Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 14

Environmental chemistry of colloids and surfaces

PROBLEMS/SOLUTIONS
1.

Based on what you know about the structure of humic material, explain why it is more likely to have a
net negative charge rather than net positive charge.

Solution
Humic materials (humic acid and fulvic acid in particular) have many acidic functional groups which have
pKa values in the range of 4-5. In natural aqueous environments it is likely that many of these acidic sites
will be deprotonated, leaving an overall net negative charge associated with this material.

2.

Show that a cation exchange capacity (CEC) value in units of cmol (+) kg–1 is numerically the same as
one using units of meq 100 g–1. An explanation is provided in Chapter 18.

Solution
Equate cmol(+) kg-1 to

meq 100 g-1

100 g = 0.10 kg, 1 mol of (+) charge = 1 eq charge, c = 10-2, and m = 10-3
use 250 cmol (+) kg-1 as an example value and convert to units of meq 100 g-1
250 cmol (+) kg-1 x 10 mmol/cmol = 2500 mmol (+) kg-1
2500 mmol (+) kg-1 x 1 eq/mol (+) = 2500 meq kg-1
2500 meq kg-1 x 1 kg/1000 g = 2.5 meq g-1

(x 100 for 100 g) gives:

2.5 meq g-1 x 100 = 250 meq 100 g-1
Hence;

3.

250 cmol (+) kg-1 = 250 meq 100 g-1

A sediment sample containing 47% clay and 3.2% organic matter was found to have a cation exchange
capacity of 10.5 cmol (+) kg–1. Assuming that the organic matter has a CEC of 210 cmol (+) kg–1, show
that the clay could be kaolinite but not montmorillonite.

Solution
Assume that the clay and organic matter are solely responsible for the total CEC found for the sediment.
47% clay,

3.2% OM, & CEC = 10.5 cmol (+) kg -1

OM has CEC = 210 cmol (+) kg -1 but represents only 3.2 % of the sediment.
The OM accounts for:
The clay accounts for:

0.032 x 210 cmol (+) kg-1 = 6.7 cmol (+) kg-1
10.5 – 6.7 = 3.8 cmol (+) kg -1
0.47 x clay (CEC) = 3.8 cmol (+) kg-1

123



Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 14

Environmental chemistry of colloids and surfaces
clay (CEC) = 8.1 cmol (+) kg -1

The calculated CEC value for the clay matches the expected range for kaolinite (3-15 cmol (+) kg-1) but not
that of montmorillonite, which has a much higher range (80-150 cmol (+) kg-1) (see Table 14.6).

4.

A particular lake sediment sample is made up of 23% organic matter and 77% mineral fraction, the
latter containing: 42% clay minerals; 24% silt; 11% sand. The clay minerals are 90% kaolinite and
10% halloysite. Calculate an approximate cation exchange capacity (cmol (+) kg–1).

Solution
Kaolinite ranges from 3 – 8 cmol (+) kg -1 with an average value of 8 cmol (+) kg-1, Table 14.6
Halloysite ranges from 4-10 cmol (+) kg -1 with an average of 8 cmol (+) kg-1, Table 14.6
The CEC due to clay mineral fraction is calculated as shown.
For kaolinite:

CEC = 8 x 0.9 x 0.42 = 3.0 cmol (+) kg-1

For halloysite:

CEC = 8 x 0.1 x 0.42 = 0.34 cmol (+) kg-1

Total CEC for clay minerals:


= 3.3 cmol (+) kg-1

The OM has an average value of 200 cmol(+) kg-1 (Table 14.6) and represents 23% of the sample. This
gives:
200 cmol (+) kg-1 x 0.23 = 46 cmol (+) kg-1
The total CEC for this lake sediment is expected to be about:
46 + 3.3 = 49 cmol (+) kg-1
An approximate cation exchange capacity for this lake sediment is 49 cmol (+) kg-1.
5.

Among the forms of phosphorus found in sediments are:
(a) organic phosphorus;
(b) apatite ((Ca5(PO4)3(F,Cl,OH) is one representation for this mineral);
(c) adsorbed on hydrated iron(III) oxides;

(d) associated with clay minerals;
How would you expect the solubility of each of these forms to be affected by the ambient pE of the water?
Solution
Ambient pE of water and effect on solubility of:

124


Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 14

Environmental chemistry of colloids and surfaces

a) organic phosphorus species in sediment - expect phosphorus solubility to increase as organic matter
decomposes, mostly through oxidation reactions. Therefore increasing soluble phosphate with increasing

pE
b) apatite (Ca5(PO4)3(F,Cl,OH) - there are no redox-sensitive elements in this stable mineral. Therefore,
pE would have little or no effect on phosphorus release.
c) adsorbed on hydrated iron(III) oxides - under low pE conditions, some of the solid iron(III) oxide would
be reduced, releasing iron(II) into solution. Along with it, would be phosphate that had been covalently
bonded to the solid. Therefore, soluble phosphate would increase with decreasing pE.
d) associated with clay minerals - the phosphate anion is not strongly held by clay minerals, and the
binding would be little affected by changing pE conditions.

6.

(a) Use Fig. 10.P.1 to suggest how phosphorus release could occur from buried iron-rich sediments.
(b) In laboratory experiments, it has been shown that phosphorus is released from sediment under dark
conditions,1 but is accumulated in the sediment when it is exposed to light. Suggest an explanation.

Solution
Part a) Release of P from iron rich sediments.
Referring to Figure 10.P.1 and overlaying Figure10.6, it can be seen that aqueous Fe2+ is the most
prominent iron species derived from submerged lake sediment. Originally, phosphorus would have been
adsorbed onto Fe(OH)3. As the iron-rich sediment becomes buried, the pE decreases, as is known to happen
in the lake sediment environment, and the Fe(OH) 3 can be solubilized releasing Fe2+. Any phosphate that
was adsorbed onto this insoluble iron hydroxide, will also become solubilized and thus release the
phosphorus back into the aqueous system.
Part b) Photosynthetic activity / biological uptake of nutrients require sunlight.
Phosphate is one of the required nutrients. In the dark, photosynthetic activity ceases and the reverse
process involving respiration results in biological decay and with it, a net release of phosphate.

7. The behaviour of lead in soil / water systems has been described in terms of Langmuir adsorption. From
laboratory experiments using ‘pure’ solids, the following values for Langmuir parameters have been
estimated.


For clay

b  3200 L mol –1
Csm  4  10 5 mol g –1

For organic matter b  10 000 L mol –1
Csm  4  10 – 4 mol g –1
1

Moore, P.A. Jr., K.R. Reddy, and D.A. Graetz, Phosphorus geochemistry in the sediment–water
column of a hypereutrophic lake, J. Environ. Qual., 20 (1991), 869–75.

125


Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 14

Environmental chemistry of colloids and surfaces

(a) Use these values to calculate the concentration (in mg L–1 or ppb) of water-soluble lead in
equilibrium with clay containing 5 ppm adsorbed lead, and with organic matter also containing 5
ppm adsorbed lead. Both soil concentrations are those found at equilibrium.
(b) Use the information from this question to predict the relative mobility of lead (released from
gasoline combustion) deposited on clay-rich versus organic rich roadside soils.
Solution
Use the equation for the Langmuir relation (Equation 14.4)
Cs
bCsm

---- = -------------------- or (Equation 14.9)
Caq
1 + Caqb

1
----- =
Cs

1
1
------- + -----------Csm
bCaqCsm

Since values given are equilibrium values, we can assume no further uptake of lead by the clay or the
organic matter will occur. The aqueous concentration of lead will be calculated separately for the clay and
organic matter.
For the clay soil:
Cs = 5 µg g-1 (convert to mol g-1, M.M. Pb = 207.2 g mol-1)
= 2.41 x 10-8 mol g-1
1
1
1
----- = ------- + -----------Cs
Csm
bCaqCsm
insert values and solve for Caq:
1
4.14 x 107 = 2.5 x 104 + -----------------------------3200 x Caq x 4 x 10-5
4.14 x 107 x 0.128 = Caq-1
Caq = 1.9 x 10-7 mol L-1 (soluble Pb from the clay)

For the organic matter soil:
Cs = 5 µg g-1 (convert to mol g-1, M.M. Pb = 207.2 g mol-1)
= 2.41 x 10-8 mol g-1
1
4.14 x 107 = 2.5 x 103 + -------------------------------10000 x Caq x 4 x 10-4
4.14 x 107 x 4.0 = Caq-1
Caq = 6.0 x 10-9 mol L-1 (soluble Pb from the organic matter)
Convert to units of ppb: for clay
1.9x10-7 mol L-1 x 207.2 g mol-1 x 106 µg g-1 = 39 µg L-1

(or ppb)

126


Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 14

Environmental chemistry of colloids and surfaces

Convert to units of ppb: for organic matter
6.0x10-9 mol L-1 x 207.2 g mol-1 x 106 µg g-1 = 1.2 µg L-1

(or ppb)

It is clear from the above calculations that lead mobility would be restricted when deposited on organic-rich
soils compared with lead that is deposited on clay-rich soils.

8.


Under what conditions do the Langmuir and Freundlich equations reduce to having the same form?

Solution
The Freundlich relation will begin to resemble that of the Langmuir relation when the concentration of the
species under consideration begins to become very high, and further adsorption does not continue. In other
words, the system has attained a maximum absorbable quantity, which is the basis for the description of the
Langmuir relation. The Freundlich relation is useful for small molecules present in low concentration.

9.

The KOW function is frequently used to describe the movement of pesticides in the surface soil pore
water, but it is used less frequently to describe the behaviour of an organic contaminant in
groundwater. Explain.

Solution
The KOW function describes the distribution of an organic compound between water and n-octanol under
equilibrium conditions. The octanol is amphiphilic and has properties in some ways similar to natural
organic matter such as that found in soil. It has often been found that the organic fraction is largely
responsible for interactions between soil/sediment and hydrophobic organic compounds such as many
pesticides. For this reason, this parameter can be used to predict mobility in a surface soil.
On the other hand, ground water, that is located in the phreatic zone, is present in a soil matrix that is
usually devoid of solid organic matter. At depth, the soil matrix is almost exclusively inorganic, and any
retention of organic compounds can only be described in terms of (usually limited) interactions with these
materials. The KOW is not useful for this purpose.

10. To determine the value of KOW, one method is the shake-flask procedure whereby the molecule of
interest is placed in a flask containing octanol and water and shaken to establish equilibrium and then
the concentration is measured in both phases. This direct method has practical problems associated
with it. As an alternative, the KOW can be determined by relating it to retention times in reverse phase
liquid chromatography. Explain the basis of this method.

Solution
Reverse phase liquid chromatography contains both a non-polar and polar phase similar to the octanol /
water mixture. A very typical reverse phase HPLC system could consist of a bonded C-18 functional group
to a silica particle (3-5 μm) as the stationary phase while a solvent, such as a methanol/water mixture would
be the mobile phase. As a molecule travels through this system, those with a larger KOW value will tend to

127


Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 14

Environmental chemistry of colloids and surfaces

take longer exit from the column. In this case a number of molecules with known KOW values could be used
to create a calibration, and an unknown KOW could be determined.

11. The correlations between KOW and KOC imply that organic carbon in all soils has similar properties.
Other relations that have been suggested include correlations between KOC and the ratio (N + O) / C of
soil organic matter and between KOC and percentage aromaticity in the organic fraction of the soil.
Suggest situations where these correlations might be appropriate.
Solution
Using a relation between KOC and (N + O)/C implies that it is the N and O heteroatoms in soil organic
matter that are primarily responsible for retention of a particular organic species. Figure 12.5 shows an
example of where the carbamate pesticide carbaryl is bound to humic material via hydrogen bonds
involving both O and N atoms. Where this is the principal retention mechanism, the suggested relation
might be a good one to describe how values of KOC are influenced by soil organic matter properties.
Similarly, Figure 12.7 illustrates a possible mechanism of binding of DDT to humic material, via
hydrophobic interactions. The aromatic component of OM can play an important role in such interactions.


12.

The solubility in water (S) and log KOW values for several pesticides are as follows:
S / mg L-1

log KOW

DDT

0.0028

6.0

Aldrin

0.08

5.8

Parathion

19

3.7

Atrazine

38

2.6


Carbaryl

73

2.4

Using your understanding of the basis of the KOW coefficient, comment on the relationship between the
trends shown here.
Solution
The solubility trend (increasing) from DDT to carbaryl is the opposite of the log K OW trend (decreasing)
from DDT to carbaryl. As molecules become more hydrophobic, less soluble, they tend to partition more
strongly towards being associated with the organic phase of octanol.

13. In Atlantic Canada, like elsewhere around the world, aquaculture, specifically salmon farming, has
established itself as a viable industry. However, there are concerns that the high fat content of this fish
leaves it susceptible to significant bioaccumulation or biomagnification of persistent organic
pollutants from nutritional feed stocks used in the industry. Calculate the concentration of PCBs that
would bioconcentrate into the fish if the total concentration of PCBs = 0.05 pg L-1 in the ocean water
(density 1.015 kg L-1) where the fish are raised. Use the value of log K OW = 7.11 for the PCBs. (note
the use of the different terms above; refer to the definitions given in Section 14.5)

128


Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 14

Environmental chemistry of colloids and surfaces


Solution
Assuming octanol is a good representation of the fatty tissues of the salmon fish we can use the following
simple relationship to estimate an approximate value;
KOW = CO / Caq (Equation 14.24)
Log KOW = 7.11
KOW = 1.29x107 = CO / 0.05 pg L-1
CO = 0.645 µg L-1
Clearly, even if this is a rough approximation, there is a significant increase in the amount of contaminant
contained in the fish relative to the concentration in it’s environment.
14. Is the log KOW value used in the previous question reasonable? Refer to the webpage ‘LOGKOW ©
(ICUS/CODATA) a databank of evaluated octanol water partition coefficients’ found at
accessed October, 2016, for specific log KOW values for
various PCB congeners (and many other compounds).
For classroom discussion, no solution provided. This is meant to provide students the opportunity to
research various literature values and see the range (and difficulty) of choosing a value where no one single
value may exist. A brief classroom discussion could compare results and the students can decide if the
value chosen was reasonable.

15. The average concentration of pentachlorophenol (PCP, formula C 6HCl5O) in rainbow trout, raised in a
Northern Ontario, Canada, fish hatchery was found to be 2.7 ng L-1. Assume the only process involved
is bioconcentration and calculate the PCP concentration in the fish habitat. Refer to the webpage given
in the previous question.
Solution
Refer to the webpage ‘LOGKOW© (ICUS/CODATA) found at />(or other reliable resource material) accessed October, 2016, where a log KOW value of 5.18 is
recommended for pentachlorophenol (PCP, formula C6HCl5O).
KOW = CO / Caq (Equation 14.24)
Log KOW = 5.18
KOW = 1.51x105 = 2.7 ng L-1 / Caq
Caq


= 0.0178 pg L-1

The fish habitat would contain the small equilibrium amount of 0.0178 pg L-1 of pentachlorophenol, a
common wood preservative.

129