Page 255 - Soil and water contamination, 2nd edition

242 Soil and Water Contamination

Box 13.I Determination of the ratio between solid mass and solution volume

In the saturated zone (groundwater), the mass of the solids divided by the solution
volume can be calculated by:
M 1 ( n)
s s b (13.Ia)
V n n
3
where M = the mass of the solids [M], V = the solution volume [L ], n = the water-filled
s
-3
porosity of the sediment [-], ρ = the sediment particle density (approximately 2.65 g cm
s
-3
for mineral sediments, much smaller for organic sediments) [M L ], and ρ = the dry
b
bulk density of the bed sediment (defined as weight of dry solids divided by the volume
-3
of the bulk soil or sediment, so ρ = (1 – n) ⋅ ρ [M L ].
b s
In the unsaturated zone in soil, the mass of the solids divided by the solution volume is
given by:
M s b w
(13.Ib)
V w
-3
where θ = the soil volumetric moisture content [-], ρ = density of water (≈ 1.00 g cm
w
for fresh water), and w = the soil gravimetric moisture content.
If the solids are in suspension in surface water, the mass of the solids divided by the
solution volume represents the concentration of suspended matter:
M
s SS (13.Ic)
V
-3
where SS = the suspended solids concentration [M L ]. For exchange between surface
water and bed sediments it is usual to assume an active top layer in which the sediment
mass is fully able to interact with the overlying water:
(
M A d 1 n) d
s
s b (13.Id)
V A( H n d) H n d
Where A = a standard surface area over which the exchange takes place, d = the depth
of the active top sediment layer (usually in the order of between 1 and 5 mm) [L], and
H = the water depth [L].

-1
Table 13.1 Relationships between K oc and K ow (K oc and K ow in l kg ) (source: Lymann et al., 1990).
Class of organic compounds Regression equation
Aromatics, polynuclear aromatics, triazines, and dinitroaniline log K oc = 0.937 log K ow – 0.006
herbicides
Mostly aromatic or polynuclear aromatics; two chlorinated log K oc = 1.00 log K ow – 0.21
s-Triazines and dinitroaniline herbicides log K oc = 0.94 log K ow + 0.02
Variety of pesticides log K oc = 1.029 log K ow – 0.18
Wide variety of organic compounds, mainly pesticides log K oc = 0.544 log K ow +1.377
Note: units of K oc and K ow are in l kg -1

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