The Chemistry Of Natural Waters

In this chapter, the following introductory chemistry topics are used:

■ Concepts of oxidation and reduction as electron loss/gain; half-reactions; redox reactions; oxidizing and reducing agents; electrode potentials ;

■ Solubility product and weak acid/weak base calculations; water constant Kw

■ Oxidation numbers and the balancing of redox reactions (reviewed in Box 13-1)

Background from previous chapters used in this chapter:

■ Equlibria involving gases dissolved in water: Henry's law (Chapter 3)


All life forms on Earth depend on water. Each human being needs to consume several liters of fresh water daily to sustain life. Much more water is used for other domestic activities: Typical daily usages for showering/bathing, washing, and toilets each amount to about 50 L, in addition to about 20 L for dishwashing and 10 L for cooking. (A hose delivers 10 L or more of water per minute, so watering gardens and lawns can easily double average domestic consumption.) Vastly larger amounts per capita are used by industry and especially for irrigation in agriculture. For example, thousands of liters of fresh water are required to produce one kilogram of beef or cotton or even of rice.

However, fresh water is at a premium. Over 97% of the world's water is seawater, unsuitable for drinking and for most agricultural purposes. Three-quarters of the fresh water is trapped in glaciers and ice caps. Lakes and rivers are one of the main sources of drinking water, even though taken together they constitute less than 0.01% of the total water on the planet. About half

FIGURE 13-1 Global pools and fluxes of water on Earth, showing the size of groundwater storage relative to other major water sources and fluxes. All pool volumes (green) are in cubic kilometers, and all fluxes (black) are in cubic kilometers per year. [Source: W. H. Schlesinger, Biogeochemistry—An Analysis of Global Change, 2nd ed. (San Diego: Academic Press, 1997), Chapter 10.]

Precipitation on land 111,000

Soil water v a

Atmosphere 13,000

Net transport to land • 40,000

Evapotranspiration from land 71,000

River flow to oceans 40,000

Soil water v a

Groundwater 15,300,000

Precipitation on oceans 385,000

Evaporation from oceans 425,000

Precipitation on oceans 385,000

Evaporation from oceans 425,000

Evaporation From Oceans

of drinking water is obtained from groundwater, the fresh water that lies underground and that is discussed in detail in Chapter 14. The annual fluxes of water between its global pools in the oceans, the air, and beneath the ground are shown in Figure 13-1, along with the sizes of each pool.

Humanity currently consumes, mostly for agriculture, about one-fifth of the accessible runoff water that travels through rivers to the seas; this fraction is predicted to rise to about three-quarters by 2025. Runoff water is highly variable in both location and time in terms of its availability unless storage and transport are available. Although only 10% of the world's population in 2000 lived under conditions of water stress or scarcity, this figure is expected to rise to 38% by 2025 (Figure 13-2).

It is important to understand the types of chemical activities that prevail in natural waters and how the science and application of chemistry can be employed to purify water intended for drinking purposes. Although some discussion of pollution problems is contained in this chapter, the remediation of contaminated water is considered in detail in Chapter 14-

It will be convenient to divide our considerations of water chemistry in this chapter into the two common reaction categories: acid-base reactions and oxidation-reduction (redox) reactions. Acid-base and solubility phenomena control the concentrations of dissolved inorganic ions such as carbonate in waters, whereas the organic content of water is dominated by redox reactions. The pH and concentrations of the principal ions in most natural water systems are controlled by the dissolution of atmospheric carbon dioxide


Total population: 6 billion

2025 (medium projection)

Total population: 7.9 billion

Environmental Chemistry

FIGURE 13-2 Global water supply in 2000 and projection for 2025. [Source: R. Engelman et al., People in the Balance (Washington, DC: Popular Action International, 2000) as reproduced in Nature 422 (2003): 252.]

Percentage of world population with

| stress H scarcity relative sufficiency


Total population: 6 billion

2025 (medium projection)

Total population: 7.9 billion

Percentage of world population with

| stress H scarcity relative sufficiency

FIGURE 13-2 Global water supply in 2000 and projection for 2025. [Source: R. Engelman et al., People in the Balance (Washington, DC: Popular Action International, 2000) as reproduced in Nature 422 (2003): 252.]

and soi[-bound carbonate ions; such reactions are considered in detail later in the chapter. Before considering these acid-base processes, we consider some important redox processes, especially those involving dissolved oxygen. For clarity, the phase (aq) for ions and molecules dissolved in aqueous solution will not be shown in equations but simply assumed.

Oxidation-Reduction Chemistry in Natural Waters Dissolved Oxygen

By far the most important oxidizing agent (i.e., substance that extracts electrons from other species) in natural waters is dissolved molecular oxygen, 02. Upon reaction, each of the oxygen atoms in 02 is reduced from the zero oxidation number to — 2, in H20 or OH . (The concept and calculation of oxidation numbers is reviewed in Box 13-1, as is the balancing of redox equations.) The half-reaction that occurs in acidic solution is

02 + 4 H+ + 4e"-> 2 HzO

whereas that which occurs in basic aqueous solution is

02 + 2 H20 + 4e"-> 4 OH"

The concentration of dissolved oxygen in water is small and therefore precarious from the ecological point of view. As discussed in Chapter 3, for the dissolution of a gas in water such as the process

BOX 13-1

Redox Equation Balancing Reviewed

Assigning Oxidation Numbers

A simple way to determine the extent (if any) to which an element is oxidized or reduced in a reaction is to deduce the change in its oxidation number, O.N., in the product as compared to that in the reactant. The oxidation number of the elements in most compounds and ions can be determined by applying, in sequence, the following set of rules, keeping in mind that the sum of all the oxidation numbers in a substance must equal its net charge. The rules are listed in terms of priority so that, e.g., if for a compound rule (iv) is inconsistent with rule (iii), then rule (iii) takes precedence since it is higher in the order.

(i) Elements appearing in the free, unbonded form have an O.N. equal to their ionic charge, which is zero if the element is uncharged.

(ii) Fluorine has an O.N. of —1 in compounds. Group I and II metals have O.N. values corresponding to their ionic charges +1 and +2, respectively, and Al is +3.

(iii) Hydrogen has an O.N. of +1, except when bonded to a metal, where it is — 1.

(iv) Oxygen has an O.N. of —2 (except when overridden by a rule higher in the sequence, as an example below illustrates).

(v) Chlorine, bromine, and iodine have O.N.'s of -1 (except when overridden by a rule higher in the sequence, as an example below illustrates).

Some examples:

HF: F is — 1 (rule ii) and H is +1 (rule iii); the sum is zero, as required.

HjOj: H is +1 (rule iii), but O here cannot be -2 (rule iv) since the sum of charges would be 2 (+1) + 2 (—2) = —2; the charges must add up to zero for the molecule as a whole. Since rule (iii) takes precedence, each H must be +1, so each O here must be — 1 in order that the sum be zero.

C102 : Each O is —2 (rule iv), for a total of —4, so CI here cannot be — 1 (rule v) since the sum of charges would then be —1 + 2 (—2) = —5, as compared to the actual net charge of —1. Since O.N.q + 2 X (-2) = -1, it follows that = +3 here.

As an example of the use of oxidation numbers in reactions, consider the half-reaction in which nitrate ion is converted into nitrous oxide;

the appropriate equilibrium constant is the Henry's law constant KH, which for oxygen at 25°C has the value 1.3 X 1CT3 mol L *1 atm ':

Kh = [02(aq)]/Po2 = 1-3 X 1CT3 mol L-1 atm"1 at 25°C

Since in dry air at sea level the partial pressure, Pq,, of oxygen is 0.21 atmospheres (atm), it follows that the solubility of 02 is 8.7 milligrams per liter of water (see Problem 13-1). This value can also be stated as 8.7 ppm since, as discussed in Chapter 10, ppm concentrations for condensed phases are based

Since in nitrate ion, each O is —2, the sum of which is —6, and the charge on the ion is only — 1, it follows that N here is +5.

In nitrous oxide, the O is — 2, and the sum of oxidation numbers is zero, so each N must be +1.

Keeping in mind that the reaction requires 2 nitrate ions to supply enough nitrogen for one nitrous oxide molecule, we see that 2+5 N's, total +10, here become 2+1 N's, total +2. Thus the half-reaction must be a (10 — 2 =) 8-electron reduction:

If it is required to know the amounts of water molecules and H+ or OH" ions involved, the detailed balancing scheme discussed below must be employed.

Balancing Redox Equations

There are many equivalent schemes to completely balance redox half-reactions and overall reactions, of which the following is one example,

• To balance a half-reaction, first deduce the number of electrons involved in the process, as in the scheme above, by balancing atoms other than H and O.

• Next, to balance charge, add sufficient H+ ions to the side having excess negative charge; note that only real charges on ions and electrons are considered here, not oxidation numbers.

• Finally, balance the number of oxygen atoms by adding H20 molecules to the side deficient in oxygen.

Consider, for example, the nitrate to nitrous oxide example given above:

The actual charge on the left-hand side is 2 X (-1) + (-8) = -10, but that on the right side is zero. Thus we should add 10 positive charges, each in the form of H+, to the left side, so that its charge also becomes zero:

Finally, since we now have 2 X 3 = 6 O atoms on the left side and only one on the right side, we need to add 5 O each in the form of H20 molecules to the right side:

2 N03~ + 8 e~ + 10 H+->NzO + 5 H20

The half-reaction is now balanced.

on mass rather than moles. (Note that for simplicity, molar concentrations rather than activities are used in all equilibrium calculations in this book, since in general we are considering only very dilute solutions.)

Because the solubilities of gases increase with decreasing temperature, the amount of 02 that dissolves at 0°C (14.7 ppm) is greater than the amount that dissolves at 35°C (7.0 ppm). The median concentration of oxygen found in natural, unpolluted surface waters in the United States is about 10 ppm.


Confirm by calculation the value of 8.7 mg/L for the solubility of oxygen in water at 25 °C,


Given the solubility quoted above for 02 at 0°C, calculate the value of KH for it at this temperature.

River or lake water that has been artificially warmed can be considered to have undergone thermal pollution in the sense that, at equilibrium, it will contain less oxygen than colder water because of the decrease in gas solubility with increasing temperature. To sustain their lives, most fish species require water containing at least 5 ppm of dissolved oxygen; consequently, their survival in warmed water can be problematic. Thermal pollution often occurs in the region of electric power plants (whether fossil fuel, nuclear, or solar), since they draw cold water from a river or lake, use it for cooling purposes, and then return the warmed water to its source.

Oxygen Demand

The most common substance oxidized by dissolved oxygen in water is organic matter having a biological origin, such as dead plant matter and animal wastes. If, for the sake of simplicity, the organic matter is assumed to be entirely polymerized carbohydrate (e.g., plant fiber) with an approximate empirical formula of CHzO, the oxidation reaction would be

CH20(aq) + 02(aq)->C02(g) + H,0(aq)


Dissolved oxygen in water is also consumed by the oxidation of dissolved ammonia, NH3, and ammonium ion, NH^—substances that, like organic matter, are present in water as a result of biological activity—eventually to nitrate ion, N03_ (see Problem 13-4).


Show that I L of water saturated with oxygen at 25°C is capable of oxidizing 8.2 mg of polymeric CHzO.


Determine the balanced redox reaction for the oxidation of ammonia to nitrate ion by 02 in alkaline (basic) solution. Does this reaction make the water more alkaline or less? [Hint: Recall the redox balancing procedure in

Water that is aerated by flowing in shallow streams and rivers is constantly replenished with oxygen. However, stagnant water or that near the bottom of a deep lake is usually almost completely depleted of oxygen because of its reaction with organic matter and the lack of any mechanism to replenish it quickly, diffusion being a slow process and turbulent mixing being absent.

The capacity of the organic and biological matter in a sample of natural water to consume oxygen, a process catalyzed by bacteria present, is called its biochemical oxygen demand, BOD. It is evaluated experimentally by determining the concentration of dissolved 02 at the beginning and at the end of a period in which a sealed water sample seeded with bacteria is maintained in the dark at a constant temperature, usually either 20°C or 25 °C. A neutral pH is maintained by use of a buffer consisting of two ions of phosphoric acid, namely H2P04 and HP042~:

H2P04- HP042- + H+

The BOD equals the amount of oxygen consumed as a result of the oxidation of dissolved organic matter in the sample. The oxidation reactions are catalyzed in the sample by the action of microorganisms present in the natural water. If it is suspected that the sample will have a high BOD, it is first diluted with pure, oxygen-saturated water so that sufficient 02 will be available overall to oxidize all the organic matter; the results are corrected for this dilution. Usually the reaction is allowed to proceed for five days before the residual oxygen is determined. The oxygen demand determined from such a test, often designated BOD5, corresponds to about 80% of that which would be determined if the experiment were allowed to proceed for a very long time—which of course is not very practical. The median BOD for unpolluted surface water in the United States is about 0.7 mg 02 per liter, which is considerably less than the maximum solubility of 02 in water (of 8.7 mg/L at 25°C). In contrast, the BOD values for sewage are typically several hundreds of milligrams of oxygen per liter.

A faster determination of oxygen demand of a water sample can be made by evaluating its chemical oxygen demand, COD. Dichromate ion, Cr2072^, can be dissolved as one of its salts, such as K2Cr207, in sulfuric acid: The result is a powerful oxidizing agent. It is this solution, rather than 02, that is used to ascertain COD values. The reduction half-reaction for dichromate when it oxidizes the organic matter is

Crz072~ + 14 H+ + 6 e^->2 Cr3+ + 7 H20

(In practice, excess dichromate is added to the sample and the resulting solution is back-titrated with FeZ+ to the end point.) The number of moles of 02

that the sample would have consumed in accomplishing the oxidation of the same material equals 6/4 (= 1.5) times the number of moles of dichromate, since the latter accepts six electrons per ion whereas 02 accepts only four:

02 + 4 H+ + 4e~-»■ 2 H20

Thus the moles of 02 required for the oxidation is 1.5 times the number of moles of dichromate actually used. (See Problems 13-5 and 13-6.)


A 25-mL sample of river water was titrated with 0.0010 M K2Cr207 and required 8.3 mL to reach the end point. What is the chemical oxygen demand, in milligrams of 02 per liter, of the sample?


The COD of a water sample is found to be 30 mg of Oz per liter. What volume of 0.0020 M K2Cr20? will be required to titrate a 50-mL sample of the water?

The difficulty with the COD index as a measure of oxygen demand is that acidified dichromate is such a strong oxidant that it oxidizes substances that are very slow to consume oxygen in natural waters and that therefore pose no real threat to their dissolved oxygen content. In other words, dichromate oxidizes substances that would not be oxidized by 02 in the determination of the BOD. Because of this excess oxidation, namely of stable organic matter such as cellulose to C02, and of CP to Cl2, the COD value for a water sample as a rule is slightly greater than its BOD value. Neither method of analysis oxidizes aromatic hydrocarbons or many alkanes, which, in any event, resist degradation, and therefore oxygen consumption, in natural waters.

It is not uncommon for water polluted by organic substances associated with animal or food waste or sewage to have an oxygen demand that exceeds the maximum equilibrium solubility of dissolved oxygen. Under such circumstances, unless the water is continuously aerated, it will soon be depleted of its oxygen, and fish living in the water will die. The treatment of wastewater to reduce its BOD is discussed in Chapter 14.

Finally, we note that there are two other measures used for the amount of organic substances present in natural waters. The total organic carbon, TOC, is used to characterize the dissolved and suspended organic matter in raw water. For example, the TOC usually has a value of approximately 1 milligram per liter, i.e., 1 ppm carbon, for groundwater. The parameter dissolved organic carbon, DOC, is used to characterize only organic material that is actually dissolved, not suspended. For surface waters, the DOC averages about 5 ppm, although bogs arid swamps can have DOC values that are ten times this amount, and untreated sewage typically has a DOC value of hundreds of ppm. The largest component of organic carbon in natural waters is usually carbohydrates, but many other types including proteins and low-molecular-weight aldehydes, ketones, and carboxylic acids are also present. The humic materials in water are discussed in Chapter 16.

Green Chemistry: Enzymatic Preparation of Cotton Textiles

Globally over 40 billion pounds (20 million kilograms) of cotton are produced each year. Even with the invasion of synthetic fibers such as nylon, polyester, and acrylics, cotton still holds over 50% of the market share for apparel and home furnishings that are sold in the United States. In preparing raw cotton for use as a fiber, several steps—including desizing, scouring, and bleaching—are required, leaving a fiber that is 99% cellulose. These steps use copious amounts of chemicals, water, and energy, and they produce millions of pounds of aqueous waste that is high in BOD and COD.

Raw cotton is composed of several concentric layers. The outermost layer is composed of fats, waxes, and pectin, while the inner layers consist primarily of cellulose. The fats and waxes make the raw cotton fiber waterproof, and the pectin acts as a powerful glue to hold the layers together. In order to prepare cotton for use as a fiber for bleaching and dyeing, the outer layer must be removed. This process, which is known as scouring, has traditionally been done by immersing the cotton in 18-25% aqueous sodium hydroxide solution at elevated temperatures. This results in hydrolysis of the fats (saponification) and waxes, which solubilizes the components of the outer layer so that they can simply be rinsed away. Scouring results in fibers of even and high wettability. In addition to sodium hydroxide, chelating agents (Chapter 14) and emulsifiers are added during the scouring process. To end the process, the mixture is neutralized with acetic acid and rinsed several times.

The scouring process is estimated by the U.S. EPA to account for about half of the total BOD produced in the preparation of cotton fibers. The BOD in wastewater from cotton production generally exceeds 1100 mg/L, which is several times that of raw sewage. In addition to the large amounts of chemicals, energy, and water that are used, and the concomitant pollution that is produced, another disadvantage of this process is the unintended weakening and loss of some of the cellulose fibers.

An alternative to the traditional scouring process, known as Bio-Preparation, was developed by Novozymes-North America Inc. and won a Presidential Green Chemistry Challenge Award in 2001. BioPreparation employs an enzyme (a pectin lyase) that selectively degrades pectin at ambient temperatures. Since pectin acts as a glue to hold the outer layer of the cotton fiber together, destruction of the pectin results in disintegration of this layer. Because the lyase is selective for only pectin, its use is much less aggressive than the typical scouring process and removes much less organic material (including cellulose) from the cotton. Since the dissolved organic materials are what contribute to the high BOD and COD of the wastewater, this enzymatic treatment lowers the BOD by 20% and the COD by 50%.

In addition to these environmental advantages, BioPreparation eliminates the use of sodium hydroxide solutions at elevated temperatures, which in turn

• lowers the pH of the wastewater,

• eliminates the need for neutralization with acetic acid and the concomitant wastes,

• lowers energy requirements, and

• lowers rinsing requirements, reducing water consumption by 30—50%.

Elimination of the use of sodium hydroxide also reduces the risk to workers, and the reduced degradation of cellulose provides for more robust fibers and higher yield.

Decomposition of Organic Matter in Water

Dissolved organic matter will decompose in water under anaerobic (oxygen-free) conditions if appropriate bacteria are present. Anaerobic conditions occur naturally in stagnant water such as swamps and at the bottom of deep lakes. The bacteria operate on carbon to disproportionate it; that is, some carbon is oxidized to carbon dioxide, C02, and the rest is reduced to methane, CH4:


2 CH20 —'-* CH4 + C02

organic matter

C oxidation number 0 —4-1-4

This is an example of a fermentation reaction, which in chemistry is defined as one in which both the oxidizing and the reducing agents are organic materials. Since methane is almost insoluble in water, it forms bubbles that can be seen rising to the surface in swamps and sometimes catches fire; indeed, methane was originally called marsh or swamp gas. The same chemical reaction shown above occurs in digestor units used by rural inhabitants in semi-tropical developing countries (India, for instance) to convert animal wastes into methane gas that can be used as a fuel. The reaction also occurs in landfills, as discussed in Chapter 16.

Since anaerobic conditions are reducing conditions in the chemical sense, insoluble Fe ' * compounds that are present in sediments at the bottom

of lakes are converted into soluble Fe compounds, which then dissolve into the lake water:

Fe3+ + e insoluble Fe(IIl) soluble Fe(II)

It is not uncommon to find aerobic and anaerobic conditions in different parts of the same lake at the same time, particularly in the summertime when a stable stratification of distinct layers often occurs (see Figure 13-3). Water at the top of the lake is warmed by the absorption of sunshine by biological materials, while that below the level of penetration of sunlight remains cold. Since warm water is less dense than cold (at temperatures above 4°C), the warm upper layer "floats" on the cold layer below, and little transfer between them occurs. The top layer, called the epilimnium, usually contains near-saturation levels of dissolved oxygen, due both to its contact with air and to the Oz produced in photosynthesis by algae. Since conditions in the top layer are aerobic, elements exist there in their most oxidized forms:

• carbon, with an oxidation number (O.N.) of +4, as COz or H2C03 orHC03~;

• sulfur, O.N. of+6, as S042";

• nitrogen, O.N. of +5, as N03~; and

• iron, as Fe(IlI), in the form of insoluble Fe(OH)3.

Near the bottom, in the hypolimnium, the water is oxygen-depleted since it has no contact with air and since 02 is consumed when biological material, such as the dead algae that have sunk to these depths, decomposes. Under such anaerobic conditions, elements exist in their most reduced forms:

• carbon, with an O.N. of —4, as CH4;

• nitrogen, O.N. of -3, as NH3 and NH4+; and

• iron, as Fe(II), in the form of soluble Fe2+.

Anaerobic conditions usually do not last indefinitely. In the fall and winter, the top layer of water is cooled by cold air passing over it, so that eventually the oxygen-rich water at the top becomes more dense than that below it and gravity induces mixing between the layers. Thus in the winter and early spring the environment near the bottom of a lake usually is aerobic.

Aerobic conditions (warm water)


h2co3 no3~


Anaerobic conditions




(cold water)



FIGURE 13-3 The stratification of a lake in the summer, showing the typical forms of the major elements it contains at different levels.

FIGURE 13-3 The stratification of a lake in the summer, showing the typical forms of the major elements it contains at different levels.

TABLt .13-1

TABLt .13-1

Increasing Levels of Sulfur Oxidation

Increasing Levels of Sulfur Oxidation

Aqueous solution and salts H2s h2so3 h2so4


so? soi


Gas phase Molecular solids h7s

Sulfur Compounds in Natural Waters

The common inorganic oxidation numbers in which sulfur is encountered in the environment are illustrated in Table 13-1; they range from the highly reduced — 2 form that is found in hydrogen sulfide gas, H2S, and insoluble minerals containing the sulfide ion, S' to the highly oxidized +6 form that is encountered in sulfuric acid, H2S04, and in salts containing the sulfate 1011, S04 . In organic and bioorganic molecules such as amino acids, intermediate levels of sulfur oxidation are present. When such molecules decompose anaerobically, hydrogen sulfide and other gases such as CH3SH and CH3SSCH3 containing in highly reduced forms are released, thereby giving swamps their unpleasant odor. The occurrence of such gases as air pollutants was mentioned in Chapter 3.

As discussed in Chapter 3, hydrogen sulfide is oxidized in air first to sulfur dioxide, S02, and then fully to sulfuric acid or a salt containing the sulfate ion. Similarly, hydrogen sulfide dissolved in water can be oxidized by certain bacteria to elemental sulfur or more completely to sulfate. Overall the complete oxidation reactions correspond to

Some anaerobic bacteria are able to use sulfate ion as the oxidizing agent to convert, organic matter, such as polymeric CH20. to carbon dioxide when the concentration of oxygen in the water is very low; the S042" ions are reduced in the process to elemental sulfur or even to hydrogen sulfide:

2 S042^ + 3 CH,© + 4 H+-» 2 S + 3 C02 + 5 H-O

Such reactions are especially important in seawater, for which the sulfate ion concentration is much higher than the average for fresh-water systems.

Add Mine Drainage

One characteristic reaction of groundwater, which by definition is not well-aerated since it has spent much time not exposed to air, is that when it reaches the surface and 02 has an opportunity to dissolve in it, its rather high level of soluble Fe2' is converted to insoluble Fe3+, and an orange-brown deposit of Fe(OH)3 is formed.

4 Fez+ + 02 + 2 H20-> 4 Fe3+ + 4 OHT

4 [Fe3+ + 3 OFT-» Fe(OH)3<s)]

The overall reaction is

4 Fe2+ + 02 + 2 H20 + 8 OFT->4 Fe(OH)3(s)

An analogous reaction occurs in some underground coal and metal (especially copper) mines, especially abandoned ones, and in piles of mined coal left open to the environment. Normally FeS2, called iron pyrites, or fool's gold, is a stable, insoluble component of underground rocks as long as it does not come into contact with air. However, as a result of the mining of coal and certain ores—and especially after underground mines have been abandoned and spontaneously fill with groundwater—some of it is exposed to water, oxygen, and certain bacteria and becomes partially solubilized as a result of its oxidation. The disulfide ion, S22^, in which sulfur has an average oxidation number of — 1, is oxidized to sulfate ion, S042 which contains sulfur in the +6 form:

sz2 + 8 H;0-> 2 S042" + 16 H+ + 14 e~

The main oxidizing agent acting on the sulfur is atmospheric 02:

7 [02 + 4 H+ + 4 e"-► 2 H20]

When this (28 e~) half-reaction is added to twice the (14 e~) oxidation half-reaction, the net redox reaction is obtained:

2 S22" + 7 02 + 2 H,0-> 4 SO/" + 4 H+

Since the sulfate salt of the ferrous ion, Fe2+, is soluble in water, the iron pyrites are effectively solubilized by the reaction. More importantly, the reaction produces a large amount of concentrated acid (note the H+ product), only a portion of which is consumed by the air oxidation of Fe2 + to Fe3+ that accompanies the process:

4 Fe2" + 02 + 4 H+-»4 Fe3''' + 2 H20

In acidic environments, this reaction is catalyzed by bacteria (Thiobacillus ferrooxidans); the resulting Fe3+ can oxidize various metal sulfides, liberating the metal ions.

Combining the last two reactions in the correct ratio, i.e., 2:1, we obtain the overall reaction for the oxidation of both the iron and the sulfur:

4 FeS, + 15 02 + 2 H,0-»4 Fe3+ + 8 S042^ + 4 H+

i.e., 2 Fe2(S04)3 + 2 H2S04

In other words, the oxidation of the fool's gold produces soluble iron (III) sulfate (also called ferrous sulfate), Fe2(S04)3, and sulfuric acid. The Fe3+ ion is soluble in the highly acidic water that is first produced, the pH of which can be less than zero, with the usual range being 0 to 2. However, once the drainage from the highly acidic mine water becomes diluted and its pH consequently rises, a yellowish-brown precipitate of Fe(OH)3 forms from Fe3+, discoloring the water and waterway and smothering plant and animal life (including fish) in it (see Problem 13-7). Thus the pollution associated with acid mine drainage is characterized in the first instance by the seeping from the mine of copious amounts of both acidified water and a rust-colored solid. Unfortunately, the concentrated acid can liberate toxic heavy metals— especially zinc, copper, and nickel, but also lead, arsenic, manganese, and aluminum—from their ores in the rock of the mine, further adding to the pollution of the waterway.

Interestingly, the oxidation of disulfide ion to sulfate ion in the above process is accomplished to some extent by the action of Fe3+ as the oxidizing agent, rather than by 02:

Sz2~ + 14 Fe3+ + 8 H20-» 2 S042~ + 14 Fe2 r + 16 H+

The phenomenon of acid drainage is currently of particular importance in the many abandoned mines in the mountains of Colorado. However, the most acidic water in the world comes from the Richmond Mine at Iron Mountain, California. There the pH reaches as low as —3.6 because the high temperatures (up to 47°C) of the mine water cause much of it to evaporate, thus concentrating the acid. By comparison, the most acidic natural waters occur near the Ebeko volcano in Russia, with a pH as low as —1.7; the acidity is due to hydrochloric and sulfuric acids in the hot springs water.

The acid produced by acid mine drainage is spontaneously neutralized if the soil contains limestone, in the same way we encountered for acid rain in Chapter 4. For example, some of the coal mines in Pennsylvania discharge water that is acidic (pH < 5), whereas others discharge alkaline water resulting from dissolution of limestone. Powdered or chipped limestone can also be added to water exiting the mine, although the insoluble calcium sulfate that forms on the surface of the limestone particles prevents full reaction. Calcium oxide or hydroxide can also be used to neutralize the acid. Raising the pH precipitates most of the liberated heavy metals as their insoluble hydroxides in a sludge that can be removed from the water. An alternative method of remediation is the introduction into the waters of anaerobic bacteria that reverse the oxidation of sulfate ion back to sulfide, thereby precipitating the heavy metals as insoluble sulfides as well as raising the pH. In some instances, the sulfides are rich enough in metals to be used as ores. Some abandoned mines have been sealed to prevent further intrusions of water and oxygen, but this means of preventing further production of acid is sometimes unsuccessful.


The Ksp values for Fe(OH)2 and Fe(OH) , are 7-9 X 1(T15 and 6.3 X 1CT38, respectively. Calculate the solubilities of Fe2+ and Fe3+ at a pH of 8, assuming they are controlled by their hydroxides. Also calculate the pH value at which the ion solubilities reach 100 ppm.

The pE Scale

Environmental scientists sometimes use the concept of pE to characterize the extent to which natural waters are chemically reducing in nature, by analogy to the way in which pH is used to characterize their acidity. In particular, pE is defined as the negative base-10 logarithm of the effective concentration— i.e., of the activity—of electrons in water, notwithstanding the fact that free electrons do not exist in solution (any more than do bare protons, H+ ions). pE values are dimensionless numbers; like pH, they have no units.

• Low pE values signify that electrons are readily available from substances dissolved in the water, so the medium is very reducing in nature.

• High pE values signify that the dominant dissolved substances are oxidizing agents, so that few electrons are available for reduction purposes.

When several acids or bases are present in a water sample, one of them usually makes a dominant contribution to the hydrogen or hydroxide ion concentration. In such situations, the position of equilibrium for the other, less dominant weak acids or bases is determined by the H+ or OH " level set by the dominant process. In a similar way, in natural waters one or another redox equilibrium reaction is dominant, and it determines the electron availability for the other redox reactions that occur simultaneously. If we know the position of equilibrium for the dominant process, we can calculate the pE and from it the position of equilibrium—and hence the dominant species—in the other reactions.

When a significant amount of 02 is dissolved in water, the reduction of the oxygen to water is the dominant reaction determining overall electron availability:

^02 + H+ + e~ ^^ \ H20

In such circumstances, the pE of the water is related to the acidity and to the partial pressure of oxygen and its acidity by the following equation, the origin of which is discussed subsequently:

pE = 20.75 + log ([H+] P021/4) = 20.75 -pH + i log (P0J

For a neutral sample of water that is saturated by oxygen from air, i.e., when Pq2 = 0.21 atm and is free of carbon dioxide so that its pH = 7, the pE value is calculated from this equation to be 13.9. If the concentration of dissolved oxygen is less than the equilibrium amount, then the equivalent partial pressure of atmospheric oxygen is less than 0.21 atm, so the pE value is smaller than 13.9 and in some cases even negative.

The pE expression given above looks very much like the Nemst equations encountered in the study of electrochemistry. Indeed the pE value for a water sample is simply the electrode potential, E, for whatever process determines electron availability, but divided by RT/F (where R is the gas constant, T the absolute temperature, and F the Faraday constant), which at 25°C has the value 0.0591:

Thus the pE expression for any half-reaction in water can be obtained from its standard electrode potential E°, corrected by the usual concentration and/or pressure terms and evaluated for a Qne'electron reduction. For example, for the half-reaction Unking the reduction of nitrate ion to ammonium ion, we first write the process as a (balanced) one-electron reduction:

|N03~ + JH+ + +|H20

For this reaction, E° = + 0.836 volts (from standard tables), so pE° = E°/0.0591 = + 14.15. The equation for pE involves the subtraction from the standard pE° of the logarithm of the ratio of concentrations of products to reactants, each raised to its coefficient in the one-electron half-reaction:

pE = pE° - log ([NH4+]1/8/[NCV]i/8 [trf4) = 14.15 -fpH-Jlog ([NH4+]/[N03-])

(Here we have used the properties of logarithms that log ax = x log a, and that log (l/b) = —log b.) As usual, the concentration of water does not appear in the expression as its effect is already included in pE°.


Deduce the equilibrium ratio of NIi4~ to N03~ at a pH of 6.0 (a) for aerobic water having a pE value of +11, and (b) for anaerobic water for which the pE value is 3.

Returning to the subject of the dominant reaction that determines pE in natural waters, we note that low values of dissolved oxygen usually are caused by the operation of microorganism-catalyzed organic decomposition reactions, and their dissolved products, rather than 02, can determine electron availability. For example, in cases of low oxygen availability, the pE of water may be determined by ions such as nitrate or sulfate. In the extreme case of the anaerobic conditions found at the bottoms of lakes in the summer and in swamps and rice paddies, the electron availability is determined by the ratio of dissolved methane, a reducing agent, to dissolved carbon dioxide, an oxidizing agent, both of which are produced by the fermentation of organic matter, discussed above. They are connected in the redox sense by the half-reaction

|COz + H+ + + H20

The pE value for water controlled by this half-reaction is pE = 2.87 -- pH + | log (PCo2/Pch4)

For example, if the partial pressures of the two gases are equal and the water is neutral, the pE value is —4.1. Thus the lower levels of a stratified lake are characterized by negative pE values, whereas the oxygenated top layer has a substantially positive pE.

The pE concept is useful in predicting the ratio of oxidized to reduced forms of an element in a water body when we know how the electron availability is controlled by another species. Consider, for example, the equilibrium between the two common ions of iron:

For this reaction, pE = 13.2 + log ([Fe3+]/[Fe2+])

If the pE is determined by another redox process and its value is known, the ratio of Fe3+ to Fe2+ can be deduced. Thus for the oxygen-depleted water discussed above that has a pE value of —4-1,

-4-1 = 13.2 + log ([Fe3+]/[Fe2+])

and hence

[Fe3+]/[Fe2+] = 5 X 10~18

In contrast, for a sample of aerobic water that has a pE of 13,9, the calculated ratio is 5:1 in favor of the Fe3+ ion. The transition between dominance of the two forms occurs when their concentrations are equal:


Find the pE value for acidic water at which the ratio of concentrations of Fe3+ to Fe2+ is 100:1.


Assuming that dissolved oxygen determines the electron availability in an aqueous solution and that the partial pressure equivalent to the amount dissolved is 0.10 atm, deduce the ratio of dissolved C02 to dissolved CH4 at a pH of 4.

pE-pH Diagrams

A visual representation of the zones of dominance for the various oxidation states in water of an element can be displayed in a pE-pH diagram, as shown in Figure 13-4 for iron. It is clear from the diagram that the situation is more complicated than that included in our example above, since, in moderately acidic or alkaline environments, the solid hydroxides Fe(OH)2 and Fe(OH)3 also come into play in the equilibria. The solid lines in the diagrams indicate combinations of pE and pH values where the concentrations of the two species listed on either side of the line are equal. Thus we see from the top left side of Figure 13-4 that equilibrium between dissolved Fe2+ and Fe3+ is important only for pH < 3. Equality in the concentrations of these two dissolved forms corresponds to the short horizontal line in the figure. As expected from our calculations above, the transition occurs at pE = 13.2 regardless of pH; hence the line is horizontal.

If iron is in the 3 + state at higher pH, it exists predominantly as the solid Fe(OH)3, whereas solutions containing iron in the 2+ state are not mainly precipitated until the solution becomes basic (Figure 13-4).

Diagram Source Hydrogen
FIGURE 13-4 ThepE-pH diagram for the iron system at 10~5 M concentration. [Source: Adapted from S. E. Manahan, Environmental Chemistry4th ed. (Boston, MA: Wi I lard Grant Press/PVVS Publishers, 1984).J

The shaded regions at the top right and bottom left side of the pE-pH diagram represent extreme conditions, under which water itself is unstable to decomposition, being oxidized or reduced to yield 02 or H2, respectively:

ZH20-> 02 + 4 H+ + 4e~

2 H70 + 2 e~->H7 + 2 0H"

Nitrogen Compounds in Natural Waters

In some natural waters, nitrogen occurs in inorganic and organic forms that are of concern with respect to human health. As discussed in Chapter 6, there are several environmentally important forms of nitrogen that differ in the extent of oxidation of the nitrogen atom.

TABLE 13-2

Common Oxidation Numbers for Nitrogen

TABLE 13-2

Common Oxidation Numbers for Nitrogen

Oxidation Number of N

Increasing Level

J of Nitrogen


-3 0 +1

+2 +3

+4 +5

Aqueous solution




and salts


Gas phase

nh3 n2 n2o



The common oxidation numbers of nitrogen, along with the most important examples for each, are illustrated in Table 13-2. The most reduced forms all have the —3 oxidation number, as occurs in ammonia, NH3, and its conjugate acid, the ammonium ion, NH4+. The most oxidized form, having an oxidation number of +5, occurs as the nitrate ion, N03^, which exists in salts, aqueous solutions, and nitric acid, HN03. In solution, the most important intermediates between these extremes are the nitrite ion, N02~, and molecular nitrogen, N2.

The pE-pH diagram for the existence of these forms in aqueous solution is shown in Figure 13-5. Notice the relatively small field of dominance (the small triangle on the right side of the diagram) for the nitrite ion, N02~, in which nitrogen's oxidation number has the intermediate value of +3. In particular, it is the predominant species only under alkaline conditions that are intermediate in oxygen content (small positive pE values).

The equilibrium between the most highly reduced and the most highly oxidized forms of nitrogen is given by the half-reaction

NH4+ + 3 H20 N03~ + 10 H+ + 8 e^

Previously we derived the equation relating pE to pH for this reaction written as a reduction; this equation defines the diagonal line in Figure 13-4 that separates these two ions when their concentrations are equal, so that pE = 14.15 — | pH — | log (1) = 14.15 — | pH

We might conclude from this equation, and from the slope of the diagonal line in Figure 13-5 separating NH4+ and N03~~, that since the oxidation of ammonium ion is highly pH dependent, it would not occur under highly acidic conditions. However, electron availability from the reduction of dissolved oxygen also decreases as the pH is lowered, so the pE value in such

Pressure Enthalpy Diagram R22 Dupont
FIGURE 13-5 pE-pH diagram for inorganic nitrogen in an aqueous system. [Source: Adapted from C. N. Sawyer, P. L. McCarty, and C. F. Parkin, Chemistry for Environmental Engineering, 4th ed. (New York: McGraw-Hill, 1994).]

water is quite high, and nitrate still predominates. For example, the pE value calculated for oxygen in water at pH of 1 is about 20, so nitrate still predominates.

Recall from Chapter 6 that in the microorganism-catalyzed process of nitrification, ammonia and ammonium ion are oxidized to nitrate, whereas in the corresponding denitrification process, nitrate and nitrite are reduced to molecular nitrogen. Both processes are important in soils and in natural waters. In aerobic environments such as the surface of lakes, nitrogen exists as the fully oxidized nitrate, whereas in anaerobic environments such as the bottom of stratified lakes, nitrogen exists as the fully reduced forms ammonia and ammonium ion (Figure 13-3). Nitrite ion occurs in anaerobic environments such as waterlogged soils that are not sufficiently reducing to convert the nitrogen all the way to ammonia. Most plants can absorb nitrogen only in the form of nitrate ion, so any ammonia or ammonium ion used as fertilizer must first be oxidized via microorganisms before it is useful to such plant life.


Consider the reduction of nitrate ion to nitrite ion in a natural water system.

(a) Write the balanced one-electron half-reaction for the process if it occurs in acidic media.

(b) Given that for this reaction, E° = + 0.881 volts, calculate pE°.

(c) From your answer to (a), deduce the expression relating pE to pE n d ion concentrations.

(d) From your result in part (c), obtain an equation relating the pE and pH conditions under which the ratio of nitrate to nitrite is 100:1.

(e) From your result in part (c), deduce the ratio of nitrite to nitrate under conditions of pE = 12, pH = 5.

Acid-Base Chemistry in Natural Waters: The Carbonate System

Natural waters, even when "pure," contain significant quantities of dissolved carbon dioxide and of the anions it produces, as well as cations of calcium and magnesium. In addition, the pH of such natural water is rarely equal to exactly 7.0, the value expected for pure water. In this section, the natural processes that involve these substances in natural waters are analyzed.

The C02-Carbonate System

The acid—base chemistry of many natural water systems, including both rivers and lakes, is dominated by the interaction of the carbonate ion, C032^, a moderately strong base, with the weak acid H2C03, carbonic acid. Loss of one hydrogen ion from the acid produces the bicarbonate ion, HC03~ (also called hydrogen carbonate ion):

H2CO3 H+ + HCO, ( 1 )

The acid dissociation constant for this process, Kj, is numerically much, greater than K2, the constant for the second stage of ionization, which produces carbonate ion:

HC03~ H+ + C032~ (2)

In order to discover the dominant form at any given pH, it is instructive to consider a species diagram for the C02-bicarbonate-carbonate system in water, such as that shown in Figure 13-6. In it, the fraction of the total inorganic carbon that is present in each of the three forms is shown as a function of the master variable, the pH of the solution. Clearly, carbonic acid is the dominant species at low pH (< 5), carbonate is dominant at high pH (> 12), and bicarbonate is the predominant—but not the only—species present in the pH range of most natural waters, i.e., from 7 to 10. At the pH of natural rainwater, 5.6, most of the dissolved carbon dioxide exists as carbonic acid, but a measurable fraction is bicarbonate ion (Figure 13-6).

The curves in Figure 13-6 were constructed by solving for the three unknowns [H2C03], [HC03_], and [C032"], relative to their total concentration, C, from individual equilibrium constant expressions, as detailed in Box 13-2.

[H2co3]/c = [H+]2/D [HC03-]/C = K, [H+]/D

Manahan Environmental Chemistry

II 12

FIGURE 13-6 Species diagram for the aqueous carbon dioxide-bicarbonate ion-carbonate system. (Source: S. E. Manahan, Environmental Chemistry, 6th ed. (Boca Raton, FL: Lewis Publishers, 2000), Figure 3.3, p. 54.]

II 12

[C032~]/C = K{ Kz/D

where the common denominator D = [H+]2 4- K^H"1"] + K^ K2.

These expressions clearly show that H2C03 is the dominant species under conditions of high acidity and determine the other conditions as discussed above.

The carbonic acid in natural waters results from the dissolution of carbon dioxide gas in water, the gas originating either in the air or from the decomposition of organic matter in the water. The gas in the air and the acid in water in contact with the surface usually are at equilibrium:

FIGURE 13-6 Species diagram for the aqueous carbon dioxide-bicarbonate ion-carbonate system. (Source: S. E. Manahan, Environmental Chemistry, 6th ed. (Boca Raton, FL: Lewis Publishers, 2000), Figure 3.3, p. 54.]

C02(g) + H20(aq) H2C03(aq)

The relevant equilibrium constant for this reaction is the Henry's law constant, Kh, for C02. [In fact, much of the dissolved carbon dioxide exists as C02 (aq) rather than as H2C03 (aq), but following conventional practice we collect the two forms together and represent it all as the latter.] The pH of pure water in equilibrium with the current level of atmospheric C02 is 5.6, according to the methods discussed in Chapter 3 (see Problem 3-10).

The sequestration of carbon dioxide, as such, into oceans would result in an increase in the acidity of the surrounding waters since the resulting increase in H2C03 concentration [reaction (3)] would give rise to further ionization [reaction (1)], which is called ocean acidic for that reason (Chapter 7). The resulting decrease in pH could affect biological life in the vicinity. Indeed, the increase in the atmospheric concentration of C02 that has already occurred has resulted in a drop of about 0.1 pH unit in surface ocean water worldwide.

The predominant source of the carbonate ion in natural waters is limestone rocks, which are largely made up of calcium carbonate, CaC03.

Derivation of the Equations for Species Diagram Curves

BOX 13-2

'TPhe final equations relating the concentra-1 tions of carbonic acid, bicarbonate ion, and carbonate ion to the pH, the equilibrium constants, and their total concentration C were obtained by algebraic manipulation of the three simultaneous equations.

Mass balance:

Continue reading here: K2 [co32 ih[hco3t

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    Where is the most neutral natural water found?
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    How does pe compare to ph in chemistry?
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    What species determines the pe value in natural water?
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    How to construct an EhpH diagrams for nitrogen?
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    How can caculate pE value of water?
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    How to find pE in acidic water ratio is 100:1?
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