March 29, 2024, 02:55:47 AM
Forum Rules: Read This Before Posting


Topic: Solubility in cold water vs. diethyl ether  (Read 10666 times)

0 Members and 1 Guest are viewing this topic.

Offline noiseordinance

  • Regular Member
  • ***
  • Posts: 66
  • Mole Snacks: +0/-0
Solubility in cold water vs. diethyl ether
« on: July 10, 2011, 08:29:22 PM »
Hello there,

I'm answering a few questions for a lab and am stuck. Here's the question. "The pain reliever phenacetin is soluble in cold water to the extent of 1.0 g/1310 mL and soluble in diethyl ether to the extent of 1.0 g/90 mL.

a) Determine the approximate distribution coefficient for phenacetin in these two solvents.
b) If 50 mg of phenacetin were dissolved in 100 mL of water, how much ether would be required to extract 90% of the phenacetin in a single extraction?
c) What percent of the phenacetin would be extracted from the aqueous solution in b) by two 25-mL portions of ether?

I believe I have a) solved. K = [phenacetin in di. ether]/[phenacetin in water] = (1.1 / 100mL)/(7.6x10-2 / 100mL) = 15.

I'm stuck on b), however. This has gotta be easy and I'm just not seeing it. Can anyone offer any direction? I imagine I can start out by figuring out how much diethyl ether is required to extract 100% first, but I don't even know how to get at that...

Offline Babcock_Hall

  • Chemist
  • Sr. Member
  • *
  • Posts: 5592
  • Mole Snacks: +319/-22
Re: Solubility in cold water vs. diethyl ether
« Reply #1 on: July 11, 2011, 04:23:44 PM »
I just have time for a quick reply right now, but your idea about calculating what it would take for 100% extraction does not look promising.  This is basically an equilibrium problem, with 90% in one phase and 10% in another.

Offline noiseordinance

  • Regular Member
  • ***
  • Posts: 66
  • Mole Snacks: +0/-0
Re: Solubility in cold water vs. diethyl ether
« Reply #2 on: July 11, 2011, 06:26:25 PM »
I just have time for a quick reply right now, but your idea about calculating what it would take for 100% extraction does not look promising.  This is basically an equilibrium problem, with 90% in one phase and 10% in another.

Thanks for the reply. This is one idea I've had... can someone verify?

K = [solv 2]/[solv 1] = (0.045g / x mL)/(0.005g / 100mL), x = 60mL

Any thoughts?

Offline Dan

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 4716
  • Mole Snacks: +469/-72
  • Gender: Male
  • Organic Chemist
    • My research
Re: Solubility in cold water vs. diethyl ether
« Reply #3 on: July 11, 2011, 06:49:18 PM »
60 mL looks good to me
My research: Google Scholar and Researchgate

Offline noiseordinance

  • Regular Member
  • ***
  • Posts: 66
  • Mole Snacks: +0/-0
Re: Solubility in cold water vs. diethyl ether
« Reply #4 on: July 11, 2011, 06:58:36 PM »
Sweetness. Thanks much!!

Sponsored Links