CHAPTER 2 SOLUTIONS
1. Note that the figure showing conductivity versus time is not
completely symmetrical; instead, it exhibits tailing. Thus, the
solutions below are approximate. The asymmetry may occur due to
several reasons. First, the cloud of salt is continuing to disperse as it
passes Site 1. Second, there may exist in the stream some zones where
flow is partially isolated from the main streamflow. Third, an
imperfect injection technique may have been used. A plot of
conductivity (concentration) versus distance ideally would approach
a Gaussian shape after the chemical mass has traveled a sufficient
distance for mixing to become uniform. A plot of conductivity would
also be approximately Gaussian with respect to time if the time for the
peak to pass Site 1 were small compared to the total travel time since
the salt was injected.
a. To estimate the average stream velocity in this reach, measure the
2. a. The minimum possible flux density of downward Fickian oxygen
transport is due to molecular diffusion. Using Eq. (1.3), an
b. Turbulence would increase the diffusion coefficient Dsignificantly.
3. a. As in exercise (2), the minimum possible flux density is due to
molecular diffusion. Given a thermocline of 5 m earlier in the
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c. If the arsenic that enters the lake via streamflow is entirely in the
4. To solve this problem in terms of the measured dissolved
concentration, define:
Given that TSS is all organic carbon in this problem, the total mass
of sorbed and dissolved chemical per liter of water can be written:
Converting the TSS concentration of 20 mg/liter to 210
5
kg/liter:
5. To estimate the loss rate from the spilled pool of gasoline, Eq. (2.39) is
appropriate:
e21CHAPTER 2 SOLUTIONS
Assuming an air temperature of 15 C, and using the gas constant,
the air concentration can be estimated:
6. Given that gasoline has dissolved in the lake, an appropriate model to
use for volatilization is the thin film model with water-side control.
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For octane, a piston velocity can be estimated using Eq. (2.32) and
the piston velocity for pentane:
Therefore, the volatilization flux for octane is:
These volatilization fluxes can be thought of as the change in
chemical mass over time for a unit area:
The dimensions of (k/depth) are [T
1
]. The solution to
this first-order equation is Eq. (1.20):
For pentane, the equation describing its concentration
versus time is:
e23CHAPTER 2 SOLUTIONS
7. a. To estimate the flux density of TCE to the atmosphere over the
puddle, use Eqs. (2.36), (2.38), and (2.39), and the vapor pressure
and molecular weight of TCE from Table 1.3:
b. The puddle will grow until the rate of dripping from the tank truck
equals the rate of volatilization from the puddle:
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8. Referring to the solubility data for 2,2,4-trimethylpentane presented
in exercise (5), it is evident that the chemical is present in the
wastewater at nearly its solubility limit. Note that the system is not
9. a. The maximum concentration of 1,1,1-TCA will occur immediately
downstream, before any degradation or volatilization has
b. Although 1,1,1-TCA is denser than water (1.34 g/cm
3
, from
is much greater than its maximum stream concentration.
c. 1,1,1-TCA is quite volatile (K
H
¼1.810
2
atm m
3
/mol, from
e25CHAPTER 2 SOLUTIONS
10. To estimate how long it will take for the n-octane to volatilize, first
calculate the concentration of n-octane at the NAPL-air interface.
11. a. Given that a slick of hexane has formed on the lake, volatilization is
occurring from a pure phase liquid. The vapor pressure of hexane
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Converting the vapor pressure into an air concentration using
Eq. (2.38):
Given the wind speed of 1 m/sec and using Eq. (2.36) (or,
alternatively, Eq. 2.35), a piston velocity for turbulent diffusion
into the air above the slick is approximately 1100 cm/hr.
b. If chloroform runs into the lake, a slick will not form, due to the
density of chloroform (1.48 g/cm
3
in Table 1.3). Given its solubility
12. To determine if the inspector is in time to collect any evidence from
the road, the mass of solvents that could volatilize in 20 min must be
compared to the mass of solvent spread on the road.
e27CHAPTER 2 SOLUTIONS
Using Eq. (2.36) (or, alternatively, Eq. 2.35), the piston velocity in
air for the solvents is estimated as:
Using Eq. (2.38) to estimate the concentration of solvents in air
immediately above the pure phase on the road:
In 20 min, the mass of solvents that could volatilize per unit area is:
The mass of solvents present on the road per unit area, given a
uniform layer 2 mm thick, is:
13. Given a (þI) oxidation state for hydrogen:
a. 2xðÞþ6þIðÞ¼0
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14. a. The relevant reaction for O
2
consumption is:
b. The jug will become anaerobic if there is not sufficient O
2
to
degrade the mouse. Calculate the moles of CH
2
O and O
2
present:
c. After the O
2
has been used up in an aerobic reaction, 0.226 mol
CH
2
O remains. The next process to occur will be denitrification
(Eq. 2.53):
The next process to occur will be sulfate reduction (Eq. 2.55):
e29CHAPTER 2 SOLUTIONS
15. a. Given a (þI) oxidation state for hydrogen:
b.
c. If TNT is detonated in the absence of significant amounts of O
2
,
there are no external electron acceptors or donors readily
available. Assume that oxygen and hydrogen in the products are
16. a. Bubbly waters that smell of sulfide most likely contain bacteria
h
b. If small fish are swimming around in the water, oxidizing
c. Highly chlorinated PCBs slowly degrade through the loss of
d. Water leaves brown stains in the sink when it contains Fe
2þ
that
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17. Toxaphene is incorporated at a concentration of 0.2 ppm. Because
some toxaphene will be dissolved in the water and some will be
sorbed to the organic-rich lake sediments, the concentration will be
regarded as mass of toxaphene per mass of sediment-water mixture:
Given that the mixture of sediment and water is bubbled with air,
the initial concentration of O
2
in the water would be at its solubility
e31CHAPTER 2 SOLUTIONS
c. From Table 2.6:
As oxygen disappears, water appears.
18. a. Using Fick’s first law (Eq. 1.3), the diffusive flux of sulfate into the
sediment is:
sec 3600 sec
hr 24 hr
day 365 day
year 1 mol
106mmol ¼4:3104mol
b. The lake volume is:
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c. Using a mass balance expression and assuming steady state:
19. a. In writing a half-reaction, the following steps should be
performed:
iii. Write the number of electrons lost or gained from one species
to the other species, according to the oxidation states
iv. Add H
2
O to the side of the equation which loses electrons. The
number of moles of H
O added must balance the number of
e33CHAPTER 2 SOLUTIONS
v. Add H
þ
on the side of the equation which gains electrons.
þ
vi. By convention, the half-reaction is expressed in terms of the
transfer of one electron. Therefore, divide the above reaction
b. The given reaction can be written as follows:
Given a pH of 7 and equal concentrations of {H
SeO
}and{H
Se}:
20. To make the provided equation comparable to the redox half-
reactions presented in Table 2.6, it must be normalized to the transfer
of one electron:
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21. According to Table 2.11, the half-life of methyl parathion due to direct
photolysis is 30 days. The significance of this rate depends on the rates
at which other degradation processes occur.
22. a. If the lake remains stratified for 4 months (120 days) and mass
b. The lowest possible value of Dwould be approximately 10
5
cm
2
/
23. When the lake stratifies in late spring, the water contains 10.8 mg/liter
O
2
. (From Table 2.4, the water temperature must be about 12 C at that
time.) Assume that after stratification, a negligible flux of O
2
occurs
across the thermocline.
a. Given that the 6 m deep hypolimnion is fairly well mixed,
calculations can be made on a per-sediment-area basis. Initially,
the amount of available DO is:
e35CHAPTER 2 SOLUTIONS
b. A cold-water fishery cannot be supported because the hypolimnion
24. Chlorobenzene is present in lake water at a concentration of 2 ppb
(2 mg/liter). To estimate the chlorobenzene concentration in trout in
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25. First estimate the intake rate of pesticides in the osprey’s diet:
1 mg pesticide
1 kg minnows 0:2 kg minnows
day ¼0:2 mg pesticide=day
e37CHAPTER 2 SOLUTIONS
26. a. First estimate the intake rate of PCBs in the eagle’s diet:
0:1 mg PCBs
1 kg trout 0:5 kg trout
day ¼0:05 mg PCBs=day
b. Two plausible explanations as to why the PCBs bioaccumulated in
the eagle have a higher ratio of highly chlorinated to less
chlorinated congeners compared to industrial discharges are:
27. a. Given a half-life of about 3 weeks in the lab, the rate of photolysis
of quinoline (from Eq. 1.21) is approximately:
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