MITES
Summer 2010
Biochemistry
CLASS READING #1
Prokaryotes – single cellular, simple organisms (e.g. bacteria)
Eukaryotes – complex organisms whose cells contain nuclear envelopes and often
organelles (e.g. fungi, plants, animals)
Plasma Membrane – the thin layer that surrounds cells
Cytoplasm – aqueous suspension inside the plasma membrane.
Ribosomes – 18- to 22-nm particles that are composed of over 50 RNA and protein
molecules. Ribosomes are the sites of protein synthesis
Genome – the complete set of a cell’s chromosomal DNA
Plasmid – a small circular DNA that is independent of chromosomal DNA
Nucleus – the location inside a eukaryotic cell where the genome is contained.
Surrounded by a membrane called the nuclear envelope.
Nucleoid – the location inside a prokaryotic cell where genomic DNA is localized.
There is no barrier between the nucleoid and the cytoplasm
For excellent review checkout:
http://www.microscopy.fsu.edu/cells/bacteriacell.html
http://www.microscopy.fsu.edu/cells/animalcell.html
CHEMICAL BONDING
Covalent bond – bonds formed when two atoms share electrons. Can exist as single, double,
and triple bonds.
Examples: By sharing electrons, the atoms below are able to fill their outermost electron shells
(with two electrons in the case of hydrogen, eight in the case of oxygen).
O
H
H
H
H
Hydrogen (H2)
Water (H2O)
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MITES
Biochemistry
Ionic bond – attraction between oppositely charged molecules
Example: Sodium contains one electron in its outer shell, whereas chlorine is missing one
electron from its outer shell. Sodium donates its outer electron to chlorine, leaving both atoms
with full outermost electron shells. These oppositely charged ions are attracted to one another.
Na
Cl
Sodium
Chlorine
Na+
Cl–
Sodium Chloride
Strength of bonds
Strength
Type of bond (kJ/mole)
Ionic
20-40
Single Covalent 200-500
Double Covalent 500-700
Triple Covalent 800-900
POLARITY OF BONDS
Electronegativity – the tendency of an atom or molecule to draw electrons to itself in a chemical
bond. A table of electronegativities for the first seventeen elements is shown below (the noble
gases He, Ne, and Ar are omitted since they do not participate in bonds). Notice that fluorine is
the most electronegative atom, followed by oxygen.
1
H
2.20
3
Li
0.98
11
Na
0.93
Atomic number
Symbol
Electronegativity
4
5
6
7
8
9
Be
B
C
N
O
F
1.57 2.04 2.55 3.04 3.44 3.98
12
13
14
15
16
17
Mg
Al
Si
P
S
Cl
1.31 1.61 1.90 2.19 2.58 3.16
Polar – may refer to either a bond or a molecule. A polar bond is one in which the electrons are
more closely associated with one atom than the other. A polar molecule is one that contains one
or more polar bonds.
Nonpolar – may refer to either a bond or a molecule. A nonpolar bond is one in which the
electrons are more closely associated with neither of the atoms. A nonpolar molecule is one that
contains no polar bonds.
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MITES
Biochemistry
Dipole moment (δ) – a measure of the polarity of a chemical bond. Polar bonds and molecules
have a dipole moment, whereas nonpolar bonds and molecules have no dipole moment.
Not all bonds can be classified as fully covalent or fully ionic
Most real bonds are somewhere between covalent and ionic bonds and cannot truly be classified
as being in either category. Many molecules encountered in biochemistry contain polar covalent
bonds in which there is a partial transfer of electron density, but not enough for the bond to be
ionic. These bonds fall somewhere in the middle of the polarity spectrum.
Electronegativity difference
The difference in the electronegativities of two atoms allows us to estimate the degree of polarity
that would exist in a bond between those two atoms
Example: The bond in H2 has electronegativity difference of 2.20 – 2.20 = 0.00. This bond is
truly nonpolar since neither atom will draw electrons more closely to itself.
HYDROGEN BONDING
Hydrogen bonding – an intermolecular attraction by which a hydrogen atom bound to an
electronegative atom interacts with the lone electron pair of another nearby electronegative atom.
Hydrogen bonds are represented as dashed lines. The general form for a hydrogen bond is
X−H- - -Y. Since hydrogen bonding typically only occurs with the most highly electronegative
elements, X and Y are usually N, O, or F. X and Y can be the same or different species.
Example 1: Hydrogen bonding in water. The
Example 2: Hydrogen bonding in ammonia
oxygen atom in water has two lone electron
and hydrogen fluoride.
H
H
pairs (represented by • •), giving it a
H F H F
negative dipole moment (δ –). As a result,
H N H N H
each hydrogen atom has a positive dipole
H
moment (δ +). These opposing dipoles
Hydrogen Fluoride (HF)
Ammonia (NH3)
attract one another to form a hydrogen bond.
!–
!–
!+
!+
Example 3: Hydrogen bonding between
O
H
O
ethanol and pyrimidine. Note that X and Y
H
are different species (X = O, Y = N)
H
!+
Hydrogen
Bond
H
!+
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MITES
Biochemistry
C C
CH3 CH2 O H
C
N
C N
Ethanol
Pyrimidine
Since fluorine is typically not seen in biochemistry, all of the hydrogen bonds in this class will
involve either oxygen or nitrogen (and hydrogen of course!). See Figure 4-3 in the reading for
more examples of hydrogen bonds involving only nitrogen and oxygen.
Hydrogen atoms bound to carbon do not participate in hydrogen bonding
As seen on the previous page, C-H bonds are highly nonpolar and have a very small dipole
moment. This prohibits C atoms from participating in hydrogen bonding.
Hydrogen bonding contributes to solubility
Compounds that can form hydrogen bonds with water are generally more soluble in water than
those that cannot form hydrogen bonds.
Example 4: Consider hexane, 3-hexanol, and 2-hexanone (shown below). Hexane contains only
nonpolar C-C and C-H bonds, will not participate in hydrogen bonding, and is poorly soluble in
water. On the other hand, 3-hexanol and 2-hexanone contain polar O-H, C-O, and C=O bonds,
will participate in hydrogen bonding, and are 1,000 times more soluble in water than hexane.
H
H O
H
O H
O H
H
H
O
O
CH3 CH2 CH2 CH2 CH2 CH3
CH3 CH2 CH2 CH CH2 CH3
Hexane
Hydrogen Bonding between
Water and 3-Hexanol
CH3 CH2 CH2 C CH2 CH3
Hydrogen Bonding between
Water and 2-Hexanone
IONIZATION OF WATER
Conjugate acid/base pair – An acid/base pair that differ by a single hydrogen ion, e.g.
CH3COOH / CH3COO¯, HCl / Cl¯, H2CO3 / HCO3¯
Buffer – an aqueous system containing both a weak acid and its conjugate weak base. Such systems will
resist changes in pH and are commonly used in both natural and laboratory biological systems.
Monoprotic acid – an acid that can donate only one H+ ion, e.g. HCl, CH3COOH, HF
Polyprotic acid – an acid that can donate multiple H+ ions, e.g. H2SO4, H3PO4, H2CO3. Diprotic and
triprotic acids are types of polyprotic acids.
Protonate/Deprotonate – to add or remove a H+ ion. Remember that a H+ ion is just a proton.
Ka – the ionization constant of an acid
Kb – the ionization constant of a base
KW – the ionization constant of water = 10-14
Le Chatelier’s Principle
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Biochemistry
When a system at equilibrium is subjected to a stress, the system will respond in such a way to counteract
the stress. Consider the reaction:
A + B
C + D
If the system is at equilibrium and additional B is injected into it, the system will counteract this stress by
consuming B. This reaction will also consume A and produce more C and D until a new equilibrium is
reached.
Equations involving pH, pOH, and KW
pH = -log [H3O+]
pOH = -log [OH−]
KW = [H3O+][OH−] @25°C
= 10-14
-log(KW) = -log([H3O+][OH−]) = -log([H3O+]) - log([OH−]) = pH + pOH @25°C
= 14
Equations involving pKa and pKb
HA + H2O
A− + H3O+
[A ! ][H3 O + ]
Ka =
[HA]
pKa = -log Ka
A− + H2O
HA + OH−
[HA][OH ! ]
[A ! ]
pKb = -log Kb
Kb =
Note that in the equilibrium expressions above [H2O] is excluded since it is highly abundant and
remains relatively unchanged throughout the reactions. Notice also that when pH = pKa,
[H3O+] = Ka and [A−] = [HA]. Similarly, when pOH = pKb, [OH−] = Kb and [A−] = [HA].
[A ! ][H3 O + ] [HA][OH ! ]
·
= [H3O+][OH−] = KW
!
[HA]
[A ]
@25°C
-log(KW) = -log(KaKb) = -log(Ka) - log(Kb) = pKa + pKb = 14
KaKb =
A Note on Naming Acids and Bases
Acids often end with the suffix –ic acid, and their conjugate bases often end with the suffix –ate
Examples: acetic acid / acetate, benzoic acid / benzoate, formic acid / formate
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MITES
Biochemistry
TABLE OF Ka AND pKa VALUES (25°C)
Acid
Hydrochloric
Sulfuric (1)
HCl
H2SO4
Hydronium Ion
H3 O
Sulfuric (2)
HSO4
Phosphoric (1)
H3PO4
Citric (1)
Hydrofluoric
A-
HA
Ka
-
Cl
HSO4
+
SO4
1
0
1.2 x 10
H2PO4
C6 H8 O 7
C6 H7 O 7
F
HF
-3
-
7.25 x 10
-4
-
7.41 x 10
-
Acetic
CH3COOH
CH3COO
Citric (2)
C6 H7 O 7
Carbonic (1)
H2CO3
C6 H6 O 7
H2PO4
-
-
C6 H6 O 7
HCO3
2-
2-
3.75
4.19
-5
1.74 x 10
-5
1.74 x 10
-7
3.98 x 10
-7
1.38 x 10
ClO
HClO
TRIS**
Ammonium Ion
NH4
Carbonic (2)
HCO3
Phosphoric (3)
HPO4
+
-
NH3
CO3
2-
PO4
OH
3-
3.98 x 10
6.40
6.86
7.53
-9
8.08
9.25
-11
6.31 x 10
6.37
7.15
3.0 x 10
2-
4.76
-8
8.32 x 10
-10
5.62 x 10
-
4.76
-8
7.08 x 10
MOPS*
3.13
3.18
4.3 x 10
3-
2.14
-4
1.77 x 10
-5
6.46 x 10
-7
-
C6 H5 O 7
HPO4
2-
-
1.92
-4
6.6 x 10
-
HCOO
C6H5COO
Hypochlorous
~-7
~-2
-2
2-
HCOOH
C6H5COOH
Phosphoric (2)
~10
2
~10
H2 O
-
Formic
Benzoic
Citric (3)
pKa
7
-13
10.2
12.4
-14
1.0 x 10
H2 O
Water
14.0
*MOPS stands for (3-N-Morpholino)propanesulfonic acid
**TRIS stands for Tris(hydroxymethyl)aminomethane
Table assembled from Data in:
Oxtoby, D.W. and N.H.Nachtrieb. Principles of Modern Chemistry. 2nd Ed. Saunders College Publishing, 1990.
Nelson, D.L. and M.M.Cox. Lehninger Principles of Biochemistry. 3rd Ed. Worth Publishers, 2000.
Voet, D. and J.G.Voet. Biochemistry. 2nd Ed. John Wiley & Sons, Inc, 1995.
The Significance of pKa
[A ! ][H3 O + ]
Ka =
[HA]
When pH < pKa, [H3O+] > Ka,
[A ! ]
−
When pH = pKa, [H3O ] = Ka,
= 1 ⇒ [A ] = [HA]
[A ! ]
[HA]
< 1 ⇒ [A−] < [HA]
[HA]
[A ! ]
When pH > pKa, [H3O+] < Ka,
> 1 ⇒ [A−] > [HA]
[HA]
+
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MITES
Biochemistry
(At pH near pKa, the acid and base are found at similar
concentrations)
(At pH above pKa, base dominates)
16
(At pH below pKa, acid dominates)
MITES
Biochemistry
TITRATION CURVES
Titration Curve - A curve generated by adding small amounts of a strong base to a weak acid solution.
After each addition, the system is allowed to come to equilibrium and the pH is measured. The titration
can be used to determine the amount of the weak acid as well as its pKa(s).
Equivalent – An equal number of moles.
Example 1: In the titration of 1 mole of acetic acid, 200 mL of a 5 M NaOH solution (1 mole NaOH)
would be one equivalent.
Midpoint – A point in a titration at which one of the weak acid species is exactly half-protonated. At the
midpoint, the concentration of this weak acid species will equal the concentration of its conjugate base. A
monoprotic acid will have one midpoint, a diprotic acid will have two midpoints, etc… Midpoints will
occur at 0.5, 1.5, 2.5, … equivalents.
Endpoint – A point in a titration at which the number of equivalents is a whole number (1, 2, 3, …
equivalents). A monoprotic acid will have one endpoint, a diprotic acid will have two endpoints, etc…
Example 2: 1 L of a 1-M acetic acid solution is titrated using a
concentrated sodium hydroxide solution. The curve to the right is
generated. Notice that the curve is fairly flat in the middle, i.e.
increasing addition of NaOH has little effect on the pH in this
region. This resistance to pH change is what allows weak acids to
act as buffers. The relatively flat part of the curve is called the
buffering region, and for acetic acid occurs approximately between
pH 4 and 6. At the starting point, there is roughly 1 mole of acetic
acid. Upon addition of NaOH, the following reaction occurs:
CH3COOH + NaOH
CH3COO¯ + Na+ + H2O
Titration Titration
Curve (100
mL -ofAcetic
1 M Acetic
Curve
Acid Acid)
9
Endpoint
CH3COO- Dominates
8
7
6
pH
Midpoint
[CH3COOH] = [CH3COO-]
pH = pKa = 4.76
5
4
3
2
Starting Point
CH3COOH Dominates
Buffering
Region
1
At the midpoint, 0.5 moles (or 0.5 equivalents) of NaOH have
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
been added. The reaction above has consumed 0.5 moles of acetic
Equivalents of NaOH
acid and produced 0.5 moles of acetate.
Therefore, the
concentrations of acetic acid and acetate
Titration
Curve
AcidAcid )
Titration Curve
(100
mL of- Phosphoric
1 M Phosphoric
are equal. At the endpoint, 1 mole (or 1
16
Third Endpoint
equivalent) of NaOH has been added.
3Third Midpoint PO4 Dominates
23The above reaction has converted almost
[HPO4 ] = [PO4 ]
14
pH = pKa3 = 12.4
all of the acetic acid to acetate.
12
Second Endpoint
HPO42- Dominates
10
Buffering
Region 3
Second Midpoint
[H2PO4-] = [HPO42-]
pH = pKa2 = 6.86
pH 8
Buffering
Region 2
6
First Endpoint
H2PO4- Dominates
4
First Midpoint
[H3PO4] = [H2PO4-]
pH = pKa1 = 2.14
2
Buffering
Region 1
Starting Point
H3PO4 Dominates
0
0
0.5
1
1.5
2
Equivalents of NaOH
2.5
17
3
Example 3: 100 mL of a 1-M phosphoric
acid solution is titrated using a sodium
hydroxide solution. The curve to the left
is generated. Phosphoric acid (H3PO4) is a
triprotic acid. Therefore, the titration curve
has three midpoints, three endpoints, and
three buffering regions. Phosphoric acid
has three protons to donate, and will
release them one at a time, based on the
pH. Notice that each proton has its own
pKa, which determines the pH at which it
will be released.
STRONG ACIDS AND BASES
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MITES
Biochemistry
Calculating the pH of a Strong Acid/Base Solution
To solve these problems, apply the definitions of pH or pOH as given below:
pH = -log [H3O+](Eq1)
pOH = -log [OH−] (Eq2)
pH + pOH = -log(KW) = 14 (Eq3)
Example 1: What is the pH of a 0.01-M HCl solution?
Reaction:
HCl + H2O  Cl−
+ H3O+ (Rxn1)
Since HCl is a strong acid, this reaction goes to completion. Therefore, [H3O+] = 0.01 M.
pH = -log (0.01 M) = -(-2.00) = 2.00 (Applying Eq1)
Example 2: What is the pH of a 0.02-M NaOH solution?
Reaction:
NaOH  Na+ + OH− (Rxn2)
Since NaOH is a strong base, this reaction goes to completion. Therefore, [OH−] = 0.02 M.
pOH = -log (0.02 M) = -(-1.70) = 1.70 (Applying Eq2)
pH = 14 – pOH = 14 – 1.70 = 12.30 (Applying Eq3)
Example 3: What is the pH of a solution made by mixing equal volumes of a 0.01-M HCl and a
0.02-M NaOH solution.
Reactions:
HCl + H2O  Cl− + H3O+ (Rxn1)
NaOH  Na+ + OH− (Rxn2)
H3O+ + OH−  2 H2O
(Rxn3)
The first two reactions go to completion, as described above. The third reaction will also go to
completion since both [H3O+] and [OH−] are large. Since H3O+ is the limiting reagent, it will
be completely consumed, while some OH− will remain.
Let V = the volume of each solution added (in liters)
Then 2V = the total volume of the solution
Moles of HCl added = (0.01 M)(V liters) = 0.01V moles
Moles of NaOH added = (0.02 M)(V liters) = 0.02V moles
Only 0.01V moles of OH− will be consumed in Rxn3 above, with another 0.01V moles
remaining.
0.01V moles
= 0.005 M
Final concentration of OH¯ =
2V liters
So, this mixture will have the same pH as a 0.005-M NaOH solution.
pOH = -log(0.005 M) = -(-2.30) = 2.30 (Applying Eq2)
pH = 14 – pOH = 14 – 2.30 = 11.70 (Applying Eq3)
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MITES
Biochemistry
THE HENDERSON-HASSELBALCH EQUATION – BUFFER PREPARATION
The Henderson-Hasselbalch Equation relates the pH to the equilibrium concentrations of a weak
conjugate acid/base pair.
pH = pK a + log
[A ! ]
[HA]
where A− and HA are the conjugate acid/base pair from the reaction:
A− + H3O+
HA + H2O
This equation is useful for calculations in buffer preparation. Given a desired pH and the pKa of the
buffer species, the Henderson-Hasselbalch Equation allows us to calculate the ratio of each species
needed to prepare the buffer.
Example 1: You would like to prepare an acetic acid buffer (pKa = 4.76) using sodium acetate
and acetic acid. What acetate:acetic acid concentration ratio is required to produce a buffer with
pH = 4.3?
Reaction:
CH3COOH + H2O
CH3COO− + H3O+
[CH3COO ! ] [ A ! ]
acetate
=
=
acetic acid [CH3COOH] [HA ]
Rearranging the Henderson-Hasselbalch Equation gives
[A ! ]
= 10 (pH !pK a ) = 10 (4.3!4.76 ) = 0.35
[HA]
For every mole of acetic acid used, 0.35 moles of acetate must be added to reach pH 4.3.
When preparing buffers, the total concentration (CTot) of acid + base is usually given. This requires
simultaneous solution of two equations as shown below.
Example 2: You would like to prepare a 0.1-M benzoic acid buffer (pKa = 4.19) using sodium
benzoate and benzoic acid. What concentrations of each species are needed to produce a buffer
with pH = 4.3?
Reaction:
C6H5COOH + H2O
C6H5COO− + H3O+
The two equations that must be solved are:
[A ! ]
= 10 (pH !pK a ) (Eq1)
[HA]
and
−
[A ] + [HA] = CTot (Eq2)
Rearranging Eq1 and substituting into Eq2 gives
[HA] 10 (pH ! pK a ) + [HA] = CTot
[HA] =
Applying Eq2 gives
(10
C Tot
(pH - pK a )
=
) (10
+1
0.1 M
(4.3 - 4.19 )
)
+1
= 0.044 M (benzoic acid)
[A−] = CTot - [HA] = 0.1 M - 0.044 M = 0.056 M (sodium benzoate)
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MITES
Biochemistry
THE HENDERSON-HASSELBALCH EQUATION – BUFFER pH CALCULATIONS
Although technically the Henderson-Hasselbalch Equation only applies to systems at
equilibrium, it usually provides a good estimate for systems that are not at equilibrium.
Example 1: A buffer is prepared by combining 0.2 moles hypochlorous acid (HClO) and 0.1
moles of its conjugate base (ClO−) in 1 L of water. The pKa for HClO is 7.53. What is the pH of
the buffer?
Reaction:
HClO + H2O
ClO− + H3O+
We know the initial amounts of the weak acid/base pair, but we do not know the
equilibrium concentrations of each species. Does this mean that we can’t use the HendersonHasselbalch Equation? It turns out that we can apply the H-H Equation with confidence in this
case, because the equilibrium concentrations will be very similar to the initial concentrations.
Although the solution may not be exactly correct, it will be very close.
[A ! ]
0.1 M
pH = pK a + log
= 7.53 + log
= 7.23
[HA]
0.2 M
The Henderson-Hasselbalch Equation gives a good approximation when [HA] ≅ [A− ]
When [HA] = [A−], the Henderson-Hasselbalch Equation gives an exact solution as shown in the
example below. As [HA] and [A−] move farther apart, the H-H Equation becomes less accurate.
However, if [HA] and [A−] are close, the H-H Equation is still a very good approximation.
Example 2: A buffer is prepared by combining 0.3 moles acetic acid (pKa = 4.76) and 0.3 moles
sodium acetate in 1 L water. What is the pH of the buffer?
Reaction:
CH3COO− + H3O+
CH3COOH + H2O
This problem can be solved without using any calculations. Since, [HA] = [A−], we know that
pH = pKa = 4.76. However, we can apply the Henderson-Hasselbalch Equation to demonstrate
its accuracy:
[A ! ]
0.3 M
pH = pK a + log
= 4.76 + log
= 4.76 + log (1) = 4.76
[HA]
0.3 M
In the case where [HA] = [A−] the H-H Equation gives an answer that is exactly correct.
When [HA] and [A−] are very different, you cannot apply the Henderson-Hasselbalch Equation
and would have to use “ICE” or some other algebraic method. You should already be familiar
with problems like this, so we will not consider them in this course. For our purposes, the H-H
Equation can be applied when
[A ! ]
.001 <
< 1000
[HA ]
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MITES
Biochemistry
WEAK + STRONG (THE OTHER WAY TO MAKE BUFFERS)
So far, we have examined only buffer solutions prepared by combining both a weak acid and its
conjugate weak base. Buffers can also be prepared by adding a strong acid to a weak base, or by
adding a strong base to a weak acid. The resulting solution will be the same—a weak acid and its
conjugate weak base—but the starting materials will be different.
Example 1: A buffer is prepared by combining 0.3 moles sodium acetate and 0.1 moles HCl into
1 L of water.
Reactions:
HCl + H2O  Cl− + H3O+ (Rxn1)
CH3COO− + H3O+  CH3COOH + H2O (Rxn2)
CH3COO− + H2O
CH3COOH + OH− (Rxn3)
Since HCl is a strong acid, it will completely dissociate into Cl− and H3O+ (Rxn1). The H3O+
generated from this reaction will combine with acetate to form acetic acid (Rxn2). In this case,
H3O+ will be the limiting reagent. Therefore, only 0.1 moles of acetic acid will be generated
and 0.2 moles acetate will remain. The pH for this buffer will be the same as that of a buffer
made by combining 0.1 moles acetic acid and 0.2 moles acetate in 1 L water. Rxn3 will be the
equilibrium buffering reaction. Applying the Henderson-Hasselbalch Equation gives
[A ! ]
0.2 M
pH = pK a + log
= 4.76 + log
= 4.76 + log(2) = 4.76 + 0.30 = 5.06
[HA]
0.1 M
Example 2: A buffer is prepared by combining 0.8 moles benzoic acid and 0.6 moles NaOH into
2 L of water.
Reactions:
NaOH  Na+ + OH¯ (Rxn1)
C6H5COOH + OH¯  C6H5COO¯ + H2O (Rxn2)
C6H5COOH + H2O
C6H5COO¯ + H3O+ (Rxn3)
Since NaOH is a strong base, it will completely dissociate as in Rxn1. The OH¯ generated
will react as in Rxn2 to convert 0.6 moles benzoic acid to benzoate. The final solution will
consist of the remaining 0.2 moles benzoic acid and 0.6 moles benzoate. These components
will form an equilibrium system described by Rxn3. The pH of this solution can be calculated
by applying the Henderson-Hasselbalch Equation, with 0.1 M benzoic acid and 0.3 M
benzoate.
[A ! ]
0.3 M
pH = pK a + log
= 4.19 + log
= 4.19 + log(3) = 4.19 + 0.48 = 4.67
[HA]
0.1 M
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MITES
Biochemistry
POLYPROTIC ACIDS
Polyprotic Acids – acids that can donate multiple protons.
Examples: sulfuric acid (H2SO4), carbonic acid (H2CO3), and phosphoric acid (H3PO4)
To this point, we have mainly considered monoprotic acids such as acetic acid and benzoic acid.
However, many biological buffers are polyprotic acids. Problems with these acids can have
solutions that are much more complex than those discussed previously. Fortunately, the
solutions can often be simplified to be no different than those we have already examined.
Example 1: We would like to prepare a phosphoric acid buffer at pH 7.0. What two phosphoric
acid components would you add to make this buffer and in what proportion?
Reactions: H3PO4 + H2O
H2PO4− + H3O+ (Rxn1)
pKa1 = 2.14
2−
+
−
H2PO4 + H2O
HPO4 + H3O (Rxn2)
pKa2 = 6.86
2−
3−
+
HPO4 + H2O
PO4
+ H3O (Rxn3)
pKa3 = 12.4
Because there are three reactions that occur involving phosphoric acid, an exact solution would
require solving the three equilibrium equations simultaneously. Typically, we can neglect
some of these reactions based on their pKa values. Since pKa1 <<pH, the basic species in Rxn1
will dominate.
[H2PO4−] >> [H3PO4]
Similarly, pKa3 >> pH, so the acidic species in Rxn3 will dominate.
[HPO42−] >> [PO43−]
Therefore, we can neglect the species H3PO4 and PO43−. If we consider only the remaining
two species, we can narrow our scope to Rxn2. Applying the Henderson-Hasselbalch
Equation, we find
[HPO 24! ] [A ! ]
=
= 10 (pH !pK a ) = 10 (7.0!6.86 ) = 1.38
!
[H2PO 4 ] [HA]
To make this buffer, you would add 1.38 moles of HPO42− for every mole of H2PO4−.
The H-H Equation can also be applied to determine the pH of polyprotic buffers (the inverse of
the above). While some polyprotic acids may seem very different, they can often be simplified
to the familiar one-reaction problems as in the above example.
Example 2: What will be the pH of the buffers made by adding the following to 1 L water
a) 0.3 moles H2PO4−, 0.4 moles HPO42−.
This one is easy. We can limit ourselves to Rxn2 above and use the H-H Equation:
[HPO 24! ]
[A ! ]
0.4 M
= 6.98
pH = pK a + log
= pK a2 + log
= 6.86 + log
!
[HA ]
0.3 M
[H2PO 4 ]
b) 0.5 moles H2PO4−, 0.2 moles PO43−.
This one is a little more difficult. Do we have to consider both Rxn2 and Rxn3 above? The
answer is no. By recognizing that these two components will participate in the following
reaction, the problem can be simplified.
0.2 moles H2PO4− + 0.2 moles PO43−  0.4 moles HPO42−
Once this reaction occurs, the buffer will contain the remaining 0.3 moles H2PO4− and the 0.4
moles HPO42− generated by the above reaction. This is the same composition as in part (a),
therefore the answer is the same: pH = 6.98
22
MITES
Biochemistry
TAKE THE EASY WAY OUT
Some problems can be solved without doing any calculations at all. The trick is to recognize
them. Here are some examples to help you figure out when to take the easy way out.
Example 1: What is the pH of a buffer made by combing 0.055 moles formic acid and 0.055
moles formate in 920 mL water?
Whenever you see a problem with equal concentrations of a conjugate acid and base, you
should recognize this as an easy problem. Remember that when [A−] = [HA], pH = pKa. In
this case, since [HCOO−] = [HCOOH], pH = pKa = 3.75.
Example 2: What is the pH of a buffer prepared by combining 0.5 mole acetic acid and 0.25
moles NaOH in 1 L water?
In this problem, 0.5 equivalents of NaOH are added to acetic acid. Again, no calculations
need to be done. Simply recognize that half of the acetic acid will be converted to acetate by
the following reaction
CH3COOH + NaOH
CH3COO− + Na+ + H2O
In the end, the concentrations of acetic acid and acetate will
be equal, [CH3COOH] = [CH3COO−]. Therefore, pH =
pKa = 4.76. It may also help to sketch out a titration
Titration Titration
Curve (100
mL -ofAcetic
1 M Acetic
Curve
Acid Acid)
9
curve of acetic acid to convince yourself that this is the
midpoint of the titration.
Example 3: What is the pH of a buffer prepared by
combining 0.8 moles TRIS base and 0.4 moles HCl in
1 L water?
This is very similar to Example 2. The strong acid,
HCl, will convert half of the TRIS base to TRIS acid.
The concentrations of acid and base will be equal,
producing a buffer with pH = pKa = 8.08.
The two previous examples illustrate that calculations
with monoprotic species become much simpler with
0.5 equivalents of strong base or acid. For polyprotic
acids, 1.5 and 2.5 equivalents also simplify things.
Endpoint
CH3COO- Dominates
8
Midpoint
[CH3COOH] = [CH3COO-]
pH = pKa = 4.76
7
6
pH
5
4
3
Starting Point
CH3COOH Dominates
2
1
0
0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Equivalents of NaOH
1
Titration
Curve
AcidAcid )
Titration Curve
(100
mL of- Phosphoric
1 M Phosphoric
16
Third Midpoint
[HPO42-] = [PO43-]
pH = pKa3 = 12.4
14
12
Second Midpoint
[H PO ] = [HPO ]
10
Example 4: What is the pH of a buffer prepared by
pH = pK = 6.86
combining 0.6 moles phosphoric acid (H3PO4) to 0.9
pH 8
moles NaOH in 1 L water?
First Midpoint
6
In this problem, 1.5 equivalents of NaOH are added
[H PO ] = [H PO ]
pH = pK = 2.14
4
to phosphoric acid. The equilibrated system will
come to the second midpoint of the titration. At this
2
point, pH = pKa2 = 6.86. Another way to think of
0
this problem is that 1 equivalent of NaOH will
0
0.5
1
1.5
2
−
Equivalents of NaOH
convert all of the H3PO4 to H2PO4 . The remaining
0.5 equivalents will convert only half of that
H2PO4− to HPO42−, generating a buffer where [H2PO4−] = [HPO42−] and pH = pKa2.
4
2
24
a2
3
4
2
4
a1
23
2.5
3
MITES
Biochemistry
SUMMARY OF ACID/BASE CALCULATIONS
The previous pages have covered how to solve common acid/base problems for one- and twocomponent systems. The tables below summarize the techniques that should be used in each
case:
Component 1
Strong Acid
Weak Acid
Strong Acid
Strong Base
One-Component Systems
Strong Acid Def'n of pH p. 14
Strong Base Def'n of pOH p. 14
Weak Acid
ICE
p. 17
Two-Component
Weak
Base
ICESystems
p. 18
Component 2
Strong Base Def'n of pH/pOH
if [Acid]![Base]
Weak Base
H-H Eqn
Weak Base
H-H Eqn
Weak Acid
H-H Eqn
24
p. 14
pp. 15&16
p. 17
p. 17