Fundamentals of Analytical
Chemistry, 10e
Chapter 13: Complex Acid-Base
Systems
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1
Chapter Objectives (1 of 2)
By the end of this chapter, you should know:
• How to find the pH of mixtures of strong and weak acids and strong
and weak bases.
• How to construct titration curves of mixtures with NaOH.
• How to calculate the pH in solutions of polyfunctional acids and
bases, such as phosphoric and carbonic acid.
• How to calculate the pH of buffer solutions made from
polyfunctional acids.
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2
Chapter Objectives (2 of 2)
• How to determine the pH of amphiprotic salts.
• How to construct titration curves of polyfunctional acids and bases.
• How to find the pH of sulfuric acid solutions.
• How to make titration curves for amphiprotic species such as
amino acids.
• How to compute alpha values for polyprotic acid solutions.
• How to construct and use logarithmic concentration diagrams.
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3
Important Equations
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4
13A Mixtures of Strong and Weak Acids or
Strong and Weak Bases (1 of 3)
• Each of the components in a mixture containing a strong acid and
a weak acid (or a strong base and a weak base) can be
determined provided that the concentrations of the two are of the
same order of magnitude and that the dissociation constant for the
weak acid or base is somewhat less than about 104.
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5
Example 13-1 (1 of 3)
Calculate the pH of a mixture that is 0.1200 M in hydrochloric acid and 0.0800 M in the weak acid
HA(K a  1.00  10 4 ) during its titration with 0.1000 M KOH. Compute results for additions of the
following volumes of base: (a) 0.00 mL and (b) 5.00 mL to 25.00 mL of the mixture.
Solution
(a) 0.00 mL KOH
The molar hydronium ion concentration in this mixture is equal to the concentration of HCl plus the
concentration of hydronium ions that results from dissociation of HA and H2O. In the presence of
the two acids, however, the concentration of hydronium ions from the dissociation of water is
extremely small. We, therefore, need to take into account only the other two sources of protons.
Thus,
0
 H 3O    cHCL
  A    0.1200   A  
Note that [A  ] is equal to the concentration of hydronium ions from the dissociation of HA.
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6
Example 13-1 (2 of 3)
Now, assume that the presence of the strong acid so represses the dissociation of HA that
[A  ]  0.1200 M; then,
H3O+   0.1200M, and the pH is 0.92
To check this assumption, the provisional value for [H3O ] is substituted into the dissociationconstant expression for HA. When this expression is rearranged, we obtain
 A  
Ka
1.00  104
4



8.33

10
0.1200
 HA   H3O+ 
This expression can be rearranged to
 HA 
  A   /  8.33  104 
From the concentration of the weak acid, the mass-balance expression is
0
cHA
  HA    A    0.0800 M
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7
Example 13-1 (3 of 3)
Substituting the value of [HA] from the previous equation gives
 A  
 8.33
 10 4

  A   
1.20

 103  A    0.0800 M
 A    6.7  10 5 M
Note that
 A  
is indeed much smaller than 0.1200 M, as assumed.
b) 5.00 mL KOH
cHCl 
and we may write
25.00  0.1200  5.00  0.100
 0.0833 M
25.00  5.00
H3O   0.0833   A    0.0833
pH  1.08
To determine whether the assumption is still valid, compute  A   as we did in part (a), knowing that
the concentration of HA is now 0.0800 × 25.00/30.00 = 0.0667, and find
 A    8.0  10 5 M
which is still much smaller than 0.0833 M.
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8
Example 13-2 (1 of 3)
Calculate the pH of the resulting solution after the addition of 29.00 mL of 0.1000 M NaOH to
25.00 mL of the solution described in Example 13-1.
Solution
In this case,
cHCL 
25.00  0.1200  29.00  0.1000
 1.85  103 M
25.00  29.00
cHA 
25.00  0.0800
 3.70  10 2 M
54.00

3
As in the previous example, a provisional result based on the assumption that [H3O ]  1.85  10 M
yields a value of 1.90  10 3 M for [A  ]. Note that [A  ] is no longer much smaller than
[H3O ], and thus
 H 3O +   cHCL   A    1.85  10 3   A  
(13-1)
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9
Example 13-2 (2 of 3)
In addition, from mass-balance considerations, we know that
 HA 
  A    cHA  3.70  10 2
(13-2)
Rearrange the acid dissociation-constant expression for HA and obtain
 HA 
 H 3O +   A  

1.00  104
Substitution of this expression into Equation 13-2 yields
 H 3O +   A  
  A    3.70  10 2
4
1.00  10
3.70  106
 A  
 H 3O +   1.00  104

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10
Example 13-2 (3 of 3)
Substitution for [A  ] and cHCI in Equation 13-1 yields
6
3.70

10
 H 3O +   1.85  103 
 H 3O +   1.00  104
Multiplying through to clear the denominator and collecting terms gives
 H 3O 
+
2
 1.75  10 3   H 3O +   3.885  10 6  0
Solving the quadratic equation gives
 H 3O +   3.03  10 3 M
pH = 2.52
Note that the contributions to the hydronium ion concentration from
HCl 1.85 × 10 3 M and HA  3.03 × 10 3 M  1.85 × 10 3 M  are of comparable
magnitude. Hence, we cannot make the assumption made in Example 13-1.
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11
13A Mixtures of Strong and Weak Acids or
Strong and Weak Bases (2 of 3)
•
Figure 13-1 shows curves for the titration
of strong/weak acid mixtures with 0.1000 M
NaOH. Each titration curve is for 25.00 mL
of a solution that is 0.1200 M in HCl and
0.0800 M in the weak acid HA.
•
The rise in pH at the first equivalence point
is small or essentially nonexistent when the
weak acid has a relatively large
dissociation constant.
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12
13A Mixtures of Strong and Weak Acids or
Strong and Weak Bases (3 of 3)
•
For titrations such as these, only the total
number of millimoles of weak and strong
acid can be determined accurately.
•
When the weak acid has a very small
dissociation constant, only the strong acid
content can be determined.
•
For weak acids of indeterminate strength,
there are usually two useful end points.
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13
13B Polyfunctional Acids and Bases
• Species with two or more acidic or basic functional groups exhibit
polyfunctional acidic or basic behavior.
• These usually exhibit multiple end points in a neutralization titration.
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14
13B-1 The Phosphoric Acid System (1 of 3)
• Phosphoric acid is a polyfunctional acid that undergoes the
following three dissociation reactions:
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15
13B-1 The Phosphoric Acid System (2 of 3)
• The equilibrium constant for combining the first two dissociation
equilibria for H3PO4 is:
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16
13B-1 The Phosphoric Acid System (3 of 3)
• The equation below shows how to calculate the equilibrium
constant for the overall reaction that results from combining all
three dissociation reactions of phosphoric acid.
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17
13B-2 The Carbon Dioxide/Carbonic Acid
System (1 of 2)
• Equations 13-3, 13-4, and 13-5 describe the reactions that occur
when carbon dioxide is dissolved in water.
(13-3)
(13-4)
(13-5)
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18
13B-2 The Carbon Dioxide/Carbonic Acid
System (2 of 2)
•
Combining Equations 13-3 and 13-4 gives Equations 13-6 and 13-7.
(13-6)
(13-7)
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19
Example 13-3 (1 of 2)
Calculate the pH of a solution that is 0.02500 M CO2 .
Solution
The mass-balance expression for CO2 -containing species is
0

2



cCO

0.02500

CO
aq

H
CO

HCO

CO






2
3
3 
 2


 3 
2
The small magnitude of K hyd , K1, and K 2 (see Equations 13-3, 13-4, and 13-5) suggests that
 H CO 
2
and
3
  HCO3   CO32 

 CO 2  aq  
0
 0.02500 M
CO 2  aq    cCO
2
The charge-balance equation is
 H 3O +    HCO3   2 CO32   OH  
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20
Example 13-3 (2 of 2)
Then assume that
2 CO32   OH     HCO3 
Therefore,
 H 3O +    HCO3 
Substituting these approximations in Equation 13-6 leads to
2
 H 3O + 
 K a1  4.2  10 7
0.02500
 H 3O +  
0.02500  4.2  10 7  1.02  10 4 M
pH   log 1.02  10 4   3.99
Calculating values for H2CO3 , CO3 2   , and OH  indicates that the assumptions were valid.
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21
13C Buffer Solutions Involving Polyprotic
Acids (1 of 3)
• Two buffer systems can be prepared from a weak dibasic acid and
its salts.
 One contains free acid H2 A and its conjugate base NaHA.
 The second contains the acid NaHA and its conjugate base Na2 A.
•
The pH of the NaHA Na2 A system is higher than that of the
H2 A NaHA system because the acid dissociation constant for HA 
is always less than that for H2 A.
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22
Example 13-4 (1 of 2)
Calculate the hydronium ion concentration for a buffer solution that is 2.00 M in
phosphoric acid and 1.50 M in potassium dihydrogen phosphate.
Solution
The principal equilibrium in this solution is the dissociation of H3PO4 .
Assume that the dissociation of H2PO4  is negligible, that is,
 H3PO4 
 cH0 3PO4  2.00 M
0
 H 2 PO 4    cKH
 1.50 M
2 PO 4
7.11  103  2.00
+
 H 3O  
 9.48  10 3 M
1.50
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23
Example 13-4 (2 of 2)
Now use the equilibrium-constant expression for K a2 to see if the assumption was valid.
K a2  6.32  10
8
 H 3O    HPO 4 2 
9.48  103  HPO 4 2 



1.50
 H 2 PO 4 
Solving this equation yields
 HPO 4 2   1.00  105 M
Since this concentration is much smaller than the concentrations of the major
species, H3PO4 and H2PO 4  , the assumption is valid. Note that [PO43 ]
is even smaller than [HPO4 2  ].
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24
13C Buffer Solutions Involving Polyprotic
Acids (2 of 3)
•
For a buffer prepared from NaHa and Na2 A, the second dissociation
usually predominates.
•
The concentration of H2 A is negligible compared with that of HA or A 2  .
•
The hydronium ion concentration can be calculated from the second
dissociation constant by the techniques for a simple buffer solution.
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25
Example 13-5 (1 of 4)
Calculate the hydronium ion concentration of a buffer that is 0.0500 M in
potassium hydrogen phthalate (KHP) and 0.150 M in potassium phthalate (K 2P).
Solution
Assume that the concentration of H2P is negligible in this solution.
Therefore,
0
 HP    cKHP
 0.0500 M
 P 2   cK0 2 P  0.150 M
6
3.91

10
 0.0500
 H 3O   
 1.30  10 6 M
0.150
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26
Example 13-5 (2 of 4)
To check the first assumption, an approximate value for [H2P] is calculated
by substituting numerical values for [H3O] and [HP ] into the K a1 expression:
K a1
 H 3O    HP  

 1.12  103 
H2P
H2P
1.30  10   0.0500 
6
H2P
 6  10 5 M
Since [H2P]  [HP ] and [P2  ], the assumption that the reaction of HP to form
OH is negligible is justified.
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27
Example 13-5 (3 of 4)
Calculate the hydronium ion concentration of a buffer that is 0.0500 M in
potassium hydrogen phthalate (KHP) and 0.150 M in potassium phthalate (K 2P).
Solution
Assume that the concentration of H2P is negligible in this solution.
Therefore,
0
 HP    cKHP
 0.0500 M
 P 2   cK0 2 P  0.150 M
3.91106  0.0500

 H 3O  
 1.30  10 6 M
0.150
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28
Example 13-5 (4 of 4)
To check the first assumption, an approximate value for [H2P] is calculated
by substituting numerical values for [H3O] and [HP ] into the K a1 expression:
K a1
 H 3O    HP  

 1.12 103 
H2P
H2P
1.30 10   0.0500 
6
H2P
 6 105 M
Since [H2P]  [HP ] and [P2  ], the assumption that the reaction of HP to form
OH is negligible is justified.
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29
13C Buffer Solutions Involving Polyprotic
Acids (3 of 3)
• Although the assumption of a single principal equilibrium is often
satisfactory to estimate the pH of buffer mixtures derived from
polybasic acids, appreciable errors occur under two circumstances:
 The concentration of the acid or salt is very low.
 The two dissociation constants are numerically close.
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30
13D Calculation of the pH of Solutions of
NaHA (1 of 6)
• When 1 mol of NaOH is added to a solution containing 1 mol of the
acid H2 A, 1 mole of NaHA is formed, and the pH is determined by
two equilibria.
•
If the first reaction predominates, the reaction will be acidic. If the
second reaction predominates, the reaction will be basic.
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31
13D Calculation of the pH of Solutions of
NaHA (2 of 6)
•
Equations 13-8 and 13-9 calculate the equilibrium constants for these
two equilibria.
K a2
K b2 
 H 3O    A 2 

 HA  
KW

K a1
 H 2 A  OH  
(13-8)
(13-9)
 HA  
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32
13D Calculation of the pH of Solutions of
NaHA (3 of 6)
•
Equation 13-10 is the mass-balance expression.
cNaHA   HA     H 2 A    A 2 
•
(13-10)
The charge-balance equation is
 Na     H 3O     HA    2  A 2   OH  
•
The charge-balance equation can be rewritten as Equation 13-11.
cNaHA   H 3O     HA    2  A 2   OH  
(13-11)
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33
13D Calculation of the pH of Solutions of
NaHA (4 of 6)
• Equation 13-12 is produced by subtracting the mass-balance
equation from the charge-balance equation.
cNaHA   H 3O     HA    2  A 2   OH   charge balance
cNaHA   H 2 A    HA     A 2 
mass balance
 H 3O     A 2   OH     H 2 A 
(13-12)
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34
13D Calculation of the pH of Solutions of
NaHA (5 of 6)
• Equation 13-15 is produced by rearranging the acid-dissociation
constant expressions for H2 A and HA  , substituting these
expressions and that for the ion-product constant for water (K w )
into Equation 13-12, rearranging, and using the approximation
given by Equation 13-14.
 HA    cNaHA
 H 3O  

K a2 cNaHA  K W
1  cNaHA / K a1
(13-14)
(13-15)
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35
13D Calculation of the pH of Solutions of
NaHA (6 of 6)
•
When the ratio cNaHA K a1 is much larger than unity in the denominator of
Equation 13-15 and K a2cNaHA is considerably greater than K w in the
numerator, Equation 13-15 simplifies to Equation 13-16.
 H 3O   
K a1 K a2
(13-16)
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36
Example 13-6 (1 of 2)
Calculate the hydronium ion concentration of a 1.00  10 3 M Na2HPO4 solution.
Solution
The pertinent dissociation constants are K a2 and K a3 , which both contain [HPO4 2 ].
Their values are K a2  6.32  10 8 and K a3  4.5  10 13. In the case of a
Na2HPO 4 solution, Equation 13-15 can be written as
 H 3O   
K a3cNaHA  K W
1  cNaHA / K a2
Note that we have used K a3 in place of K a2 in Equation 13-15 and K a2 in place of
K a1 since these are the appropriate dissociation constants when Na2HPO4
is the salt.
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37
Example 13-6 (2 of 2)
Checking the assumptions that led to Equation 13-16, the term
cNaHA K a2  (1.0  10 3 ) (6.32  10 8 ) is much larger than 1 so that the denominator
can be simplified. In the numerator, however, K a3cNaHA  4.5  1013  1.00  103
is comparable to K w so that no simplification can be made there. Therefore, use a
partially simplified version of Equation 13-15:
 H 3O   
K a3cNaHA  K W
cNaHA
 4.5  10 1.00  10   1.00  10
1.00  10  /  6.32  10 
13

3
3
8
14
 8.1  1010 M
The simplified Equation 13-16 gave 1.7  10 10 M, which is in error by a large
amount.
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38
Example 13-7
Find the hydronium ion concentration of a 0.0100 M NaH2PO4 solution.
Solution
2
[H
PO
The two dissociation constants of importance (those containing 2 4 ]
are K a1  7.11  10 3 and K a2  6.32  10 8 ). A test shows that the
denominator of Equation 13-15 cannot be simplified, but the numerator
reduces to K a2cNaH2PO4 . Thus, Equation 13-15 becomes,
 6.32  10 1.00  10 
1  1.00  10  /  7.11  10  


8
 H 3O   
2
2
3
 1.62  10 5 M
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39
13E Titration Curves for Polyfunctional Acids
(1 of 3)
•
Figure 13-2 shows the titration curve for
a diprotic acid (20.00 mL of 0.1000 M H2 A
with 0.1000 M NaOH).
•
The K a1 K a2 ratio is significantly greater
than 103 , so the curve (except for the
first equivalence point) can be calculated
as though it contained a single
monoprotic acid with a dissociation
3
constant of K a1  1.00  10 .
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40
Example 13-9 (1 of 19)
Construct a curve for the titration of 25.00 mL of 0.1000 M maleic acid,
with 0.1000 M NaOH.
Write the two dissociation equilibria as
Because the ratio K a1 K a2 is large (2  10 4 ), proceed using the techniques just
described. Note that the two equivalence points should occur at 25.00 mL and
50.00 mL of NaOH.
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41
Example 13-9 (2 of 19)
Solution
Initial pH
Initially, the solution is 0.1000 M H2M. At this point, only the first dissociation
makes an appreciable contribution to [H3O ]; thus,
 H 3O     HM  
Mass balance requires that
cH0 2 M   H 2 M    HM     M 2   0.1000 M
Since the second dissociation is negligible, [M2  ] is very small so that
cH0 2 M   H 2 M    HM    0.1000 M
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42
Example 13-9 (3 of 19)
or
H2M
 0.1000   HM    0.1000   H 3O  
Substituting these relationships into the expression for K a1 gives
2
K a1  1.3  102
 H 3O +   HM  
 H 3O  


0.1000   H 3O  
H2M
Rearranging yields
 H 3O 

2
 1.3  102  H 3O    1.3  10 3  0
Because K a1 for maleic acid is relatively large, we must solve the quadratic
equation or find [H3O ] by successive approximations. Doing so, gives
 H 3O  
2
 3.01  10 2 M
pH =  log  3.01  102   2  log 3.01  1.52
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43
Example 13-9 (4 of 19)
First Buffer Region
The addition of base, for example 5.00 mL, results in the formation of 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, the solution can be treated as a
simple buffer system. Thus, applying Equations 7-27 and 7-28 gives
5.00  0.1000
 1.67  10 2 M
30.00
25.00  0.1000  5.00  0.1000

 6.67  10 2 M
30.00
cNaHM   HM   
cH2M   H 2 M 
Substitution of these values into the equilibrium-constant expression for K a1
yields a tentative value of 5.2  10 2 M for [H3O  ]. It is clear, however, that the
approximation H3O+  << cH M or cHM is not valid; therefore, Equations 7-25 and 7-26
must be used, and
2

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44
Example 13-9 (5 of 19)
 HM    1.67  102   H 3O    OH  
H2M
 6.67  102   H 3O    OH  
Because the solution is quite acidic, the approximation that [OH ] is very small is
surely justified. Substitution of these expressions into the dissociation-constant
relationship gives

2

K a1

 H 3O  1.67  10   H 3O 

6.67  102   H 3O  
 H 3O  
2

 2.97  10  H O
2
3


 1.3  10 2
  8.67  10 4  0
 H 3O    1.81  102 M
pH =  log 1.81  102   1.74
Additional points in the first buffer region are computed in a similar way until just
prior to the first equivalence point.
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45
Example 13-9 (6 of 19)
Just Prior to First Equivalence Point
Just prior to the first equivalence point, the concentration of H2M is so small that it
becomes comparable to the concentration of M2 , and the second equilibrium
must also be considered. Within approximately 0.1 mL of the first equivalence
point, we have a solution of primarily HM with a small amount of H2M remaining
and a small amount of M2 formed. For example, at 24.90 mL of NaOH added,
24.90  0.1000
 HM   cNaHM 
 4.99  10 2 M
49.90
25.00  0.1000
24.90  0.1000
cH2M 

 2.00  10 4 M
49.90
49.90

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46
Example 13-9 (7 of 19)
Mass balance gives
cH2 M  cNaHM   H 2 M    HM     M 2 
Charge balance gives
 H 3O     Na +    HM    2  M 2   OH  
Since the solution consists primarily of the acid HM at the first equivalence point, we
can safely neglect [OH ] in the previous equation and replace [Na ] with cNaHM.
After rearranging, we obtain
cNaHM   HM    2  M 2     H 3O  
Substituting this equation into the mass-balance expression and solving for [H3O ] give
 H 3O    cH2 M   M 2    H 2 M 
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47
Example 13-9 (8 of 19)
If we express M2  and H2M in terms of HM  and H3O+  , the result is
 H 3O    cH2 M 
K a 2  HM  
 H 3O  
 H 3O    HM  

K a1

Multiplying through by [H3O ] gives, after rearrangement,
  HM   
   c  H O +   K  HM    0
 H 3O   1  
H2M  3
a2 



K a1 


2
Substituting HM   4.99  102 , cH M  2.00  104 , and the values for K a1 and K a2 leads to
2
4.838  H 3O + 
2
 2.00  10 4  H 3O +   2.94  10 8  0
The solution to this equation is
 H 3O +   1.014  10 4 M
pH = 3.99
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48
Example 13-9 (9 of 19)
The same reasoning applies at 24.99 mL of titrant, where we find
H3O+  = 8.01 × 10 5 M
pH = 4.10
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49
Example 13-9 (10 of 19)
First Equivalence Point
At the first equivalence point,
 HM   cNaHM

25.00  0.1000

 5.00  10 2 M
50.00
Simplification of the numerator in Equation 13-15 is certainly justified. On the
other hand, the second term in the denominator is not <<1. Hence,
 H 3O  

K a2 cNaHM

1  cNaHM / K a1
5.9  107  5.00  10 2
1   5.00  102  / 1.3  10 2 
 7.80  105 M
pH =  log  7.80  105 M   4.11
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50
Example 13-9 (11 of 19)
Just After the First Equivalence Point
Prior to the second equivalence point, we can obtain the analytical concentrations of
NaHM and Na2M from the titration stoichiometry. At 25.01 mL, for example, the values are
cNaHM
cNa 2 M
mmol NaHM formed   mmol NaOH added  mmol NaHM formed 
total volume of solution
25.00  0.1000   25.01  25.00   0.1000

 0.04997 M
50.01
 mmol NaOH added  mmol NaHM formed   1.9996  105 M

total volume of solution

In the region a few tenths of a milliliter beyond the first equivalence point, the
solution is primarily HM with some M2 formed as a result of the titration. The
mass balance at 25.01 mL added is
cNa 2 M  cNaHM   H 2 M    HM     M 2   0.04997  1.9996  10 5
 0.04999 M
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51
Example 13-9 (12 of 19)
and the charge balance is
 H 3O     Na     HM    2  M 2   OH  
Again, the solution should be acidic, and so, we can neglect OH as an important
species. The Na concentration equals the number of millimoles of NaOH added divided
by the total volume, or
 Na   
25.01  0.1000
 0.05001 M
50.01
Subtracting the mass balance from the charge balance and solving for [H3O ] gives

 H 3O +    M 2    H 2 M   cNa 2 M  cNaHM

  Na + 
Expressing the [M2 ] and [H2M] in terms of the predominant species HM , we have
 H 3O  
+
K a2  HM  
 H 3O + 
 H 3O +   HM  

 cNa 2 M  cNaHM
K a1


  Na  
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52
Example 13-9 (13 of 19)
Since [HM ]  cNaHM  0.04997 M. Therefore, if we substitute this value and numerical
values for cNa2M  cNaHM and [Na+ ] into the previous equation, we have, after
rearranging, the following quadratic equation:
 H 3O   
K a1  H 3O + 
 K a1
2
K a2  0.04997 
 H 3O + 
 H 3O +   0.04997 

 1.9996  10 5
K a1
 0.04997 K a1 K a2  0.04997  H 3O + 
 0.04997   H 3O + 
2
2
 1.9996  10 5 K a1  H 3O  
 1.9996  10 5 K a1  H 3O +   0.04997 K a1 K a2  0
This equation can then be solved for [H3O ].
 H 3O +   7.60  10 5 M
pH = 4.12
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53
Example 13-9 (14 of 19)
Second Buffer Region

2
Further additions of base to the solution create a new buffer system consisting of HM and M .
When enough base has been added so that the reaction of HM with water to give OH can be
neglected (a few tenths of a milliliter beyond the first equivalence point), the pH of the mixture may
be calculated from K a2 . With the introduction of 25.50 mL of NaOH, for example,
 M 2   cNa 2 M 
 25.50  25.00  0.1000 
0.050

M
50.50
50.50
and the molar concentration of NaHM is
 HM   cNaHM 

 25.00
 0.1000    25.50  25.00  0.1000 
2.45

M
50.50
50.50
Substituting these values into the expression for K a2 gives
K a2

 H 3O   0.050 / 50.50
 H 3O    M 2 


 HM  
2.45 / 50.50

 5.9  107
 H 3O +   2.89  10 5 M
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54
Example 13-9 (15 of 19)
The assumption that [H3O ] is small relative to cHM and cM2 is valid, and pH = 4.54.
The other values in the second buffer region are calculated in a similar manner.
Just Prior to Second Equivalence Point
Just prior to the second equivalence point (49.90 mL and more), the ratio
[M2  ] [HM ] becomes large, and the simple buffer equation no longer applies.
At 49.90 mL, cHM  1.335  10 4 M, and cM2  0.03324 M. The primary
equilibrium is now
Write the equilibrium constant as
K b1
OH    HM  
KW


K a2
 M 2 

OH   1.335  104  OH  

0.03324  OH  



1.00  1014
8


1.69

10
5.9  107
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55
Example 13-9 (16 of 19)


In this case, it is easier to solve for [OH ] than for [H3O ]. Solving the resulting
quadratic equation gives
OH    4.10  106 M
pOH = 5.39
pH = 14.00  pOH = 8.61
The same reasoning for 49.99 mL leads to [OH ]  1.80  105 M, and pH = 9.26.
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 and the only one that we need to take
into account. Thus,
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56
Example 13-9 (17 of 19)
pH Just Beyond Second Equivalence Point
In the region just beyond the second equivalence point (50.01 mL, for example), we still
need to take into account the reaction of M2 with water to give OH since not enough
OH has been added in excess to suppress this reaction. The analytical concentration of
M2 is the number of millimoles of M2 produced divided by the total solution volume:
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57
Example 13-9 (18 of 19)
cM 2 
25.00  0.1000
 0.03333 M
75.01

The OH now comes from the reaction of M2  with water and from the excess OH added as
titrant. The number of millimoles of excess OH is then the number of millimoles of NaOH added
minus the number required to reach the second equivalence point. The concentration of this
excess is the number of millimoles of excess OH divided by the total solution volume, or
50.01  50.00   0.1000


OH 

 1.333  105 M
excess
75.01
The concentration of HM can now be found from K b1.
 M 2   cM 2   HM    0.03333   HM  
OH    1.3333  105   HM  
K b1

 HM   1.3333  105   HM  
 HM   OH  


2
 M 
0.03333   HM  

 1.69  108
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58
Example 13-9 (19 of 19)
Solving the quadratic equation for [HM ] gives
 HM    1.807  105 M
and
OH    1.3333  105   HM    1.33  10 5  1.807  10 5  3.14  105 M
pOH  4.50, and pH  14.00  pOH  9.50
The same reasoning applies to 50.10 mL where the calculations give pH = 10.14.
pH Beyond the Second Equivalence Point
Addition of more than a few tenths of a milliliter of NaOH beyond the second equivalence point gives
enough excess OH to repress the basic dissociation of M2 . The pH is then calculated from the
concentration of NaOH added in excess of that required for the complete neutralization of H2M.
Thus, when 51.00 mL of NaOH have been added, we have 1.00-mL excess of 0.1000 M NaOH, and
1.00  0.100
 1.32  10 3 M
76.00
pOH   log 1.32  103   2.88
OH   
pH  14.00  pOH  11.12
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59
13E Titration Curves for Polyfunctional Acids
(2 of 3)
•
Figure 13-3 shows the titration curve for
25.00 mL of 0.1000 M maleic acid, H2M,
titrated with 0.1000 M NaOH.
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60
13E Titration Curves for Polyfunctional Acids
(3 of 3)
•
Figure 13-4 shows the titration curves for
three polyprotic acids all titrated using a
0.1000 M NaOH solution.
•
Curve A: 25.00 mL of 0.1000 M H3PO4
•
Curve B: 25.00 mL of 0.1000 M oxalic acid
•
Curve C: 25.00 mL of 0.1000 M H2SO4 .
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Feature 13-1: The Dissociation of Sulfuric
Acid (1 of 3)
Sulfuric acid is unusual in that one of its protons behaves as a strong acid in water and
the other as a weak acid (K a2  1.02  102 ). Let us consider how the hydronium ion
concentration of sulfuric acid solutions is computed using a 0.0400 M solution as an
example.
First assume that the dissociation of HSO4  is negligible because of the large excess of
H3O resulting from the complete dissociation of H2SO4 . Therefore,
 H 3O     HSO 4    0.0400 M
2
An estimate of SO 4  based on this approximation and the expression for K a2
reveals that
0.0400 SO 4 2 
 1.02  102
0.0400
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62
Feature 13-1: The Dissociation of Sulfuric
Acid (2 of 3)
Note that SO 4 2  is not small relative to  HSO 4   , and a more rigorous solution is
required.
From stoichiometric considerations, it is necessary that
 H 3O    0.0400  SO 4 2  
The first term on the right is the concentration of H3O resulting from dissociation of the
H2SO4 to HSO 4  . The second term is the contribution of the dissociation of HSO4  .
Rearrangement yields
SO 4 2    H 3O    0.0400
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63
Feature 13-1: The Dissociation of Sulfuric
Acid (3 of 3)
Mass-balance considerations require that
cH2SO4  0.0400   HSO 4   + SO 4 2 
Combining the last two equations and rearranging yield
 HSO 4   = 0.0800   H 3O + 
2

By introducing these equations for SO 4  and HSO 4 into the expression for K a2 , we find
that

 H 3O +   H 3O +   0.0400
0.0800   H 3O + 

 1.02  102
Solving the quadratic equation for [H3O ] yields
 H 3O +   0.0471 M and pH  1.33
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64
Exercises in Excel Activity (1 of 4)
In Chapter 8 of Applications of Microsoft® Excel® in Analytical
Chemistry, 4th ed., we extend the treatment of neutralization titration
curves to polyfunctional acids. Both a stoichiometric approach and a
master equation approach are used for the titration of maleic acid with
sodium hydroxide.
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65
13F Titration Curves for Polyfunctional Bases
(1 of 2)
• The same principles used for constructing titration curves for
polyfunctional acids can be applied to titration curves for
polyfunctional bases.
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66
13F Titration Curves for Polyfunctional Bases
(2 of 2)
•
Figure 13-5 illustrates that two end
points appear in the titration of
sodium carbonate.
•
25.00 mL of 0.1000 M Na2CO3
is titrated with 0.1000 M HCl.
•
The second end point is sharper than
the first, suggesting that the individual
components in mixtures of sodium
carbonate and sodium hydrogen
carbonate can be determined by
neutralization methods.
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67
Exercises in Excel Activity (2 of 4)
The titration curve for a difunctional base being titrated with strong
acid is developed in Chapter 8 of Applications of Microsoft® Excel® in
Analytical Chemistry, 4th ed. In the example studied, ethylene
diamine is titrated with hydrochloric acid. A master equation approach
is explored, and the spreadsheet is used to plot pH versus fraction
titrated.
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68
13G Titration Curves for Amphiprotic Species
• An amphiprotic species when dissolved in a suitable solvent
behaves both as a weak acid and as a weak base.
• If either of its acidic or basic characters predominates, titration of
the substance with a strong acid or base may be feasible.
• Amino acids are amphiprotic.
• A zwitterion is an ionic species that has both a positive and a
negative charge.
• The isoelectric point of a species is the pH at which no net
migration occurs in an electric field.
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69
Feature 13-2: Acid-Base Behavior of Amino
Acids (1 of 5)
The simple amino acids are an important class of amphiprotic compounds that
contain both a weak acid and a weak base functional group. In an aqueous
solution of a typical amino acid, such as glycine, three important equilibria
operate:
(13-17)
(13-18)
(13-19)
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70
Feature 13-2: Acid-Base Behavior of Amino
Acids (2 of 5)
The first equilibrium constitutes a kind of internal acid-base reaction and is analogous to
the reaction one would observe between a carboxylic acid and an amine:
(13-20)
The typical aliphatic amine has a base dissociation constant of 10 4 to 10 5 (see
Appendix 3), while many carboxylic acids have acid dissociation constants of about the
same magnitude. As a result, both Reactions 13-18 and 13-19 proceed far to the right,
with the product or products being the predominant species in the solution.
The amino acid species in Reaction 13-17, which bears both a positive and a negative
charge, is called a zwitterion. As shown by Reactions 13-18 and 13-19, the zwitterion of
glycine is stronger as an acid than as a base. Thus, an aqueous solution of glycine is
somewhat acidic.
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71
Feature 13-2: Acid-Base Behavior of Amino
Acids (3 of 5)
The zwitterion of an amino acid, which contains both a positive and a negative charge,
has no tendency to migrate in an electric field, although the singly charged anionic and
cationic species are attracted to electrodes of opposite polarity. No net migration of the
amino acid occurs in an electric field when the pH of the solvent is such that the
concentrations of the anionic and cationic forms are identical. The pH at which no net
migration occurs is called the isoelectric point and 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,
NH2CH2COO   H3O+ 
Ka =
NH3+CH2COO  
NH3+CH2COOH OH 
Kb =
NH3+CH2COO  
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72
Feature 13-2: Acid-Base Behavior of Amino
Acids (4 of 5)
At the isoelectric point,
 NH 2 CH 2 COO     NH 3 CH 2COOH 
Therefore, if we divide K a by K b and substitute this relationship, we obtain for the
isoelectric point
 H 3O    NH 2 CH 2 COO  
 H 3O  
Ka




Kb
OH   NH 3 CH 2 COOH 
OH  
If we substitute Kw [H3O ] for [OH ] and rearrange, we get
Ka K w
Kb
 H 3O   
The isoelectric point for glycine occurs at a pH of 6.0, that is
 2  10 1  10 
10
 H 3O  

2  10
14
12
 1  10 6 M
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73
Feature 13-2: Acid-Base Behavior of Amino
Acids (5 of 5)
For simple amino acids, K a and K b are generally so small that their determination
by direct neutralization is impossible. Addition of formaldehyde removes the
amine functional group, however, and leaves the carboxylic acid available for
titration with a standard base. For example, with glycine,
NH3+CH2COO + CH2O  CH2C = NCH2COOH + H2O
The titration curve for the product is that of a typical carboxylic acid.
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74
Exercises in Excel Activity (3 of 4)
The final exercise in Chapter 8 of Applications of Microsoft® Excel® in
Analytical Chemistry, 4th ed., considers the titration of an amphiprotic
species, phenylalanine. A spreadsheet is developed to plot the
titration curve of this amino acid, and the isoelectric pH is calculated.
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75
13H Composition of Polyprotic Solutions as a
Function of pH (1 of 5)
• Alpha values are useful in considering properties of polyfunctional
acids and bases.
•
If cT is the sum of the molar concentrations of the maleate-containing species
in the solution throughout the titration described in Example 13-9, then the
equation shown describes the alpha value for the free acid α0 .
α0 =
H2M
cT
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76
13H Composition of Polyprotic Solutions as a
Function of pH (2 of 5)
•
Equation 13-21 calculates cT .
cT = H2M + HM  + M2 
•
(13-21)
The alpha values for HM and M2  are given by the two equations below.
HM 
α1 =
cT
M2 
α2 =
cT
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77
13H Composition of Polyprotic Solutions as a
Function of pH (3 of 5)
•
The sum of the alpha values of a system must equal 1 as shown.
 0  1   2  1
•
Equations 13-22, 13-23, and 13-24 show how to obtain α0 , α1, and α2 for the
maleic acid system.
[H 3O  ]2

[H 3O  ]2  K a1[H 3O  ]  K a1 K a 2
(13-22)
K a1[H 3O  ]
1 
[H 3O  ]2  K a1[H 3O  ]  K a1 K a 2
(13-23)
K a1 K a2
[H 3O  ]2  K a1[H 3O  ]  K a1 K a 2
(13-24)
0
2 
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78
Feature 13-3: A General Expression for
Alpha Values
For the weak acid Hn A, the denominator D in all alpha-value expressions takes
the form:
D  [H 3O  ]n  K a1[H 3O  ]
n 1
 K a1 K a2 [H 3O  ]
n  2

K a1 K a2
K an
The numerator for α0 is the first term in the denominator, and for α1, it is the
n 1
second term, and so forth. Thus, α 0 = [H3O+ ]n D , and α1 = K a1 [H3O+ ]  D .
Alpha values for polyfunctional bases are generated in an analogous way, with
the equations being written in terms of base dissociation constants and [OH ].
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79
13H Composition of Polyprotic Solutions as a
Function of pH (4 of 5)
•
Figure 13-6 shows the alpha values
for each maleate-containing species
of H2M solutions as a function of pH.
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80
13H Composition of Polyprotic Solutions as a
Function of pH (5 of 5)
•
Figure 13-7 shows the titration of
25.00 mL of 0.1000 M maleic acid
with 0.1000 M NaOH.
•
Solid curves are plots of alpha
values as a function of titrant
volume.
•
The broken curve is the titration
curve of pH as a function of
volume.
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81
Exercises in Excel Activity (4 of 4)
In the first exercise in Chapter 8 of Applications of Microsoft® Excel® in
Analytical Chemistry, 4th ed., we investigate the calculation of distribution
diagrams for polyfunctional acids and bases. The alpha values are plotted
as a function of pH. The plots are used to find concentrations at a given pH
and to infer which species can be neglected in more extensive calculations.
A logarithmic concentration diagram is constructed. The diagram is used to
estimate concentrations at a given pH and to find the pH for various starting
conditions with a weak acid system.
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82
Analytical Chemistry Online Activity
Search for an Excel tutorial on polyprotic acid titration curves. Learn
how to construct first and second derivative curves and to plot them
on the same axis as a pH titration curve. One possible website is
https://www.youtube.com/watch?v=l2Z8gK4adqk. Find a different
website that discusses distribution diagrams of polyprotic acids.
Describe how pK a values can be obtained from distribution diagrams.
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83
Key Terms Activity
•
Alpha values
•
Polyfunctional acids
•
Amino acid titrations
•
Polyfunctional bases
•
Amphiprotic species
•
Sulfuric acid dissociations
•
Isoelectric point
•
Titrations of polyfunctional acids
•
Logarithmic concentration diagrams
•
Zwitterion
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Assessments: Discussion Questions
13-1. As its name implies, NaHA is an “acid salt” because it has a proton available
to donate to a base. Briefly explain why a pH calculation for a solution of NaHA
differs from that for a weak acid of the type HA.
13-2. Explain the origin and significance of each of the terms on the right side of
Equation 13-12. Does the equation make intuitive sense? Why or why not?
13-3. Briefly explain why Equation 13-15 can only be used to calculate the
hydronium ion concentration of solutions in which NaHA is the only solute that
determines the pH.
13-4. Why is it impossible to titrate all three protons of phosphoric acid in aqueous
solution?
Skoog, West, Holler, and Crouch, Fundamentals of Analytical Chemistry, 10e. © 2022 Cengage. All Rights Reserved. May
not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
85
Challenge Problem (1 of 2)
13-38. Challenge Problem:
(a) Plot logarithmic concentration diagrams for 0.10000 M solutions of each
of the acids in Problem 13-36.
(b) For phthalic acid, find the concentrations of all species at pH 4.8.
(c) For tartaric acid, find the concentrations of all species at pH 4.3.
(d) From the log concentration diagram, find the pH of a 0.1000 M solution
of phthalic acid, H2P. Find the pH of a 0.100 M solution of HP .
Skoog, West, Holler, and Crouch, Fundamentals of Analytical Chemistry, 10e. © 2022 Cengage. All Rights Reserved. May
not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
86
Challenge Problem (2 of 2)
e) Discuss how you might modify the log concentration diagram for
phthalic acid so that it shows the pH in terms of the hydrogen ion
activity, aH , instead of the hydrogen ion concentration
pH   log aH , instead of pH   log cH . Be specific in your
discussion and show what the difficulties might be.
+
+

Skoog, West, Holler, and Crouch, Fundamentals of Analytical Chemistry, 10e. © 2022 Cengage. All Rights Reserved. May
not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
87