Question

Volatile chemicals are often extracted into a water immiscible phase for analysis using gas chromatography. Consider benzene, which is mildly soluble (for an aromatic compound) in a water sample in a sealed 100-mL bottle containing 60 mL of water. The benzene has reached equilibrium between the two phases. Benzene has a molecular weight of 78 g/mol, liquid density of 0.88 g/mL, vapor pressure of 0.125 atm. (which is equivalent to 405 mg/L), liquid activity coefficient of 2.4 × 103 (25 °C and 1 atm), solubility of 1780 mg/L and an octanol-water coefficient of log KBow=2.0.

a. Gas analysis shows that the headspace contains 1.95 mg/L of benzene. What is the concentration of benzene (mg/L) in the water?

b. What is the total mass of benzene (mg) in the sealed bottle?

c. One mL of octanol is now added to the bottle. What is the resulting concentration of benzene in the water after equilibrium with the octanol?

Answer 1

The distribution ratio K_d can be determined by the formula X_G / X_L where X_G is the concentration of benzene in gaseous state while X_L is the concentration of benzene in water.

                                              

For neutral compounds like benzene, we can relate Henry's law constant with the vapour pressure,and solubility of benzene by the following expression.

                                              

where P_vp is in atm and S is in mol/m³ is liquid activity coefficients

So, now H can be modified to H', Henry's dimensionless constant which is same as K_d

                                                

ehre S is in mg/L and P_vp is in mmHg

          Since H' = K_d

          By rearranging the two equations, we get

                                                  

where S is in mg/L.

Since [X]_G is given and all other values are known, we can calculate [X]_L

                                                                  

                                               

                                                [X]_L = 3.6 mg/L

(a) So, concentration of benzene in water = 3.6 mg/L

         Amount of benzene in water = [X]_L x volume of water

                                                        = 3.6 x 60 x 10^-3

                                                        = 0.216 mg

       Amount of benzene in headspace= [X]_G x volume of headspace

                                                       = 1.95 x 40 x 10^-3

                                                       = 0.078 mg

(b) So, Total Amount of benzene = 0.294 mg