Why can't we apply Kd directly to a equilibrium solution of CCl4 and H2O,with CH3COOH dissolved?

Do CH3COOH form polymers in CCl4? or are their single molecules of CH3COOH in CCl4?

5 answers

I think it's because the CH3COOH is partially ionized in H2O and the percent ionization changes with the concentration of the acid.I was unsuccessful at finding the solubility of acetic acid in CCl4.
So you think this has nothing to do with the reason (2nd statement)?
Not really. The CCl4 solution will be single molecules, I think, Also, I know CH3COOH forms dimers (especially in the solid state) and in the liquid state the formula mass is somewhere between 60 and 90 indicating that some do and some don't. But I think whether dimers are formed or not formed or what is actually in CCl4 solution is not the question.
The discussion at this site talks about this.
https://en.wikipedia.org/wiki/Partition_coefficient

I believe the bottom line is that distribution coefficients work best for unionized solutes AND I suspect the overriding reason why this doesn't work very well is because CH3COOH is miscible with H2O. In such a solution, the organic solvent has almost no chance at pulling the solute into the organic phase.
I understand... Thank you!
what is the ph of 50ml buffer solution which is 2M in CH3COOH and 2M in CH3CooNa?
1 Initial PH before the addition of acids and base?
2 What is the new PH after 2ml of 6.00M HCl is added to this buffer?
3 what is the new PH after 2.00ml of 6.00M NaoH is added to the original buffer?