
pKa = -log(Ka)
(-) log Ka = (-)log [H+] [A-] / [HA]

6. (-) log Ka = (-)log [H+] [A-] / [HA] If pH = pKa, then pH = -log[H+] = -log [H+][A-] / [HA] [H+] = [H+] [A-] / [HA] 1 = [A-] / [HA] or [HA] = [A-] If pH=pKa, conc of acid is same as base

7. pKa = - log (Ka) ACID-BASE EQUILIBRIA: Strength of an Acid
- Acids vary greatly in strength, resulting in a very large range of Ka values. It is often useful to express their acidity as pKa values:
pKa = - log (Ka)
Strong Acid
HCl H+ +
Cl-
Ka > 103
pKa < -3
Weak Acid
CH3COOH H+ +
CH3COO-
Ka = 1.8 x 10-5
pKa = 4.7
Water is itself a very weak acid
H2O H+ +
OH-
Kw = 1.0 x 10-14
pKw = 14
Source: Stephen Kawai

8. Addition of HCl to water or sodium acetate

9. Henderson-Hasselbalch equation
Ka = [H+] [A-] / [HA] log Ka = log [H+] + log [A-] / [HA] pH = pKa + log [A-] / [HA] [A-] / [HA] = 10 pH-pKa

10. 1 mM phosphate, change pH from 7. 4 to 7
1 mM phosphate, change pH from 7.4 to 7.3, how much acid do you have to add?
Absence of buffer, pH7.4= = 5 x 10-8
pH7.3= = 4 x10-8
Difference 1 x 10-8

11. 1 mM phosphate, change pH from 7. 4 to 7
1 mM phosphate, change pH from 7.4 to 7.3, how much acid do you have to add?
[A-] / [HA] = 10 pH-pKa
[HPO42-] / [H2PO4-] = = =1.55
[HPO42-] = 1.55 x [H2PO4-]
Total concentration is 1mM, so [HPO42-] + [H2PO4-] = 1
[H2PO4-] = 1 – [HPO42-]

12. 1 mM phosphate, change pH from 7. 4 to 7
1 mM phosphate, change pH from 7.4 to 7.3, how much acid do you have to add?
[A-] / [HA] = 10 pH-pKa
[HPO42-] / [H2PO4-] = = =1.55
[HPO42-] = 1.55 x [H2PO4-]
Total concentration is 1mM, so [HPO42-] + [H2PO4-] = 1
[H2PO4-] = 1 – [HPO42-]

13. 1 mM phosphate, change pH from 7. 4 to 7
1 mM phosphate, change pH from 7.4 to 7.3, how much acid do you have to add?
[HPO42-] = 1.55 x (1-[HPO42-]) =
1.55 – 1.55 x [HPO42-]
2. 55 x [HPO42-] = 1.55
[HPO42-] = 1.55 / 2.55 = (mM)
[H2PO4-] = 1 – = 0.392

14. 1 mM phosphate, change pH from 7. 4 to 7
1 mM phosphate, change pH from 7.4 to 7.3, how much acid do you have to add?
Same procedure for pH 7.3 gives [HPO42-] = and [H2PO4-] = 0.448
0.668 – = HPO42- is converted into H2PO4-, using up mM H+ (acid)
With buffer mM
without buffer 1 x 10-8
5600 times more acid consumed.

15. Back to amino acids and peptides

18. Amino acids are joined by peptide bonds

19. N- and C-termini of polypeptides

20. The polypeptide backbone
Main covalent bonding between amino acids

21. Covalent bonding between cysteines

22. Primary structure of bovine insulin

23. Planar character of the peptide bond

24. Explanation for planar character

25. Typical bond lengths within a peptide unit

26. Peptide bonds can be “cis” or “trans”
trans predominates

27. Proline is an exception

28. Rotation is possible around Ca bonds

29. “Dihedral” or “torsion” angles define the degrees of rotation

30. Only some conformations are allowed

31. Amino acid sequence defines 2o and 3o structures

32. For the tutorial quiz week 1:
-be able to do pH problems (simple ones)
-ionization of groups on amino acids
-amino acids-be able to a) recognize structures; b) know names, 1 letter code and 3 letter code; c) know properties of R groups
For the laboratory week 1:
-read the intro to the lab manual
-come with required supplies and clothing
-prepare a “pre-lab” (see page 5 of the manual)