Chapter 5: Acid-Base Reactions


5.1 Introduction

In the previous chapters, we have learned a general modeling approach that begins with mass balances on the medium and or constituents of interest. We have been limited to non-reacting systems up to this point. Before we incorporate the generic concept of component interactions in our models, we will learn about specific types of reactions that are common in environmental systems: acid-base reactions. Acid-base chemistry governs many important phenomena in environmental engineering processes. An examination of acid-base reactions and equilibrium is necessary for understanding the relationships among the various constituents of natural and contaminated waters, and for quantitatively assessing the effects of treatment processes. This chapter will examine acid-base reactions through specific examples relevant to environmental engineering. We will later use them im more elaborate models of environmental processes.

 

5.2 Concepts: Acids and Bases, pH

J.N. BrØnsted and T.M. Lowry proposed the following definition of acids and bases in 1923: an acid is a proton donor and base is a proton acceptor . In the general reaction,

[5.1]

a proton (H+) is transferred from HA to the anion, B-. HA is considered an acid because it can donate a proton while B- is a base as it can accept a proton. Because the reaction is reversible, HB and A- form a new acid and base, respectively. The HA / A- and the HB / B- combinations which differ by only a proton are called conjugate acid-base pairs. Water acts as both an acid and a base as demonstrated in the following two reactions. In the first reaction,

[5.2]

H2O is the base which accepts a proton from phenol, C6H5OH, to become H3O+. While in the second reaction, H2O is an acid which donates a proton to ammonia, NH3, to

[5.3]

become OH-. The conjugate acid-base pairs are H3O+ / H2O and C6H5OH / C6H5O - in equation [5.2] and H3O+ / H2O and NH4+ / NH3 in equation [5.3].

What enables water to act as either an acid or base is its ability to reversibly ionize as depicted in the following reaction:

[5.4]

Because the proton is bonded so strongly to a water molecule it is considered more accurate to use H3O+ (the hydronium ion) rather than H+ in equations. However the shorthand H+ for H3O+ will be adapted throughout the rest of the text because of its ease of use. Thus equation [5.4] is rewritten in the shorthand as

[5.5]

An equilibrium constant for this reaction can be written as

[5.6]

where the [ ] designation represents concentrations expressed in moles per liter. A convention in acid-base chemistry is to treat the molar concentration of water as a constant. The molar concentration of water is its density (1000 g/L) divided by its molar mass (18 g/mol), or 55.6 mol /L. Because water ionizes or dissociates only slightly, the molar concentration of water is essentially a constant that can be incorporated into the equilibrium constant . Thus equation [5.6] is rewritten as

[5.7]

KW is called the dissociation constant for water. Equation [5.7] must always be satisfied when solving acid-base equilibria problems no matter what the source of hydrogen ions (H+) or hydroxide ions (OH-) dissolved in water.

Because the [H+] or [OH-] in solution is so small, a convention (established by the Danish biochemist SØren SØrensen in 1909) of taking the negative logarithm of these concentrations is followed. To express the hydrogen ion concentrations, the pH scale is followed, where

[5.8]

or

[5.9]

Similarly to express the hydroxide ion concentration, the pOH scale can be used, where

[5.10]

or

[5.11]

From equation [5.7] it follows that

[5.12]

An aqueous solution near pH = 7 is considered neutral since [H+] = [OH-]. An acidic solution has a pH less than 7, while a basic solution has a pH greater than 7. The conventional pH scale ranges from 0 to 14. This does not mean the pH cannot be less zero; as an example, a solution with 10 moles per liter of H+ has a pH of -1, since [H+] = 10+1.

 

Figure 1 The pH scale.

 

Example 5.1 Find the hydrogen ion concentration and the hydroxide ion concentration of rain water which has a pH of 5.6.

 

5.3 Strong Acids and Bases, Weak Acids and Bases

The strength of an acid is determined by the extent to which it donates its proton in the presence of a common base such as water. Consider the generic reaction of an acid (HA) in water,

[5.13]

which in the shorthand notation is written as

[5.14]

The equilibrium constant for the above reaction is

[5.15]

with KA called the dissociation constant. Strong acids essentially dissociate (lose their protons) completely and have relatively large dissociation constants, whereas weak acids dissociate only slightly and have relatively small dissociation constants.

Likewise the strength of a base is determined by the extent to which it accepts a proton in the presence of a common acid such as water. Accepting a proton is similar to donating OH- since OH- than accepts the H+ (remember equation [5.7] always holds). So we can think of bases as OH- donors similarly to how acids are H+ donors. Consider the generic reaction of a base (BOH) in water,

[5.16]

The equilibrium constant for the above reaction is

[5.17]

Strong bases essentially dissociate (lose their hydroxide ions) completely and have relatively large dissociation constants, whereas weak bases dissociate only slightly and have relatively small dissociation constants.

 

5.4 Equilibrium Calculations

In all cases considered above a mass balance holds for the conjugate acid-base pair. Thus the reacting species are conserved. Again consider HA added to water in equations [5.13] and [5.14]. Whether HA dissociates partially or completely it must either be in the HA acid form or its conjugate base form A-. A mass balance on all species containing A is written as

[5.18]

where CT,A is the number of moles of species A per liter and is equal to the number of moles of HA added per liter.

 

Example 5.2 Nitrogen in a wastewater treatment plant is either in the ammonia form (NH3) or the ammonium ion form (NH4+) governed by the equilibrium reaction of equation [5.3]. The equilibrium constant K for equation [5.3] is 1.82 x 10^-5. The ammonia form volatilizes easily to the form NH3(gas) such that a polishing step of nitrogen removal may be accomplished through a process called ammonia stripping. The greater the proportion of the nitrogen that is in the ammonia form, the more rapidly this stripping will occur. If the total nitrogen concentration is 10^-3 moles per liter (M) what percent of the nitrogen is in the NH3 form at pH = 10.5?

 

 

Besides mass balances, another general equation that must hold true for all acid-base equilibrium problems is the charge balance or electroneutrality. An aquatic solution must be electrically neutral. The total number of positive charges in the form of positive ions must be equal to the total number of negative charges in the form of negative ions. [H+] and [OH-] always appear in the charge balance since they are always present in solution. For example if sodium chloride (NaCl) is added to pure water, we must have

[5.19]

This same equation must also apply if we add hydrochloric acid (HCl) to a solution of sodium hydroxide (NaOH), since the same set of ions is present.

Care must be taken when accounting for divalent ions, since the molar concentration must be multiplied by the valence to give the concentration of charge. For example, if Na2SO4 is added to pure water, the charge balance is

[5.20]

A final general equation that always describes equilibrium acid-base relationships is the proton condition. The proton condition is a special type of mass balance equation on protons that is a useful component of equilibrium problem solving if either H+ or OH- are involved in the equilibria. The proton mass balance is established with reference to a "zero level" or a "reference level" for protons (the proton reference level, PRL). The PRL is established as the species from which the solution was prepared. In other words, every increase in a species concentration that contains more protons than those in the initial species must be equalled by a decrease in concentration of some species with fewer protons. The species that have protons in excess of the PRL are then equated with the species having fewer protons than the PRL.

A simple diagram helps illustrate how the proton condition is determined, although with practice the equation can be written directly. The conjugate acid-base pairs are listed vertically, with the form containing the extra proton (the acid) on the left. The H+ / H2O pair is always on the top of the diagram and the H2O / OH- pair is always on the bottom. Conjugate acid-base pairs of interest are placed in between. The PRL species are then determined; these always include water and the species added to water. By drawing a line connecting the PRL species, the proton condition may be determined by equating species on opposite sides of the line. This is illustrated in example 5.3.

Example 5.3 Sodium hypochlorite (NaOCl) is a commonly used disinfectant for the disinfection of water and wastewater. The chemical reactions of NaOCl are

 

NaOCl -> Na+ + OCl- ; assume that this dissociation is essentially complete

OCl- + H2O <--> HOCl + OH-

 

Determine the mass balance, charge balance, and proton condition.

 

There are 5 species in solution Na+, OCl-, HOCl, and H+, OH-. In every aquatic chemistry problem H+ and OH- are always present.

 

The mass balance for the conjugate acid-base pair : CT,OCl = [HOCl] + [OCl-]

The charge balance: [Na+] + [H+] = [OH-] + [OCl-]

The proton condition is determined by first drawing the proton balance diagram. The conjugate acid-base pairs are H+ / H2O, HOCl / OCl-, and H2O / OH- as shown below. The species that provide the PRL are then determined. Water is always one of these since it is initially present. In this example, NaOCl is also added but its dissociation into Na+ and OCl- is not an acid-base reaction (since neither H+ nor OH- is involved). However, this provides OCl- which is part of the HOCl / OCl- conjugate pair, so it is a PRL species.

 

 

Proton condition: [H+] + [HOCl] = [OH-]

 

Note that we could also write a mass balance on the dissociation of NaOCl to form Na+, OCl-, and (by protonation) HOCl. Conservation of sodium and hypochlorite allows us to state

[Na+] = [HOCl] + [OCl-]

[Na+] = CT,OCl

 

This equation, however, can also be obtained by algebraic manipulation of the previous three equations. There are thus four equations, but only three can be used in providing independent information. (This can be any three of the four).

Example 5.4 Determine the mass balances, charge balance, and proton condition of a solution prepared with 10^-3M ammonium chloride (NH4Cl) and 10^-2M sodium acetate (CH3COONa) which can be abbreviated NaAc.

 

There are 8 species in solution Na+, Cl-, NH4+, NH3, HAc, Ac-, OH-, H+.

 

The mass balance for the conjugate acid-base pairs :

CT,Ac = [HAc] + [Ac-] = 10^-2 M

CT,NH3 = [NH4+] + [NH3] = 10^-3 M

 

The mass balance for the sodium and chloride ions:

CT,Na = [Na+] = 10^-2 M

CT,Cl = [Cl-] = 10^-3 M

 

The charge balance: [Na+] + [NH4+] + [H+] = [Ac-] + [Cl-]+ [OH-]

The proton condition is determined by first drawing the proton balance diagram. The conjugate acid-base pairs are H+ / H2O, HAc / Ac-, NH4+ / NH3, and H2O / OH- as shown below. The species that provide the PRL are H2O, NH4+, and Ac- as these are the species used to prepare the solution.

Proton condition: [H+] + [HAc] = [NH3 ] + [OH-]

 

5.5 Calculating pH - An Analytical Approach

Strong Acids

Strong acids dissociate completely such that if HA in equation [5.14] is a strong acid, the reaction goes completely to the right and [HA] essentially equals zero. Some common strong acids are listed in the table below. The pKa value is the negative logarithm of the dissociation constant:

[5.21]

As you will recall, the stronger the acid, the higher the KA value. Thus, the stronger the acid, the lower the pKA value will be.

Strong Acid Reaction pKA

Nitric HNO3 HNO3 ' NO3- + H+ 0

Sulfuric H2SO4 H2SO4 ' HSO4- + H+ -3

Bisulfate ion HSO4- HSO4- ' SO42- + H+ 2

Hydrochloric HCl HCl ' Cl- + H+ -3

Perchloric HClO4 HClO4 ' ClO4- + H+ -7

 

Table 1 Common Strong Acids.

 

 

Example 5.5 Confirm that when nitric acid is added to water, the concentration of HNO3 at equilibrium is negligibly small. Assume that the pH of the acidified water to be 2.0 or greater.

 

Example 5.6 If we add 0.005 moles sulfuric acid to one liter of pure water, calculate the pH. If we only added 5 x 10^-8 moles to one liter of pure water, what would be the resulting pH?

Thus from the stoichiometric equation for every mole of sulfuric acid added, 2 moles of H+ are created. Beginning with a charge balance :

From the dissociation of water (eq[5.7]) the above becomes

This example illustrates that as the concentration of the strong acid decreases the importance of the [OH-] term in the charge balance increases. The [OH-] manifests itself as the 4 (10^-14) term in the radical of the quadratic formula. The other term under the radical in the quadratic formula is dependent on the strong acid concentration. By inspection we can see that as long as the strong acid concentration is greater than 10^-6 the 4 (10^-14) term will be insignificant. Thus the pH of a 10^-4 M solution of HCl is 4, the pH of a 10^-3.58 M solution of HClO4 is 3.58, and the pH of a 10^-4.3 M solution of H2SO4 is 4.

 

Strong Bases

Strong bases dissociate completely such that if BOH in equation [5.16] is a strong base, the reaction goes completely to the right and [BOH] essentially equals zero (this can be demonstrated in a similar manner to example 5.5). The development of pH calculations for strong bases is very similar to that of strong acids. Thus, by analogy to the calculations shown in example 5.6, strong bases at a concentration greater than 10-6 M allow the [H+] to be neglected in the charge balance. Thus the pOH of a 10^-4 M solution of NaOH is 4, the pOH of a 10^-3.58 M solution of KOH is 3.58, and the pOH of a 10^-4.3 M solution of Ca(OH)2 is 4.

 

Weak Acids

Weak acids dissociate only slightly such that if HA in equation [5.14] is a weak acid, the reaction proceeds only partially to the right and [HA] is still a significant species in solution. A simple equation can be derived for determining the pH of a weak acid added to water by making use of equation [5.14] and equation [5.15], combined with a mass balance (equation [5.18]).

 

Algebraically manipulating equation [5.15] and equation [5.18], while utilizing the stoichiometry of equation [5.14] results in the following equation:

[5.22]

 

[5.23]

 

Note that the above simplifying assumption can - and should - be checked once the value of [H+] is determined. If the final result is not valid, the final result can be used in the preceding solution (equation[5.22]) of [H+] and iterated to obtain an improved solution. This iteration of successive approximation is generally quicker than trying to solve the quadratic equation.

 

Example 5.7 Acetic acid (H3CCOOH) is produced in anaerobic sludge digesters. Typically, digesters contain 60,000 mg/L or 0.1M acetic acid. What is the pH of a 0.1 M solution of acetic acid? Acetic acid has a KA of 1.8 x 10^-5. Ignore all species that might be in the digester except acetic acid and water.

 

Weak Bases

The development of an analytical equation for pH determination for a weak bases is similar to that for a weak acid. Let us consider the weak base B in the following reaction:

[5.24]

The dissociation constant of the base would be

[5.25]

Note the dissociation constant for the conjugate acid of base B would be determined by adding the acid [BH+ ] on the right-hand side of equation [5.24] to water. Thus, the dissociation constant of the conjugate acid BH+ would be

[5.26]

By multiplying equation [5.25] and equation [5.26] and utilizing equation [5.7] an important relationship between the acid and base dissociation constants of a conjugate pair can be determined

[5.27]

Utilizing equations [5.25] and [5.27], and emulating the derivation in the weak acids section results in the following equations for weak bases:

[5.28]

[5.29]

 

Example 5.8 What is the pH of a 0.005 M solution sodium acetate? Sodium acetate (NaH3CCOO) is the salt of the weak acid, acetic acid (H3CCOOH). Generally, the salt of a weak acid is basic (alkaline) in character and behaves like a base.

 

5.6 Calculating pH and Equilibrium Relationships - A General Approach

While the analytical approach is useful for simplistic situations, a more rigorous approach must be utilized for complex problems. This method will apply to all acid-base problems and utilizes the equilibrium relationships discussed in Section 5.4. Generally, more information than just the pH of the solution is required. Furthermore, the analytical approach falls short of determining the pH for simple acid-base equilibria such as polyprotic acids. Polyprotic acids are acids that can donate more than one proton, as demonstrated for the general polyprotic acid H3A:

[5.30]

Carbonic acid (H2CO3) is a polyprotic acid that we will examine in great detail because of its major significance in environmental engineering. Carbonic acid along with its ionization products bicarbonate (HCO3-) and carbonate (CO32-) comprise the carbonate system. The carbonate system is the most important acid-base system in natural waters, as it usually controls the pH. Carbon dioxide in the atmosphere dissolves in water to form carbonic acid

[5.31]

The following are the important equilibria relationships governing the carbonate system

The carbonate system's species allow natural waters to resist changes in pH. Waters being resistant to changes in pH are said to be buffered. A buffer is a species in solution that offers resistance to pH changes upon acid or base addition. The bases HCO3-, CO32-, and OH- give natural waters the ability to resist changes in pH when a strong acid is added. Acids such as H2CO3, HCO3-, H+ provide buffering against the addition of strong bases. This buffering capacity of natural waters is extremely important because of the narrow pH range tolerable to aquatic life. Before proceeding further let us look at a few examples utilizing the general approach.

 

Example 5.9 Calculate the pH of a solution containing 50 mg/L of carbonic acid.

With the general approach, we write down all the equations that must hold for the solution under consideration. These include the equilibrium reactions and their equilibrium constants, the dissociation of water, mass balances of the interacting species, and either the proton condition or charge balance as appropriate. Generally, this will lead to a system of equations that is ready solvable by making realistic approximations or through the use of computer software such as Mathematica or Maple.

Without access to such powerful software you might proceed as follows:

 

1) Assume OH- << H+ as it is a solution of carbonic acid.

2) The dominant species of the diprotic weak acid are H2CO3 and HCO3- such that CO32- is negligible.

 

After getting the pH the original assumptions must be checked to insure accuracy. This has been omitted since the Mathematica® solved solution is the same.

Example 5.10 In the year 2005, you find yourself employed by a consulting firm that is designing a water treatment facility. You've been asked to characterize some of the chemical processes to be employed. The facility is buffered by the carbonate system, with CT = 10^-5 M. You measure the pH to be 7.65. Under these conditions what are the concentrations of [H+], [OH-], [H2CO3], [HCO3-], [CO32-] in mol /L?

 

 

The general approach also enables the pH to be determined for systems of multiple components. Most waters of interest are mixtures of various acid and base containing components. For these more complex situations the general approach must be employed to determine the equilibria of the system.

 

Example 5.11 To control acid rain, sulfur dioxide (SO2) must be removed from power plant emissions. Although the generation of acid rain is a complex process the following equations have been proposed to represent its formation from sulfur containing fossil fuels:

H2SO4 is sulfuric acid which we have seen is a strong acid.

Beginning in 1995, the 110 largest coal-fired electric utilities in the country were each required to reduce sulfur dioxide emissions by 10 million tons compared with 1980 levels. One of these utilities, Delmarva Power & Light Co., added scrubbing equipment to their Millsboro, DE plant in order to meet the federal guidelines. Following scrubbing of the stack (or flue) gases at this Millsboro, DE plant, the flue gases are sprayed with a lime slurry, Ca(OH)2, to form calcium sulfate, Ca(SO4)2, suspended in water. If a sulfuric acid solution of pH 2.4 results after scrubbing how much lime must be added during spraying to produce a slurry of pH 8.0?

 

Example 5.12 A solution of hydrogen sulfide (H2S) was added to NaOH. The mixture was analyzed and found to contain 0.006583 M H2S and 0.003155 M NaOH; its pH was measured as 6.416 with a glass electrode. Find pKA,1 for H2S.

Equations of interest:

Assume the dominant species of the diprotic weak acid are H2S and HS- such that S2- is negligible.

 

Alkalinity

As mentioned earlier the carbonate system buffers natural waters. A concept closely related to buffering is alkalinity. Alkalinity is a measure of a water's capacity to neutralize acid. Alkalinity is attributable to all the bases present in a water. In natural waters as previously stated this capacity is usually attributable to HCO3-, CO32-, and OH-. The alkalinity of a water is operationally determined by measuring the addition of a strong acid (such as sulfuric acid) that will lower the solution pH to 4.5. This process is called a titration. Mathematically alkalinity can be defined as

[5.32]

Equation [5.32] assumes that only the carbonate system contributes to the alkalinity.

 

Ionization Fractions

Equation [5.32] can be put in terms of one variable by defining a set of dimensionless ionization fractions, a. Each such a represents the fraction of the total species concentrations, CT, present as the particular species. The ionization fractions can be determined from the mass balance and equilibrium equations. Using the carbonate system, the ionization fractions for a diprotic acid will be determined. The following terms are defined:

Now all the different species can be represented as a function of only [H+] if CT is known.

Similar expressions can be developed for a1 and a2.

 

 

Using the above results the alkalinity equation [5.32] can be puts in terms of just H+ if CT is known.

 

For a monoprotic acid, similar expressions for a0 and a1 can be derived and there is no a2 (or a2 = 0). They are simply given here as

[5.36]

[5.37]

 

Example 5.13 An industrial wastewater is composed 0.01 M ammonium cyanide (NH4CN). What is the pH of this wastewater? Use the following chemical reactions combined with ionization fractions to simplify the problem.

The assumption was a reasonable one, because the concentration of the concerned species (0.01M) was considerably higher than the expected H+ or OH- concentration.

 

Example 5.14 If the Delmarva Power and Light Co. sprayed the scrubbed stack gases at their Millsboro, DE plant with groundwater having a total alkalinity of 0.002 eq/L and a pH of 7.9, what is the minimum dilution ratio (volume wastewater / volume groundwater) that must be maintained to raise the pH of the wastewater to 6.3? At this pH the plant is permitted to discharge its wastewater to a receiving stream. Assume that only the carbonate system contributes to the total alkalinity.

 

5.7 Calculating pH and Equilibrium Relationships - A Graphical Approach

 

Example 5.15 Use a pC-pH diagram to confirm the answer in example 5.2.

 

Example 5.16 What is the pH of a 10^-4 M aqueous boric acid (H3BO4) solution? Assume a pKA of 9.15 and solve graphically.

 

Example 5.17 Sodium bicarbonate (NaHCO3) is added to water at a concentration of 10 mg/L. Draw a pC-pH graph of the carbonate system and use this to determine the pH of the system. Repeat this for initial concentrations of 100 and 1000 mg/L.

 

Example 5.18 By means of Logarithmic diagrams (pC-pH graphs), find pH in solutions containing (a) 0.1M HA, 0.01M HB, (b) 0.1M NaA, 0.01 M HB, (c) 0.1 M HA, 0.01M NaB, (d) 0.1 M NaA, 0.01 M NaB, if pKa is 5.0 for HA and 7.2 for HB, pKw is 14.