1. 100 mL of aqueous solution containing 2, 4, 6-trinitrotoluene (TNT) at an ini
ID: 970411 • Letter: 1
Question
1. 100 mL of aqueous solution containing 2, 4, 6-trinitrotoluene (TNT) at an initial concentration of 0.05 ppm is shaken with 10 mL hexane. After phase separation, it was determined that the organic phase has 0.4 ppm. What is the partition coefficient and what is the percent extracted? If the equilibrium concentration (rather than initial concentration) is 0.05 ppm, this would simplify the calculation. What would be then partition coefficient and percent extracted?
2. Two organic compounds (a and b) are sepreated on a GC column with retention time of 20.2 and 24.3 min, and peak widths (at base) 0.76 and 0.89 min for a and b, respectively. An unretained air peak occurs at 3.6 min. Please calculate the (a) retention factor, (b) separation factor, (c) resolution, (d) average plate number.
Explanation / Answer
ANSWER : partition coefficient is given by = concentration of TNT in aqueous layer
concentration of TNT in organic layer
Partition coefficient is given by Kd = (0.05 / 100) / (0.4 / 10)
Kd = 0.0005 / 0.004 = 0.0125 ppm
Therefore partition coefficient = 0.0125 ppm
total volume = 100ml + 10ml = 110ml
For 110ml the Kd = 0.0125
for 100ml ---------- (0.0125 x 100) / 110 = 12.5 / 110 =0.113 % (percent extracted)
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