Max Conjugate Base: Find pH w/ Sum of pKa's

In summary, there is a method used to find the pH at which the concentration of a conjugate base is at a maximum by adding up the 2 pKa values and dividing by 2. However, this method only works for diprotic acids and for other cases, it results in a polynomial. It is not clear where this method came from or what approximation it requires. There is also a question about the accuracy of this method and under what conditions it tends to work. It has been observed that the approximation consistently underestimates the Ka3 value. Further investigation has shown that the accuracy of the approximation increases as the [H+] for maximal [H2X-] grows larger. However, it is not clear how this can
  • #1
Big-Daddy
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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?
 
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  • #2
Big-Daddy said:
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?

I don't know where you saw what things thrown around, you would have to quote. You quoted results just for diprotic acids.

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 n-protic 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.
 
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  • #3
epenguin said:
I don't know where you saw what things thrown around, you would have to quote. You quoted results just for diprotic acids.

Well I can quote a problem I recently did:
"It is known that in the solution of citric acid H3X, the maximum concentration of H2X- is at pH = 3.95; the maximum concentration of HX2- is at pH = 5.57; the concentrations of H2X- and HX2- are equal at pH = 4.76. Determine the acidity constants Ka1, Ka2, Ka3 of citric acid."

The method expected was to write (Ka1*Ka2)1/2=10-3.95, (Ka2*Ka3)1/2=10-5.57, Ka2 = 10-4.76. But surely neither of these two is justified, since we are dealing with a triprotic acid here? The results found were Ka1 = 1.24 * 10-4, Ka2 = 1.14 * 10-5, Ka3 = 4.17 * 10-7.

Yet the result is highly accurate. e.g. for Ka3 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 [H2X-] grows larger, our approximation for Ka3 grows more accurate; 2) the approximation consistently underestimates Ka3. 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 Ka1 value compared to Ka2 and Ka3, the more we can ignore it in the calculation and treat the acid as if only Ka2 and Ka3 existed, but also that the approximation will be decent unless Ka1 is exceedingly close.

epenguin said:
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.

Can I ask whether you were taught this/found it somewhere, or worked it out yourself? If the former, I would love if you could suggest where I can find more tricky issues like this.

epenguin said:
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 n-protic 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.

It is indeed noticeable that when [H+]=(Ka(k)*Ka(k+1))1/2, the approximation discussed in this thread (the suggestion being that the maximum concentration of [Hn-kA] is reached at approximately this pH), as a result we always have [Hn-k-1A]=[Hn-k+1A] at such points - in fact this relies only on equilibrium constant expressions. But I don't see that this can be extended to tell us anything about any forms of the acid except [Hn-k-1A], [Hn-k+1A] and thus about the acid overall?
 
  • #4
Big-Daddy said:
The method expected was to write (Ka1*Ka2)1/2=10-3.95, (Ka2*Ka3)1/2=10-5.57, Ka2 = 10-4.76. But surely neither of these two is justified, since we are dealing with a triprotic acid here? The results found were Ka1 = 1.24 * 10-4, Ka2 = 1.14 * 10-5, Ka3 = 4.17 * 10-7.

Yet the result is highly accurate. e.g. for Ka3 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 [H2X-] grows larger, our approximation for Ka3 grows more accurate; 2) the approximation consistently underestimates Ka3. The approximation is typically highly accurate.

OK, they almost give you Ka2 but please check as I get not far but significantly different from you , 1.74*10-5 .

I don't know how you did your 'exact calculation' for Ka3 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 Ka and pKa otherwise a reader is having to do calculations just to try and follow what you are saying. You give two widely different values for Ka2. You give a Ka1 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 pKa3. 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.

Big-Daddy said:
Can I ask whether you were taught this/found it somewhere, or worked it out yourself? If the former, I would love if you could suggest where I can find more tricky issues like this.

I had known about that little theorem but never had any use for it before now. You could I think use it in your exact calculations. I looked it up and the way it is presented in my book it is easier and you are far better off to derive it yourself! Which I did. Not difficult - try it for diprotic then triprotic acids.

There might some small slips in calculations but you seem to be approaching this in a right way.
 
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  • #5
epenguin said:
OK, they almost give you Ka2 but please check as I get not far but significantly different from you , 1.74*10-5 .

I don't know how you did your 'exact calculation' for Ka3 so please give it.

See below. Note that Ka2 = 10-4.76 = 1.74 * 10-5 to 3 sig figs, pKa=4.76 is known exactly as per the question.

epenguin said:
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.

I don't want to show the entire differentiation process as it's quite long. I took the concentration of [H2A-] as a function of all Ka values and [H+] and differentiated with respect to [H+], set equal to 0 and got a polynomial for concentration of [H+] at which [H2A-] would be maximal. I did the same for [HA2-]. This gives two equations, in which we have a total of 2 variables, Ka1 and Ka3, since Ka2 is known to be 10-4.76 (pKa=4.76) and the concentration of [H+] at which each of these two conjugate base forms is maximal is also known. I rearranged one equation for Ka1, substituted into the other and rearranged for Ka3. I can provide the final result if you'd like.

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.

epenguin said:
You give a Ka1 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.

I haven't provided any exact calculation values for Ka1. The values that came as a result of the approximation discussed were given in my last post.

epenguin said:
Qualitatively in e.g. your first maximum you are more than 2 pH units away from pKa3. 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.

But we hadn't calculated pKa3 yet at the time?

Perhaps if we define symbolically the information given, it will make things easier to discuss. Let us say that [H+] at which concentration of [H2A-] is maximal is defined as [H+]k=1 and [H+] at which concentration of [HA2-] 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, pHk=1, pHk=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 twice-dissociated 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!

epenguin said:
I had known about that little theorem but never had any use for it before now. You could I think use it in your exact calculations. I looked it up and the way it is presented in my book it is easier and you are far better off to derive it yourself! Which I did. Not difficult - try it for diprotic then triprotic acids.

I see. It seems like a slightly niche thing - my analytical chemistry textbook does not even broach the question ("what pH is the concentration of a given form maximal?") - so I was wondering if you knew a link or textbook I could look at which would contain more of these rarer issues.

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 (pKa1 + pKa2) = pH equation (i.e. [H+]=(Ka1*Ka2)1/2).
 
  • #6
I'll answer the easiest points now, may make additional ones later.
Big-Daddy said:
See below. Note that Ka2 = 10-4.76 = 1.74 * 10-5 to 3 sig figs, pKa=4.76 is known exactly as per the question.
Misunderstanding due to a typo
Big-Daddy said:
Ka2 = 1.14 * 10-5,


Big-Daddy said:
This gives two equations, in which we have a total of 2 variables, Ka1 and Ka3, since Ka2 is known to be 10-4.76 (pKa=4.76) and the concentration of [H+] at which each of these two conjugate base forms is maximal is also known. I rearranged one equation for Ka1, substituted into the other and rearranged for Ka3.
IOW you solved two simultaneous linear equations in two unknowns. :biggrin:

Big-Daddy said:
I can provide the final result if you'd like.
Why not? In particular did you have a zero coefficient in each polynomial? If not I think we need to look again. However
Big-Daddy said:
I took the concentration of [H2A-] as a function of all Ka values and [H+] and differentiated with respect to [H+], set equal to 0 and got a polynomial for concentration of [H+] at which [H2A-] would be maximal. I did the same for [HA2-].


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 (pKa1 + pKa2) = pH equation (i.e. [H+]=(Ka1*Ka2)1/2).

the last result is telling me you did get a zero in the central term of the quadratic.


Big-Daddy said:
the sheer length of differentiating them etc.

The things that students tossed off routinely a few years back when it was called Question 6 of Exercise 1 of Chapter 3, Calculus tend to get considered, if not insoluble, an undertaking of the most extravagant labour when it is called
Biophysics. :biggrin:


Big-Daddy said:
I haven't provided any exact calculation values for Ka1. The values that came as a result of the approximation discussed were given in my last post.

My point was even the approximate value you gave seems grossly wrong, perhaps due to another typo. Your Ka1 gives me pKa1 of 3.96, very close to and on the wrong side of your pH of maximum, 3.95. Instead the latter is supposed to be the mean of pKa1 and 4.76, from which I make pKa1 to be 3.14.

If you pull your Profs up on tiny inaccuracies but make a huge one yourself they may take revenge! :eek:

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.


Big-Daddy said:
But we hadn't calculated pKa3 yet at the time?

Very acute. :approve: It sounded better how I said it :biggrin: but I think we could have got from other stuff we had that it was a reasonable at least rough approximation. When you are 1 pH unit away from something its effect is 10% or less. The results fully confirm my intuition! waves hands Indeed in this field you may often find what may at first sound like circular arguments - but it is pretty indicative when the circles close.


Big-Daddy said:
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!

I thought you said you had done the first. I did the second in a previous post

Big-Daddy said:
I see. It seems like a slightly niche thing - my analytical chemistry textbook does not even broach the question ("what pH is the concentration of a given form maximal?") - so I was wondering if you knew a link or textbook I could look at which would contain more of these rarer issues.

Yes it is slightly - but it sounds you are making it your niche. For most biochemistry and biophysics approximate solutions are sufficìent and the point, and exact ones useless and even irritating to hear about. In fact they are so thrown out that you'll likely forget them youself, so I recommend you to make good notes while this is fresh, including what I said about setting out and tabulating this problem. Where this could have application would be if you get into e.g. metal complexes of acids, amino acids, peptides etc. More mainstream would be in co-operative phenomena in proteins and other biomolecules. Probs are slightly different - your examples are called negative co-operativity - a proton dissociating makes it harder for the next one to do so - and so the K's are nicely separated. Instead the thing of most biophysical interst is the opposite, one thing making the next easier. That makes it almost impossible to determine the K's! But the practice you are giving yourself will stand you in good stead. Just remember experimetalists will be concerned with big effects they can measure, not tiny differences they can't. A mathematical/physical book about this is J. Wyman & S.Gill "Binding and Linkage" USB publ. and the result we mentioned is on p.72.
 
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  • #7
epenguin said:
Why not? In particular did you have a zero coefficient in each polynomial? If not I think we need to look again. However


the last result is telling me you did get a zero in the central term of the quadratic.

Yes I did. They were actually cubics since it's a triprotic acid, but each had one 0 coefficient. And of course I didn't have to solve the cubic equations as such because I wasn't solving for the leading term.

epenguin said:
The things that students tossed off routinely a few years back when it was called Question 6 of Exercise 1 of Chapter 3, Calculus tend to get considered, if not insoluble, an undertaking of the most extravagant labour when it is called
Biophysics. :biggrin:

Is there a book where this would be found as Question 6 of Exercise 1 of Chapter 3?

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

epenguin said:
My point was even the approximate value you gave seems grossly wrong, perhaps due to another typo. Your Ka1 gives me pKa1 of 3.96, very close to and on the wrong side of your pH of maximum, 3.95. Instead the latter is supposed to be the mean of pKa1 and 4.76, from which I make pKa1 to be 3.14.

You're right, I think there is a typo in the solutions, because the approximate method gives me Ka1 = 7.24 * 10-4, pKa1 = 3.14.

epenguin said:
I thought you said you had done the first. I did the second in a previous post

Thanks, I think I've understood this issue now. At pH when a certain form is maximal, so long as the equilibria that form is involved in have K values not too closely surrounded by the K values of adjacent equilibria, it can be assumed that that form and the two adjacent to it will take up a vast majority of the total concentration of the acid and can neglect the rest of the acid forms, and neglect the presence of all the other equilibria except the two which this form is involved in. This does indeed lead by the same procedure of differentiation to the approximate result we've been discussing.

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 Ka1 → ∞ (pKa1 → -∞), the maxima for the twice-dissociated 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 learned anything new from this bit of maths. :P

epenguin said:
Yes it is slightly - but it sounds you are making it your niche. For most biochemistry and biophysics approximate solutions are sufficìent and the point, and exact ones useless and even irritating to hear about. In fact they are so thrown out that you'll likely forget them youself, so I recommend you to make good notes while this is fresh, including what I said about setting out and tabulating this problem. Where this could have application would be if you get into e.g. metal complexes of acids, amino acids, peptides etc. More mainstream would be in co-operative phenomena in proteins and other biomolecules. Probs are slightly different - your examples are called negative co-operativity - a proton dissociating makes it harder for the next one to do so - and so the K's are nicely separated. Instead the thing of most biophysical interst is the opposite, one thing making the next easier. That makes it almost impossible to determine the K's! But the practice you are giving yourself will stand you in good stead. Just remember experimetalists will be concerned with big effects they can measure, not tiny differences they can't. A mathematical/physical book about this is J. Wyman & S.Gill "Binding and Linkage" USB publ. and the result we mentioned is on p.72.

Thanks for all the advice. My main purpose in this investigation was less to show my determination to use exact solutions and more to try and find out why and when the approximation is likely to be good.

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.
 

What is Max Conjugate Base?

Max Conjugate Base refers to the maximum amount of a conjugate base that can be present in a solution before the solution becomes basic.

How do you calculate the pH using the sum of pKa's?

The pH can be calculated using the Henderson-Hasselbalch equation, which is pH = pKa + log([conjugate base]/[weak acid]). The sum of pKa's refers to the total pKa values of all weak acids in the solution.

What does the sum of pKa's tell us about a solution?

The sum of pKa's can tell us the overall acidity or basicity of a solution. A higher sum of pKa's indicates a more basic solution, while a lower sum of pKa's indicates a more acidic solution.

What factors can affect the sum of pKa's in a solution?

The sum of pKa's can be affected by the concentration and strength of weak acids and their corresponding conjugate bases in the solution. Temperature and ionic strength can also have an impact.

How can the sum of pKa's be used in titrations?

The sum of pKa's can be used to determine the equivalence point in a titration, which is when the moles of acid equal the moles of base. This can help in calculating the concentration of an unknown solution.

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