1. CHAPTER 2 Water, pH, and Ionic Equilibria to accompany
Biochemistry, 2/e by
Reginald Garrett and Charles Grisham
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2. Outline 2.1 Properties of Water 2.2 pH 2.3 Buffers
2.4 Water's Unique Role in the Fitness of the Environment

3. Properties of Water
High b.p., m.p., heat of vaporization, surface tension
Bent structure makes it polar
Non-tetrahedral bond angles
H-bond donor and acceptor
Potential to form four H-bonds per water

5. Comparison of Ice and Water
Issues: H-bonds and Motion
Ice: 4 H-bonds per water molecule
Water: 2.3 H-bonds per water molecule
Ice: H-bond lifetime - about 10 microsec
Water: H-bond lifetime - about 10 psec
(10 psec = sec)
Thats "one times ten to the minus eleven second"!

8. Solvent Properties of Water
Ions are always hydrated in water and carry around a "hydration shell"
Water forms H-bonds with polar solutes
Hydrophobic interactions - a "secret of life"

10. Hydrophobic Interactions
A nonpolar solute "organizes" water
The H-bond network of water reorganizes to accommodate the nonpolar solute
This is an increase in "order" of water
This is a decrease in ENTROPY

12. Amphiphilic Molecules
Also called "amphipathic"
Refers to molecules that contain both polar and nonpolar groups
Equivalently - to molecules that are attracted to both polar and nonpolar environments
Good examples - fatty acids

16. Acid-base Equilibria The pH Scale
A convenient means of writing small concentrations:
pH = -log10 [H+]
Sørensen (Denmark)
If [H+] = 1 x M
Then pH = 7

20. Dissociation of Weak Electrolytes
Consider a weak acid, HA
The acid dissociation constant is given by:
HA  H+ + A-
Ka = [ H + ] [ A - ]
____________________ [HA]

21. The Henderson-Hasselbalch Equation
Know this! You'll use it constantly.
For any acid HA, the relationship between the pKa, the concentrations existing at equilibrium and the solution pH is given by:
pH = pKa + log10 [A¯ ]
¯¯¯¯¯¯¯¯¯¯
[HA]

23. Consider the Dissociation of Acetic Acid
Assume 0.1 eq base has been added to a fully protonated solution of acetic acid
The Henderson-Hasselbalch equation can be used to calculate the pH of the solution:
With 0.1 eq OH¯ added:
pH = pKa + log10 [0.1 ]
¯¯¯¯¯¯¯¯¯¯
[0.9]
pH = (-0.95)
pH = 3.81

24. Consider the Dissociation of Acetic Acid
Another case....
What happens if exactly 0.5 eq of base is added to a solution of the fully protonated acetic acid?
With 0.5 eq OH¯ added:
pH = pKa + log10 [0.5 ]
¯¯¯¯¯¯¯¯¯¯
[0.5]
pH =
pH = 4.76 = pKa

25. Consider the Dissociation of Acetic Acid
A final case to consider....
What is the pH if 0.9 eq of base is added to a solution of the fully protonated acid?
With 0.9 eq OH¯ added:
pH = pKa + log10 [0.9 ]
¯¯¯¯¯¯¯¯¯¯
[0.1]
pH =
pH = 5.71

28. Buffers
Buffers are solutions that resist changes in pH as acid and base are added
Most buffers consist of a weak acid and its conjugate base
Note in Figure 2.15 how the plot of pH versus base added is flat near the pKa
Buffers can only be used reliably within a pH unit of their pKa