A truck carrying the herbicide linuron has crashed into a lake Although no one w

Question

A truck carrying the herbicide linuron has crashed into a lake Although no one was injured, the truck's tank sprung a leak and the linuron is now spilling into the lake. You've been tasked to evaluate the risk the spill poses. Use the tools in this class to determine the fraction of the compound in air, water, and sediment at equilibrium. Assume a ratio of 10 L:1 L:0.1 kg for air water sediment. From prior measurements, you know the pH of the water is 7.87 and the f_om of the sediment is 0.05. Water solubility (mg/L) 81 Henry's Law Constant (atm-L/mol) 1.97 times 10^6 Log K_ow = 3.00 Use the following relationships to estimate the partitioning coefficient between water and sediment (K_D): Log K_om = + 0.82Log K_ow + 0.14; K_D f_omK_om; K_D = C_sed/C_w Note that in this formulation, C_sed is the concentration of the contaminant per mass of solid phase (mol/kg and C_w is the concentration of the contaminant per liter of water (mol/L), so K_D has units of L/kg.

Explanation / Answer

Partition (or distribution) coefficient:

Sorbed phase conc cs (mass contaminant per mass solid) divided by dissolved phase conc cd (mass contaminant per volume solvent) in equilibrium.

Kp = cs / cd

Kp depends on contaminant, its concentration, concentration of organic matter, redox potential, etc.

In sediment bed Mass of dissolved contaminant/volume = theta * cd

Mass of dissolved contaminant/volume = pscs (1- theta) = psKpcd (1- theta)

Ratio of sorbed to total mass:

F = psKp (1- theta) / psKp (1- theta) + theta = Kpp / Kpp +1

P = ps(1 – theta) / theta = pb,d / theta ----Solid-water phase ratio

Pb,d = ps (1 – theta ) --- Bulk (dry) sediment density

Pb,d = ps (1-theta ) + theta --- Bulk (wet) sediment density

For Superficail sediments:

s ~ 1.5 - 2.5 g/cm3 ; theta ~ 0.6 – 0.8 =>

b,d ~ 0.3-1 g/cm3 , pb,w ~ 1.1-1.6 g/cm3 , ~ 0.4-1.7 g/cm3

Therefore f = Kpp / Kpp + 1

Where ~ 1 => for Kp >> 1 most contamination is sorbed to particles

While most of the contaminant is associated with solids, the dissolved phase is very important because it is more bio-available and amenable to sediment-water exchange Sediment quality criteria often derived from target dissolved phase concentrations assuming equilibrium partitioning.

In water column porosity = 1 so p = ps (1 – theta) / theta = [TSS]

[TSS] = 1 to 100 mg/L (10-4 to 10-6 g/cm3)

F = Kpp / Kpp + 1 = Kp[TSS] / Kp[TSS]+1 = (10-4 to 10-6)Kp / (10-4 to 10-6) Kp + 1

Point to be noted is that for hydrophobic contaminants (Kp ~ 104 to 105) concentrations in sorbed and dissolved phases can be comparable. Hydrophilic contaminants are mostly in dissolved phase.

For the values we have in hand:Log Kom = +0.82Log Kow +0.14; KD = fomKom; KD = Csed / Cw

We know that fine particles usually most important usuallyt

RESULTS

The experimental values for Kow for herbicide and the corresponding log Kow are shown:

Pesticide fate models are being used in regional assessments of the risk of pesticide movement by leaching, runoff, and water-eroded soil. Sorption coefficients (measures of pesticide sorption by soil) are among the most sensitive input parameters. It has been demonstrated that sorption coefficients vary within soil landscapes due to differences in soil organic carbon (SOC) content or soil pH between slope positions. Establishing methods to account for this variation could reduce uncertainties in regional scale assessments of pesticide fate.

An additional aspect of soil Kd values is that there are minimum and maximum values that will have a significant effect in an assessment, depending on the time scale of the values observed. Values of Kd below about 0.1 L kg–1 are indicative of high mobility, and variation of tenfold around this value, and especially lower, will not have a large effect on assessment results because the element is essentially as mobile as water. Thus, it is not productive to create a detailed predictive model for Kd in this range. Similarly, there is an upper limit Kd beyond which the element can be considered essentially immobile for a specific assessment time frame. Thus, it is again not productive to create detailed predictive models for high values, in the order of 104 L kg–1 and above. Usually Kd values below and above these limits are reported, but with less emphasis on predictive equations. Where possible, regression equations were developed to enable prediction of best-estimate Kd values for a specific set of soil properties. The dependent variable was log10(Kd). The definition of the dependent variable and a series of potential independent variables are seen. The process begins by including the one independent variable that explains the largest portion of the variance in log Kd. It is included only if it’s corresponding regression coefficient is statistically significant at P < 0.05. In the next step, the next most useful independent variable is included, again only if it’s corresponding regression coefficient is statistically significant. This proceeds until no further independent variables significantly contribute to explaining the variance in log Kd. At each step, if the inclusion of a new independent variable renders the coefficient of a previously included variable non-significant at P < 0.05, the non-significant variable is removed from the regression – effectively it has been made redundant. In the end, the regression equation may contain anywhere from none to all the independent variables, with only those that are statistically significant remaining in the equation. Interactions among independent variables are not implicitly included in the stepwise regression process. For soil Kd, pH and clay content are often the most influential variables, and there is an inherent interaction between them in that acidic soils are often low clay content soils (sands). In the above case it is

Log(Kd) = 4.62 – 1.40 where the coefficient –1.40 was significant at P < 0.0001.

Compound Kow Log Kow Herbicide 38.2 1.58

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