WATERSHEDS - Water Quality Engineering in Natural Systems - page 233

Estimate the annual phosphorus loading from sediment-
laden runoff into hillsdale Lake in 2002.

weakly adsorbed onto soil particles; hence, they have
higher bioavailability and leach more easily into
groundwater. The mobility of an organic compound in
soils and sediments is related to the octanol-water par-
titioning coefficient, K ow . Sorption equilibria are quanti-
fied by the isotherms described in Section 5.5.1. The
Langmuir, Freundlich, and linear isotherms are all
widely used and accepted, although the linear isotherm
is the most used. In cases where a fraction of the con-
taminant is adsorbed, the total concentration of the
contaminant in the soil, c T (ML −3 ), is equal to the sum
of the aqueous (dissolved) concentration, c aq (ML −3 ),
and adsorbed particulate concentration, c p (ML −3 );
hence,

Solution

From the data given, S s = 0.71 g/kg, ER = 2.4, and
Y s = 21.9 × 10 6 kg/yr. The potency factor, p , for phos-
phorus is given by Equation (6.33) as

p S =

s ER

⋅ = ( . 0 71 ) ( . ) ⋅ 2 4 = 1 7 . g/kg

and the annual phosphorus loading, Y , on hillsdale
Lake is given by Equation (6.32) as

Y pY = = ( . )( 1 7 21 9 10 . ×

6

) = 37 2 10 . ×

6

g P/year

s

c = θ c + c = θ c + F ρ (6.35)

= 37 200 , kg P/year

T aq p aq b

where θ is the water content of the soil (dimensionless),
F is the adsorbed concentration of the contaminant
(MM −1 ), and ρ b is the bulk density of the soil (ML −3 ).
If adsorption is described by a linear isotherm, the
adsorbed concentration, F , and aqueous concentration,
c aq , are related by

This phosphorus loading (37,200 kg/yr in 2002) can be
associated with the phosphorus concentration in the
lake to assess whether reductions in phosphorus loading
are needed.

Pollutants exist in soils, sediments, and sediment-
laden water in several phases. They can be precipitated
and/or strongly adsorbed to particles (solid phase), be
dissolved or dissociated in water or soil moisture (liquid
phase), volatilize, or be gasified. The capacity of soils
and sediments to adsorb and retain contaminants
depends on their composition. For organic micropollut-
ants, the most important component in soils is the
organic particulate matter, which has the strongest
binding capacity. For inorganic contaminants such as
toxic metals, the adsorbing capacity of both organic and
inorganic soil sediment particulates should be consid-
ered. The adsorbing capacity is related to the surface
area of the particles, and hence small particles, such as
clay minerals, have the highest adsorbing capacity. The
important capacity-controlling parameters are the soil
organic matter content and the cation exchange capac-
ity (CEC), which is determined by the surface area of
the particles. Most clays and organic matter have a net
negative charge, making them effective in holding posi-
tively charged particles (cations), such as Al, Fe, h + , and
Ca. The CEC in milliequivalents per 100 g of soil can be
estimated using the relation

F K c
d

=

aq

(6.36)

where K d is the distribution coefficient (L 3 M −1 ). Com-
bining Equations (6.35) and (6.36) gives the following
relationship between the aqueous concentration, c aq ,
and the total concentration, c T :

1

c = c

(6.37)

aq T

θ + K ρ

d b

EXAMPLE 6.6

A 1-m 3 sample of soil is found to contain 53 g of atra-
zine (a herbicide) and to have a water content of 0.15
and a bulk (dry) density of 1610 kg/m 3 . If the soil is 1%
organic carbon and has an estimated distribution coef-
ficient of 1.6 mL/g, estimate the concentration of atra-
zine in the pore water. The solubility of atrazine is
33 mg/L.

Solution

From the data given, c T = 53 g/m 3 = 53 mg/L, θ = 0.15,
ρ b = 1610 kg/m 3 , and K d = 1.6 mL/g = 1.6 × 10 −3 m 3 /kg.
Substituting these data into Equation (6.37) yields

CEC = 2 5 . × (% organic matter ) + 0 57 . × (% clay ) (6.34)

Adsorption or complexation of the contaminant into
the particulate form also immobilizes contaminants
and makes them biologically unavailable in most cases.
Generally, water soluble (hydrophilic) compounds are

1 1

c = c =

) (

53 )

aq T

− 3

θ + K ρ 0 15 . + ( . 1 6 10 × )( 1610

d b

= 19 mg/L