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Acid-Base Equilibrium

Module by: John S. Hutchinson

Foundation

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.

Goals

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.

Observation 1: Strong Acids and Weak Acids

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.

Observation 2: Percent Ionization in 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.

Observation 3: Autoionization of Water

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.

Observation 4: Base Ionization, Neutralization and Hydrolysis of Salts

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