Partition coefficient of chlorpromazine in DMSO & pentane

[Schwarzschild90]

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.

 

[Borek]

What is the exact definition of the partition coefficient?

 

[Schwarzschild90]

In the physical sciences, a partition-coefficient(P) or distribution-coefficient(D) is the ratio of concentrations of a compound in a mixture of two immisciblephases atequilibrium. These coefficients are a measure of the difference in solubility of the compound in these two phases.

 

[Borek]

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?

 

[Schwarzschild90]

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

 

[Borek]

So, to calculate partition coefficient you need to know concentrations. You measure absorbance. Can you use the absorbance to find the concentration?

 

[Schwarzschild90]

Yes and using Lambert-Beer's law

 

[Borek]

So where is the problem?

 

[Schwarzschild90]

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?

P = \frac{C_0}{(C_0-C_1)}
C_0 = initial concentration of CPZ in DMSO
C_1 = concentration of CPZ in DMSO determined by UV-vis

 

[Borek]

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.

 

[Schwarzschild90]

So, I must use the volume of the pentane phase in calculating concentrations?

 

[Borek]

Yes.

 

[Schwarzschild90]

The concentration of the pentane phase is then
(m_{DMSO}-m_{pentane})/v_{pentane}

 

[Borek]

Not sure what you mean by m_DMSO.

 

[Schwarzschild90]

Initial mass of CPZ (it should read this)

 

[Borek]

Doesn't make it good.

Try you express the total number of moles of CPZ at equilibrium using concentrations and volumes of both phases.

 

[Schwarzschild90]

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

 

[Borek]

n = M*V

Can't you do the same for both phases?

I don't know how well CPZ partitions into the pentane phase. I wish to use UV-vis to determine this

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).

 

[Schwarzschild90]

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

 

[Borek]

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.

 

[Schwarzschild90]

And I measure the concentration of CPZ left in DMSO using UV-vis absorbance, then getting the concentration from absorbance, right?

 

[Borek]

Yes.

 

[Schwarzschild90]

How will the formula look? Using
1. 0,2 mM, mg, 10 ml DMSO

n = M*V -> 3,5533 mg * 10 ml = 35,533 moles

1 ml pentane

 

[Borek]

I told you several posts ago - try you express the total number of moles of CPZ at equilibrium using concentrations and volumes of both phases. For each sample you know total number of moles, you know V_DMSO and V_pentane, you will measure C_DMSO, the only unknown will be C_pentane.

 

[Schwarzschild90]

Okay. Thank you.

I have the following samples prepared
10 ml DMSO+CPZ dissolved in 1 ml pentane
10 ml DMSO+CPZ dissolved in 5 ml pentane
10 ml DMSO+CPZ dissolved in 10 ml pentane
10 ml DMSO+CPZ dissolved in 20 ml pentane

The phases are seperated.

I also have the following prepared
2,7 mg DMSO+Methylene blue dissolved in 2 ml pentane
2,5 mg DMSO+Methylene blue dissolved in 2 ml pentane
2,1 mg DMSO+Methylene blue dissolved in 2 ml pentane
1,7 mg DMSO+Methylene blue dissolved in 2 ml pentane
0,3 mg DMSO+Methylene blue dissolved in 2 ml pentane

For this part I found out, that the fine pipette that I used, was less than accurate. I wanted to measure out 2 ml, but instead got the values 2.7, 2.5, 2.1, 1.7 and 0.3. I want to "fix" this, by mixing all the samples together, creating a ~roughly 10 mg DMSO + 10 ml pentane solution. I will then shake it. Then take ten 1 ml samples, which I will dilute by adding 5, 10, 20, 30, 50 mls of pentane, each.

Will you recommened a method for me to do spectroscopy without disrupting the phases?