1. Lehninger Principles of Biochemistry
David L. Nelson and Michael M. Cox
Lehninger Principles of Biochemistry
Fourth Edition
Chapter 2:
Water
Copyright © 2004 by W. H. Freeman & Company

2. CONTENTS [1] Weak Interactions in Aqueous Systems
(1.1) Why Study Water?
(1.2) Characteristics of Water
(1.3) Aqueous (Water-based) Solutions
(1.4) Hydrophobic Interactions
(1.5) van der Waals Interactions
(1.6) Noncovalent interactions
(1.7) Water can be part of a protein’s structure
(1.8) Solutes Affect the Colligative Properties of Aqueous Solutions
[2] Ionization of Water, Weak Acids, and Weak Bases
(2.1) Ionization of Water
(2.2) The pH Scale
(2.3) Weak Acids and Weak Bases
(2.4) pKa and Titration Curve
(2.5) Buffers
(2.6) The Henderson-Hasselbalch (H-H) Equation
(2.7) Polyprotic Acids
(2.8) Biological buffers
(2.9) Water as a Reactant

3. [1] Weak Interactions in Aqueous Systems
(1.1) Why Study Water:
Life exists in an aqueous environment.
- Intra- and extra-cellular fluids very similar to sea water.
(1.2) Characteristics of Water
 The water molecule is bent.
 O (3.5) is more electronegative than H (2.2)
 H2O is highly polar with a net dipole.
 has 33% ionic character.

4.  H2O interacts through hydrogen bonds. b.p. (oC) m.p. Water 100 0
MeOH ‒98
BuOH
Butane –0.5
1 Å (Angstrom)
= 10‒8 cm
= 0.1 nm

5. Hydrogen bond: Can a H atom bonded to C form a H-bond?
A type of dipole force.
An electrostatic attractive force between molecules. Found when a H atom is bonded to N, O or F.
Weaker than covalent bonds (23 kJ/mol in water).
C—C: 350 kJ/mol, O—H: 470 kJ/mol.
Can a H atom bonded to C form a H-bond?
The answer is No (for the moment).
C (2.5) is only slightly more electronegative than H (2.2)

6.  Hydrogen bonding in ice.
Each H2O forms 4 H-bonds.
In liquid water,
each H2O forms, on average,
3.4 H-bonds.
This is why less dense ice
floats in liquid water, and why
water expands when frozen.

7. Directionality of the H-bond:
The bond strength depends not only on the distance between
the H or X (N or O), but also the angle of the atoms in space.
Thus, Hydrogen bonds can vary in strength from
very weak (1-2 kJ mol−1) to strong (20~30 kJ mol−1).

8.  Common H-bonds in biological systems:

11. Good solvent for polar and ionic material such as salts, small polar molecules (amino acids, sugars, nucleic acids …) and the exterior of proteins.
Solubility enhanced for molecules that can form H-bonds i.e. hydroxy, keto, carboxy, amino groups, and these are termed hydrophilic (water loving) groups.
Solubility reduced for molecules that cannot form H-bonds (i.e. hydrocarbons such as alkanes, alkenes), and these are termed hydrophobic (water fearing) compounds.

12. Nonpolar
Polar
Amphipathic
phospholipids

13. (1.3) Aqueous (Water-based) Solutions
Water is a polar solvent.
Solvation of crystalline salt: Water breaks up the salt crystal
lattice by H-bonding with the individual ions. hydration
High dielectric constant () of water makes
ionic interactions weaker. (F= Q1Q2/ r2)
Entropy-driven: G < 0  H ≥ 0, S ≫ 0 (of ions)
dielectric constant : a quantity measuring the ability of a substance to store electrical energy in an electric field

14. This is why we need a water soluble O2 carrier, hemoglobin.
Non-polar gases (O2, CO2) and nonpolar compounds
(benzene, hexane) are not soluble in water.
H > 0  breaking up the water H-bonds.
S < 0  cage-like, highly ordered H2O molecules
around the hydrophobic compounds.
Overall, G > 0.
This is why we need a water soluble O2 carrier, hemoglobin.

16. Amphipathic (Amphiphilic) compounds contain
both polar (hydrophilic) & non-polar (hydrophobic) regions
(ex. fatty acids)
In aqueous solution: water hydrates the hydrophilic portion
but excludes the hydrophobic region to give micelles and
lipid bilayers in a process that reduces entropy (S)
or increases the order of the lipid molecules, but increases the
entropy of surrounding water (basis of cell membranes)

17. Formation of micelles

18. Not favorable
G > 0

19. Favorable
G < 0

20. Hydrophobic Interactions

21. (1.4) Hydrophobic Interactions
The tendency of hydrophobic (lipophilic) groups to form
intermolecular aggregates in an aqueous medium,
and analogous intramolecular interactions.
The name may be misleading, not repulsion between
water and hydrophobic compounds, per se,
but effects of the hydrophobic groups on the water-water
interactions are the driving force for the hydrophobic interactions.
Hydrophobic interaction is driven by S ≫ 0.
Per se (terminology), a Latin phrase meaning "in itself".

22. “Clathrate” structure
(Ordered water structure surrounding a compound.)
First hydration shell of MeOH/H2O solution.
Neither the negative end nor the positive end of the dipole
wants to be pointed to the non-polar molecule, hence,
the water molecules in the hydration shell tend to be oriented
with their dipole moments oriented tangentially to the non-polar,
neutral molecule, resulting in a cage-like, or 'clathrate' structure
around the non-polar group.

23. Hydrophobic Interactions
The transfer of nonpolar solutes from non-aq. solvent to water.
Note that water molecules surrounding the non-polar solutes are
highly ordered (red).

25.  Hydrophobic interactions are Important in protein folding,
biomembrane organization, and enzyme-substrate binding. 

27. Hydrophobic Interaction and Protein Folding:
Hydrophobic interactions are the most important noncovalent
force that will cause the linear polypeptide to fold into
a compact structure.
However, it is not the interactions between side chains of
hydrophobic amino acids per se (mainly van der Waals)
that induce the strong interaction, but the increase in entropy
gained by the removal of hydrophobic surface area
from ordered solvating water.
The aggregation of the hydrophobic surfaces gives
the tightly packed core of a protein.

28. (1.5) van der Waals Interactions
① Dispersion forces (London forces):
Intermolecular, attractive force arising from temporary dipoles.
b.p. of noble gases:
He ‒269°C
Ne ‒246°C
Ar ‒186°C
Original temporary
dipole
Induced
dipole

29. ② Dipole-dipole interactions:
Intermolecular, attractive force arising from permanent dipoles.
ethane
b.p K + –
fluoroethane
b.p K
Permanent dipole
③ All molecules experience dispersion forces.
④ Dipole-dipole interactions are not an alternative to
dispersion forces. - They occur in addition to them.

30. (1.6) Noncovalent interactions:
-individually weak, but their cumulative effects are critical in biological functions.

31. (1.7) Water can be part of a protein’s structure
Tightly bound water molecules (red spheres) in hemoglobin.

32. Water can play an important role in protein function.
Water chain
in cytochrome f.
Protons move across
the membrane,
probably through
“proton hopping.”

33. Water molecule found at the active site of stilbene synthase.
The trapped water molecule
is involved in the hydrolysis
reaction catalyzed by the
enzyme.
Other water molecules forms
H-bonds network that hold the amino acid residues in correct positions in space.
Active site of Stilbene Synthase

34. (1.8) Solutes Affect the Colligative Properties of Aqueous Solutions:
[H2O]eff is lower in solution than in pure water.
Colligative: depending on the number of particles (as molecules) and not on the nature of the particles
<pressure is a colligative property>
Other colligative properties
of solution:
b.p. 
m.p. 

35.  Osmosis: water moves from a region of high to low water conc.
- Effect can be measured by the van’t Hoff equation:
 =  (incn)RT
 - Osmotic pressure,
force required to resist water movement
i - van’t Hoff factor
- measure of extent to which the solute dissociates into two or more ionic species (a fraction, no units)
2 is the maximal possible value for NaCl
1 is non-ionizing
c - molar concentration of solute (M)
R - gas constant ( L•atm/K •mol)
T – absolute temperature (K)

36. Osmotic Concenttration (Osmolarity): The osmotic concentration of a solution expressed as osmoles of solute per liter of solution.
The number of osmotically active particles in solution
Osmolarity is dependent on the
number of particles in solution
but independent of the nature
of the particles.
1 mM NaCl
Osmolarity of a simple solution
= (molarity) X
(the number of particles per molecule)
1 M glucose solution = 1 OsM
1 M NaCl = 2 OsM
1 M Na2SO4 = 3 OsM
Total osmolarity = 2 mOsM

37. isotonic hypertonic hypotonic
Water moves fairly freely across the cell membrane.
Not all molecules can cross the membrane
(semi-permeable).

38. 1. Fresh water fish in salt water 2. A cell in distilled water
Examples of Osmosis
1. Fresh water fish in salt water
2. A cell in distilled water
3. Blood
4. Bacteria & salt
5. Wilted lettuce
Q: Why do the cells store energy as polysaccharides, not as individual glucose molecules?

41. [2] Ionization of Water, Weak Acids, and Weak Bases (2
[2] Ionization of Water, Weak Acids, and Weak Bases   (2.1) Ionization of Water:
Slight tendency of H2O to undergo reversible ionization to hydrogen ion (proton) and hydroxide ion: H2O ⇄ H+ + OH‒ BUT, free protons do not exist in solution: protons are hydrated to hydronium ions:
H2O + H+ ⇄ H3O+ (very fast due to proton hopping)

43. The ionization is expressed by an equilibrium constant (Keq)
H2O ⇄ H+ + OH‒
[H+][OH‒]
Keq =
[H2O]
At 25 °C, [H2O] = 55.5 M.
∴ (55.5 M)(Keq) = [H+][OH‒] = Kw (ion product of water)
Keq is 1.8 x M as determined by electro-conductivity exp.
Kw = (55.5 M)(1.8 x 10‒16 M) = [H+][OH‒] = 1.0 x 10‒14 M2
At neutral pH, [H+] =[OH‒] and [H+][OH-] = [H+]2
 [H+] = 1.0 x 10‒7 M in pure water at 25°C.
Since Kw is constant,
when [H+] > 10-7 M ⇒[OH‒] < 10-7 M, and
when [H+] < 10-7 M ⇒[OH‒] > 10-7 M.

44. (2.2) The pH Scale 1 pH = log [H+] = ‒ log [H+] pH 7 is neutral,
[ H+] = [OH‒]
pH < 7 is acidic,
[ H+] > [OH‒]
pH > 7 is basic,
[OH‒] > [ H+]
pH + pOH = 14

45. The pH must be controlled in
an organism; where the breakdown
in pH regulation can lead to serious
metabolic disturbances:
The pH of blood is normally kept within 7.35~7.45.
Outside the narrow range, the organism can not function.
The pH of the cytosol of most cells is ~ 7.4, however, in the lysosomal organelles the pH is ~ 5.0. This is the pH at which the degradative enzymes (proteases) of the lysosome function best, and they are actually inactive at cytosolic pH!

46. Negative pH ?
YES
Most substances have a pH in the range 0 to 14, although extremely acidic or extremely basic substances may have pH less than 0 or greater than 14.
An example is acid mine runoff, with a pH = –3.6

47. What is the pH of 0.1 M NaOH solution ? (p.62)
NaOH is a strong base and completely ionized in dilute aq. solution.
[OH‒] = 0.1 M.
From Kw = [H+][OH‒] = 1.0  10‒14 M2,
[H+] = M2/0.1 M = 10‒13 M.  pH = ‒log10‒13 = 13.
Or, from [OH‒] = 0.1 M, pOH = ‒log 10‒1 = 1.
Since pH + pOH = 14, pH of the solution is 13.
What is [OH‒] in a solution with [H+] of 1.3 x 10‒4 M?
[H+] = Kw/[OH‒] = 10‒14 M2/1.3  10-4 M
=  10‒10 M
= 7.7  10‒11 M
(Notice the number of significant figures.)
pH meter: History and How it works

48. (2.3) Weak Acids and Weak Bases
Acids – proton (H+) donors; Bases – proton acceptors
HA ⇄ H+ + A‒ (A‒ : conjugate base)
[H+] [A‒]
Keq = Ka (dissociation constant) =
[HA]
Stronger acids  larger Ka  lower pKa (‒ logKa)
⇒ phosphoric acid (H3PO4): pKa = 2.34;
monohydrogen phosphate (HPO4‒2): pKa = 12.4
Conjugate base of a strong acid is a weak base.

50. (2.4) pKa and Titration Curve
- How do pH values of an acetic acid solution
vary with added [OH‒]?
a titration curve
- Constructed by:
a) experiment
b) H-H equation
midpoint pH of titration = pKa of corresponding acid:
b/c pH = pKa
when [HA] = [A‒]

51. * Slope lower near midpoint
When [HA] = [A‒],
pH is relatively insensitive to
addition of strong acid or base i.e. buffered solution
Buffering capacity is maximal when pH = pKa.
The useful range of a buffer is within one pH unit of its pKa.
Above this, the pH will change rapidly.
Buffer = Pka ± 1

52. * Slope lower near midpoint
When [HA] = [A‒],
pH is relatively insensitive to
addition of strong acid or base i.e. buffered solution.
Buffering capacity is maximal when pH = pKa.
The useful range of a buffer is within one pH unit of its pKa.
Above this, the pH will change rapidly.

53. Titration Curve for a Weak Acid
Living Graph
Titration Curve for a Weak Acid

54. (2.5) Buffers Buffer: A system whose pH changes only slightly
when small amounts of acid or base is added.
A buffer ordinarily consists of a weak acid and
its conjugate base, present in roughly equal amounts
(at pH = pKa of the acid)
Used to control the pH within a system

55. [buffer] >> added [H+]
How buffer works? (Fig 2-19)
Buffer works because the added H+ or OH‒ ions
are consumed and do not directly affect the pH.
HAc + OH‒ ⇄ H2O + Ac‒
Ac‒ + H+ ⇄ HAc
[buffer] >> added [H+]
or [OH‒]

56. (2.6) The Henderson-Hasselbalch (H-H) Equation
The pH of a solution, and the concentration of an acid and its conjugate base are related by the H-H equation:
[A‒]
pH = pKa + log
[HA]
When the molar concentration of an acid (HA) and its conjugate base (A‒) are equal ([A‒] = [HA]),
[A‒]/[HA] = 1; and log[A‒]/[HA] = log1 = 0
So the pH of the solution simply equals the pKa of the acid.
When [A‒] > [HA], pH > pKa.
When [A‒] < [HA], pH < pKa.

57. Derivation of the Henderson-Hasselbalch (H-H) Equation
From: Ka = [H+][A‒]/[HA], Solve for [H+], [H+] = Ka [HA]/[A‒]
Negative log of each side:
–log [H+] = –log Ka – log([HA]/[A‒])
Convert to p scale: pH = pKa – log([HA]/[A‒])
Invert log: pH = pKa + log([A‒]/[HA])
[proton acceptor]
pH = pKa + log
[proton donor]
Calculate the pH of a 2 L solution containing 10 mL of 5 M acetic acid (CH3COOH) and 10 mL of 1 M sodium acetate (CH3COONa). pKa of CH3COOH = 4.76.

58. Henderson-Hasselbalch Equation (Box 2.3)
Living Graph
Henderson-Hasselbalch Equation (Box 2.3)

59. (2.7) Polyprotic Acids:
Substances that have more than one acid/base group.
H3PO4 ⇄ H2PO4‒ + H+ ⇄ HPO42‒ + H+ ⇄ PO43‒ + H+
pKa1 = pKa2 = pKa3 = 12.4
Example: 1.00 mole of phosphoric acid (H3PO4) and 1.75 moles of NaOH are added to 1 L of water. Calculate the pH.
Step 1: mol of H3PO mol OH- 
1 mol H2PO mol H2O mol OH-
Step 2: 1 mol of H2PO mol OH- 
0.25 mol H2PO mol HPO mol H2O
Step 3: In the end, we have 0.25 moles of H2PO4- and 0.75 moles of HPO42-,
we can calculate the pH using the H-H equation:
Step 4: Look up the pKa of the reaction:
pKa for H2PO4- ⇄ HPO42- + H+, is 6.86
Step 5: Calculate [HA] = [H2PO4-]: 0.25 mol/1 L = 0.25 M
Step 6: calculate [A-] = [HPO42-] : 0.75 mol/1 L = M
Therefore: pH = pKa+ log[A-]/[HA]
= log((0.75 M)/(0.25 M)) = 7.34

60. Practice Questions
At 37oC, ion product for water (Kw) has a value
of 2.4 x
What is the pH of pure water at 37oC? (6.81)
2. Hydroxide ions were released during an enzyme reaction performed at pH 6.8.
Circle the buffer you would select for the enzyme assay.
CH3COOH/CH3COO- pKa = 4.76
H2PO4-/HPO42- pKa = 7.21
HCO3-/CO32- pKa = 10.0

61. Example: The addition of a 0
Example: The addition of a 0.01 mL drop of 1 M HCl to 1 L of water will change the pH from 7 to 5. A small concentration of buffer can alter this so that there is virtually no change in the pH, even with much larger amounts of acid added!
Q13, p.74.
A buffer contains mol of lactic acid (pKa=3.86) and mol of sodium lactate per liter.
Calculate the pH of the buffer.
Calculate the change in pH when 5 mL of 0.5 M HCl is added to 1 L of the buffer.
What pH change would you expect if you added the same quantity of HCl to 1 L of pure water (pH=7).
Work on Q1, 2, 3, 5(c), 8, 9, 10, 11, 12, 14, 15, 16 (p.73-4).

62. (2.8) Biological buffers:
HCO3‒ : H2CO3 pKa = 6.35
HPO42‒ : H2PO4‒ pKa = 6.86
pKa of amino acid, histidine = 6.0

63. Common buffers used in lab: Buffer pKa (25oC) Effective pH Range
succinate (pK1)
acetate
citrate (pK2)
malate (pK2)
succinate (pK2)
MES
carbonate (pK1)
citrate (pK3)
imidazole
MOPS
phosphate (pK2)
HEPES
Trizma (Tris)
glycine (pK2)
carbonate (pK2)

64. Bicarbonate Buffer System in Blood and Lung

65. * Slope lower near midpoint
When [HA] = [A‒],
pH is relatively insensitive to
addition of strong acid or base i.e. buffered solution
Buffering capacity is maximal when pH = pKa.
The useful range of a buffer is within one pH unit of its pKa.
Above this, the pH will change rapidly.

66. Why does the titration curve of a week acid look the way it looks?
Using the H-H equation, let’s follow the change of pH as we increase the ratio of [acid]/[base].
[CH3COO]
pH = pKa + log
[CH3COOH]
[CH3COO]/[CH3COOH] log([CH3COO]/[CH3COOH]) pH
1/ log(1/10) = log10 =   1 =
1/ log(1/4) = log4 =   = 4.16
1/ log(1/2) = log2 =   = 4.46
3/ log(3/4) =   = 4.64
log =
4/ log(4/3) = = 4.89
2/ log = = 5.06
4/1 log = = 5.36
10/1 log10 = =
and, so on....
From the H-H eq. we know that the pH is related to the ratio of [acid] and [conjugate base].
Since the relationship is logarithmic,
pH decreases as the ratio approaches to 1.0 (titration midpoint), and
(2) the titration curve is symmetrical on both sides of the midpoint.
Conclusion: A solution made of a weak acid and its conjugate base has its highest buffer
capacity at pH equal to its pKa.

67. (2.9) Water as a Reactant Hydrolysis (exergonic)
Hydrolysis (exergonic)
are catalyzed by hydrolases
( Condensation)

68. Assignment Derive the Henderson-Hasselbalch (H-H) Equation
Draw titration curve of Glutamate and label it
What is importance of Hydrophobic interactions?
Note: prepare hand written assignment on assignment pages