# Need help on solving pkas, concentrationDecember 17, 2008 4:00 PM   Subscribe

I apologize for posting such a specific question: I have an ionization scheme with 3 different pKas. I can figure out, using Henderson Hasselbalch, the hypothetical concentration of two of them. But my research adviser wants to know the concentration of all the structures in solution. And I don't have the math skills for that.

So here's my basic ionization scheme for Serine, but I'm trying to be as general as possible.

[a] --> [b] --> [c] --> [d]
(1) (2) (3)

pk1 (btw. [a] and [b] = 2.2
pk 2 (btw. [b] and [c]= 9.2
pk 3 (btw. [c] and [d]= 13
a, b, c, d are differently protonated in solution. At a certain pH, I need to find out how much is in solution. Can Henderson Hasselbalch handle that?

here's what I've got:
Using Henderson
The original concentration of the sample is 1.2M, so I set it up as so, with pH at 13.431. And the solution exists mostly between C and D:

13.431 = 13 - log((1.2 - [c])/[c])
-.431 = log (1.2-b)/b
d= .87552
c = 1.2 - b = .32458

So my question is, is it possible for me to solve for A and B, in order to have a more realistic model of what's going on in my solution? I've looked, but I guess I'm not sure how to incorporate that into Henderson Hasselbalch. Thanks!
posted by geronimo's folly to Science & Nature (3 answers total)

Well, you can write statements that express the fraction of each form (like, from 0 to 1), is that what you're after?

Since I think this sounds like homework, I'm not going to exactly spell it out for you, and this might not be what you want anyway.

For a diprotic system (e.g. glycine, three forms), it goes something like:

f(+1)= [[H+]^2)/(([H+]^2)+([H+]*Ka1)+(Ka1*Ka2))

f(zwitterion)= [[H+]*Ka1)/(([H+]^2)+([H+]*Ka1)+(Ka1*Ka2))

f(-1)= (Ka1*Ka2)/(([H+]^2)+([H+]*Ka1)+(Ka1*Ka2))

I leave it as an exercise for the reader to extrapolate that to describe a triprotic system.
Anyway, each statement will give you the fraction of each form present at the given [H+], and you can apply that directly to the concentration of the amino acid to find the concentration of each form. If you graphed f vs. pH it, two lines would cross at f=0.5 at the points where pH=pKa.
posted by pullayup at 5:32 PM on December 17, 2008

If you can get your hands on an analytical chemistry text, it'll probably spell this out more comprehensibly than I can. You can look at the acids/bases chapter, or sometimes there's a section on EDTA (which has six! dissociable protons) that will really blow your mind.
posted by pullayup at 5:38 PM on December 17, 2008

Here we go. Sorry, I don't remember having learned the pKa for serine--I'll just assume the hydroxyl group has a neutral charge at physiological pH and gets protonated at high pH. If I'm wrong, the only thing that changes would be the charge notation for the different ionic species, but the solution will still apply. It's been a while.

[Ser1+] --- pK1 = 2.2 ----> [Ser0] --- pK2 = 9.2 ---> [Ser1-] --- pK3 = 13 ---> [Ser2-]

At pH 13, assume that the carboxyl group is fully unprotonated and that the alpha-amino group is fully unprotonated. The ionic species depends mainly on the side chain (pK3 = 13).

pH = pK2 + log ([Ser2-]/[Ser1-])
[Ser2-] + [Ser1-] = 1.2M
You know pH = 13.431 and pK3 = 13, solve system of two equations to get [Ser1-] and [Ser2-]

Henderson-Hasselbalch equation, for the second time, to calculate [Ser1-] and [Ser0]:

pH = pK2 + log ([Ser1-]/[Ser0])
pH is still 7, pK2 = 9.2 and we already found [Ser1]. Straightforward.

Henderson Hasselbalch equation for the third time, to find [Ser0] and [Ser1+]:

pH = pK1 + log([Ser0]/[Ser1+])
[Ser1+] is the only unknown here, so this is easy, too!

This gives you all the concentrations of the ionic species; make sure that they all add up to 1.2M.
posted by halogen at 6:16 PM on December 17, 2008

« Older propaganda says what?   |   Can one order up a miracle? Newer »
This thread is closed to new comments.