November 29, 2024, 05:47:34 PM
Forum Rules: Read This Before Posting


Topic: Position of an equilibirum in a proton transfer step  (Read 2241 times)

0 Members and 1 Guest are viewing this topic.

Offline davidenarb

  • Full Member
  • ****
  • Posts: 133
  • Mole Snacks: +0/-1
Position of an equilibirum in a proton transfer step
« on: August 16, 2014, 07:47:15 PM »
Hi all,

I would like to understand why the equilibrium will favor the compound that has the larger pka (the basic compound) ?

please I am looking just for a short and logic explanation

Offline rwiew

  • Regular Member
  • ***
  • Posts: 88
  • Mole Snacks: +10/-6
  • Gender: Male
  • MChem, PhD Researcher in Chemical Biology
Re: Position of an equilibirum in a proton transfer step
« Reply #1 on: August 16, 2014, 09:30:13 PM »
OK, let's try to get you to think about this. What is pKa defined is, what does it mean?

Offline davidenarb

  • Full Member
  • ****
  • Posts: 133
  • Mole Snacks: +0/-1
Re: Position of an equilibirum in a proton transfer step
« Reply #2 on: August 17, 2014, 10:29:30 AM »
OK, let's try to get you to think about this. What is pKa defined is, what does it mean?

pka is a value to tell us the strength of an acid (the shorter the pka, the stronger the acid)

Offline rwiew

  • Regular Member
  • ***
  • Posts: 88
  • Mole Snacks: +10/-6
  • Gender: Male
  • MChem, PhD Researcher in Chemical Biology
Re: Position of an equilibirum in a proton transfer step
« Reply #3 on: August 17, 2014, 10:53:50 AM »
Perfect, so now we have two acids - water and butane. You said the smaller the pKa, the stronger the acid - so you can see that water is a much stronger acid than butane, it wants to give the proton away much more than butane does hence the right side of the equilibrium is much favoured. I think some confusion might be coming from the fact that in your first post you said "compound that has the larger pka (the basic compound)" - a large pKa does not mean that the compound itself is basic, but that it's conjugate base (in this case the carbanion) is a strong one - hence it wants to get protonated. Is this making sense to you now?

Offline davidenarb

  • Full Member
  • ****
  • Posts: 133
  • Mole Snacks: +0/-1
Re: Position of an equilibirum in a proton transfer step
« Reply #4 on: August 17, 2014, 05:28:15 PM »
Perfect, so now we have two acids - water and butane. You said the smaller the pKa, the stronger the acid - so you can see that water is a much stronger acid than butane, it wants to give the proton away much more than butane does hence the right side of the equilibrium is much favoured. I think some confusion might be coming from the fact that in your first post you said "compound that has the larger pka (the basic compound)" - a large pKa does not mean that the compound itself is basic, but that it's conjugate base (in this case the carbanion) is a strong one - hence it wants to get protonated. Is this making sense to you now?

Perfect !!! thanks

Sponsored Links