Benzene is an aromatic organic compound which is present in crude oil and is a k

Question

Benzene is an aromatic organic compound which is present in crude oil and is a known carcinogen. It has a log K_OW of 2.13 You are performing an octanol-water partition experiment to confirm the low K_ow Value of benzene. For your experiment, you add 5 mg of benzene to a separatory funnel containing 0.75 L of octanol and 0.25 L of water. What concentration of benzene do you expect to measure in the water at equilibrium? In a lake contaminated with benzene, the benzene concentration in the water is 2.1 mug/L and the benzene concentration in the sediment is 8.4 mug/g. What is the organic carbon content of the soil? If you took 2 g of contaminated soil from the lake described in part (b), and placed it in 25 L of pure water, what would the benzene concentration be on the soil and in the water once the system reached equilibrium?

Explanation / Answer

Ans-

a) Kow is calculated using the followinf formula:

Kow= concentration of benzene in octanol saturated with water/ cocentration of benzene in water saturated with octanol

Given: log Kow of benzene =2.13, therefore taking the antilog,

Kow= 134.89

mass of benzene =5 mg = 5/1000 g

Thus, Concentration of benzene in octanol saturated with water = Mass/ volume= (5/1000)g / 0.75 L

                                                                                            =0.0066 g/L

now, using the formula for Kow, concentration for benzene in watera t equilibrium can be calculated .

Concentration of benzene in water at equilibrium=concentration of Benzene in octanol/ Kow

                                                                     =(0.0066 g/L)/ 134.81 = 4.940 * 10^-5 g/L.

b) % organic carbon content in soil is calculated using the formula:

Koc= (Kd*100)/ %organic carbon content

Distribution coefficient, Kd=Concentration of chemical in soil/ concentration of chemical in water

Therefore, Kd=(8.4ug/L)/(2.1 ug/L) = 4.0

now, Koc is related to Kow by following equation:

log Koc=0.937 log Kow - 0.006

calculating,

log Koc= 1.9898

Thus Koc= 97.6787

coming back to % carbon content in soil, using the formula we get,

% Organic carbon content = 4*100/ 97.6787

                                       =4.095 %

c) After adding 2 g of contaminated soil of benzene concentration = 8.4ug/g into 25 ml of pure water, this benzene concentration will distribute itself b/w two phses.

suppose the benzene concentration in the soil after equilibrium be x

so benzene concencentration in water after eqm b (8.4-x)

so Kow= x/(8.4-x)

Kow= 134.89 ( calculated in part a)

Therefore ,

134.89= x/(8.4-x)

on calculating we get,

x=8.338

Thus concentration of benzene in soil after eqm = 8.338 ug/g

and concentration of benzene in water after eqm = 0.062 ug/L

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