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.
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.
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.
Table 3: % 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.
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.
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 Table 6, we have also calculated the value
of the function
KbKb
defined as:
Kb=[NH4+][OH-][NH3]
Kb
[NH4+]
[OH-]
[NH3]
(9)
Table 6: Equilibrium Concentrations and Kb for Ammonia
|
c0c0
(M) |
[OH-][OH-] |
KbKb |
pH |
| 0.50 |
3.2×10-33.2-3 |
2.0×10-52.0-5 |
11.5 |
| 0.20 |
2.0×10-32.0-3 |
2.0×10-52.0-5 |
11.3 |
| 0.10 |
1.4×10-31.4-3 |
2.0×10-52.0-5 |
11.1 |
| 0.050 |
9.7×10-49.7-4 |
1.9×10-51.9-5 |
11.0 |
| 0.020 |
6.0×10-46.0-4 |
1.9×10-51.9-5 |
10.8 |
| 0.010 |
4.2×10-44.2-4 |
1.9×10-51.9-5 |
10.6 |
| 0.005 |
3.0×10-43.0-4 |
1.9×10-51.9-5 |
10.5 |
| 0.001 |
1.3×10-41.3-4 |
1.8×10-51.8-5 |
10.1 |
| 0.0005 |
8.7×10-58.7-5 |
1.8×10-51.8-5 |
9.9 |
Given that we have dissolved a base in pure
water, we might be surprised to discover the presence of positive
hydrogen ions,
H3O+H3O+,
in solution, but a measurement of the pH for each of the solutions
reveals small amounts. The pH for each solution is also listed in
Table 6. The source of these
H3O+H3O+
ions must be the autoionization of water. Note, however, that in
each case in basic solution, the concentration of
H3O+H3O+
ions is less than that in pure water. Hence, the presence of the
base in solution has suppressed the autoionization. Because of
this, in each case the pH of a basic solution is greater than
7.
Base ionization is therefore quite analogous
to acid ionization observed earlier. We now consider a comparison
of the strength of an acid to the strength of a base. To do so, we
consider a class of reactions called "neutralization
reactions" which occur when we mix an acid solution with a
base solution. Since the acid donates protons and the base accepts
protons, we might expect, when mixing acid and base, to achieve a
solution which is no longer acidic or basic. For example, if we mix
together equal volumes of 0.1M
HCl(aq)HCl(aq)
and 0.1M
NaOH(aq)NaOH(aq),
the following reaction occurs:
HCl(aq)+
NaOH(aq)→
Na+(aq)+
Cl-(aq)+
H2O(l)
HCl(aq)+
NaOH(aq)→
Na+(aq)+
Cl-(aq)+
H2O(l)
(10)
The resultant solution is simply a salt
solution with
NaClNaCl
dissolved in water. This solution has neither acidic nor basic
properties, and the pH is 7; hence the acid and base have
neutralized each other. In this case, we have mixed together a
strong acid with a strong base. Since both are strong and since we
mixed equal molar quantities of each, the neutralization reaction
is essentially complete.
We next consider mixing together a weak acid
solution with a strong base solution, again with equal molar
quantities of acid and base. As an example, we mix 100ml of 0.1M
acetic acid
(HAHA)
solution with 100ml of 0.1M sodium hydroxide. In this discussion,
we will abbreviate the acetic acid molecular formula
CH3COOHCH3COOH
as
HAHA
and the acetate ion
CH3COO-CH3COO-
as
A-A-.
The reaction of
HAHA
and
NaOHNaOH
is:
HA(aq)+
NaOH(aq)→
Na+(aq)+
A-(aq)+
H2O(l)
HA(aq)+
NaOH(aq)→
Na+(aq)+
A-(aq)+
H2O(l)
(11)
A-(aq)A-(aq)
is the acetate ion in solution, formed when an acetic acid molecule
donates the positive hydrogen ion. We have thus created a salt
solution again, in this case of sodium acetate in water. Note that
the volume of the combined solution is 200ml, so the concentration
of sodium acetate
(NaANaA)
in solution is 0.050M.
Unlike our previous
NaClNaCl
salt solution, a measurement in this case reveals that the pH of
the product salt solution is 9.4, so the solution is basic. Thus,
mixing equal molar quantities of strong base with weak acid
produces a basic solution. In essence, the weak acid does not fully
neutralize the strong base. To understand this, we examine the
behavior of sodium acetate in solution. Since the pH is greater
than 7, then there is an excess of
OH-OH-
ions in solution relative to pure water. These ions must have come
from the reaction of sodium acetate with the water. Therefore, the
negative acetate ions in solution must behave as a base, accepting
positive hydrogen ions:
A-(aq)+
H2O(aq)→
HA(aq)+
OH-(l)
A-(aq)+
H2O(aq)→
HA(aq)+
OH-(l)
(12)
The reaction of an ion with water to form
either an acid or a base solution is referred to as
hydrolysis. From this example, the salt of a weak acid
behaves as a base in water, resulting in a pH greater than
7.
To understand the extent to which the
hydrolysis of the negative ion occurs, we need to know the
equilibrium constant for this reaction. This turns out to be
determined by the acid ionization constant for
HAHA.
To see this, we write the equilibrium constant for the hydrolysis
of
A-A-
as
Kh=[HA][OH-][A-]
Kh
[HA]
[OH-]
[A-]
(13)
Multiplying numerator and denominator by
[H3O+][H3O+],
we find that
Kh=[HA][OH-][A-][H3O+][H3O+]=KwKa
Kh
[HA]
[OH-]
[A-]
[H3O+]
[H3O+]
Kw
Ka
(14)
Therefore, for the hydrolysis of acetate ions
in solution,
Kh=5.8×10-10Kh5.8-10.
This is fairly small, so the acetate ion is a very weak
base.
We now have a fairly complete quantitative
description of acid-base equilibrium. To complete our understanding
of acid-base equilibrium, we need a predictive model which relates
acid strength or base strength to molecular properties. In general,
we expect that the strength of an acid is related either to the
relative ease by which it can donate a hydrogen ion or by the
relative stability of the remaining negative ion formed after the
departure of the hydrogen ion.
To begin, we note that there are three basic
categories of acids which we have examined in this study. First,
there are simple
binary acids: • HFHF; • HClHCl; • HBrHBr; • HIHI.
Second, there are acids formed from main group elements combined
with one or more oxygen atoms, such
H2SO4H2SO4
or
HNO3HNO3.
These are called
oxyacids. Third, there are the
carboxylic acids, organic molecules which contain the
carboxylic functional group in Figure 1.
We consider first the simple binary acids.
HClHCl,
HBrHBr,
and
HIHI
are all strong acids, whereas
HFHF
is a weak acid. In comparing the experimental values of
pKapKa
values in Table 7, we note that
the acid strength increases in the order
HF<HCl<HBr<HIHFHClHBrHI.
This means that the hydrogen ion can more readily separate from the
covalent bond with the halogen atom (X) as we move down the
periodic table. This is reasonable, because the strength of the H-X
bond also decreases as we move down the periodic table, as shown in
Table 7.
Table 7: H-X Bond Strengths and pKa
| |
pKapKa |
Bond Energy
(kJmolkJmol) |
|
HFHF |
3.1 |
567.7 |
|
HClHCl |
-6.0 |
431.6 |
|
HBrHBr |
-9.0 |
365.9 |
|
HIHI |
-9.5 |
298.0 |
The decreasing strength of the H-X bond is
primarily due to the increase is the size of the X atom as we move
down the periodic table. We conclude that one factor which
influences acidity is the strength of the H-X bond: a weaker bond
produces a stronger acid, and vice versa.
In the acids in the other two categories, the
hydrogen atom which ionizes is attached directly to an oxygen atom.
Thus, to understand acidity in these molecules, we must examine
what the oxygen atom is in turn bonded to. It is very interesting
to note that, in examining compounds like R-O-H, where R is an atom
or group of atoms, we can get either acidic or basic properties.
For examples,
NaOHNaOH
is a strong base, whereas
HOClHOCl
is a weak acid. This means that, when
NaOHNaOH
ionizes in solution, the Na-O linkage ionizes, whereas when
HOClHOCl
ionizes in solution, the H-O bond ionizes.
To understand this behavior, we compare the
strength of the simple oxyacids
HOIHOI,
HOBrHOBr,
and
HOClHOCl.
The
pKapKa's
for these acids are found experimentally to be, respectively, 10.6,
8.6, and 7.5. The acid strength for
HOXHOX
increases as we move up the periodic table in the halogen group.
This means that the H-O bond ionizes more readily when the oxygen
atom is bonded to a more electronegative atom.
We can add to this observation by comparing
the strengths of the acids
HOClHOCl,
HOClOHOClO,
HOClO2HOClO2,
and
HOClO3HOClO3.
(Note that the molecular formulae are more commonly written as
HClOHClO,
HClO2HClO2,
HClO3HClO3,
and
HClO4HClO4.
We have written them instead to emphasize the molecular structure.)
The
pKapKa's
of these acids are, respectively, 7.5, 2.0, -2.7, and -8.0.
In each case, the molecule with more oxygen atoms on the central Cl
atom is the stronger acid:
HOClOHOClO
is more acidic than
HOClHOCl,
etc. A similar result is found in comparing the
oxyacids of nitrogen.
HONO2HONO2,
nitric acid, is one of the strong acids, whereas
HONOHONO,
nitrous acid, is a weak acid. Since oxygen atoms are very strongly
electronegative, these trends add to our observation that
increasing electronegativity of the attached atoms increases the
ionization of the O-H bond.
Why would electronegativity play a role in
acid strength? There are two conclusions we might draw. First, a
greater electronegativity of the atom or atoms attached to the H-O
in the oxyacid apparently results in a weaker H-O bond, which is
thus more readily ionized. We know that an electronegative atom
polarizes bonds by drawing the electrons in the molecule towards
it. In this case, the Cl in
HOClHOCl
and the Br in
HOBrHOBr
must polarize the H-O bond, weakening it and facilitating the
ionization of the hydrogen. In comparing
HOClHOCl
to
HOClOHOClO,
the added oxygen atom must increase the polarization of the H-O
bond, thus weakening the bond further and increasing the extent of
ionization.
A second conclusion has to do with the ion
created by the acid ionization. The negative ion produced has a
surplus electron, and the relative energy of this ion will depend
on how readily that extra electron is attracted to the atoms of
ion. The more electronegative those atoms are, the stronger is the
attraction. Therefore, the
OCl-OCl-
ion can more readily accommodate the negative charge than can the
OBr-OBr-
ion. And the
OClO-OClO-
ion can more readily accommodate the negative charge than can the
OCl-OCl-
ion.
We conclude that the presence of strongly
electronegative atoms in an acid increases the polarization of the
H-O bond, thus facilitating ionization of the acid, and increases
the attraction of the extra electron to the negative ion, thus
stabilizing the negative ion. Both of these factors increase the
acid strength. Chemists commonly use both of these conclusions in
understanding and predicting relative acid strength.
The relative acidity of carbon compounds is a
major subject of organic chemistry, which we can only visit briefly
here. In each of the carboxylic acids, the H-O group is attached to
a carbonyl C=O group, which is in turn bonded to other atoms. The
comparison we observe here is between carboxylic acid molecules,
denoted as
RCOOHRCOOH,
and other organic molecules containing the H-O group, such as
alcohols denoted as
ROHROH.
(R is simply an atom or group of atoms attached to the functional
group.) The former are obviously acids whereas the latter group
contains molecules which are generally extremely weak acids. One
interesting comparison is for the acid and alcohol when R is the
benzene ring,
C6H5C6H5.
Benzoic acid,
C6H5COOHC6H5COOH,
has
pKa=4.2pKa4.2,
whereas phenol,
C6H5OHC6H5OH,
has
pKa=9.9pKa9.9.
Thus, the presence of the doubly bonded oxygen atom on the carbon
atom adjacent to the O-H clearly increases the acidity of the
molecule, and thus increases ionization of the O-H bond.
This observation is quite reasonable in the
context of our previous conclusion. Adding an electronegative
oxygen atom in near proximity to the O-H bond both increases the
polarization of the O-H bond and stabilizes the negative ion
produced by the acid ionization. In addition to the
electronegativity effect, carboxylate anions,
RCOO-RCOO-,
exhibit resonance stabilization, as seen in Figure 2.
The resonance results in a sharing of the
negative charge over several atoms, thus stabilizing the negative
ion. This is a major contributing factor in the acidity of
carboxylic acids versus alcohols.
Strong acids have a higher percent ionization
than do weak acids. Why don't we use percent ionization as a
measure of acid strength, rather than
KaKa?
Using the data in Table 4 for nitrous acid, plot
[H3O+][H3O+]
versus
c0c0,
the initial concentration of the acid, and versus
[HNO2][HNO2]
the equilibrium concentration of the acid. On a second graph, plot
[H3O+]2[H3O+]2
versus
c0c0,
the initial concentration of the acid, and versus
[HNO2][HNO2]
the equilibrium concentration of the acid. Which of these results
gives a straight line? Using the equilibrium constant expression,
explain your answer.
Using Le Châtelier's principle,
explain why the concentration of
[OH-][OH-]
is much lower in acidic solution than it is in neutral
solution.
We considered mixing a strong base with a weak
acid, but we did not consider mixing a strong acid with a weak
acid. Consider mixing 0.1M
HNO3HNO3
and 0.1M
HNO2HNO2.
Predict the pH of the solution and the percent ionization of the
nitrous acid. Rationalize your prediction using Le
Châtelier's principle.
Imagine taking a 0.5M solution of nitrous acid
and slowing adding water to it. Looking at Table 3, we see that, as the concentration
of nitrous acid decreases, the percent ionization increases. By
contrast,
[H3O+][H3O+]
decreases. Rationalize these results using Le
Châtelier's principle.
We observed that mixing a strong acid and a
strong base, in equal amounts and concentrations, produces a
neutral solution, and that mixing a strong base with a weak acid,
in equal amounts and concentrations, produces a basic solution.
Imagine mixing a weak acid and a weak base, in equal amounts and
concentrations. Predict whether the resulting solution will be
acidic, basic, or neutral, and explain your prediction.
Using the electronegativity arguments
presented above,
explain why, in general, compounds like M-O-H are bases rather than
acids, when M is a metal atom. Predict the relationship between the
properties of the metal atom M and the strength of the base
MOHMOH.
Ionization of sulfuric acid
H2SO4H2SO4
produces
HSO4-HSO4-,
which is also an acid. However,
HSO4-HSO4-
is a much weaker acid than
H2SO4H2SO4.
Using the conclusions from above, explain why
HSO4-HSO4-
is a much weaker acid.
Predict and explain the relative acid
strengths of
H2SH2S
and
HClHCl.
Predict and explain the relative acid strengths of
H3PO4H3PO4
and
H3AsO4H3AsO4.