Hurricane Harvey may also have released polychlorinated biphenyls (PCBs), which

Question

Hurricane Harvey may also have released polychlorinated biphenyls (PCBs), which are reported to be carcinogenic and bioaccumulative. PCBs are a group of chemicals that includes 209 different congeners. PCB4 (C_12H_8Cl_2) is one common representative PCB. Assume dissolved PCB4 is present in a pond. a. One study suggests that PCB4 has a Henry's constant of 0.84 [mol/L-atm]. Using this value, calculate the concentration of PCB4 in the air above the pond, assuming equilibrium, when PCB4 in water is fixed at 48 ppb. b. Another study suggests that PCB4 has a Henry's constant of 0.02 [unitless, air/water]. Using this value, calculate the concentration of PCB4 in the air above the pond, assuming equilibrium, when PCB4 in water is present at 48 ppb.

Explanation / Answer

According to henry's law the amount of dissolved gas is proportional to its partial pressure in the gas phase

a. Assuming PCB4 dissolved in a pond

let concentration of PCB4 in air be c

then c = p [ where p is partial pressure of PCB4)]

so using henry's law

H = c'/p

c' = concentration inside the pond = 48 parts per billion

for 1 billion molecules of water, there are 48 molecules of PCB4

1 mole of PCB4 is present in 6.022*10^23/48 billion molecules of water

mass of 6.022*10^23/48 billion molecules of water = 18*10^9/48 grams = 375000 kg

volume V = 375000 lit

so, c' = 1/375000 = 2.666*10^-6 mol/L

Given H = 0.84 mol/L atm

p = c'/H = 2.6668106*10^-6/0.84 atm = 3.1746*10^-6 atm

so concentration of PCB4 in air = partial pressure of PCB4/atmospheric pressure = 3.1764*10^-6

b. Assuming PCB4 dissolved in a pond

let concentration of PCB4 in air be c

then c = p [ where p is partial pressure of PCB4)]

so using henry's law

H = Ca/Cg

c' = concentration inside the pond = 48 parts per billion

for 1 billion molecules of water, there are 48 molecules of PCB4

1 mole of PCB4 is present in 6.022*10^23/48 billion molecules of water

mass of 6.022*10^23/48 billion molecules of water = 18*10^9/48 grams = 375000 kg

volume V = 375000 lit

so, ca = 1/375000 = 2.666*10^-6 mol/L

Given H = 0.002

Cg = Ca/H = 2.666*10^-6 / 0.002

Cg = 0.001583 mol / lit

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