Question

A 10-mg quantity of phenanthrene (MW = 178) is added to a 1-L flash containing 500 mL of air and 500 mL of water. Predict the aqueous phase equilibrium concentration of phenanthrene at 25*C if the values for this compound are Cs = 1.1 mg/L, P^v = 1.6 x 10^-7 atm, KH = 4.0 x 10^-5 atm*m^3/mol, and logKow = 4.52.

Answer 1

Outline to solve the problem:

(P)_phenanthrene = K_H * [Phenanthrene]_{in water}

Here, K_H = Henry's constant = 4.0 x 10^-5 atm*m³/mol = 4*10^-5 atm*10³ L/mol and the partical pressure of phenanthrene, i.e. (P)_phenanthrene = 1.6 x 10^-7 atm

(Note: 1 m = 10² cm, 1 cm³ = 1 mL)

Now, [Phenanthrene]_{in water} = 1.6 x 10^-7 atm / (4*10^-5 atm*10³ L/mol)

= 4*10^-6 mol/L

Given that the solublle concentration of phenanthrene = 1.1 mg/L = (1.1*10^-3 g / 178 g mol^-1) / L = 6.18*10^-6 mol/L

Therefore, the concentration in ocatnol phase = (6.18 - 4)*10^-6 = 2.18*10^-6 mol/L

Given that Log Kow = equilibrium concentration in octanol phase/equilibrium concentration in aqueous phase = 4.52

Therefore, the equilibrium concentration of phenanthrene in aqueous phase = 2.18*10^-6 / 4.52

= 4.82*10^-7 mol/L