November 24, 2024, 05:53:02 PM
Forum Rules: Read This Before Posting


Topic: computation question  (Read 3927 times)

0 Members and 1 Guest are viewing this topic.

Offline jezelee

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
computation question
« on: December 03, 2006, 11:55:32 PM »
A tablet is crushed/suspened in 50mL .1M NaCl.  After filtering to remove the insoluable solids, aqueous solution was put through sep. funnel. Add 2mL .1M HCl to convert interferent to less soluable form.  Acid form has K[d] for an ether:h2o extraction equilibrium 7.5.  How many 10 mL portions of ether must be used to ensure 99.5% interferent is removed from aq. solution?  (aq. layer is 51mL).  I am not sure how to go about computing this.  Could someone please help me?

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27861
  • Mole Snacks: +1813/-412
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: computation question
« Reply #1 on: December 04, 2006, 04:11:30 AM »
form has K[d] for an ether:h2o extraction equilibrium 7.5.  How many 10 mL portions of ether must be used to ensure 99.5% interferent is removed from aq. solution?  (aq. layer is 51mL).

That's the whole question, everything else is there just to confuse you.

Kd is a ratio of concentrations of the extracted substance in both liquid phases staying in contact.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline jezelee

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
Re: computation question
« Reply #2 on: December 04, 2006, 01:04:23 PM »
 ??? Well....I'm still confused?  If the K[d] is a ratio of concentrations, is there some formula or something that lets me compute the # of 10 mL portions needed?  We did not discuss this type of extraction in my class and it is not in my textbook.

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27861
  • Mole Snacks: +1813/-412
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: computation question
« Reply #3 on: December 04, 2006, 01:10:20 PM »
It is not that hard to do simply by trial and error - check how much will be left after first extraction, after second extraction and so on. Try to derive a formula for amount of substance left after one extraction and use it several times or just raise it to power.

For fast calculations you should use logarithms (hint: if y=xn then log(y) = n log(x), you can solve for n).
« Last Edit: December 04, 2006, 01:15:44 PM by Borek »
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Sponsored Links