Transcript Chapter 2B Lecture

Chap. 2. Water (Part B)
Topics
• Weak interactions in aqueous systems
• Ionization of water, weak acids, and weak bases
• Buffering against pH changes in biological systems
• Water as a reactant
• Fitness of the aqueous environment
for living organisms
Fig. 2-2. Hydrogen bonding in ice.
Ionization of Water
Water molecules have a very slight tendency to ionize to a
hydrogen ion and a hydroxide ion
H2O  H+ + OH-.
Actually, free protons (H+) do not exist in solution, and instead,
hydrogen ions formed in water are immediately hydrated to
form hydronium ions (H3O+).
Through proton hopping (Fig. 2-14), the
short hops of protons between a series
of hydrogen-bonded water molecules
results in a very rapid net movement of a
proton over a long distance. Thus the
movement of hydronium (and hydroxide)
ions in an electric field is extremely fast
compared to the movement of ions such
as Na+ and Cl-.
Equilibrium Constant for H2O Ionization (I)
The equilibrium constant for the reversible ionization of water is
Keq = [H+] [OH-] / [H2O].
In pure water at 25˚C, the concentration of H2O is 55.5 M. On
rearranging, the equation becomes
(55.5 M) (Keq) = [H+] [OH-] = KW
where KW designates the product (55.5 M) (Keq) and is the ion
product of water. Since the value of Keq, determined by
electrical conductivity measurements is 1.8 x 10-16 M, the
equation becomes
KW = [H+] [OH-] = (55.5 M) ( 1.8 x 10-16 M)
or
KW = [H+] [OH-] = 1.0 x 10-14 M2.
Equilibrium Constant for H2O Ionization (II)
The above calculations indicate that the product [H+] [OH-] in
aqueous solution at 25˚C always equals 1 x 10-14 M2. Thus,
when the concentrations of H+ and OH- in solution are equal (a
neutral solution at neutral pH, see below)
[H+] = [OH-] = √KW = √1 x 10-14 M2 = 1 x 10-7 M.
Worked Examples 2-3 and 2-4 from the text will be covered in
class to show how the concentrations of H+ and OH- in solution
are determined using the ion product of water equation
KW = [H+] [OH-] = 1.0 x 10-14 M2.
Worked Example 2-3. Calculation of [H+]
Worked Example 2-4. Calculation of [OH-]
The pH Scale
The ion product of water, KW,
is the basis for the pH scale
(Table 2-6). The pH scale is a
convenient way of designating
the concentrations of H+ and
OH- in solution in the range
between 1.0 M H+ and 1.0 M
OH-. The pH is defined by the
equation
pH = -log [H+].
A precisely neutral solution
occurs when [H+] = [OH-], and
thus when pH = 7.0. Solutions
in which [H+] > [OH-] are acidic
and have pH values less than
7.0. Solutions in which [H+] <
[OH-] are basic and have pH
values greater than 7.0.
pH Values of Some Common Liquids
The pH values of some common
aqueous fluids are shown in Fig.
2-15. Remember that the pH
scale is logarithmic, and solutions
that differ by one pH unit
actually differ by 10-fold in
[H+]. The [H+] of a solution of
known pH is calculated by
rearranging the pH equation to
its exponential form:
[H+] = 10-pH.
The pH of biological solutions is
carefully controlled because pH
strongly affects the structure
and activity of biological
macromolecules, e.g., enzymes.
Dissociation Constants of Weak Acids and
Bases (I)
Unlike strong acids (HCl, H2SO4) and strong bases (NaOH,
KOH), weak acids and bases (e.g., acetic acid (CH3COOH) and
amino acids) are not completely ionized when dissolved in water.
Instead, depending upon the pH, both the proton donor species
(conjugate acid) and proton acceptor species (the conjugate
base) occur together in solution. Each weak acid has a
characteristic tendency to lose its proton in aqueous solution.
The stronger the acid, the greater the tendency to ionize. The
tendency of any conjugate acid (HA) to lose a proton and form
its conjugate base (A-) is defined by the equilibrium constant
(Keq) for the reversible reaction
HA  H+ + AFor which
Keq = [H+][A-]/[HA] = Ka.
Equilibrium constants for ionization reactions are usually called
acid dissociation constants, Ka. (Continued on the next slide)
Dissociation Constants of Weak Acids and
Bases (II)
The higher the value of Ka, the stronger the acid. Biochemists
also commonly convert Ka values to their logarithmic forms, pKa.
pKa = -log Ka.
The lower the value of the pKa, the stronger the acid. pKas
commonly are determined by titration of weak acids. Also note
that the Ka of a weak acid can be calculated from the pKa by
using the exponential form of the pKa equation, namely
Ka = 10-pKa.
Conjugate Acid-base Pairs of Weak Acids
and Bases
The dissociation reactions for a number of common mono-, di-,
and triprotic weak acids and bases are shown in Fig. 2-16. The
pKa values for the dissociation reactions are shown where they
occur along a pH gradient.
Titration Curves and the pKa (I)
Titrations are used to
determine the amount of
weak acid in a solution.
Titrations also yield the pKa
value for a weak acid. In a
titration, the conjugate acid
form (HA) of a weak acid is
stoichiometrically converted
to its conjugate base form
(A-) by the addition of a
strong base. The titration
curve for acetic acid is
shown in Fig. 2-17. As
discussed below, the pKa
value of the acid is
equivalent to the pH
obtained at the midpoint of
the titration curve where
50% of the acid has been
neutralized.
Titration Curves and the pKa (II)
Two reversible equilibria are involved in the titration of a weak acid
such as acetic acid (HAc).
H2O  H+ + OHHAc  H+ + Ac-.
These equilibria simultaneously obey their characteristic equilibrium
equations
KW = [H+] [OH-] = 1.0 x 10-14 M2
Ka = [H+] [Ac-] / [HAc] = 1.74 x 10-5 M.
As OH- is added, it combines with H+ to form water. As free H+ is
removed, HAc subsequently dissociates to satisfy its own
equilibrium equation. The net result is that during the titration,
more and more HAc is converted to Ac-. When an amount of base
stoichiometrically equivalent to the amount of HAc originally in
solution has been added, the titration is complete.
Titration Curves and the pKa (III)
The titration curves of all
weak acids have the same
shape (Fig. 2-18). They only
differ with respect to where
the midpoints of the curves
(pKas) intersect the pH axis
of the graphs. The region in
which buffering ability is the
highest for each weak acid
corresponds to a pH range
where approximately 10% to
90% of the proton donor
species has been neutralized.
This is the same as the pH
range ±1 pH unit above and
below the pKa.
Buffer Systems
Buffers are aqueous solutions that tend to resist changes in pH
when small amounts of strong acid (H+) or base (OH-) are added.
An optimum buffer consists of equal amounts of the conjugate
acid and conjugate base forms of a weak acid in solution. As
shown from titration curves, pH changes are the smallest at the
midpoint of the titration curve (plateau region) where [HA] = [A-]
on addition of acid or base. At this point the pH is equivalent to
the pKa of the weak acid.
Because buffer systems have
both HA and A- species, they
are capable of absorbing
either strong base or strong
acid as illustrated for acetic
acid in Fig. 2-19. Whenever
acid or base is added to the
buffer there is a small
change in the ratio of the
conjugate base and conjugate
acid, and thus a small change
in the pH. The sum of the
buffer components in solution
does not change, only their
ratio changes.
The Henderson-Hasselbalch Equation (I)
The shape of the titration curves for all weak acids is described
by the Henderson-Hasselbalch equation. This equation simply
restates the expression for the ionization of a weak acid, HA, as
shown in the following derivation:
Ka = [H+] [A-] / [HA]
After solving for [H+], the equation becomes
[H+] = Ka [HA] /[A-].
Then take the negative logarithm of both sides
-log [H+] = -log Ka - log [HA] / [A-]
After substituting pH and pKa into the equation, it becomes
pH = pKa - log [HA] / [A-]
Finally, the Henderson-Hasselbach equation results after
mathematical manipulation of the log term on the right, namely
pH = pKa + log [A-] / [HA]
The Henderson-Hasselbalch Equation (II)
Examination of this equation shows that
[A-]/[HA] = 10/1 when pH = pKa + 1
[A-]/[HA] = 1/1 when pH = pKa (midpoint of a titration curve)
[A-]/[HA] = 1/10 when pH = pKa - 1
Remember A- is the conjugate base of HA, and HA is its
conjugate acid. The Henderson-Hasselbalch equation allows us
to calculate (1) pKa, given pH and the molar ratio of proton
acceptor and donor in solution; (2) pH, given pKa and the molar
ratio of proton acceptor and donor; and (3) the molar ratio of
proton acceptor and donor, given pH and pKa.
The Histidine R Group Buffer System
Cells and extracellular fluids contain a high concentration of
proteins. Histidine is an amino acid that occurs in most proteins.
The R group (side-chain) of histidine (see Chap. 3) has an
imidazole functional group that undergoes reversible protonation
(Fig. 2-20). Because the pKa of the imidazole R group is 6.0,
histidine residues in proteins help buffer the pH of the cell
cytosol and extracellular fluids around neutrality. The cell cytosol
also is buffered by phosphate, which has a pKa of 6.86 (see
below). The blood also is buffered by the bicarbonate system.
Worked Example 2-6. Phosphate Buffers
Worked Example 2-6. Phosphate Buffers
Bicarbonate Buffer System (I)
The bicarbonate buffer system is an effective physiological buffer
near pH 7.4. As shown below, this is because H2CO3 of blood
plasma is in equilibrium with the large reserve capacity of CO2(g)
in the air space of the lungs. This buffer system involves three
reversible equilibria (below), in this case between gaseous CO2 in
the lungs and bicarbonate (HCO3-) in the blood plasma (Fig. 2-21).
H2CO3  H+ + HCO3-
K1 = [H+][HCO3-]/[H2CO3]
CO2(d) + H2O  H2CO3
K2 = [H2CO3]/[CO2(d)][H2O]
CO2(g)  CO2(d)
K3 = [CO2(d)]/[CO2(g)]
Bicarbonate Buffer System (II)
When H+ is added to blood plasma as it passes through peripheral
tissues, Reaction 1 in Fig. 2-21 shifts towards the production of
H2CO3. This in turn increases CO2(d) in the blood (Reaction 2) and
thus increases the partial pressure of CO2(g) in the air space of
the lungs (Reaction 3). The extra CO2 is then exhaled. When H+ is
lost from the blood plasma, the opposite events occurs. More
H2CO3 dissociates into HCO3- and H+ and thus more CO2(g) from
the lungs dissolves in the blood. The rate of inhaling and exhaling,
which is controlled by the brain stem in response to pH changes in
the blood plasma, can quickly adjust these equilibria to keep the
pH of blood nearly constant. The partial pressure of CO2 in the
gas phase is denoted as pCO2.
Bicarbonate Buffer System (III)
At the pH of the blood plasma (7.4) very little H2CO3 is present
in comparison with HCO3- (Note that the pKa for H2CO3 is 3.57
at 37˚C). So why is the HCO3-/H2CO3 buffer system effective at
buffering the blood near pH 7.4? The answer is that there is a
very large reservoir of CO2(d) in the blood which is in rapid
equilibrium with H2CO3 via Reaction 2 in Fig. 2-21.
In clinical medicine it is common to refer to CO2(d) as the
conjugate acid of the bicarbonate buffer system. Further a
simple expression analogous to the HH equation can be derived
(see text) to calculate blood pH from the amounts of CO2(d) and
HCO3- in blood:
pH = 6.1 + log [HCO3-]/(0.23 x pCO2)
In this equation, pCO2 is expressed in kilopascals (kPa; typically
pCO2 is 4.6-6.7 kPa). 0.23 is the corresponding solubility
coefficient for CO2 in aqueous solution. Thus, the term 0.23 x
pCO2 ~ 1.2 kPa. Plasma [HCO3-] is normally about 24 mM keeping
blood plasma pH around 7.4.
The pH Optima of Enzymes
Enzymes typically show maximal
catalytic activity at a
characteristic pH called the pH
optimum (Fig. 2-22). Above and
below the optimum pH, activity
usually declines. For this reason,
it is important that the pH of
biological fluids such as the
cytosol and blood be controlled
within a narrow range. For
example, in untreated diabetes
mellitus, high concentrations of
ketones (weak acids) accumulate
in the blood, reducing its pH.
Severe ketoacidosis leads to
headache, drowsiness, nausea,
vomiting, and diarrhea, followed
by stupor, coma, and convulsions.
These symptoms are caused by
the disruption of proper enzyme
function at low pH.
Water as a Reactant
Water is a direct participant in many biochemical reactions.
When a biomolecule is split apart by water, the reaction is
called a hydrolysis reaction. When a biomolecule is formed from
two components with the elimination of water, the reaction is
called a condensation reaction. The condensation reaction in
which ATP is formed from ADP and inorganic phosphate (HPO42-)
is illustrated in Fig. 2-23. Biopolymers such as proteins,
polysaccharides, and nucleic acids are formed and broken down
by condensation and hydrolysis reactions, respectively.