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Maximum conjugate base 
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#1
Dec2613, 09:08 PM

P: 343

I have seen it thrown around a lot that the pH at which concentration of a conjugate base is at a maximum can be found by adding up the 2 pKa's whose reactions that base is involved in and dividing by 2.
But I tried differentiating and this only appears to be the case for HA^{} maximum concentration (reached when pH=pKa1+pKa2, for a diprotic acid). Any other case, and we reach a polynomial... so where does this simple method for finding the pH for maximum concentration come from? What approximation does it require? 


#2
Dec2813, 08:27 PM

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P: 1,985

In other cases you do have a higher polynomial. I am not aware of any particular simplification  even though the polynomials always do contain one zero coefficient. So nothing very nice there on the face of it, unless you can quote us something. Somewhat nice though, since you seem interested in these things, is that these maxima occur where a nprotic acid has bound overall 1, 2,... (n  1) protons per acid molecule (moles/mole)  you might show this to yourself and us. It is not limited to proton binding of course but applies to any ligand that can multiply bind to another molecule. 


#3
Dec2913, 03:26 PM

P: 343

"It is known that in the solution of citric acid H_{3}X, the maximum concentration of H_{2}X^{} is at pH = 3.95; the maximum concentration of HX^{2} is at pH = 5.57; the concentrations of H_{2}X^{} and HX^{2} are equal at pH = 4.76. Determine the acidity constants K_{a1}, K_{a2}, K_{a3} of citric acid." The method expected was to write (K_{a1}*K_{a2})^{1/2}=10^{3.95}, (K_{a2}*K_{a3})^{1/2}=10^{5.57}, K_{a2} = 10^{4.76}. But surely neither of these two is justified, since we are dealing with a triprotic acid here? The results found were K_{a1} = 1.24 * 10^{4}, K_{a2} = 1.14 * 10^{5}, K_{a3} = 4.17 * 10^{7}. Yet the result is highly accurate. e.g. for K_{a3} my exact calculation gave 4.2 * 10^{7}. So the question returns: given that this seems to be a decent approximation often, how do we explain it, and under what conditions will it tend to work? I've done some investigating of this now. Any suggestions on what to interpret from this would be most welcome. What I've found is that (1) it can be algebraically shown and numerically checked that as the value of as the [H^{+}] for maximal [H_{2}X^{}] grows larger, our approximation for K_{a3} grows more accurate; 2) the approximation consistently underestimates K_{a3}. The approximation is typically highly accurate. Can we use this to make any generalizations about this approximation? It does indeed look to me like the larger the K_{a1} value compared to K_{a2} and K_{a3}, the more we can ignore it in the calculation and treat the acid as if only K_{a2} and K_{a3} existed, but also that the approximation will be decent unless K_{a1} is exceedingly close. 


#4
Dec2913, 08:58 PM

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P: 1,985

Maximum conjugate base
I don't know how you did your 'exact calculation' for K_{a3} so please give it. Now I see what you meant by your original question. I gather from your 'the method expected' you have been told somewhere that this is the method. You'd have to show the calculations you made in its justification if you want any comment on them. Please in future give all values as both K_{a} and pK_{a} otherwise a reader is having to do calculations just to try and follow what you are saying. You give two widely different values for K_{a2}. You give a K_{a1} which seems to correspond to a pH close to your maximum so I don't see how you can possibly have been using means. If you do not set out calculations then any errors, yours or mine, make what you did incomprehensible. Qualitatively in e.g. your first maximum you are more than 2 pH units away from pK_{a3}. Then the 3^{} form should be less than 1% of all, so looks safe to ignore and the formulation they gave you right to a decent approximation. There might some small slips in calculations but you seem to be approaching this in a right way. 


#5
Dec2913, 09:49 PM

P: 343

As for "the method expected", I originally didn't know how to approach it because I knew that HA^{} was the only form for which we have this simple exact result (and at the time the sheer length of differentiating them etc. as above made me think it probably wasn't what they wanted). I looked at the given solutions and they used this approximate method. Then I used the above differentiation method to calculate exact values and it seems that they are in pretty close agreement, which is what I want to find out about most. Perhaps if we define symbolically the information given, it will make things easier to discuss. Let us say that [H+] at which concentration of [H_{2}A^{}] is maximal is defined as [H+]_{k=1} and [H+] at which concentration of [HA^{2}] is maximal is defined as [H+]_{k=2}. All we have to work with in the original question is Ka2, [H+]_{k=1}, [H+]_{k=2} (or if you prefer, pKa2, pH_{k=1}, pH_{k=2}). To get the rest, we have to choose our method  exact, or approximate, and if the latter is likely to give us a good result let's go for that! More broadly, how do we decide if the (Ka2*Ka3)^{1/2} formula (alternatively (1/2)(pKa2+pKa3) equation) for maximal concentration of the twicedissociated form will give us a good, close result? Seems to require that Ka1 and I presume Ka4 (if the acid were 4+protic), or pKa1 and pKa4, are not too close to Ka2 and Ka3 (pKa2 and pKa3). So then, if they are indeed not too close to Ka2 and Ka3 (pKa2 and pKa3), does it mean we are calling them negligible? In the problem, according to the approximate method (which is pretty close to the exact method) we got pKa1=3.91, pKa2=4.76, pKa3=6.38 (Ka1 = 1.24 * 10^{4}, Ka2 = 1.14 * 10^{5}, Ka3 = 4.17 * 10^{7}). So it looks like the 'interfering' equilibrium constant has to be within a fraction of a pKa unit of the 'important' ones in order for the approximate method to give poor results? In other words, a very good approximation most of the time. I'll try to see if I can find some ways to prove the approximate equations from the exact procedure and get back to you. In the meanwhile any help you can give with qualitative interpretations of why the approximate method works will be most appreciated! My "exact calculation" method is to differentiate the mole fraction of the form with respect to [H^{+}] and set equal to 0. For HA^{} form from a diprotic acid, we get the neat (and exact) 1/2 (pK_{a1} + pK_{a2}) = pH equation (i.e. [H+]=(Ka1*Ka2)^{1/2}). 


#6
Dec3013, 11:50 AM

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P: 1,985

I'll answer the easiest points now, may make additional ones later.
Biophysics. If you pull your Profs up on tiny inaccuracies but make a huge one yourself they may take revenge! And I think you would do well (also for yourself later) if you set out all relevant data and results in a table (K's and pK's !) otherwise no one will understand this, nor you later. 


#7
Dec3013, 03:12 PM

P: 343

I was quite happy to go for the differentiation. It just occurred to me that it couldn't possibly be expected for the problem  that's why the idea only came afterwards :P The approximate result can also be found by taking the appropriate limit on the exact expression. The limit taken is tantamount to saying that as K_{a1} → ∞ (pK_{a1} → ∞), the maxima for the twicedissociated form converges to the value our approximation predicts. Since this is the same thing as we've said above  that the surrounding Ka values are far enough away from the too important ones that we can call them negligible  we haven't learnt anything new from this bit of maths. :P Interestingly, with the exact approach I don't think it matters whether the dissociation constants get larger or smaller. The calculation relies on more pure maths starting from a mass balance and the equilibrium expressions, and that's it. But I see why the approximate approach would get screwed up if one thing makes the next easier  because at the maximum of a given form, it is helping to produce more and more of the next form onwards and so forth. We thus cannot assume that only the form in question and the two adjacent to it are present significantly at this maximum, because however much there is of these will promote even more of the later forms to be produced. (Maybe I'm explaining my intuition badly!) But I checked the approximate vs. exact calculations and think you're right to say the approximate method won't work there. Thanks again for the help. 


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