pKa/b and Ka/b

[amikhchi]

I read through this stuff so many times it all seems like gibberish to me. I think i understand the basic concept of what these values mean... but can someone dumb it down and explain it, i thought the Ka was just the Keq for an acid, so where does the "it will/won't be protonated" stuff come in?

Thanks,

-Amir

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[MundaneMD]

Note: I stole the following stuff from http://library.tedankara.k12.tr/carey/pka.html

pKa = pH + log [HA] / [A-]

This tells us that when the pH = pKa then log [HA] / [A-] = 0 therefore [HA] = [A-] ie equal amounts of the two forms.

If we make the solution more acidic, ie lower the pH, then pH < pKa and log [HA] / [A-] has to be > 0 so [HA] > [A-]. This makes sense as it tells us that the protonated form dominates in an acidic medium.

If instead we make the solution more basic, ie raise the pH, then pH > pKa and log [HA] / [A-] has to be < 0 so [HA] <[A-]. This makes sense as it tells us that the deprotonated form dominates in the basic medium.

 

[werd]

you're correct in that Ka is just the Keq for a reaction in which an acid loses a proton. genericaly...
HA <--> (H+) + (A-). the Keq = Ka = [H+][A-] / [HA]

now, the "p" of anything is defined as the (-log) of the value. pH is the negative log of the H concentration, pKa is the negative log of the Ka value. now, doing some math and reviewing the properties for adding logs,

-log Ka = -log( [H+][A-] / [HA] )
pKa = -log(H+) - log (A-/HA)
pKa = pH - log (A-/HA)
pKa = pH + log (HA/A-)

the whole protonated/deprotonated stuff comes from working with this equation... when pH = pKa, then log(HA/A-) = 0 and so [HA]/[A-] = 1, that is the protonated and deprotonated forms are equal in concentration. you can remember this equation each time or you can remember the pattern derived from it.... when pH is above pKa the species is mostly deprotonated (A-); when pH is below pKa it's mostly protonated (HA). specifically, the ratio changes by an order of magnitude for each pH point away from pKa... so at 1 pH point above pKa it's 91% A- and 9.1% HA.

hopefully this helps.

 

[DocWalken]

Ka is the acid dissociation constant, or measure of how readily an acid loses a proton:

HA + H2o <=> A- + H3o+ (where HA is the protonated acid and A- is the deprotonated form or conjugate base)

Ka=([A-][h3o+])/(HA)

Kb is the base dissociation constant, or measure of how readily A- will accept a proton.

B + h2o <=> BH+ + oH- (where B is base or A- in this case and BH+ is HA)

kb=([oh-][Bh+])/(B)


pka is the negative log of the ka. This tells you essentially at what pH an acid will be deprotonated:

ex. Acetic acid's pKa is ~4.7, at a pH of 6, Acetic acid will be deprotonated. At a pH of 3, however, it will be protonated.

hope this helps and I haven't made any errors.

 

[amikhchi]

thanks a lot guys, its all making sense now. wow, nearly a week to go...