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BIOC 462a
General Chemistry Review: Chemical Equilibrium, Ionization of H2O, pH, pKa
General Equilibrium:
What are the UNITS of Keq?
Example reactions:
A --> B
units of Keq ?
_________
A --> B + C
units of Keq ?
_________
A + B --> C
units of Keq ?
_________
A + B --> C + D
units of Keq ?
_________
2A --> B
units of Keq ?
_________
2A + B --> C + D
units of Keq ?
_________
Units of Keq depend on the reaction!
Dissociation of H2O:
Bioc462a General Chemistry Review (2007)
p. 1
Table 2-6, from Nelson & Cox: Lehninger Principles of Biochemistry, 4th edition (2004)
NOTE:
+
–7
•If pH > 7, [H ] < 10 M
(solution is "basic")
+
–7
•If pH < 7, [H ] > 10 M
(solution is "acidic")
+
–7
•If pH = 7, [H ] = 10 M
(solution is "neutral")
•pH scale is logarithmic:
•If change in pH (ΔpH) = 1,
+
1
Δ[H ] = 10 (10-fold).
•If change in pH (ΔpH) = 2,
+
2
Δ[H ] = 10 (100-fold).
Bioc462a General Chemistry Review (2007)
p. 2
Ionization of weak acids and bases:
Brοnsted definitions:
acid = a proton donor
base = a proton acceptor
Definition of pKa:
Below is a table of some weak acids, with their pKa values (25° C).
Explain the differences in pKa within the 3 groups below. Why does substitution of each successive H
in the methyl group of acetic acid with a Cl decrease the pKa? Why does substitution of each
additional methyl or methylene group for the H in formic acid or in methylamine increase the pKa?
Why does addition of the OH group to ethylamine to make ethanolamine decrease the pKa?
Bioc462a General Chemistry Review (2007)
p. 3
Acid
acetic acid
CH3-COOH
chloroacetic acid
ClCH2-COOH
dichloroacetic acid
(Cl)2CH-COOH
trichloroacetic acid
(Cl)3C-COOH
pKa
4.76
formic acid
H-COOH
acetic acid
CH3-COOH
propionic acid
CH3-CH2-COOH
3.75
2.85
1.48
0.70
Acid
ammonium ion
+
NH4
methylamine
+
CH3-NH3
ethylamine
+
CH3-CH2-NH3
ethanolamine
+
HO-CH2-CH2-NH3
pKa
9.25
10.6
10.8
9.5
4.76
4.87
Fig. 2-16: Proton dissociation reactions of some mono-, di-, and triprotic acids, with their pKas
arranged on a pH scale (from Nelson & Cox: Lehninger Principles of Biochemistry, 4th edition (2004)
Bioc462a General Chemistry Review (2007)
p. 4
Quantitative relationship between pKa and pH:
Look again at the equilibrium dissociation reaction and remember LeChatelier's Principle:
+
–
HA
H
+ A
+
The higher the [H ] (i.e., the lower the pH), the more the equilibrium shifts to the left, so the more
+
conjugate acid will be present. Conversely, the lower the [H ] (i.e., the higher the pH), the more the
equilibrium shifts to the right, so the more conjugate base will be present. The exact ratio of base to acid
depends on the pH and the pKa (Henderson-Hasselbalch Equation).
Note that when pH = pKa, [A–] = [HA], so [base] /[ acid] ratio = 1/1 ([base] = [acid]). When pH =
pKa, the “buffering capacity” of the mixture will be the greatest. That is, a given change in concentration
+
of base (OH–) or acid (H ) will result in the smallest change in pH. If the pH is ABOVE the pKa, there
will be more base than acid in the solution, i.e., the [base] /[ acid] ratio will be greater than 1/1. If the
pH is BELOW the pKa, there will be more acid than base, i.e., the [base] /[ acid] ratio will be less than
1/1. When you do calculations, always stop at the end to think about whether your answer makes sense.
Work the posted practice problems at the end of this review, on pH, pKa, and [base]/[acid] ratio, and
learn how to convert a [base]/[acid] ratio to the FRACTION of the total ([base] + [acid]) that's in the
form of the conjugate base (or the conjugate acid).
Bioc462a General Chemistry Review (2007)
p. 5
Titration Curves:
•add measured volume of solution of strong base (NaOH) of known concentration
•measure the pH, e.g. with a pH meter
•can be used to determine
–
°amount of acid in the solution (from mols of OH required to –completely deprotonate each acidic
group), which permits conversion of volume of standard OH added (which was measured) to
equivalents OH – added (mols OH –/mol acidic group)
°pKa values of acidic groups (from pH at which an acidic group is exactly half titrated)
•example: titration of acetic acid, CH3-COOH
°involves 2 reversible equilibria:
Fig. 2-17: Titration curve of acetic acid (from Nelson &
Cox: Lehninger Principles of Biochemistry, 4th edition
(2004). When no OH– has been added, predominant form is
HAc; at the endpoint, when 1 equivalent OH– has been
added, all the HAc has been converted to Ac –. When
exactly 0.5 equivalent of OH– has been added, exactly
half the original HAc has dissociated, so [Ac –] = [HAc] ,
i.e., [base] = [acid]), and pH = pKa. Shaded zone shows
the pH range of useful buffering power, from about 1 pH
unit below to about 1 pH unit above the pKa, i.e. between
about 10% and 90% titration of the weak acid.
Bioc462a General Chemistry Review (2007)
Fig. 2-18: Titration curves of acetic
acid, H2PO4–, and ammonium ion
(from Nelson & Cox: Lehninger
Principles of Biochemistry, 4th edition
(2004).
p. 6
Fig. 2.13 from Garrett &
Grisham, Biochemistry,
3rd ed., 2004. For a
polyprotic acid (more
than one acidic group, so
more than one pKa on
titration curve), it takes 1
equivalent of OH– to
titrate each acidic group,
so a triprotic acid like
phosphoric acid requires
3 equivalents. The pKa
values are pH values at
each half equivalent
added.
Summary:
+
1) H2O and acids in aqueous solution dissociate to yield protons (H ), which in actuality would be
+
hydrated to form H3O .
2) The tendency of an acid to donate its proton to H2O (dissociate the proton) can be quantitatively
described by its dissociation equilibrium constant Ka (or its pKa, which = –logKa).
3) pKa values can be very accurately measured by titration curves, as the pH at half equivalence points.
4) Relationship between pH, pKa, and ratio of [base]/[acid] can be described by the HendersonHasselbalch Equation:
Buffers:
•homeostasis: the maintenance of constant conditions in internal environment
•Fluids in living systems have a characteristic, almost constant, pH.
•How is pH controlled in living systems? By buffer systems.
Buffer: aqueous system that tends to resist changes in pH when small amounts of acid or base are added.
Buffer system: aqueous solution of a weak acid and its conjugate base
Buffer range of a weak acid: near its pKa, about ±1 pH unit from the pKa. Maximum buffering
capacity is at the pKa.
Look again at the equilibrium dissociation reaction and remember LeChatelier's Principle:
HA
H
+
+
–
A
Bioc462a General Chemistry Review (2007)
p. 7
+
The higher the [H ] (i.e., the lower the pH), the more the equilibrium shifts to the left, so the more
+
conjugate acid will be present. Conversely, the lower the [H ] (i.e., the higher the pH), the more the
equilibrium shifts to the right, so the more conjugate base will be present. The exact ratio of base to acid
depends on the pH and the pKa (Henderson-Hasselbalch Equation).
Remember that when pH = pKa, [A–] = [HA], so [base] /[ acid] ratio = 1/1 ([base] = [acid]).
When pH = pKa, the “buffering capacity” of the mixture will be the greatest. That is, a given change in
+
concentration of base (OH–) or acid (H ) will result in the smallest change in pH.
2 physiologically important buffer systems:
1) INTRACELLULAR:
–
inorganic phosphate ion (H2PO4
–
phosphomonoesters (R-OPO2OH
2–
HPO4 , pKa ~ 7) and organic phosphates, e.g.
2–
R-OPO3 , pKa ~ 7)
2) EXTRACELLULAR (blood plasma of mammals):
+
–
bicarbonate buffer system (CO2 + H2O
H2CO3 <--> H + HCO3 )
3 linked equilibria:
–
•How can the [HCO3 ] / [H2CO3] system effectively buffer the blood plasma at pH ~7.4, with a
pKa well below 7? Hint: see Box 2-1 in textbook, and look at the Biology Project website (below)
•clinical considerations of bicarbonate buffer system:
+
a) What happens physiologically if the blood pH↓ (i.e. [H ]↑? (acidosis)
You exhale excess CO2, so [CO2] ↓, so (Le Chatelier's Principle)
+
[H ]↓, i.e. pH goes back up.
+
b) What happens physiologically if the blood pH↑ (i.e. [H ]↓? (alkalosis)
You exhale less CO2, so [CO2] ↑, so (Le Chatelier's Principle)
+
[H ]↑, i.e. pH goes back down.
c) What if you have difficulty exhaling (respiratory failure)?
+
Blood [CO2] ↑, so [H ]↑ (acidosis) and you can't get rid of the excess [CO2] by exhaling.
+
–
Treatment: inject bicarbonate (HCO3 ) to convert some of the excess H to H2CO3 , which
would bring pH back up -- and get the person breathing again to get rid of the excess CO2!
For a little online quiz on acids, bases, pH & pKa, and buffers, try the Biology Project website at
http://www.biology.arizona.edu/biochemistry/problem_sets/ph/ph.html
Click on "tutorial" at the bottom of each question for the tutorial that goes with that question.
Bioc462a General Chemistry Review (2007)
p. 8
For a little online quiz on clinical correlates of blood pH, try the Biology Project website at
http://www.biology.arizona.edu/biochemistry/problem_sets/medph/medph.html
Click on "tutorial" at the bottom of each question for the tutorial that goes with that question.
PRACTICE PROBLEMS
Work these without looking at the answers, which are on the last page.)
NOTE ESPECIALLY PROBLEM 2B, how to go from the ratio of base/acid (Henderson-Hasselbalch
calculation) to the fraction of the total (acid + base) that's in the form of the acid (or what fraction is
in the form of the base). In fact, anytime you calculate a RATIO and get a number, STATE the
(understood) "1" in the denominator. If ratio is 0.4, state it as 0.4 / 1. Also, start by writing chemical
equations to describe acid dissociation reactions, with CHARGE BALANCE.
+
+
–7
1. For a solution whose pH is 6.0, what is [H ]? If [H ] is 5 x 10
scientific calculator if necessary to do log problems.)
M, what is the pH? (Use a simple
2. A. For a weak acid such as the R group carboxyl group of Glu or Asp in a protein, if the pKa of that
specific residue's carboxyl group is exactly 4.0, in an environment in which the pH = 5.0, what
–
would be the ratio of base to acid ( [COO ]/[COOH] )?
B. For that same carboxyl group, what fraction, or what percentage, of the total (population of all the
molecules in solution) is present in the form of the ACID (COOH) at pH 5.0? What fraction of the
–
total is present in the form of the BASE (COO ) at pH 5?
3. Suppose that about 1% of the molecules of a particular protein in solution have the imidazole group of a
specific His residue (say it's residue #20 in the amino acid sequence of that protein) in the uncharged
(neutral) form at pH 4.5. What is the pKa of that specific His residue in that protein?
(You need to know the ionization properties of the functional group of His to answer this question -look at the ionization equilibrium for His in Fig. 3-12, p. 84, of Lehninger Principles of Biochemistry,
4th ed. All that concerns you is the ionization of the R group, the imidazole group, which for this
amino acid is the middle pKa. The pKa of the imidazole group of the free amino acid histidine is about
6.0, but the exact pKa values of amino acid functional groups in proteins vary somewhat, depending on
the specific environment around that residue in that protein. You're being asked to calculate this
specific His residue's pKa.)
Bioc462a General Chemistry Review (2007)
p. 9
SOLUTIONS TO PRACTICE PROBLEMS
+
1. pH = – log[H ] (Remember, the log is the exponent.)
+
If pH = 6.0, [H ] =10
–pH
= 10
–6.0
M
–7
+
(Use a calculator for this.) If [H ] = 5 x 10
–7
M, pH = – log(5 x 10 ) = 6.3
–
+
2. acid dissocation reaction for a carboxyl group: (R)-COOH
(R)-COO + H
A. Use the Henderson-Hasselbalch Equation whenever you see a problem involving the relationship of
pH, pKa, and base/acid ratio (or fraction of the total group that's in form of the base, or of the acid).
pH = pKa + log([base]/[acid])
log([base]/[acid]) = pH – pKa = 5.0-4.0 = 1.0
([base]/[acid]) = 101 = 10/1
(There's 10 times as much of the carboxyl group in the form of the base as in the form of the acid
when the pH is 1 unit above the pKa.)
–
B. At pH 5.0, the COO /COOH ratio is 10/1. We can do this problem most simply by using
–
proportions; the total of the carboxyl group is COOH + COO = 1 + 10 = 11. The fraction of the
total in the form of the acid (COOH) is
–
(COOH) / total = (COOH) / (COOH + COO ) = 1/11 = 0.091
9.1% of the molecules in solution have that carboxyl group in its acid form.
–
–
–
COO / total = COO / (COOH + COO ) = 10/11 = 0.909
Alternatively, if you've already calculated that acid/total = 0.091, then
base/total = 1 – (acid/total) = 1 – (0.091) = 0.909. Either way,
90.9% of the molecules in solution have that carboxyl group in its base form.
3. We'll abbreviate the imidazole group in question as Im, so neutral form is Im, the conjugate base, and
+
charged form is Im , the conjugate acid.
+
+
Im <--> Im + H
Use the Henderson-Hasselbalch Equation again.
"About 1% neutral" means that base (Im) / total = about 0.01 or 1/100. Thus base/acid ratio would be
base/(total – base) = 1 / (100 – 1) = 1/99.
pH = pKa + log([base]/[acid])
pKa = pH – log([base]/[acid]) = 4.5 – log (1/99) = 4.5 – (– 2.0) = 4.5 + 2.0 = 6.5
The pKa of that specific His residue in that protein must be about 6.5. (Note that this is slightly
different from the pKa of the R group of the free amino acid histidine.)
Bioc462a General Chemistry Review (2007)
p. 10