Question

Atrazine is an herbicide widely used for corn and is a common groundwater pollutant in the corn-producing regions of the United States. Use EPIWIN to find Koc for atrazine. Calculate the fraction of total atrazine that will be adsorbed to the soil given that the soil has an organic carbon content of 2.5%. The bulk density of the soil is 1.25 g/cm3 ; this means that each cubic centimeter of soil (soil plus water) contains 1.25 g of soil particles. The porosity of the soil is 0.4.

Answer 1

Answer:

0.94

Explanation:

Let's assume that the total sample of the soil is 1 cm³; which has a porosity of 0.4 cm³; then the original sample of the soil will be 0.6 cm³

The mass  of the soil particles in 1 cm³ of soil + water = 0.75 g

The fraction of the total atrazine can be calculated by the expression;

Total atrazine =$C_w(\frac{g}{mL})*V_w(mL)+C_{(soil)}(\frac{g}{soil})*M_{(soil)}(gsoil)$

Also; $C_{(soil)}(\frac{g}{soil})= K_{oc}(\frac{mL}{gOC})*f_{oc} (\frac{gOC}{gsoil})*C_w(\frac{g}{mL})$

NOTE THAT: $K_{oc}( \frac{mL}{gOC})=K_{oc}\frac{L}{kgOC}$

Total atrazine can now be calculated as:

$C_w*0.4+K_{oc}f_{oc}C_w*0.75$ = $\frac{C_{soil}*M_{soil}}{Total atrazine}$

$=\frac{K_{oc}f_{oc}C_w*M_{soil}}{C_w*0.4+K_{oc}f_{oc}C_w*0.75} = \frac{0.75*K_{oc}f_{oc}}{0.4+K_{oc}f_{oc}*0.75}$

$= \frac{10^{2.52}*0.025*0.75}{0.4+10^{2.52}*0.025*0.75}$

= 0.94

Thus,the fraction of total atrazine that will be adsorbed to the soil = 0.94.