Introduction to
Analytical Chemistry
CHAPTER 8
TITRATING
POLYFUNCTIONAL
ACIDS AND BASES
8A Polyfunctional Acids
 With this acid, as with other polyprotic acids, Ka1＞Ka2
＞Ka3.
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8B Describing Polyfunctional
Bases
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8B Describing Polyfunctional
Bases
 The overall basic dissociation reaction of sodium carbonate is
described by the equations
 Solving the several simultaneous equations that are involved can
be difficult and time consuming. Fortunately, simplifying
assumptions can be invoked when the successive equilibrium
constants for the acid (or base) differ by a factor of about 10³ (or
more).
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8C Finding the pH of Solutions
of Amphiprotic Salts
 The solution could be acidic because of
or basic because of
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8C Finding the pH of Solutions
of Amphiprotic Salts
 Whether a solution of NaHA is acidic or basic depends
on the relative magnitude of the equilibrium constants
for these processes:
(8-1)
(8-2)
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8C Finding the pH of Solutions
of Amphiprotic Salts
 As we can see in Feature 8-2, solution of these
equations yields an approximate value of [H₃O⁺] that is
given by the equation
(8-3)
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8C Finding the pH of Solutions
of Amphiprotic Salts
 Frequently, the ratio cNaHA / Ka1 is much larger than
unity in the denominator of Equation 8-3 and Ka2cNaHA is
considerably greater than Kw in the numerator.
(8-4)
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Example 8-1
 Calculate the hydronium ion concentration of a 0.100 M NaHCO₃
solution.
 We first examine the assumptions leading to Equation 8-4. The
dissociation constants for H₂CO₃ are Ka1 = 1.5 × 10⁻⁴ and Ka2 =
4.69 × 10⁻¹¹. Clearly, cNaHA / Ka1 in the denominator is much larger
than unity; in addition, Ka2cNaHA has a value of 4.69 × 10¯¹², which
is substantially greater than Kw . Thus, Equation 8-4 applies and
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Example 8-2
 Calculate the hydronium ion concentration of a 1.00×10¯³ M
Na₂HPO₄ solution. The pertinent dissociation constants are Ka2
and Ka3 , which both contain [HPO₄2–]. Their values are Ka2 =
6.32 × 10¯⁸ and Ka3 = 4.5 × 10⁻¹³. Considering again the
assumptions that led to Equation 8-4, we find that the ratio
(1.00 × 10¯³)/(6.32 × 10¯⁸) is much larger than 1, so the
denominator can be simplified. The product Ka3cNa₂HPO₄ is by no
means much larger than Kw , however. We therefore use a
partially simplified version of Equation 8-3:
 Use of Equation 8-4 yields a value of 1.7 × 10¯¹⁰M.
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Figure 8-1
Figure 8-1 Titration of 20.00 mL
of 0.1000 M H2A with 0.1000 M
NaOH. For H2A, Ka1 = 1.00 10–3
and Ka2 = 1.00  10–7. The
method of pH calculation is
shown for several points and
regions on the titration curve.
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Example 8-4
 Construct a curve for the titration of 25.00 mL of
0.1000 M maleic acid,
with
0.1000 M NaOH. Consider the initial pH, the first buffer
region, the first equivalence point, the second buffer
region, the second equivalence point, and the region
beyond the second equivalence point. The calculations
are done manually here.
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Example 8-4
 Because the ratio Ka1/Ka2 is large (2 × 10⁴), we proceed
as just described for constructing Figure 8-1.
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Example 8-4
 Initial pH
 Substituting these relationships into the expression for
Ka1 gives
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Example 8-4
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Example 8-4
 First Buffer Region
 A buffer consisting of the weak acid H₂M and its
conjugate base HM¯. To the extent that dissociation of
HM¯ to give M²¯ is negligible,
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Example 8-4
 Substitution of these values into the equilibrium-
constant expression for Ka1 yields a tentative value of
5.2 × 10¯² M for [H₃O⁺]. It is clear, however, that the
approximation [H₃O⁺] << cH₂M or cHM¯ is not valid;
therefore, Equations 7-6 and 7-7 must be used, and
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Example 8-4
 Because the solution is quite acidic, the approximation
that [OH¯] is very small is surely justified. Substitution
of these expressions into the dissociationconstant
relationship gives
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Example 8-4
 First Equivalence Point
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Example 8-4
 Second Buffer Region
 Further additions of base to the solution create a new
buffer system consisting of HM¯ and M²¯.
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Example 8-4
 Second Equivalence Point
 After the addition of 50.00 mL of 0.1000 M sodium
hydroxide, the solution is 0.0333 M in Na2M (2.5
mmol/75.00 mL). Reaction of the base M²¯ with water
is the predominant equilibrium in the system
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Example 8-4
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Example 8-4
 pH Beyond the Second Equivalence Point
 When 51.00 mL of NaOH have been added,
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8D Constructing Titration Curves
for Polyfunctional Acids
 Figure 8-3 shows titration curves for three other
polyprotic acids.
 The ratio Ka1/Ka2 for oxalic acid (curve B) is
approximately 1000.
 The magnitude of the pH change is too small to permit
precise location of equivalence with an indicator.
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8D Constructing Titration Curves
for Polyfunctional Acids
 Curve A in Figure 8-3 is the theoretical titration curve
for triprotic phosphoric acid. Here, the ratio Ka1/Ka2 is
approximately 10⁵, as is Ka2/Ka3.
 Curve C is the titration curve for sulfuric acid, a
substance that has one fully dissociated proton and one
that is dissociated to a relatively large extent (Ka2 =
1.02 × 10¯²).
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Figure 8-3
Figure 8-3 Curves for the titration of
polyprotic acids. A 0.1000 M NaOH
solution is used to titrate 25.00 mL
of 0.1000 M H3PO4 (curve A), 0.1000
M oxalic acid (curve B), and 0.1000
M H2SO4 (curve C).
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8E Drawing Titration Curves for
Polyfunctional Bases
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Figure 8-4
Figure 8-4 Curve for the titration of
25.00 mL of 0.1000 M Na2CO3 with
0.1000 M HCl.
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Feature 8-4 Titration Curves for
Amino Acids
 The internal proton transfer from the carboxyl group to
the amine group of an amino acid is very favorable in
aqueous solution so that amino acids exist as
zwitterions as shown below for glycine.
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Feature 8-4 Titration Curves for
Amino Acids
 The conjugate acid of the zwitterion can be treated as a
polyprotic acid.
 The zwitterion can act as an acid because of the second
proton transfer or as a base because of
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Feature 8-4 Titration Curves for
Amino Acids
 The pH at which no net migration occurs is called the
isoelectric point; this point is an important physical
constant for characterizing amino acids. The isoelectric
point is readily related to the ionization constants for
the species. Thus, for glycine,
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Feature 8-4 Titration Curves for
Amino Acids
 At the isoelectric point,
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Feature 8-4 Titration Curves for
Amino Acids
 If we substitute Kw/[H₃O⁺] for [OH¯] and rearrange, we
get
 The isoelectric point for glycine occurs at a pH of 6.0.
That is,
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8F The Composition of Polyprotic Acid
Solutions as A Function of pH
(8-11)
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8F The Composition of Polyprotic Acid
Solutions as A Function of pH
(8-12)
(8-13)
(8-14)
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Figure 8-6
Figure 8-6 Titration of 25.00
mL of 0.1000 M maleic acid
with 0.1000 M NaOH. The
solid curves are plots of alpha
values as a function of volume.
The broken curve is the
titration curve of pH as a
function of volume.
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THE END
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