Chromatography Forum

Correction!

Peter wrote:

that a weak acid will be predominantly ionised or not depending on the pH of the solution in which it finds itself

But it is not so. On the contrary, the weaker acid will always be forced to accept a proton from the stronger acid

And in case it wasn’t clear enough – what I wrote to Hans, was a kind of irony “I didn’t realize that water was a stronger acid than the benzoic acidâ€

so

according to Peter:

all the shoulder/splitting is due to that multiple phases are formed when the plug of analyte solution is injected into the column instead of due to the formation of ion/non-ion forms.

I have no problem with that physicists are proud of being a physicist just as chemists do as long as the physicists always have right answers for physical problems.

ym3142 wrote/asked:

so according to Peter:
all the shoulder/splitting is due to that multiple phases are formed when the plug of analyte solution is injected into the column instead of due to the formation of ion/non-ion forms.

It is just reformulated/told in other words, but the reason is the same: In the front fraction of the peak the environment is short of protons (the counter ion here is sodium) whilst there are enough protons in the end fraction of the peak to suppress ionization of the analyte. And this is only possible if the environment’s pH is equal to the analyte’s pH plus/minus 1. So the pH through the peak is not precisely the same and thereby the similarity to the “strong solvent effectâ€

I thought that Peter was very clear:

everywhere in a uniform solution the pH is the same and the equilibrium between the ionized form and the non-ionized form of the analyte is the same. This applies to the front fraction of the peak as well as the tail section of a peak, and there is no difference in the ratio of the ionized form and the non-ionized form from the front to the tail of the peak in isocratic chromatography.

Dear Uwe,

Firstly, a chromatographic peak is not a uniform solution. Ideally there would be 2 (3) different concentrations in a symmetrical peak - low concentration in the front, high concentration in the middle and finally at the end, equally low concentration as the front fraction.

Secondly, there is this vector forward motion – nothing goes backwards in order to create equilibrium.

Thirdly, the people, defending the assumption that the analyte being at pH = pKa doesn’t behave differently than at pH values away from It’s pKa, are just defending their believe dogmatically not feeling that they have to provide any evidence or just examples.

Best Regards

The mobile phase is the uniform solution. Please explain, why in this uniform solution a low concentration of the analyte should sometimes be more ionic (in the front of the peak) and sometimes be less ionic (e.g. in the back of the peak.

And, as a practical experiment that you can do yourself: The UV spectra of many substances are different in their ionized form and in their non-ionic form. You can measure the UV spectrum at the front of the peak and compare it to the one on the back of the peak.

As my teacher used to say: don't think, measure!

Uwe wrote/asked:

low concentration of the analyte should sometimes be more ionic (in the front of the peak) and sometimes be less ionic (e.g. in the back of the peak.

Note that I wrote ideally! That is when the chromatography works as intended. We have been discussing a situation of “less idealâ€

I'll be glad to send to anybody who is interested my HPLC Troubleshooting collection, which experimentally proves in section 29 my point that you have no problem with a peak shape at a mobile phase pH at the pKa of an analyte, if your mobile phase is properly buffered.

If you or anybody else wants a copy of this booklet, feel free to contact me at my e-mail address: Uwe.Neue@prodigy.net.

Hi Uwe,

I think (not quite sure though) I came across your guide. It is published on Waters website – right?
So I just checked section 29. But I can’t relate the material to the present discussion. So maybe that’s what causes all the fuss. You are talking about buffer capacity and that is what the example in the guide deals with. I on the other hand am talking about pH (buffer) = pKa (analyte) and that is what the present discussion was all about - from the start anyway. The example you are referring to, doesn’t even mention ibuprofen’s pKa. So how can you drag it in this discussion as an experimental proof?
I fully agree that the buffer capacity should be taken seriously and if you check other posts from me you will be more than convinced that this is indeed one of the things I’m taking seriously.

I think it is just about time for me to break my participation in this particular discussion, because the subject gets derailed all the time.

The most important thing for me is the truth and not “who said the last wordâ€

The experiment was set up at the pKa of ibuprofen in water, i.e. the pH was set up to match this pKa. In this experiment, the ibuprofen (pKa 4.3 or 4.4, depending on where you look) is roughly 50% ionized and 50% not ionized. In a proper buffer at this pH, the peak shape is completely symmetrical. Without proper buffer, the peak is tailing.

The story is that in the presence of a proper buffer, the peak shape is symmetrical, and the peak is uniform. This can be determined via the consistency of the UV spectrum from the front to the tail of the peak. I did not do this in this experiment, but there has been software around since a long, long time that does this. If there would be peak uniformity problems due to effects proposed by you, we would all know about this. Since you do not think that the experiment that I described is relevant, I encourage you to set up the same experiment and measure the peak uniformity, which will tell you that the peak in the presence of a proper buffer is uniform (unless you overload the buffer, of course).

Danko, surely there is a point when everything has been said that it is time to terminate a discussion. This has long since been reached here.
Unfortunatly, some things can not be left to stand, for instance:

"But it is not so. On the contrary, the weaker acid will always be forced to accept a proton from the stronger acid"
Peter´s statement is correct. Did you, Danko, ever hear of equilibrium?

"And in case it wasn’t clear enough – what I wrote to Hans, was a kind of irony “I didn’t realize that water was a stronger acid than the benzoic acidâ€

Not so Hans.

If you have a stronger acid (than water) in the solution then H3O+ and the acid residue (e.g. chloride or whatever) is the result.
Oh, I nearly forgot – and a lot of H2O

Over and out.

Then what is the pH of a 10^-7 molar HCl solution?

It is a long discussion.
I want to take this chance to thank all the participants, especially Uwe, Danko, Peter, Hans, and Mark. You guy’s expertise helps us learn more about the LC, both physical and chemical sides.

(only repeat what people already described) It seems that it is dominantly believed that: 1) when pH=pka, 50% is anion and 50% is neutral acid; 2) when there is enough buffer, the formation of these two forms will not cause peak to shoulder/split; 3) when there is not enough buffer, it is possible that two layers of mobile phases(these two are different) are formed in the column and sustained through the column to give peak shoulder/splitting; 4) the reason for a column to be unable to see the separation of anion/neutral acid is that proton exchange process is relatively much faster.5) further implication, that regardless the preparation of buffer, if the analyte is dissolved in a the mobile phase, there will not be any shoulder/splitting due to buffer capacity;

To my understanding, the above conclusions do not conflict from one to another and are theoretically correct. Though I do not agree with Uwe’s UV theory for supporting above conclusions. Because we know putting sodiumn benzoate and benzoic acid to any sodium acetate buffer will always give the same thing such as pH 3.2 ,both sodiumn benzoate and benzoic acid give 90% the acid, 10% the anion; 4.2 , 50% acid, 50% anion; pH 5.2 10% acid and 90% anion. In other words, what UV detector sees may have nothing to do with what happens inside a column.

Thanks