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Chapter 09
Polyprotic Acid-Base Equilibria
Proteins perform biological functions such as
structural support, catalysis of chemical
reactions, immune response to foreign
substances, transport of molecules across
membranes, and control of genetic expression.
The three-dimensional structure and function of a
protein is determined by the sequence of amino
acids from which the protein is made. The diagram
below shows how amino acids are connected to
make a polypeptide.
Polyprotic systems-acids or bases can donate or accept more than one proton.
Diprotic systems (with two acidic or basic sites).
9-1 Diprotic Acids and Bases
The amino acid building blocks of proteins have the general structure
Therefore, the nonionized form rearranges spontaneously to the zwitterion:
Diprotic acid:
H2L+ = HL + H+
HL = L- + H+
Diprotic base:
L- + H2O = HL + OHHL + H2O = H2L+ + OH-
Ka1 = K1
Ka2 = K2
(10-1)
(10-2)
Kb1
Kb2
(10-3)
(10-4)
Relations between
Ka and Kb:
That is, we would use the same procedure to find the pH of the diprotic H2A,
Where A is anything, or H2L+, where HL is leucine.
The Acidic Form, H2L+
In summary, a solution of a diprotic acid behaves like a solution of a monoprotic
acid, with Ka = Ka1.
The Basic Form, LL- + H2O = HL + OHHL + H2O = H2L+ + OH-
Kb1 = KW/Ka2 = 5.55 X 10-5
Kb2 = KW/Ka1 = 2.13 X 10-12
Kb1 tells us that L- will not hydrolyze (react with water) very much to give HL.
In summary, if there is any reasonable separation between Ka1 and Ka2 (and,
therefore, between Kb1and Kb2), the fully basic form of a diprotic acid can be
treated as monobasic, with Kb = Kb1.
BOX 9-1 Carbon Dioxide in the Air and Ocean
The Intermediate Form, HL
HL = H+ + LHL + H2O = H2L+ + OH-
Ka = Ka2 = 1.80 X 10-10
Kb = Kb2 = 2.13 X 10-12
(10-8)
(10-9)
A molecule that can both donate and accept a proton is said to be
amphiprotic.
However, the results apply to the intermediate form of any diprotic acid,
regardless of its charge.
[H+] + [H2L+] = [L-] + [OH-] or [H2L+] – [L-] + [H+] – [OH-] = 0
Intermediate form of diprotic acid:
Solution
Simplified Calculation for the Intermediate Form
or
Intermediate form of diprotic acid:
Equation 10-12 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.
Advice
When faced with the intermediate form of a diprotic acid, use Equation 10-11
to calculate the pH. The answer should be close to 1/2(pK1 + pK2).
Box 9-1 Carbon Dioxide in the Air and Ocean
Box 9-2 Successive Approximations
The method of successive approximations is a good way to deal with difficult
equations that do not have simple solutions.
[HM-]2 = F – [H2M]1 – [M2-]1
=0.001 00 – 0.000 129 – 0.000 175 = 0.000 696 M
 [H2M]2 = 8.53 X 10-5 M
[M2-]2 = 1.28 X 10-4 M
[HM-]3 = F – [H2M]2 – [M2-]2 = 0.000 786 M
[H+]3 = 4.37 X 10-5
[H+]4 = 4.35 X 10-5
Summary of Diprotic Acid Calculations
Solution of H2A
1. Treat H2A as a monoprotic acid with Ka = K1 to find [H+], [HA-], and [H2A].
K
H2A =1 H+ + HAF-x
x
x
2. Use the K2 equilibrium to solve for [A2-].
Solution of HA1. Use the approximation [HA-] ≈ F and find the pH with Equation 10-11.
The pH should be close to 1/2(pK1 + pK2).
2. With [H+] from step 1 and [HA-] ≈ F, solve for [H2A] and [A2-], using the
K1 and K2 equilibria.
Solution of A21. Treat A2- as monobasic, with Kb = Kb1 = KW/Ka2 to find [A2-], [HA-], and
[H+].
Kb1
A2- + H2O = HA- + OHF-x
x
x
2. Use the K1 equilibrium to solve for [H2A].
9-2 Diprotic Buffers
For the acid H2A, we can Write two Henderson-Hasselebalch equations, both of
which are always true.
9-3 Polyprotic Acids and Bases
H3A = H2A- + HH2A- = HA2- + H+
HA2- = A3- + H+
A3- + H2O = HA2- + OHHA2- + H2O = H2A- + OHH2A- + H2O = H3A + OH-
Ka1 = K1
Ka2 = K2
Ka3 = K3
Kb1 = KW/Ka3
Kb2 = KW/Ka2
Kb3 = KW/Ka1
1. H3A is treated as a monoprotic weak acid, with Ka = K1.
2. H2A- is treated as the intermediate form of a diprotic acid.
3. HA2- is also 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.
4. A3- is treated as monobasic, with Kb = Kb1 =KW/Ka3.
When you encounter an acid or base, decide whether you are dealing with an
acidic, basic, or intermediate form.
9-4 Which Is the Principal Species?
Benzoic acid
Speciation describes the distribution of analyte among possible species.
9-5 Fractional Composition Equations
Monoprotic Systems
Mass balance: F = [HA] + [A-]
The Fraction of molecules in the form HA is called αHA.
Fraction in the form HA:
In a similar manner, the fraction in the form A-, designated αA-, can be
obtained:
Fraction in the form A-:
Diprotic Systems
Mass balance:
For a diprotic system, we designate the fraction in the form H2A as αH A, the
fraction in the form HA- as αHA-, and the fraction in the form A2- as αA2-.
2
Fraction in the form H2A:
Fraction in the form HA-:
Fraction in the form A2-:
9-6 Isoelectric and Isoionic pH
The isoionic point (or isoionic pH) is the pH obtained when the pure,
neutral polyprotic acid HA (the neutral zwitterions) is dissolved in water.
The only ions are H2A+, A-, H+, and OH-. Most alanine is in the form HA,
and the concentrations of H2A+ and A- are not equal to each other.
The isoelectric point (or isoelectric pH) is the pH at which the average
charge of the polyprotic acid is 0. Most of the molecules are in the
uncharged form HA, and the concentrations of H2A+ and A- are equal to
each other.
When alanine is dissolved in water, the pH of the solution, by definition, is
the isoionic pH.
Isoionic point:
From [H+], K1, and K2, you could calculate [H2A+] = 1.68 X 10-5 M and
[A-] = 1.76 X 10-5 M for pure alanine in water ( the isoionic solution).
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.
Isoelectric point:
pH = 1/2(pK1 + pK2)
(10-23)
BOX 9-2 Isoelectric Focusing
At its isoelectric point, the
average charge of all forms of a
protein is 0.
This effect is the basis of a
sensitive technique of protein
separation called isoelectric
focusing.
A mixture of seven proteins( and
some impurities) was applied to a
polyacrylamide gel containing a
mixture of polyprotic compounds
called ampholytes.
The instrument that measures absorbance as a function of position along the
gel is called a densitometer.
For a protein, the isoionic pH is the pH of a solution of pure protein with no
other ions except H+ and OH-.
When the protein is subjected to intensive dialysis (Demonstration 27-1)
against pure water, the ph in the protein compartment approaches the isoionic
point if the counterions are free to pass through the semipermeable dialysis
membrane that retains the protein. The isoelectric point is the pH at which the
protein has no net charge.
The pH of zero charge is the pH at which {SiOH2+} = {SiO-} and, therefore, the
surface has no net charge. Colloidal particles (those with diameters in the range
1-100 nm) tend to remain dispersed when they are charged, but they flocculate
(come together and precipitate) near the pH of zero charge.