Expert: John Locke - 2/5/2008

Question
I've done 3 labs over the past week and I have some questions over the last one that I cannot figure out the answer to.

In the idealized case involving non-charged solutes, the equilibrium constant (Keq) for the diffusible dye in the dialysis experiment is: Keq= Dye in/Dye out. If Keq is 1, then, at equilibrium, the concentration of dyes on both sides of the membrane are equal, as expected for a passive diffusion experiment. Assuming constant pressure and temperature, answer the following:

a. Does changing the water bath alter the Keq?
b. Does changing the water bath change the amount of time it takes the system to reach equilibrium?
c. Assume the Keq of the diffusible dye in the mixed dye experiment is 10. What could account for a value greater than 1?

Also. How could the rate of dialysis of the dye be increased?

Answer
Thanks for using AllExperts, Holly. As to questions a and b, I'm not sure exactly what "changing the water bath" means. Does that mean you'd replace the water both inside and outside the dialysis membrane? Could you provide additional details of the experimental setup?

Regardless of whether you replace the water both inside and outside the dialysis membrane, Keq will not change: Keq depends only upon the conditions at equilibrium. It does not matter how you arrive at that equilibrium. It could very well change the time required to reach equilibrium, however. Consider replacing the water bath with new water that contains no dye molecules; the dye would have to re-diffuse through the dialysis membrane and reestablish its equilibrium. That would take longer than if the water bath already contained dye--in the latter case, part of the diffusion would already have taken place.

However, I would need more details of this experiment to better answer all your questions, particularly the last two. If you still would like a response, you can post a question to me with more details of this experiment. I will be happy to answer your questions then.