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Chapter 2 Water and Chemical
Bonds in Biochemistry
2-1 Water and hydrogen bond
Water
(H2O)
Due to the dipolar nature,
there is an electrostatic attraction
between the oxygen atom
of one water molecule and
the hydrogen of another.
This attraction
results in a
hydrogen bond.
In ice each water molecule
forms 4 hydrogen bonds.
In liquid water each
water molecule forms
about 3.4 hydrogen bonds.
Certain other molecules can form hydrogen bonds.
Needed is:
A hydrogen acceptor (oxygen or nitrogen with lone pair of e-).
A hydrogen donor (hydrogen atom covalently bonded to
oxygen or nitrogen)
Some biologically important hydrogen bonds
A linear hydrogen bond is the strongest
Intermolecular Forces
Table of Force Energies
Type of Force
Energy (kJ/mol)
Ionic Bond
300-600
Covalent
200-400
Hydrogen Bonding
20-40
Ion-Dipole
10-20
Dipole-Dipole
1-5
Instantaneous Dipole/
Induced Dipole
0.05-2
2-2 Hydrophilic and Hydrophobic
properties
Hydrophilic molecules
Water is a polar solvent.
Many biomolecules are charged or polar and readily
dissolve in water. They are hydrophilic.
DNA
Examples of polar groups in biomolecules
Hydrophobic molecules
Biomolecules are soluble in nonpolar solvents
such as chloroform or benzene.
They are hydrophobic.
Example: waxes, hydrocarbon chains of lipids.
Examples of nonpolar groups in biomolecules
Beeswax
Amphipathic molecules
Biomolecules that have both hydrophilic and
hydrophobic regions are termed amphipathic.
Examples of amphipathic biomolecules
Lipids self-aggregate
in aqueous solution
Lipids self-aggregate in aqueous solution
Lipids self-aggregate
in aqueous solution
Lipids self-aggregate
in aqueous solution
Enzyme – substrate interaction releases
ordered water molecules
and increases entropy.
Enzyme – substrate interaction releases ordered
water molecules and increases entropy
Solubilities of Some Gases in Water
Four types of
noncovalent
“weak”
interactions among
biomolecules in
aqueous solvent
2-3 Osmosis
Osmotic
pressure
H2O
flows in
Osmosis – water movement across a cell membrane
driven by differences in osmotic pressure.
Osmotic pressure – Force that results from water entering
a cell that is in a hypotonic solution.
Isotonic – solutions of equal osmolarity (solute concentration).
Hypertonic – A cell is in a hypertonic solution if the medium
has a higher osmolarity than the cytosol.
Water flows out of the cell, the cell shrinks.
Hypotonic - A cell is in a hypotonic solution if the medium has
a lower osmolarity than the cytosol.
Water flows into the cell, the cell swells.
How do cells compensate for living in a hypotonic solution?
1. Some cells have an outer cell wall that is rigid and prevents
cell rupture (lysis). Examples: Plants, bacteria.
2. Some organisms have a special organelle called a contractile
vacuole that serves as a water pump.
Example: Protozoans.
3. Cells can compensate by pumping sodium ions (Na+) out
of the cell. Water will leave the cell with the sodium
to maintain osmotic balance.
2-4 Ionization of water, weak
acids, and weak Bases
A. Pure Water Is Slightly Ionized
H3O+
H+
H2O
H+
OH-
B Equilibrium Constant
H2O
Equilibrium
= Keq =
constant
Ion product
of water
=
OH- + H+
[OH-] [H+]
= 1.8 x 10-16 M
[H2O]
Kw = [OH-] [H+] = 10-14 M2
[H2O]=55.5 M
Pure H2O : [H+] = [OH-] = 10-7 M
pH = - log [H+] = -log (10-7) = 7
If [H+] > 10-7 M then pH < 7 (acidic)
If [H+] < 10-7 M then pH > 7 (basic)
B Equilibrium Constant
Blood: [H+] = 4 x 10-8 M
Blood pH = 7.4
C. The pH Scale
D. Acids and Bases
Acid (HA) - something that has a proton and is willing to give it up.
Base (A-) - something that has a place to put a proton
HA
H + + A-
HA + H2O
H 3O+ + A-
Strong acids completely dissolve in H2O (H+Cl-).
Weak acids don’t completely dissolve in H2O.
[A-] [H+]
K=
O
[HA]
H2PO4(acid)
O
HPO42- + H+
(base)
P
OH
K=
[HPO42-] [H+]
[H2PO4-]
pK = - log (K) = 6.82
= 1.51 x 10-7 M
OH
E. Relationship between pH and pK
pH = pK + log (A- / HA)
(Henderson - Hasselbalch equation)
pH = pK when:
The molar concentration of acid and
conjugate base are equal.
[H2PO4-] = [HPO42-]
pH = pK = 6.8
Henderson – Hasselbalch equation
Buffers
Buffers are solutions that contain both the acidic and basic forms of a weak acid.
Buffers minimize changes in pH when strong acids (H+) and bases (OH-)are added.
H2O
+10-3 M HCl
[H+] = 10-3 M
[Cl] = 10-3 M
pH = 3.0
+10-3 M HCl
0.101 M H2PO40.099 M HPO42[H+] = 10-6.79 M
pH = 6.79
[H+] = 10-7 M
pH = 7.0
0.1 M H2PO40.1 M HPO42[H+] = 10-6.8 M
pH = 6.8
O
-O
P
OH
O
O-
+10-3 M HCl
-O
P
OH
OH
Buffering Capacity
Buffering capacity depends on the concentration of the buffering agent.
0.1 M H2PO40.1 M HPO42[H+] = 10-6.8 M
pH = 6.8
+10-3 M HCl
0.101 M H2PO40.099 M HPO42[H+] = 10-6.79 M
pH = 6.79
0.002 M H2PO40.002 M HPO42[H+] = 10-6.8 M
pH = 6.8
+10-3 M HCl
0.003 M H2PO40.001 M HPO42[H+] = 10-6.32 M
pH = 6.34
O
-O
P
OH
O
O-
+10-3 M HCl
-O
P
OH
OH
H2PO4(acid)
HPO42- + H+
(base)
O
O
[HPO42-] [H+]
Ka =
= 1.51 x 10-7 M
[H2PO4-]
P
OH
OH
[0.001 M] [H+]
= 1.51 x 10-7 M
[0.003 M]
[H+] = [(1.51 x 10-7 M)*(0.003 M)]/(0.001 M) = 4.53 x 10-7M
pH = -log [H+] = -log(4.53 x 10-7) = 6.34
K=
[HPO42-] [H+]
[H2PO4-]
K = 1.51 x 10-7 M
pK = - log (K) = 6.8
6.8
H2PO4(acid)
HPO42- + H+
(base)
O
O
P
OH
OH
Phosphate buffers act in the cytoplasm of cells,
keeping the pH inside a cell near 7.
Many biomolecules inside a cell possess ionizable groups that
are weak acids and act as intracellular buffers.
The side chain of the amino acid histidine contains an imidazole
functional group with a pK ~ 6.
Enzymes, protein molecules that serve as biological catalysts,
have a characteristic pH optimum
Physiological pH
The pH in the human body needs to remain ~7
Biological molecules are very sensitive to pH.
enzyme catalysis, protein-protein interactions, receptor binding
Acid-Base balance of the blood is maintained using a CO2 - bicarbonate buffer.
CO2 + H2O
(acid)
H2CO3
(hydrated
CO2)
H+ + HCO3(bicarbonate
base)
pKa = 6.1
There is more than 10-fold more base (HCO3-) than acid (CO2) so pH > pK (pH= 7.4)
CO2 is exhaled by the lungs H+ + HCO3-
CO2 + H2O
Breathing rate controls CO2
CO2 balance is controlled by the lungs, HCO3- by the kidneys
H2CO3
(hydrated CO2)
(acid)
H+ + HCO3(bicarbonate
base)
pKa = 6.1
If acid is produced, such as lactic acid in muscle
CH3-CH(OH)-COOH
H+ + CH3-CH(OH)-COOHCO3- + H+
H2CO3
If base is produced, such as NH3 from protein degradation
NH3 + H+
NH4
H2CO3
H+ + HCO3-
Question:
What is the pH of a solution whose
hydrogen ion concentration is 3.2 x 10-4 M?
Answer:
pH = -log[H+]
pH = -log(3.2 x 10-4)
pH = 3.5
Question:
What is the pH of a solution whose
hydroxide ion concentration is 4.0 x 10-4 M?
Answer:
Define pOH = -log [OH-]
Kw = [H+][OH-] = 10-14
log[H+] + log[OH-] = log 10-14
pH + pOH = 14
[OH-] = 4.0 x 10-4
pOH = -log [OH-] = -log(4.0 x 10-4) = 3.4
pH + pOH = 14
pH = 14 – pOH = 14 – 3.4 = 10.6
Work the problems in Box 2-3 page 67
Working in Biochemistry
Work Through Acid – Base Worksheet #1
CH3COOH  CH3COO- + H+
Acid
Base
Ka =
[CH3COO-] [H+]
[CH3COOH]
CH3COOH  CH3COO- + H+
Acid
Base
Ka =
[CH3COO-] [H+]
[CH3COOH]
Ka = 1.74 x 10-5
pKa = -log(1.74 x 10-5) = 4.76
Henderson – Hasselbalch equation
The pH of a solution of acetic acid is 6.22.
What is the ratio of base to acid?
6.22 = 4.76 + log [A-]/[HA]
6.22 – 4.76 = log [A-]/[HA]
1.46 = log [A-]/[HA]
101.46 = [A-]/[HA]
28.8 = [A-]/[HA]
[A-]/[HA] = [CH3COO-]/[CH3COOH]
The concentration of CH3COOH is 0.1 M.
The concentration of CH3COO- is 0.01 M.
What is the pH of the solution?
pH = 4.76 + log [0.01]/[0.1]
pH = 4.76 + log 0.1
pH = 4.76 + (-1)
pH = 4.76 - 1
pH = 3.76