We have developed an understanding of
equilibrium involving phase transitions and involving
reactions entirely in the gas phase. We will assume an
understanding of the principles of dynamic equilibrium, reaction
equilibrium constants, and
Le Châtelier's Principle. To understand
application of these principles to reactions in solution, we will
now assume a definition of certain classes of substances as being
either acids or bases. An acid is a substance whose molecules
donate positive hydrogen ions (protons) to other molecules or ions.
When dissolved in pure water, acid molecules will transfer a
hydrogen ion to a water molecule or to a cluster of several water
molecules. This increases the concentration of
H+H+
ions in the solution. A base is a substance whose molecules accept
hydrogen ions from other molecules. When dissolved in pure water,
base molecules will accept a hydrogen ion from a water molecule,
leaving behind an increased concentration of
OH-OH-
ions in the solution. To understand what determines acid-base
behavior, we will assume an understanding of the bonding,
structure, and properties of individual molecules.
Acids and bases are very common substances
whose properties vary greatly. Many acids are known to be quite
corrosive, with the ability to dissolve solid metals or burn flesh.
Many other acids, however, are not only benign but vital to the
processes of life. Far from destroying biological molecules, they
carry out reactions critical for organisms. Similarly, many bases
are caustic cleansers while many others are medications to calm
indigestion pains.
In this concept study, we will develop an
understanding of the characteristics of molecules which make them
either acids or bases. We will examine measurements about the
relative strengths of acids and bases, and we will use these to
develop a quantitative understanding of the relative strengths of
acids and bases. From this, we can develop a qualitative
understanding of the properties of molecules which determine
whether a molecule is a strong acid or a weak acid, a strong base
or a weak base. This understanding is valuable in predicting the
outcomes of reactions, based on the relative quantitative strengths
of acids and bases. These reactions are commonly referred to as
neutralization reactions. A surprisingly large number of reactions,
particularly in organic chemistry, can be understood as transfer of
hydrogen ions from acid molecules to base molecules.
From the definition of an acid given in the
Foundation, a typical acid can be written as
HAHA,
representing the hydrogen ion which will be donated and the rest of
the molecule which will remain as a negative ion after the
donation. The typical reaction of an acid in aqueous solution
reacting with water can be written as
HA(aq)+H2O(l)→H3O+(aq)+A-(aq)HA(aq)+H2O(l)→H3O+(aq)+A-(aq)(1)
In this reaction,
HA(aq)HA(aq)
represents an acid molecule dissolved in aqueous solution.
H3O+(aq)H3O+(aq)
is a notation to indicate that the donated proton has been
dissolved in solution. Observations indicate that the proton is
associated with several water molecules in a cluster, rather than
attached to a single molecule.
H3O+H3O+ is
a simplified notation to represent this result. Similarly, the
A-(aq)A-(aq)
ion is solvated by several water molecules. Equation 1 is referred to as
acid ionization.
Equation 1
implies that a 0.1M solution of the acid
HAHA
in water should produce
H3O+H3O+
ions in solution with a concentration of 0.1M. In fact, the
concentration of
H3O+H3O+
ions,
[H3O+][H3O+],
can be measured by a variety of techniques. Chemists commonly use a
measure of the
H3O+H3O+
ion concentration called the
pHpH, defined
by:
pH=-log[H3O+]pH[H3O+]
We now observe the concentration
[H3O+][H3O+]
produced by dissolving a variety of acids in solution at a
concentration of 0.1M, and the results are tabulated in Table 1.
H3O+ pH for 0.1M Acid Solutions
| Acid |
[H3O+][H3O+]
(M) |
pH |
|
H2SO4H2SO4 |
0.1 |
1 |
|
HNO3HNO3 |
0.1 |
1 |
|
HClHCl |
0.1 |
1 |
|
HBrHBr |
0.1 |
1 |
|
HIHI |
0.1 |
1 |
|
HClO4HClO4 |
0.1 |
1 |
|
HClO3HClO3 |
0.1 |
1 |
|
HNO2HNO2 |
6.2×10-36.2-3 |
2.2 |
|
HCNHCN |
7×10-67-6 |
5.1 |
|
HIOHIO |
1×10-61-6 |
5.8 |
|
HFHF |
5.5×10-35.5-3 |
2.3 |
|
HOCNHOCN |
5.5×10-35.5-3 |
2.3 |
|
HClO2HClO2 |
2.8×10-22.8-2 |
1.6 |
|
CH3COOHCH3COOH
(acetic acid) |
1.3×10-31.3-3 |
2.9 |
|
CH3CH2COOHCH3CH2COOH
(propionic acid) |
1.1×10-31.1-3 |
2.9 |
Note that there are several acids listed for
which
[H3O+]=0.1M[H3O+]0.1M,
and
pH=1pH1.
This shows that, for these acids, the acid ionization is complete:
essentially every acid molecule is ionized in the solution
according to Equation 1. However,
there are other acids listed for which
[H3O+][H3O+]
is considerably less than 0.1M and the pH is considerably greater
than 1. For each of these acids, therefore, not all of the acid
molecules ionize according to Equation 1. In fact, it is clear in Table 1 that in these acids the vast
majority of the acid molecules do not ionize, and only a small
percentage does ionize.
From these observations, we distinguish two
classes of acids:
strong acids and
weak acids. Strong acids are those for which nearly
100% of the acid molecules ionize, whereas weak acids are those for
which only a small percentage of molecules ionize. There are seven
strong acids listed in Table 1.
From many observations, it is possible to determine that these
seven acids are the only commonly observed strong acids. The vast
majority of all substances with acidic properties are weak acids.
We seek to characterize weak acid ionization quantitatively and to
determine what the differences in molecular properties are between
strong acids and weak acids.
Table 1
shows that the pH of 0.1M acid solutions varies from one weak acid
to another. If we dissolve 0.1 moles of acid in a 1.0L solution,
the fraction of those acid molecules which will ionize varies from
weak acid to weak acid. For a few weak acids, using the data in
Table 1 we calculate the
percentage of ionized acid molecules in 0.1M acid solutions in
Table 2.
Percent Ionization of 0.1M Acid Solutions
| Acid |
[H3O+][H3O+]
(M) |
% ionization |
|
HNO2HNO2 |
6.2×10-36.2-3 |
6.2% |
|
HCNHCN |
7×10-67-6 |
0.007% |
|
HIOHIO |
1×10-61-6 |
0.001% |
|
HFHF |
5.5×10-35.5-3 |
5.5% |
|
HOCNHOCN |
5.5×10-35.5-3 |
5.5% |
|
HClO2HClO2 |
2.8×10-22.8-2 |
28.2% |
|
CH3COOHCH3COOH
(acetic acid) |
1.3×10-31.3-3 |
1.3% |
|
CH3CH2COOHCH3CH2COOH
(propionic acid) |
1.1×10-31.1-3 |
1.1% |
We might be tempted to conclude from Table 2 that we can characterize the
strength of each acid by the percent ionization of acid molecules
in solution. However, before doing so, we observe the pH of a
single acid, nitrous acid, in solution as a function of the
concentration of the acid.
HNO2(aq)+H2O(l)→H3O+(aq)+NO2-(aq)HNO2(aq)+H2O(l)→H3O+(aq)+NO2-(aq)(2)
In this case, "concentration of the
acid" refers to the number of moles of acid that we dissolved
per liter of water. Our observations are listed in Table 3, which gives
[H3O+][H3O+],
pH, and percent ionization as a function of nitrous acid
concentration.
% Ionization of Nitrous Acid
|
c0c0
(M) |
[H3O+][H3O+] |
pH |
% Ionization |
| 0.50 |
1.7×10-21.7-2 |
1.8 |
3.3% |
| 0.20 |
1.0×10-21.0-2 |
2.0 |
5.1% |
| 0.10 |
7.0×10-37.0-3 |
2.2 |
7.0% |
| 0.050 |
4.8×10-34.8-3 |
2.3 |
9.7% |
| 0.020 |
2.9×10-32.9-3 |
2.5 |
14.7% |
| 0.010 |
2.0×10-32.0-3 |
2.7 |
20.0% |
| 0.005 |
1.3×10-31.3-3 |
2.9 |
26.7% |
| 0.001 |
4.9×10-44.9-4 |
3.3 |
49.1% |
| 0.0005 |
3.0×10-43.0-4 |
3.5 |
60.8% |
Surprisingly, perhaps, the percent ionization
varies considerably as a function of the concentration of the
nitrous acid. We recall that this means that the fraction of
molecules which ionize, according to Equation 2, depends on how many acid molecules
there are per liter of solution. Since some but not all of the acid
molecules are ionized, this means that nitrous acid molecules are
present in solution at the same time as the negative nitrite ions
and the positive hydrogen ions. Recalling our observation of
equilibrium in gas phase reactions, we can conclude that Equation 2 achieves equilibrium for each
concentration of the nitrous acid.
Since we know that gas phase reactions come to
equilibrium under conditions determined by the equilibrium
constant, we might speculate that the same is true of reactions in
aqueous solution, including acid ionization. We therefore define an
analogy to the gas phase reaction equilibrium constant. In this
case, we would not be interested in the pressures of the
components, since the reactants and products are all in solution.
Instead, we try a function composed of the equilibrium
concentrations:
K=[H3O+][NO2-][HNO2][H2O]
K
[H3O+]
[NO2-]
[HNO2]
>[H2O]
(3)
The concentrations at equilibrium can be
calculated from the data in Table 3 for nitrous acid.
[H3O+][H3O+]
is listed and
[NO2-]=[H3O+][NO2-][H3O+].
Furthermore, if
c0c0
is the initial concentration of the acid defined by the number of
moles of acid dissolved in solution per liter of solution, then
[HA]=c0-[H3O+][HA]c0[H3O+].
Note that the contribution of
[H2O(l)][H2O(l)]
to the value of the function KK is simply a
constant. This is because the "concentration" of water
in the solution is simply the molar density of water,
nH2OV=55.5MnH2OV55.5M,
which is not affected by the presence or absence of solute. All of
the relevant concentrations, along with the function in Equation 3 are calculated and tabulated in
Table 4.
Equilibrium Concentrations and K for Nitrous Acid
|
c0c0
(M) |
[H3O+][H3O+] |
[NO2-][NO2-] |
[HNO2][HNO2] |
KK |
| 0.50 |
1.7×10-21.7-2 |
1.7×10-21.7-2 |
0.48 |
1.0×10-51.0-5 |
| 0.20 |
1.0×10-21.0-2 |
1.0×10-21.0-2 |
0.19 |
9.9×10-69.9-6 |
| 0.10 |
7.0×10-37.0-3 |
7.0×10-37.0-3 |
9.3×10-29.3-2 |
9.6×10-69.6-6 |
| 0.050 |
4.8×10-34.8-3 |
4.8×10-34.8-3 |
4.5×10-24.5-2 |
9.4×10-69.4-6 |
| 0.020 |
2.9×10-32.9-3 |
2.9×10-32.9-3 |
4.5×10-24.5-2 |
9.4×10-69.4-6 |
| 0.010 |
2.0×10-32.0-3 |
2.0×10-32.0-3 |
8.0×10-38.0-3 |
8.9×10-68.9-6 |
| 0.005 |
1.3×10-31.3-3 |
1.3×10-31.3-3 |
3.6×10-33.6-3 |
8.8×10-68.8-6 |
| 0.001 |
4.9×10-44.9-4 |
4.9×10-44.9-4 |
5.1×10-45.1-4 |
8.5×10-68.5-6 |
| 0.0005 |
3.0×10-43.0-4 |
3.0×10-43.0-4 |
2.0×10-42.0-4 |
8.5×10-68.5-6 |
We note that the function
KK in Equation 3 is approximately, though only
approximately, the same for all conditions analyzed in Table 3. Variation of the concentration by
a factor of 1000 produces a change in KK of only 10%
to 15%. Hence, we can regard the function KK as a
constant which approximately describes the acid ionization
equilibrium for nitrous acid. By convention, chemists omit the
constant concentration of water from the equilibrium expression,
resulting in the
acid ionization equilibrium constant,
KaKa,
defined as:
Ka=[H3O+][NO2-][HNO2]
Ka
[H3O+]
[NO2-]
[HNO2]
(4)
From an average of the data in Table 4, we can calculate that, at
25°C for nitrous acid,
Ka=5×10-4Ka5-4.
Acid ionization constants for the other weak acids in Table 2 are listed in Table 5.
Weak Acid Ionization Constants, Ka and pKa
| Acid |
KaKa |
pKaKa |
|
HNO2HNO2 |
5×10-45-4 |
3.3 |
|
HCNHCN |
4.9×10-104.9-10 |
9.3 |
|
HIOHIO |
2.3×10-112.3-11 |
10.6 |
|
HFHF |
3.5×10-43.5-4 |
3.4 |
|
HOCNHOCN |
3.5×10-43.5-4 |
3.4 |
|
HClO2HClO2 |
1.1×10-21.1-2 |
2.0 |
|
CH3COOHCH3COOH
(acetic acid) |
1.7×10-51.7-5 |
4.8 |
|
CH3CH2COOHCH3CH2COOH
(propionic acid) |
1.4×10-51.4-5 |
4.9 |
We make two final notes about the results in
Table 5. First, it is clear the
larger the value of
KaKa,
the stronger the acid. That is, when
KaKa
is a larger number, the percent ionization of the acid is larger,
and vice versa. Second, the values of
KaKa
very over many orders of magnitude. As such, it is often convenient
to define the quanity
pKapKa,
analogous to pH, for purposes of comparing acid strengths:
pKa=-logKapKaKa(5)
The value of
pKapKa
for each acid is also listed in Table 5. Note that a small value of
pKapKa
implies a large value of
KaKa
and thus a stronger acid. Weaker acids have larger values of
pKapKa.
KaKa
and
pKapKa
thus give a simple quantitative comparison of the strength of weak
acids.
Since we have the ability to measure pH for
acid solutions, we can measure pH for pure water as well. It might
seem that this would make no sense, as we would expect
[H3O+][H3O+]
to equal zero exactly in pure water. Surprisingly, this is
incorrect: a measurement on pure water at 25°C yields
pH=7pH7,
so that
[H3O+]=1.0×10-7M[H3O+]1.0-7M.
There can be only one possible source for these ions: water
molecules. The process
H2O(l)+H2O(l)→H3O+(aq)+OH-
(aq)H2O(l)+H2O(l)→H3O+(aq)+OH-
(aq)(6)
is referred to as the
autoionization of water. Note that, in this reaction,
some water molecules behave as acid, donating protons, while other
acid molecules behave as base, accepting protons.
Since at equilibrium
[H3O+]=1.0×10-7M[H3O+]1.0-7M,
it must also be true that
[OH-]=1.0×10-7M[OH-]1.0-7M.
We can write the equilibrium constant for Equation 6, following our previous convention
of omitting the pure water from the expression, and we find that,
at 25°C,
Kw=[H3O+][OH-]=1.0×10-14MKw
[H3O+][OH-]1.0-14M(7)
(In this case, the subscript "w"
refers to "water".)
Equation 6 occurs
in pure water but must also occur when ions are dissolved in
aqueous solutions. This includes the presence of acids ionized in
solution. For example, we consider a solution of 0.1M acetic acid.
Measurements show that, in this solution
[H3O+]=1.3×10-3M[H3O+]1.3-3M
and
[OH-]=7.7×10-12M[OH-]7.7-12M.
We note two things from this observation: first, the value of
[OH-][OH-]
is considerably less than in pure water; second, the autoionization
equilibrium constant remains the same at 1.0×10-141.0-14.
From these notes, we can conclude that the autoionization
equilibrium of water occurs in acid solution, but the extent of
autoionization is suppressed by the presence of the acid in
solution.
We consider a final note on the autoionization
of water. The pH of pure water is 7 at 25°C. Adding any acid
to pure water, no matter how weak the acid, must increase
[H3O+][H3O+],
thus producing a pH below 7. As such, we can conclude that, for all
acid solutions, pH is less than 7, or on the other hand, any
solution with pH less than 7 is acidic.
We have not yet examined the behavior of base
molecules in solution, nor have we compared the relative strengths
of bases. We have defined a base molecule as one which accepts a
positive hydrogen ion from another molecule. One of the most common
examples is ammonia,
NH3NH3.
When ammonia is dissolved in aqueous solution, the following
reaction occurs:
NH3(aq)+H2O(l)→NH4+(aq)+OH-(aq)NH3(aq)+H2O(l)→NH4+(aq)+OH-(aq)(8)
Due to the lone pair of electrons on the
highly electronegative N atom,
NH3NH3
molecules will readily attach a free hydrogen ion forming the
ammonium ion
NH4+NH4+.
When we measure the concentration of
OH-OH-
for various initial concentration of
NH3NH3
in water, we observe the results in Table 6. We should anticipate that a base
ionization equilibrium constant might exist comparable to the acid
ionization equilibrium constant, and in
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