Download Water

David L. Nelson and Michael M. Cox
Lehninger Principles of
Biochemistry
Fourth Edition
Chapter 2:
Water
Copyright © 2004 by W. H. Freeman & Company
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
[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.


O
H
O
H
o
104.5 C

 H
H 
 H2O interacts through hydrogen bonds.
Water
MeOH
b.p. (oC)
100
65
BuOH
Butane
117
–0.5
1 Å (Angstrom)
= 10‒8 cm
= 0.1 nm
m.p.
0
‒98
Hydrogen 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)
 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.
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).
 Common H-bonds in biological systems:
 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.
Polar
Nonpolar
Amphipathic
phospholipids
(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)
 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.
 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)
Formation of micelles
Not favorable
G > 0
Favorable
G < 0
Hydrophobic Interactions
(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.
“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.
Hydrophobic Interactions
Hexane
O
O
O
O
O
O
O
O
O
O
O
G > 0
S < 0
O
H2O
O
O
O O O
O O
Hexane
O
G < 0
S > 0
O
O
O
O
O
O O O
O O
O
H2O
Spontaneous
The transfer of nonpolar solutes from non-aq. solvent to water.
Note that water molecules surrounding the non-polar solutes are
highly ordered (red).
O
O
O
O
O
O
O
H2O
OO
O
Add oil drops
O
O
Step 1
Less ordered bulky water solvent
Supose there are 11 water molecules,
all disordered.
O
O
O
O
O
O
O
O
O
Step 2
The rests (2 molecules) are
still disodered
O
Highly ordered water molecules
(9 molecules) surround oil drops.
When the oil drops come closer,
they will spontaneouly collapse onto
themselves to become a larger drop.
We know this from experience.
Step 3
Now, 3 previously orderd water molecules
become disordered, increasing "disorder"
in the system (S > 0), and leading to
a state of lower G (G < 0).
O
O
O
O
O
O
O
O
O
O
O
Hydrophobic interactions are Important in protein folding,
biomembrane organization, and enzyme-substrate binding.

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.
O
O
Phe
O
O
Leu
O
folding
O
O
O
S > 0
Leu Phe
O
O
O
(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
② Dipole-dipole interactions:
Intermolecular, attractive force arising from permanent dipoles.
ethane
b.p. 184.5 K
+
–
fluoroethane
b.p. 194.7 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.
(1.6) Noncovalent
interactions:
-individually weak, but
their cumulative effects
are critical in biological
functions.
(1.7) Water can be part of a protein’s structure
Tightly bound water molecules (red spheres) in hemoglobin.
Water can play an important role in protein function.
Water chain
in cytochrome f.
Protons move across
the membrane,
probably through
“proton hopping.”
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
(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. 
 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 (0.08206 L•atm/K
•mol)
T – absolute temperature (K)
Osmotic Concenttration (Osmolarity): The osmotic concentration of a
solution expressed as osmoles of solute per liter of solution.
• Osmolarity is dependent on the
number of particles in solution
but independent of the nature
of the particles.
The number of osmotically active
particles in solution
1 mM NaCl
• Osmolarity of a simple solution
= (molarity) X
(the number of particles per molecule)
• 1 M glucose solution
1 M NaCl
1 M Na2SO4
= 1 OsM
= 2 OsM
= 3 OsM
Total osmolarity = 2 mOsM
isotonic
hypertonic
hypotonic
• Water moves fairly freely across
the cell membrane.
• Not all molecules can cross the
membrane
(semi-permeable).
Examples of Osmosis
1.
2.
3.
4.
5.
Fresh water fish in salt water
A cell in distilled water
Blood
Bacteria & salt
Wilted lettuce
Q: Why do the cells store energy as polysaccharides,
not as individual glucose molecules?
[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)
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 10-16 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.
(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
• 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!
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
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+] = 10-14 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
= 0.769....  10‒10 M
= 7.7  10‒11 M
(Notice the number of significant figures.)
pH meter: History and How it works
http://center.acs.org/landmarks/landmarks/phmeter/phmeter.html
(2.3) Weak Acids and Weak Bases
 Acids – proton (H+) donors; Bases – proton acceptors
HA ⇄ H+ + A‒ (A‒ : conjugate base)
 Keq = Ka (dissociation constant) =
[H+] [A‒]
[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.
(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‒]
* 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
* 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.
Living Graph
Titration Curve for a Weak Acid
(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
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‒]
(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],
When [A‒] < [HA],
pH > pKa.
pH < pKa.
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.
Living Graph
Henderson-Hasselbalch Equation (Box 2.3)
(2.7) Polyprotic Acids:
Substances that have more than one acid/base group.
H3PO4 ⇄ H2PO4‒ + H+ ⇄ HPO42‒ + H+ ⇄ PO43‒ + H+
pKa1 = 2.14
pKa2 = 6.86
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: 1 mol of H3PO4 + 1.75 mol OH- 
1 mol H2PO4- + 1 mol H2O + 0.75 mol OHStep 2: 1 mol of H2PO4- + 0.75 mol OH- 
0.25 mol H2PO4- + 0.75 mol HPO42- + 0.75 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 = 0.75 M
Therefore: pH = pKa+ log[A-]/[HA]
= 6.86 + log((0.75 M)/(0.25 M)) = 7.34
Practice Questions
1. At 37oC, ion product for water (Kw) has a value
of 2.4 x 10-14.
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-/HPO42pKa = 7.21
HCO3-/CO32pKa = 10.0
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 0.010 mol of lactic acid (pKa=3.86) and 0.050 mol of
sodium lactate per liter.
(a) Calculate the pH of the buffer.
(b) Calculate the change in pH when 5 mL of 0.5 M HCl is added to 1 L
of the buffer.
(c) 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).
(2.8) Biological buffers:
HCO3‒ : H2CO3 pKa = 6.35
HPO42‒ : H2PO4‒ pKa = 6.86
pKa of amino acid, histidine = 6.0
Common buffers used in lab:
Buffer
pKa (25oC)
succinate (pK1) 4.21
acetate
4.76
citrate (pK2)
4.76
malate (pK2)
5.13
succinate (pK2) 5.64
MES
6.10
carbonate (pK1) 6.35
citrate (pK3)
6.40
imidazole
6.95
MOPS
7.14
phosphate (pK2) 7.20
HEPES
7.48
Trizma (Tris)
8.06
glycine (pK2)
9.78
carbonate (pK2) 10.33
3.2-5.2
3.6-5.6
5.5-6.5
6.0-8.0
5.8-8.0
6.8-8.2
9.5-11.1
Effective pH Range
3.0-6.2
4.0-6.0
5.5-6.7
5.5-7.2
6.2-7.8
6.5-7.9
7.5-9.0
8.8-10.6
Bicarbonate Buffer System in Blood and Lung
* 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.
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
1/10
log(1/10) = log10 = 1
4.76  1 =
3.76
1/4
log(1/4) = log4 = 0.602
4.76  0.602 = 4.16
1/2
log(1/2) = log2 = 0.301
4.76  0.301 = 4.46
3/4
log(3/4) =
0.125
4.76  0.125 = 4.64
1.0
log1
=
0
4.76
4/3
log(4/3) =
0.125
4.76 + 0.125 = 4.89
2/1
log2
=
0.301
4.76 + 0.301 = 5.06
4/1
log4
=
0.602
4.76 + 0.602 = 5.36
10/1
log10 =
1.0
4.76 + 1 =
5.76
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,
(1) 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.
(2.9) Water as a Reactant
Hydrolysis (exergonic)
are catalyzed by hydrolases
( Condensation)
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