Transcript Lh6Ch02Water

Chapter 2 : Water
Learning goals:
• What kind of interactions occur between
different type of molecules in water.
• Why water is a good medium for life
• Why nonpolar moieties aggregate in water
• How dissolved molecules alter properties of
water
• How weak acids and bases behave in water: to
be able to solve weak acid problems with the
Henderson-Hasslebalch equation.
• How buffers work and why we need them
• How water participates in biochemical reactions
Structure
ICE
Common H-bonds in Biochemistry
Some Biologically Important H-bonds
H-bond Strength and Alignment
Importance of Hydrogen Bonds
•
•
•
•
•
•
•
Source of unique properties of water
Structure and function of proteins
Structure and function of DNA
Structure and function of polysaccharides
Binding of substrates to enzymes
Binding of hormones to receptors
Matching of mRNA and tRNA
“I believe that as the methods of structural chemistry are further applied to
physiological problems, it will be found that the significance of the hydrogen
bond for physiology is greater than that of any other single structural feature.”
–Linus Pauling, The Nature of the Chemical Bond, 1939
Water as a Solvent
• Water is a good solvent for charged and polar
substances
– amino acids and peptides
– small alcohols
– carbohydrates
• Water is a poor solvent for nonpolar substances
– nonpolar gases
– aromatic moieties
– aliphatic chains
Solvation and Hydration Spheres
Flickering Clusters and Clathrate Cages
The Hydrophobic Effect
• Refers to the association or folding of
nonpolar molecules in the aqueous solution
• Is one of the main factors behind:
– protein folding
– protein-protein association
– formation of lipid micelles
– binding of steroid hormones to their receptors
• Does not arise because of some attractive
direct force between two nonpolar
molecules
Substrates Must Displace Water to
Bind Enzymes

Approximate Bond
Strength, kJ/mole
Distance,
nm
12-30
0.3
20
0.25
<40
-
0.4 – 4.0
0.2
Water Bound to Hemoglobin
Hb purified from water
Hb with Water Removed
Proton Hop and Hydronium
Water Bound in a Protein Channel
(Cytochrome f)
Facilitates Proton
Hopping – see
later in
Photosynthesis
Osmotic Pressure
Cell Response to Osmotic Pressures
Plants Use Osmotic Pressure
Plants Use Osmotic Pressure
Protection Against Wind
Ionization of Water
Keq = [H+][OH-] / [H2O] = 1.8 x 10-16 M
Concentration of water - one liter = 1,000g
Mole Wt Water = 18.015
[H2O] = 55.5 M
Kw = [H+][OH-] = Keq x [H2O] = 1 x 10-14 M2
for pure water [H+] = [OH-]
so, [H+] = 10-7 M
pH is negative log [H+] , for pure water = 7.0
Weak Acids
HA
↔H
+
+ A-
K e = [H+][A-] / [HA] = Ka
Henderson-Hasselbalch Equation Rearranges Ka
pH = pKa + log ( [A-] / [HA] )
when pKa = pH … [A-] = [HA]
A
Weak acids have different pKas
Enzymes have pH optima Related to their Function
Water as a Reactant
Problem 18 in Chapter 2
1 liter of 0.1 M glycine.
a. what pH’s is glycine a good buffer due to its amino
group:
Problem 18 in Chapter 2
1 liter of 0.1 M glycine.
a. What pH is glycine a good buffer due to its amino
group:
NH3+
Glycine = CH2-COOpKa’s = 2.34 and 9.6
So it would be good +/- 1.0 from each pKa
which would be from 1.3 to 3.3 and 8.6 to 10.6 for the
amino group.
Problem 18 in Chapter 2
NH3+
Glycine = CH2-COO-
I liter of 0.1 M glycine.
pKa’s = 2.34 and 9.6
b. in a 0.1 M solution, pH 9.0 what fraction has the
amino group as –NH3+ ?
Example of a Clicker Question: R-NH3+ is
A. HA
B. AC. H2O
D. H+
E. OH-
Problem 18 in Chapter 2
NH3+
Glycine = CH2-COOpKa’s = 2.34 and 9.6
I liter of 0.1 M glycine.
b. in a 0.1 M solution, pH 9.0 what fraction has the
amino group as –NH3+?
pH = pKa + log A/HA
9.0 = 9.6 + log A/HA
log A/HA = -0.6
HA + A = 0.1M
 A/HA = 0.25  0.25HA = A
so HA + 0.25HA = 0.1 M :: 1.25HA = 0.1M
so HA = 0.08 M…and that is 80% of 0.1M so 
not asked: A = 0.02 M or 20%
Problem 18 in Chapter 2
NH3+
Glycine = CH2-COOpKa’s = 2.34 and 9.6
I liter of 0.1 M glycine.
c. How much 5M KOH is needed to change pH from 9
to 10 for 1 Liter of 0.1M glycine?
Problem 18 in Chapter 2
NH3+
Glycine = CH2-COOpKa’s = 2.34 and 9.6
I liter of 0.1 M glycine.
c. How much 5M KOH is needed to change pH from 9
to 10 for 1 Liter of 0.1M glycine?
pH = pKa + log A/HA
10 = 9.6 + log A/HA
so: log A/HA = 0.4
thus A/HA = 2.5  2.5HA = A
HA + A = 0.1 M 
HA + 2.5 HA = 0.1 M  3.5HA= 0.1M
so HA at pH 10 = 0.029 moles/L
from pH 9 HA is converted to A by adding OH-, that is HA is
lowered from 0.08M to 0.029M or a change of 0.051 moles
0.051 moles/5 moles/L = 0.01 L  10 ml of 5M KOH
Problem 18 in Chapter 2
NH3+
Glycine = CH2-COO-
I liter of 0.1 M glycine.
pKa’s = 2.34 and 9.6
d. When 99% of glycine is in its –NH3+ form, what is the
pH of solution due to it’s amino group? (functionally
reworded from the text)
pH = pKa + log A/HA so this is easy HA dominates, so it will
be on the acid side of the pKa. A is only 1% or 0.01 so the
log of A/HA ≈ -2
thus pH = 9.6 – 2 = or 7.6
Things to Know and Do Before Class
1. General Chemical Properties of Water.
2. pH definition and what it means+how to calculate it.
3. Strong vs Weak Acids.
4. Henderson-Hasselbalch Equation and how to do
calculations with it.
5. Weak bonds and their relative bond strength.
6. Make sure you are able to do EOC Problems
calculating pH (2-5, 8), pH affects solubility (14)
and uptake of aspirin (15) and rest on buffers (11):
They are part of Class Clicker Questions and Case
Study (aspirin).