Homework SourseAn episode of a popular crime scene investigation television
An episode of a popular crime scene investigation television episode used a separation technique as part of the plot. Searching for clues of a crime, an extract from the vitreous humor of the eye (liquid in the eyeball) is examined for the presence of a particular drug. The procedure uses non-polar toluene to extract the drug.
When toluene, in contact with the viterous humor fluid, reached equilibrium it was determined that 0.0100 moles of the drug was found in 0.020 L of toluene and 0.0010 moles of the drug was in 0.010 L of the aqueous vitreous humor. What is the value for the partition ratio?
Using that value of KD from part \"A\" in another experiment, what fraction of the drug would remain in 0.010 L of the vitreous humor being extracted with 0.0050 L of toluene?
Solution
The partition ratio (KD), as the ratio of the solute’s total concentration in each phase measn organic phase and aqueous phase
KD =[Sorg]total / [Saq]total
Molarity = number of moles / volume in L
[Sorg]total
= 0.0100 moles /0.020 L
= 0.5 M
[Saq]total
= 0.0010 moles / 0.010L
= 0.1 M
Here [Sorg]total = 0.5 M
[Saq]total = 0.1 M
KD =[Sorg]total / [Saq]total
= 0.5/0.1
= 5
Using that value of KD from part \"A\" in another experiment, what fraction of the drug would remain in 0.010 L of the vitreous humor being extracted with 0.0050 L of toluene?
To calculate the fraction of the drug would remain in 0.010 L of the vitreous humor being extracted with 0.0050 L of toluene
[Sorg]total =5 [Saq]total
KD =[Sorg]total / [Saq]total
5 [Saq]total /0.0050 L *0.010 L /[Saq]total
10
