Page 76 - Materials Chemistry, Second Edition

Site Assessment and Remedial Investigation 59

Example 2.34: Mass Partition between Vapor and Liquid Phases
A new field technician was sent out to collect a groundwater sample from a
monitoring well. He filled only half of the 40-mL sample vial with ground-
water impacted by benzene (T = 20°C). The benzene concentration in the
collected groundwater was analyzed to be 5 mg/L.

Determine:
(a) The concentration of benzene in the head space (in ppmV) before
the vial was opened
(b) The percentage of total benzene mass in the aqueous phase of
the closed vial
(c) The true benzene concentration in the groundwater, if headspace
free sample is collected
Assume the value of the dimensionless Henry’s constant for benzene is
equal to 0.22.

Solution:
Basis: 1-liter container
(a) Concentration of benzene in the headspace
= (H)(C)
= (0.22)(5) = 1.1 mg/L = 1,100 mg/m 3
1 ppmV = (MW/24.05) mg/m = (78/24.05) mg/m = 3.24 mg/m 3
3
3
Concentration of benzene in the head space
= 1,100/3.24 = 340 ppmV
(b) Mass of benzene in the liquid phase
= (C)(volume of the liquid)
= (5)(1 × 50%) = 2.5 mg
Mass of benzene in the headspace
= (G)(volume of the air space)
= (1.1)(1 – 0.5) = 0.55 mg
Total mass of benzene
= mass in the liquid + mass in the headspace
= 2.5 + 0.55 = 3.05 mg
Percentage of total benzene mass in the aqueous phase
= 2.5/3.05 = 82%

(c) The actual liquid concentration should be
= (total mass of benzene) ÷ (volume of the liquid)
= (3.05)/(0.5) = 6.1 mg/L