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Partition coefficient of chlorpromazine in DMSO & pentane

  1. Dec 15, 2015 #1
    Determine the partition coefficient of chlorpromazine in DMSO and n-pentane

    I'm a physics student and doing a course in biophysics. I would really appreciate it, if you would take some time and provide some hints as to how to design a more concrete plan. Especialy, how do I find out the partition coefficient from system A and B?

    Solubility of CPZ in DMSO: 71 mg/mL
    Molecular weight of CPZ: 355,33 g/mol

    Dissolve 17,7665 mg CPZ in 50 mL DMSO to obtain a 1 mM solution.

    Prepare a series of samples of pentane with the following volumes (1 ml, 5 ml, 10 ml, 20 ml, 40 ml) and bring 10 ml of the CPZ+DMSO solution into contact with it.

    Shake the solutions to ensure fast equilibration and then centrifuge to separate the two phases.

    Use UV-vis to measure the absorbance of chlorpromazine in DMSO and use the absorbance to determine the final concentration of CPZ in pentane. Measure the absorbance between ~200-380 nm.

    Use LB to determine the concentrations of each solvent to determine.

    The partition coefficient of system B is defined as the difference in concentration between CPZ in DMSO in system A and the amount of CPZ that has diffused to pentane.
     
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  3. Dec 15, 2015 #2

    Borek

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    What is the exact definition of the partition coefficient?
     
  4. Dec 15, 2015 #3
  5. Dec 15, 2015 #4

    Borek

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    Great, you know how to copy and paste. Do you also know hot to apply the definition here? What you need to know to calculate the partition coefficient?
     
  6. Dec 15, 2015 #5
    P = A1/A2 = C1/C2, from LBs law

    I take a different approach: I calculate the PC by measuring UV-vis absorbance in the DMSO phase in each of the systems B
     
  7. Dec 15, 2015 #6

    Borek

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    So, to calculate partition coefficient you need to know concentrations. You measure absorbance. Can you use the absorbance to find the concentration?
     
  8. Dec 15, 2015 #7
    Yes and using Lambert-Beer's law
     
  9. Dec 15, 2015 #8

    Borek

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    So where is the problem?
     
  10. Dec 15, 2015 #9
    I prepare a series of samples of pentane with the following volumes (1 ml, 5 ml, 10 ml, 20 ml, 40 ml) and bring 10 ml of the CPZ+DMSO solution into contact with it. Centrifuge.

    Now, I transfer the solution to a skillet and let the phases separate, then I take a 1 ml sample of DMSO. Equilibrate.

    I use UV-vis on the DMSO phase to determine the amount of CPZ that diffused to the pentane phase.

    How does the partition coefficient formula then look?

    [itex]P = \frac{C_0}{(C_0-C_1)}[/itex]
    [itex]C_0[/itex] = initial concentration of CPZ in DMSO
    [itex]C_1[/itex] = concentration of CPZ in DMSO determined by UV-vis
     
  11. Dec 15, 2015 #10

    Borek

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    Partition coefficient formula always look the same. The only thing that changes is how you calculate concentrations to plug into the formula.

    You are close, but it is not exactly right. Sum of concentrations is not constant. However, total amount of CPZ is constant.
     
  12. Dec 15, 2015 #11
    So, I must use the volume of the pentane phase in calculating concentrations?
     
  13. Dec 15, 2015 #12

    Borek

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    Yes.
     
  14. Dec 15, 2015 #13
    The concentration of the pentane phase is then
    [itex](m_{DMSO}-m_{pentane})/v_{pentane}[/itex]
     
  15. Dec 15, 2015 #14

    Borek

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    Not sure what you mean by mDMSO.
     
  16. Dec 15, 2015 #15
    Initial mass of CPZ (it should read this)
     
  17. Dec 15, 2015 #16

    Borek

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    Doesn't make it good.

    Try you express the total number of moles of CPZ at equilibrium using concentrations and volumes of both phases.
     
  18. Dec 15, 2015 #17
    DMSO phase: 0,2 mM, 3,5533 mg, 10 ml DMSO.
    n = M*V -> 3,5533 mg * 10 ml = 35,533 moles

    I don't know how well CPZ partitions into the pentane phase. I wish to use UV-vis to determine this
     
  19. Dec 15, 2015 #18

    Borek

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    Can't you do the same for both phases?

    I am not convinced that's what you are told to do - the original post seems to be contradicting. It speaks about using absorbance, but it also mentions "finding the difference", which seems to be suggesting you should measure the absorbance of DMSO solution only. Knowing total amount of CPZ and volumes of both phases, it is enough to measure the concentration in one phase (at least as long as the partition coefficient is such that the concentrations are comparable and don't differ by orders of magnitude).
     
  20. Dec 15, 2015 #19
    Sample #1: 0,2 mM, 3,5533 mg, 10 ml DMSO. n = M*V -> 3,5533 mg * 10 ml = 35,533 moles
    1 ml pentane
    #2: 0,2 mM, 3,5533 mg, 10 ml DMSO. n = M*V -> 3,5533 mg * 10 ml = 35,533 moles
    5 ml pentane
    #3: 0,2 mM, 3,5533 mg, 10 ml DMSO. n = M*V -> 3,5533 mg * 10 ml = 35,533 moles
    10 ml pentane
    #4: 0,2 mM, 3,5533 mg, 10 ml DMSO. n = M*V -> 3,5533 mg * 10 ml = 35,533 moles
    20 ml pentane
    #5: 0,2 mM, 3,5533 mg, 10 ml DMSO. n = M*V -> 3,5533 mg * 10 ml = 35,533 moles
    40 ml pentane

    Do the same for both phases? For the pentane phase I would get the amount of CPZ that diffusions into the pentane phase
     
  21. Dec 15, 2015 #20

    Borek

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    What you have calculated is the initial - total - amount of CPZ, identical in each case. Good.

    After you measure the concentration of CPZ left in DMOS, you can calculate how much CPZ is left in DMSO phase. Whatever disappeared moved to the pentane phase.
     
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