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94 BASIC CONCEPTS IN MEDICINAL CHEMISTRY
[BaseForm]
[AcidForm]
[BaseForm]
[BaseForm]
m]==
Scenario 1: pH Is Equal to the pK
a
This is the easiest of the three scenarios. When the pH = pKa, then the functional group is 50% ionized and 50% unionized regardless of whether it is an acid or a base. To avoid rote memorization, the
Henderson-Hasselbalch equation is used here to verify this fact. If you understand the proof below,
it is unnecessary to use the equation in subsequent problems.
HpKlog
Because the pH is equal to the pK
=+
a
, the following is true:
a
=
og
[AcidForm]
Calculating the antilog of each side of the equation shows that the concentration of the base
form is equal to the concentration of the acid form.
[AcidForm]
or [BaseForm] [AcidFor
Therefore, regardless of the acid/base character of the functional group, it is 50% ionized and
50% unionized whenever the environmental pH is the same as the pKa.
Scenario 2: pH Is Greater Than the pK
a
When the pH is greater than the pKa, then the environment is more basic than the functional group.
This scenario is seen with the sample problem involving captopril, in which the pH is 5.9 and the pKa
is 3.7. In a basic environment, protons are less available and are primarily removed from their respective functional groups. As shown below, the equilibria for both acidic and basic functional groups
shift to the right, and the basic, or unprotonated, forms predominate.
basic functional groups are primarily unionized. Please note that this is the same conclusion that was
reached when the Henderson-Hasselbalch equation was used to solve the sample problem with captopril. Because that sample problem involved an acidic functional group, let us look at an analogous
problem involving tamoxifen, a drug molecule with a basic functional group.
tamoxifen has only one ionizable functional group, the pKa must be a property of the tertiary amine.
For the purpose of this sample problem, assume that this drug molecule resides in a solution with a
The key point here is that in a basic environment, acidic functional groups are primarily ionized and
As shown above, the structure of tamoxifen contains a functional group with a pKa of 8.9. Since
CH 4 - SOLVING pH AND pKa PROBLEMS 95
[BaseForm]
[BaseForm]
m]
[BaseForm]
[BaseForm]
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pH of 9.5. In comparing the pH and pKa, the pH is greater than the pKa; therefore, the basic functional
group is in a basic environment and is primarily unionized according to the equilibrium equation for
bases. For the sake of completeness, let us use the Henderson-Hasselbalch equation to ensure that
both methods provide the same answer.
=+
58.9 log
6log
=
[AcidFor
[AcidForm]
Because this is a qualitative problem, it is not necessary to calculate the antilog of 0.6. The
fact that the equation produces a positive value indicates that the base form (i.e., the unprotonated
form) is predominant. Since the base form, or unprotonated form, of a basic functional group is the
unionized form, this equation verifies and is consistent with the above conclusion.
Scenario 3: pH Is Less Than the pK
a
When the pH is less than the pKa, then the environment is more acidic than the functional group. In
an acidic environment, protons are more available and functional groups are primarily protonated.
The equilibria for both acidic and basic functional groups shift to the left and the acidic, or protonated, forms predominate.
The key point here is that in an acidic environment, acidic functional groups are primarily union-
ized and basic functional groups are primarily ionized. Similar to what was done in the previous scenario, it is important to verify that these conclusions are consistent with the Henderson-Hasselbalch
equation. Let us revisit two previously discussed drug molecules, captopril and tamoxifen, and alter
the pH environment in which they reside. For these examples, a gastric pH of 2.1 is used.
Using the Henderson-Hasselbalch equation for captopril, the drug molecule with the acidic
functional group, gives the following initial equation, which can then be rearranged by subtracting
3.7 from both sides of the equation.
=+
13.7 log
1.6log
=
[AcidForm]
[AcidForm]
96 BASIC CONCEPTS IN MEDICINAL CHEMISTRY
[AcidForm]
[AcidForm]
As before, this type of question simply requires you to identify if the functional group is primarily ionized or unionized. Since the number on the left has a negative value, this indicates that the
acid, or protonated, form is predominant. Because the functional group is acidic, this indicates that
it is primarily unionized in this environment. This is the same conclusion that was reached without
using the Henderson-Hasselbalch equation: acidic functional groups are primarily unionized in acidic
environments.
Similarly, the Henderson-Hasselbalch equation can be used for tamoxifen, the drug molecule
with the basic functional group.
18.9 log
=+
6.8log
=
[BaseForm]
[BaseForm]
Although the magnitude of the value is larger, the same trend is seen. When the pH is less than
, the log [Base Form]/[Acid Form] is always a negative value, and thus the acid form always
the pK
a
predominates. Again, the exact same conclusion is reached regardless of whether the HendersonHasselbalch equation is used: basic functional groups are primarily ionized in acidic environments.
Key Points Regarding All Three Scenarios
When solving qualitative pH/pKa problems, it is essential to understand that the relative difference
between the pH and the pKa values is more important than the actual pH and pKa values. This is also
applicable to the quantitative problems that are discussed next. Failure to recognize the importance
of this relative difference commonly leads to incorrect assumptions and answers. A common error
occurs when a given pH value is compared with a neutral pH value of 7 instead of the given pKa values. For example, if a pH of 4.5 is immediately considered to be an acidic environment—because it
is less than 7—without evaluating the pKa values of the functional groups, errors are likely to occur.
The same is true if a pH of 8.5 is immediately considered to be a basic environment because it is
CH 4 - SOLVING pH AND pKa PROBLEMS 97
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greater than 7. To illustrate the potential errors that can occur, let us consider meropenem and these
two pH environments.
First, let us correctly evaluate these two environments using the method previously discussed.
Relative to a pH of 4.5, the carboxylic acid (pK
= 3.5) is in an environment that is one log unit more
a
basic than its pKa. Relative to this same environment, the primary amine (pKa = 9.4) is in an environment that is approximately five log units more acidic than its pKa. Because the acidic functional
group is in a basic environment and the basic functional group is in an acidic environment, both functional groups are primarily ionized if meropenem is in an aqueous environment with a pH of 4.5. A
similar situation arises when meropenem is in an environment with a pH of 8.5. Relative to this pH,
the carboxylic acid is still in a basic environment, the primary amine is still in an acidic environment,
and both groups are still primarily ionized. Please note that in both situations, the exact same environment is viewed differently by these two functional groups. As an analogy, consider that most
individuals view the costs of items they purchase as either expensive or inexpensive depending on
their perspective (i.e., their income, the prices of similar products, and the need for the product).
What is regarded as expensive and unaffordable to one individual may be regarded as inexpensive
by another. The same is true of functional groups. The same environment, measured by its pH value,
can be acidic to one group and basic to another.
Now let us approach this same problem using some preconceived ideas and incorrect compari-
sons. Because the pH of 4.5 is less than 7, meropenem and its functional groups reside in an acidic
environment. The acidic carboxylic acid is primarily unionized in an acidic environment while the
basic primary amine is primarily ionized. Similarly, since the pH of 8.5 is greater than 7, meropenem
and its functional groups reside in a basic environment. The acidic carboxylic acid is primarily ionized
in a basic environment, while the basic primary amine is primarily unionized. These conclusions are
obviously not consistent with the previous analysis. The key error here lies in the fact that the given
pH was incorrectly compared with a standard pH and not the given pKa values.
It is strongly recommended that you approach these types of problems without any preconceived ideas or comparisons. By using the knowledge of the acidic or basic nature of the functional
group and its pKa, the given environmental pH, and the methods described above, you can correctly
answer qualitative pH/pKa problems.
Key Summary Points for Solving Qualitative pH/pKa Problems
• The initial step in solving these types of problems is to identify the acidic and basic
functional group(s) and correctly assign the given pKa value(s).
• Problems that only require the determination of whether a functional group is primar-
ily ionized or unionized can be solved by simply comparing the pH and pKa values.
• The Henderson-Hasselbalch equation is not required to solve these types of problems;
however, the answers derived are consistent with this equation.
• When the pH equals the pKa, the functional group is 50% ionized and 50% unionized.
98 BASIC CONCEPTS IN MEDICINAL CHEMISTRY
• When the pH is greater than the pKa, acidic functional groups are primarily ionized and
basic functional groups are primarily unionized.
• When the pH is less than the pKa, acidic functional groups are primarily unionized and
basic functional groups are primarily ionized.
• If a drug molecule contains more than one acidic or basic functional group, each func-
tional group and its associated pKa must be evaluated separately.
• Quaternary ammonium functional groups, such as that found within bethanechol, are
always 100% ionized regardless of the environmental pH.
• Nonelectrolytes, such as dexamethasone, are 100% unionized regardless of the physi-
ologic pH.
Solving Quantitative pH and pKa Problems
Quantitative pH/pKa problems are similar to the qualitative problems previously discussed but
require one additional step. Quantitative problems require that you calculate the percent ionization of a functional group, the pH of the environment, or the pKa of the functional group. Although
examples of each are discussed below, problems that involve the percent ionization of a functional
group are the most common type of problem that you will encounter.
Calculating the Percent Ionization of a Functional Group
Similar to the discussion of qualitative pH/pKa problems, initial discussions here involve the use of
the Henderson-Hasselbalch equation to solve these types of problems. This is then followed by a
discussion of The Rule of Nines, a quick and easy method to solve quantitative problems when the
difference between the pH and pKa is an integer.
Shown below is the structure of verapamil. Its structure contains one ionizable functional group
with a pKa of 8.9. Using this information, let us calculate the ionization of verapamil at a physiologic pH
of 7.4.
Similar to qualitative problems, the initial steps of a quantitative problem involve the identification of acidic and basic functional groups and the assignment of given pKa values to the appropriate
CH 4 - SOLVING pH AND pKa PROBLEMS 99
[BaseForm]
[AcidForm]
[BaseForm]
[AcidForm]
1
m]
[AcidForm]
==
0.032Molecules in Base Form
1%=×
1.032Total Molecules
%=×
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functional groups. In this problem, only one pKa has been given, so it can be assumed that only one
acidic or basic functional group needs to be evaluated. A structural analysis reveals the presence of a
basic tertiary amine, four ether groups, two aromatic rings, numerous aliphatic hydrocarbons, and
a nitrile. Since the tertiary amine is the only ionizable functional group, the given pKa of 8.9 must be
a property of this basic functional group.
The next step is to use the principles discussed for qualitative problems to determine if the
tertiary amine will be primarily ionized or unionized. Since the given pH of 7.4 is less than the pKa
of the tertiary amine, this basic functional group is in an acidic environment and thus is primarily
ionized. These initial steps are very important in that they provide a proof check for the subsequent
calculations. Any quantitative answer that is inconsistent with these facts serves as an alert that an
error has been made.
Because this type of problem requires a percentage calculation, the Henderson-Hasselbalch
equation needs to be used. Inserting the given pH and pKa values provides the following equation:
48.9 log
=+
1.5log
=
[AcidForm]
Taking the antilog of each side of the equation provides the following ratio:
[BaseForm]
2
or
0.032
[BaseFor
This ratio indicates that for every one molecule that contains the functional group in the acid
form, there are 0.032 molecules that contain the functional group in the base form. To calculate the
percent of molecules that contain the functional group in either the acid or base form, it is necessary to first add these numbers together since all molecules must be either in the base or acid form.
rcentofMolecules in Base Form
rcentofMolecules in AcidForm
Since it has already been determined that the pK
1.032Total Molecules
1MoleculeinAcidForm
is a property of a basic tertiary amine, the
a
100% 3.
=
100% 96.9
=
base form in these equations is the unionized form and the acid form, or conjugate acid, in these
equations is the ionized form. Thus, the tertiary amine is 96.9% ionized. Please note that this answer
is consistent with the qualitative analysis that was initially performed prior to the actual calculation.
100 BASIC CONCEPTS IN MEDICINAL CHEMISTRY
m]
[AcidForm]
[BaseForm]
[AcidForm]
[AcidForm]
1
[AcidForm]
==
A common error in solving these types of problems is the failure to correctly convert the ratio
provided by the Henderson-Hasselbalch equation into a percentage. In the problem above, the ratio
between the [Base Form]/[Acid Form] was calculated to be 0.032.
Two incorrect interpretations of this ratio are as follows:
1. The calculated value from the Henderson-Hasselbalch equation is perceived to be the percentage. Thus, for this sample problem, the basic functional group is 99.968% ionized and
0.032% unionized.
2. The calculated value from the Henderson-Hasselbalch equation is multiplied by 100%. Thus,
for this sample problem, the basic functional group is 97.8% ionized and 3.2% unionized.
Granted, both of these incorrect calculations are close to the correct answer and are consistent with the initial qualitative analysis; however, this is only because the functional group in this
example is highly ionized at the given pH and the acid form is highly predominate. To illustrate
the importance of converting the ratio given by the Henderson-Hasselbalch into a percentage
and highlight the errors that can result if this step is omitted, let us examine another sample
problem.
Shown below is the structure of zafirlukast. Its structure contains a functional group that has a pK
of 4.3. Using this information, let us calculate the percent to which this functional group is ionized at a
urinary pH of 5.
a
As previously mentioned, the initial step is to match the pKa value with its appropriate functional group. This has already been done in this example; however, you should be able to recognize
that the sulfonamide is the only ionizable functional group present in this structure. In the second
step, the pH of 5 is found to be more basic than the pKa value of 4.3. Since this pKa value is a property of the carboxylic acid, this acidic functional group is residing in a basic environment and thus is
primarily ionized. Let us now use the Henderson-Hasselbalch equation to calculate the percent ionized.
This ratio indicates that for every one molecule that contains the functional group in the acid
form, there are 5.01 molecules that contain the functional group in the base form. Because the functional group is a carboxylic acid, the acid form is the unionized form, and the base form, or conjugate
base form, is the ionized form.
04.3 log
=+
7log
=
[BaseForm]
[BaseFor
5.01
or
[BaseForm]
CH 4 - SOLVING pH AND pKa PROBLEMS 101
=×
5.01 MoleculesinIonizedForm
%
1MoleculeinUnionizedForm
%=×
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The following equations can then be used to correctly calculate the percent of the molecules that
are ionized and the percent that are unionized.
100% 83.4
rcentinIonizedForm
6.01 TotalMolecules
=
Note that the calculated quantitative answers are consistent with the initial qualitative analy-
sis; the predominant form of the carboxylic acid in a urine pH is the ionized form.
In the above solution to the problem, the value of 5.01 derived from the Henderson-Hasselbalch
equation is correctly interpreted as a ratio of 5.01:1.
Let us now revisit the two previously described incorrect interpretations of this ratio and the errors
they can cause.
1. This calculated value is actually the percentage. Thus, for this sample problem, 5.01% is in
2. This calculated value is multiplied by 100%. Thus, for this sample problem, 501% exists in
In summary, the ratio between the [Base Form]/[Acid Form] that is provided when using the
Henderson-Hasselbalch equation is not the same as the percent ionized or unionized. Failure to correctly convert this ratio into a percentage can lead to erroneous calculations.
rcentinUnionized Form
the ionized or conjugate base form, and 94.99% is in the unionized acid form. Please note
that this calculation completely contradicts the initial qualitative analysis. Conclusion: An
error has occurred somewhere in the analysis or calculation!
the ionized or conjugate base form. Conclusion: An error has occurred since the calculation
indicates that more than 100% is ionized!
6.01 TotalMolecules
100% 16.6
=
Calculating the pH of an Environment
In this type of problem, you are given a structure, the pKa values for ionizable functional groups, and
the percent to which each of these groups is ionized. These types of questions are generally not as
common as those that require the calculation of the percent ionization but require similar steps and
strategies.
As a sample problem, let us use phenobarbital, shown below, to calculate the pH required for it to
be 80% ionized. This drug molecule has a functional group with a pKa value of 7.4.
The initial step in solving this type of problem is to analyze the given structure and to match
the given pKa values with their appropriate functional groups. Please note that this is identical to the
initial step for solving percent ionization problems. In analyzing the structure of phenobarbital, you
102 BASIC CONCEPTS IN MEDICINAL CHEMISTRY
[BaseForm]
80
20
=+
=
should be able to identify the acidic imide functional group (highlighted below). Since the heterocyclic ring is symmetrical, it doesn’t matter which CO—NH—CO sequence is used.
Because this is an acidic functional group, it must reside in a basic environment for it to be pri-
marily ionized, or in this case, 80% ionized. Thus, the pH that is calculated should be greater than
7.4. Similar to the steps used in calculating percent ionization, these initial steps allow us to check
our calculations and ensure that they are consistent with the known facts. Let us once again review
the Henderson-Hasselbalch equation.
HpKlog
=+
a
[AcidForm]
The pK
value has been given and can be directly entered into the equation. The functional group
a
is acidic and is 80% ionized. This means that for every 100 molecules of this drug, 80 of the molecules exist in the ionized, conjugate base form and 20 of the molecules exist in the unionized acid
form. Thus, the [Base Form]/[Acid Form] ratio is 80:20. Inserting these values provides the following
equation:
H7.4 log
=+
Solving this equation gives a pH value of 8.0. As predicted, the calculated pH is greater than the
pKa of the functional group.
H7.4 0.6
=+
H8.0
Calculating the pKa of a Functional Group
This is probably the least common type of problem that you will encounter; however, it is no more
difficult than calculating percent ionization of a functional group or the pH of an environment. The
steps used here are identical to the previous problems. In this scenario, you are given a structure, a
highlighted functional group, a pH, and the percent to which the functional group is ionized in the
given pH. Please note that since this type of question requires a pKa calculation, a specific functional
group is highlighted. More than likely, you will still be responsible for identifying if the highlighted
functional group is acidic or basic.
As an example, let us use sulfadiazine, shown below. Its highlighted functional group is 10% ionized
in a urine pH of 5.5. Given this information and the Henderson-Hasselbalch equation, it is possible to
calculate the pKa for this functional group.
CH 4 - SOLVING pH AND pKa PROBLEMS 103
[BaseForm]
10
90
10
90
aa
11
a
=
)
a
=−
6.5)
a
=
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Similar to the previous types of problems, the initial step is to determine if the given functional
group is acidic or basic. The highlighted functional group in this example is a sulfonamide, which is
an acidic functional group. Because this sulfonamide is acidic and is only 10% ionized at the given
pH, this pH of 5.5 must be less than the pKa of the functional group. Remember, acids are primarily
unionized in acidic environments. Once again, these initial steps allow us to ensure that our calculated pKa is consistent with the given facts. The given pH can directly be entered into the HendersonHasselbalch equation.
5pKlog
=+
a
[AcidForm]
The functional group is acidic and is 10% ionized. Similar to the previous problem, this means
that for every 100 molecules of this drug molecule, 10 molecules exist in the ionized, conjugate base
form and 90 molecules exist in the unionized acid form. Thus, the [Base Form]/[Acid Form] ratio is
10:90. Inserting these values provides the following equation:
Solving this equation gives a pK
5pKlog
=+ =−
value of 6.5 for the acidic sulfonamide functional group. As
a
or pK 5.5log
initially predicted, the calculated pKa is greater than the given pH of the environment.
K5.5 –log 0.
K5.5 –( 0.96
K6.46(or
The Rule of Nines
The Rule of Nines is a quick and easy method to determine the percent ionization of acidic and basic
functional groups without directly using the Henderson-Hasselbalch equation. One important criterion must be met before this method can be used: the difference between the pH and the pKa must
be an integer (i.e., 1, 2, 3).
The Rule of Nines is based on the Henderson-Hasselbalch equation and is best explained by
way of a series of examples. For these examples, we use flurbiprofen, a nonsteroidal anti-inflammatory
drug, and ask three new questions. To what extent is this functional group ionized at a urine pH of 5.5, a
gastric pH of 2.5, and a solution pH of 7.5?
Similar to all other pH/pKa problems, the initial step requires identifying and matching pKa values to functional groups. Since we have already looked at carboxylic acids in this chapter, this step
is omitted here, and this information is shown with the structure. The functional group is acidic, so
the acid form of this functional group is the unionized form, and the conjugate base form is the ionized form. In comparing the pK
of 4.5 with the three environments, this carboxylic acid is primarily
a
unionized at the gastric pH of 2.5 and primarily ionized at the urine pH of 5.5 and solution pH of
7.5. Using this information, let us use the Henderson-Hasselbalch equation to calculate the percent
ionization of this functional group in these three environments.
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