Ch. 9: Polyprotic Acid-Base Equilibria
Outline:
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9-1 Diprotic Acids and Bases
9-2 Diprotic buffers
9-3 Polyprotic Acids and Bases
9-4 Which is the Principal Species?
9-5 Fraction Compositions
9-6 Isoelectric and Isoionic pH’s
Updated Oct. 24, 2011: new slides 1-24
Amino Acids
Amino acids are the building blocks of proteins, and always have an acidic carboxylic acid
group, a basic amino group, and a neutral substituent, R.
At low pH: both the -COO and -NH2 groups are protonated
At high pH: both the -COO and -NH2 groups are not protonated
There are separate pKa values associated with proton dissociation from all three of the
functionalities (see table next page).
AA’s can also exist in zwitterionic form. Zwitterions are species carrying both positive and
negative charges, which are stabilized, in the cases of AA’s by interactions of the -COOand -NH3+ groups with water. In fact, AA’s in the solid state almost always have their most
stable state as a zwitterion; however, in the gas phase, zwitterions rarely exist.
pKa’s of Amino Acids
Leucine
Our discussion will focus upon Leucine (HL) (i.e., R is the isobutyl group, (CH3)2CHCH2—)
The associated equilibrium constants are:
(note: sometimes the subscripts ‘a’ are omitted on Ka1 and Ka2)
Recall: relations between these coefficients involve Kw:
K a1 iK b2 = K w
K a2 iK b1 = K w
We will examine three cases, where we calculate the pH’s and compositions of separate
solutions of (i) 0.0500 M H2L+, (ii) 0.0500 M HL and (iii) 0.0500 M L-.
We note that they do not depend on the charge type of the acids and bases, meaning that
we use the same procedure to find the pH of the diprotic H2A, where A is anything, or H2L+,
where HL is leucine.
Leucine: Acidic form, H2L+
Leucine hydrochloride contains the protonated species, H2L+, which can dissociate twice.
Since K1 = 4.70 × 10−3, H2L+ is a weak acid, and HL is an even weaker acid, with K2 = 1.80 ×
10−10. We can make the approximation that a solution of H2L+ behaves as a monoprotic
acid, with Ka = K1 (since H2L+ dissociates only partly, and HL much less).
K a = K1 = 4.70 × 10 −3
x2
= K a ⇒ x = 1.32 × 10 −2 M = [HL] = [H + ]
F−x
[H + ] = x = 1.32 × 10 −2 ⇒ pH = 1.88
[H 2 L+ ] = F − x = 3.68 × 10 −2 M
For most diprotic acids, K1 is
sufficiently larger than K2 for this
approximation to be valid. Even if
K2 were just 10 times less than K1,
[H+] calculated by ignoring the
second ionization would be in
error by only 4%. The error in pH
would be only 0.01 pH unit. In
summary, a solution of a diprotic
acid behaves like a solution of a
monoprotic acid, with Ka = K1.
We have assumed that [L] is very small; nonetheless, we can still calculate it.
[H + ][L− ]
K [HL]
K2 =
⇒ [L− ] = 2 + = 1.80 × 10 −10 M
[HL]
[H ]
The approximation [H+] ≈ [HL] reduces the equation above to [L−] = K2 (good approximation!)
Leucine: Basic form, LThe species L−, found in a salt such as sodium leucinate, can be prepared by treating leucine
(HL) with an equimolar quantity of NaOH. There are two basic species with Kb1 = 5.55 × 10−5
and Kb2 = 2.13 × 10−12. We can make the approximation that a solution of L- behaves as a
monoprotic base, with Kb = Kb1 (L- does not really hydrolyze with H2O to give HL).
[H2L+] = Kb2 = 2.13 × 10−12 M, so the approximation that [H2L+] is insignificant relative to [HL]
is justified. In summary, if there is any reasonable separation between Ka1 and Ka2 (and,
therefore, between Kb1 and Kb2), the fully basic form of a diprotic acid can be treated as
monobasic, with Kb = Kb1.
CO2 in the air and ocean
Dissolved carbon dioxide is one of the most important diprotic acids in Earth’s ecosystem.
Increasing atmospheric CO2 increases the concentration of CO2 dissolved in the ocean, which
consumes carbonate and lowers the pH:
The pH of the ocean has already decreased from its preindustrial value of 8.16 to 8.04 today,
and is predicted to be 7.7 by 2100. Low [CO32-] promotes dissolution of solid CaCO3:
If CO32- in the ocean decreases enough, organisms such as
plankton and coral with CaCO3 shells or skeletons will not
survive. Calcium carbonate has two crystalline forms called
calcite and aragonite (more soluble).
When pteropods collected from the subarctic Pacific Ocean
are kept in water that is less than saturated with aragonite,
their shells begin to dissolve within 48 h. Animals such as
the pteropod lie at the base of the food chain.
Leucine: Intermediate form, HL
This is more complicated to treat, since HL may act as either an acid or a base.
The HL is said to be amphiprotic (i.e., it can accept and/or donate a proton). However, we
expect that this solution will be acidic, since Ka > Kb. But, we cannot simply ignore the
hydrolysis reaction, even if there is a fairly large difference between Ka and Kb.
Since the first reaction produces H+, it reacts with the OH- in the second reaction to form H2O,
thereby driving this reaction to the right. Hence, we need to treat this problem with the
systematic treatment of equilibrium.
We have Step 1 listed above. Step 2 is the charge balance equation:
There are a number of substitutions we can make to ultimately calculate the species
concentrations and the pH (written example). Summary for all three forms:
Simplified calculation: Intermediate form
Starting from the equation we worked out using the S.T.E., we can make a few more
simplifications (i.e., (i) K2F >> Kw and (ii) K1 >> F):
[H + ] =
K1 K 2 F + K1 K w
≈
K1 + F
K1 K 2 F
≈
K1 + F
Approx. (i)
K1 K 2 F
= K1 K 2
F
Approx. (ii)
Then, reframe the equation in terms of pH:
1
(log K1 + log K 2 )
2
1
+
− log[H ] ≈ − (log K1 + log K 2 )
2
1
pH ≈ (pK1 + pK 2 )
2
Recall:
log[H + ] ≈
This very powerful and useful equation can be used in many situations, and says that the pH
of the intermediate form of a diprotic acid is close to midway between pK1 and pK2, regardless
of the formal concentration.
Simplified calculation: Intermediate form, 2
Example:
Summary of diprotic acid calculations
Solution of H2A
Solution of HA-
Solution of HA
Treat H2A as a monoprotic acid
with Ka = K1 to find [H+], [HA−],
and [H2A].
Treat HA- as a monoprotic base
with Kb = Kb1 = Kw/Ka2 to find
[H+], [HA−], and [A2-].
Use the approximation [HA−] ≈
F and find the pH
Use the K1 equilibrium to solve
for [H2A].
With [H+] from step 1 and
[HA−] ≈ F, solve for [H2A] and
[A2−], using the K1 and K2
equilibria.
Use the K2 equilibrium to solve
for [A2−].
We must be careful, since we have not considered other equilibria (e.g., Na+ or K+ in
solutions of HA− or A2− form weak ion pairs that we have neglected).
Diprotic Buffers
A buffer made from a diprotic (or polyprotic) acid is treated in the same way as a buffer made
from a monoprotic acid. For the acid H2A, we can write two Henderson-Hasselbalch
equations, both of which are always true. If we happen to know [H2A] and [HA−], then we will
use the pK1 equation. If we know [HA−] and [A2−], we will use the pK2 equation.
[HA − ]
pH = pK1 + log
;
[H 2 A]
[A 2 − ]
pH = pK 2 + log
[HA]
Polyprotic acids and bases
The treatment of diprotic acids and bases can be extended to polyprotic systems.
1. H3A is treated as a monoprotic weak acid, with Ka = K1.
2. H2A- is treated as an intermediate form of a diprotic acid.
3. HA2- treated as the intermediate form of a diprotic acid. However, HA2− is “surrounded” by
H2A− and A3−, so the equilibrium constants to use are K2 and K3, instead of K1 and K2.
[H + ] =
K1 K 2 F + K1 K w
K1 + F
[H + ] =
4. A3− is treated as monobasic, with Kb = Kb1 = Kw/Ka3.
K 2 K 3F + K 2 K w
K2 + F
Polyprotic acids and bases, 2
Example (written solution).
We have reduced acid-base problems to just three types. When you encounter an acid or
base, decide whether you are dealing with an acidic, basic, or intermediate form. Then do the
appropriate arithmetic to answer the question at hand.
Principal Species
We sometimes must identify which species of acid, base, or intermediate predominates under
given conditions. e.g., “What is the principal form of benzoic acid at pH = 8?”
At pH = 4.20, there is a 1:1 mixture of benzoic acid (HA)
and benzoate ion (A−).
At pH = pKa + 1 = 5.20, the quotient [A−]/[HA] is 10:1.
At pH = pKa + 2 = 6.20), the quotient [A−]/[HA] is 100:1.
As pH increases, the quotient [A−]/[HA] increases further.
Hence, for a monoprotic system:
The basic species, A−, is the predominant form
when pH > pKa.
The acidic species, HA, is the predominant form
when pH < pKa.
Thus, the predominant form of benzoic acid at pH 8 is the
benzoate anion, C6H5CO2-.
Principal Species, 2
For polyprotic systems, our reasoning is similar, but there are several values of pKa. Consider
oxalic acid, H2Ox, with pK1 = 1.25 and pK2 = 4.27:
At pH = pK1, [H2Ox] = [HOx−].
At pH = pK2, [HOx−] = [Ox2−].
The chart shows the major species in each pH region.
Hence, for a polyprotic system:
The basic species, A2-, is the predominant form
when pH > pK2.
The intermediate species, HA-, is the
predominant form when pK1 < pH < pK2
The acidic species, H2A, is the predominant form
when pH < pK1.
Principal Species, 3
What is the principal form of arginine at pH 10.0?
Approximately what fraction is in this form?
What is the second most abundant form at this pH?
At pH = pK2 = 8.99,
[H2Arg+] = [HArg].
At pH = pK3 = 12.1
[HArg] = [Arg−].
At pH = 10.0, the major
species is HArg.
Since pH 10.0 is about one
pH unit higher than pK2, we
can say [HArg]/[H2Arg+] ≈
10:1. i.e., ca. 90% of
arginine is in the form HArg.
The second most important
species is H2Arg+, which
makes up ca. 10% of the
arginine.
Fractional Compositions
We now derive equations that give the fraction of each species of acid or base at a given pH.
These equations are useful for acid-base and EDTA titrations, and electrochemical equilibria.
Monoprotic systems
Goal: Find an expression for the fraction of an acid in each form (HA and A−) as a function of
pH. We can do this by combining the equilibrium constant with the mass balance:
Rearranging the MB gives [A−] = F − [HA], which is plugged into the Ka expression to give
This can be rearranged to give the concentration of HA:
The fraction of molecules in the form HA is called αHA:
Fractional Compositions, 2
If we divide the previous equation by F, we get:
α HA
[HA]
[H + ]
=
= +
F
[H ] + K a
Fraction in the form HA
α A−
[A − ]
Ka
=
= +
F
[H ] + K a
Fraction in the form A-
Here we see plots of αHA and αA- as a function of pH
for a system with a pKa of 5.00.
At low pH, almost all of the acid is in the form HA.
At high pH, almost everything is in the form A−.
Note, that: αHA + αA- = 1.
The fraction here denoted as αA- is the same as the
fraction of dissociation discussed earlier.
Fractional Compositions, 3
Diprotic Systems
The derivations for the fractional composition equations for diprotic systems is similar to that
for the monoprotic system (don’t worry about this - let’s just look at the equations):
α H2 A
[H 2 A]
[H + ]2
=
= + 2
F
[H ] + [H + ]K1 + K1K 2
α HA−
[HA − ]
K1[H + ]
=
= + 2
F
[H ] + [H + ]K1 + K1K 2
α A2−
[A 2 − ]
K1 K 2
=
= + 2
F
[H ] + [H + ]K1 + K1K 2
Note: These equations can be applied
equally to bases, where (K1 = Kw/Kb2 and
K2 = Kw/Kb1)
Fractional composition diagram for fumaric acid
(trans-butenedioic acid). At low pH, H2A is
dominant. At intermediate pH, HA− is
dominant; and, at high pH, A2− dominates.
Because pK1 and pK2 are not separated very much,
the fraction of HA− never gets very close to unity.
Isoelectronic and Isoionic pH
In biochemistry, polyprotic molecules are often discussed in terms of the isoelectric or isoionic
pH. Let’s consider a simple diprotic system: alanine.
The isoionic point (or isoionic pH) is the pH obtained when the pure, neutral polyprotic acid
HA (the neutral zwitterion) is dissolved in water (i.e. the pH of the pure acid)
Ions: H2A+, A−, H+ and OH−
Principal Species: HA, [H2A+] ≠ [A−]
The isoelectric point (or isoelectric pH) is the pH at which the average charge of the polyprotic
acid is 0.
Principal Species: uncharged form HA, [H2A+] = [A−]
Equilibria: Always some H2A+ + A− ⇌ HA.
Isoelectronic and Isoionic pH, 2
When alanine is dissolved, the pH is the isoionic pH. Since HA is the intermediate form of the
diprotic acid, H2A+, then the [H+] is given by
Isoionic point:
[H + ] =
K1 K 2 F + K1 K w
K1 + F
For 0.10 M alanine, the isoionic pH is
From [H+], K1 and K2, [H2A+] = 1.68 × 10−5 M and [A−] = 1.76 × 10−5 M for pure alanine in water
(the isoionic solution). There is a slight excess of A− because HA is a slightly stronger acid
than it is a base. It dissociates to make A− a little more than it reacts with water to make H2A+.
The isoelectric point is the pH at which [H2A+] = [A−], and, therefore, the average charge of
alanine is 0. To go from the isoionic solution (pure HA in water) to the isoelectric solution, we
could add just enough strong acid to decrease [A−] and increase [H2A+] until they are equal.
Adding acid necessarily lowers the pH. For alanine, the isoelectric pH must be lower than the
isoionic pH.
Isoelectronic and Isoionic pH, 3
Calculate the isoelectric pH by first writing expressions for [H2A+] and [A−]:
Setting [H2A+] = [A−], we find
which yields
Isoelectronic point:
1
pH = (pK1 + pK 2 )
2
For a diprotic amino acid, the isoelectric pH is halfway between the two pKa values. The
isoelectric pH of alanine is 0.5(2.34 + 9.87) = 6.10
The isoelectric and isoionic points for a polyprotic acid are almost the same. At the isoelectric
pH, the average charge of the molecule is 0; thus [H2A+] = [A−] and pH = 0.5(pK1 + pK2). At the
isoionic point, the pH is given by a more complex equation (see previous slide), and [H2A+] is
not exactly equal to [A−].
Isoelectronic Focusing
Isoelectronic focusing is a technique used for separation of proteins. At the isoelectronic
point, the average charge of all proteins is zero. The application of a strong electric field will
not result in migration of these proteins if they are at their isoelectronic pH’s.
However, if a medium is used in which there is a gradient of pH values, proteins migrate
(positive to negative poles and negative to positive poles) until they reach regions of their
isoelectronic pH’s. (i.e., Each protein is focused in one region at its isoelectric pH).
Lab-on-a-chip isoelectric focusing: 6-mm-long × 100 μm-wide
× 25-μm-deep capillary etched into silica glass.
If a molecule diffuses out of its isoelectric region, it becomes
charged and immediately migrates back to its isoelectric
zone.
(i) Fluorescent pH markers.
(ii) and (iii) Separation of fluorescence-labeled proteins:
(OVA) ovalbumin; (GFP) green fluorescent protein; (BSA)
bovine serum albumin; (Tfer) transferrin; (CA) carbonic
anhydrase; (PhB) phosphorylase B; and (Hb) hemoglobin.
[G. J. Sommer, A. K. Singh, and A.V. Hatch, “On-Chip
Isoelectric Focusing Using Photopolymerized Immobilized pH
Gradients,” Analyt. Chem. 2008, 80, 3327.]