Question

Polyprotic acids systems: thermodynamics and titration curves

Answer 1

How would you guys write up an partition function for the following system and their corresponding interaction energy between states through the thermodynamic cycle?

Answer 2

Henderson Hasselbalch is no longer enough to discribe this system assuming the interaction energy is different from zero. \(W≠0\).

Answer 3

sorry i am not able to understand or find the answer

Answer 4

It is perfectly alright, this is a hard combination of math, thermodynamics and just really clever tricks.

Answer 5

This is what I have so far:

The partition function:
\[\Large Z=\sum_{x}^{N^2} e ^{-\beta G_x^0}~e ^{\beta n_x \mu_{H^+}}\]
The titration curve of an individual site:
\[ \Large \langle x_i \rangle = \frac{ 1 }{ Z }\sum_{x}^{N^2}x_i e^{-\beta G_x^0 }~e^{\beta n_x \mu_{H^+}}\]

and pKa as the equilibrium between the states with k bound protons and k-1.
@Kainui

Answer 6

I assume (as it should be seen from the mathematics) the total deprotonated state to be the ground state (000)

Answer 7

no idea sorry :-(

Answer 8

Here's what I'm thinking, make a partition function for each site (I'm assuming they're distinguishable correct me if I'm wrong since protons are not necessarily distinguishable particles maybe that doesn't work, but maybe cause each protonation site is distinguishable then they are distinguishable?) , call them R, B and G. I'll go ahead and assume each site has the same energy but only cause I don't know any better.
\[R = e^{-\beta E_0} + e^{-\beta E_1}\]\[G = e^{-\beta E_0} + e^{-\beta E_1}\]\[B = e^{-\beta E_0} + e^{-\beta E_1}\]

I'll go ahead and set the ground state energy to 0, cause why not? \(E_0=0\). We can always shift it back up by multiplying by some factor later.
Now the total partition function is Q = RGB.

\[Q=RGB = (1+e^{-\beta E})^3\]

I'm not really sure what to do here, obviously you're using the Gibbs energy with the chemical potential and equilibrium constant somehow so I'm not sure what to do from here. But this is just like my first guess. Can you give me some hints so I can get more serious about answering this? :P

Answer 9

Yeah, for starters where does the \(W \ne 0\) condition come in? I think I don't know what 'interaction energy' is.

Answer 10

Let me just help showing how it looks for a diprotic acid system. Maybe it clears out more.

Answer 11

Sure sounds good :)

Answer 12

For a diprotic acid system there are 4 microscopic equilibrium constants (the difference between the microscopic and macroscopic constant is that the sites are indeed distinguishable :D). We can write them as:
\[\large K_{11}^{10}=\frac{ (10) \left[ H \right] }{ (11) }\]\[\large K_{11}^{01}=\frac{ (01) \left[ H \right] }{ (11) }\]\[\large K_{10}^{00}=\frac{ (00) \left[ H \right] }{ (10) }\]\[\large K_{01}^{00}=\frac{ (00) \left[ H \right] }{ (01) }\]

The microscopic pKa must then be defined as the negative decimal logarithm of the microscopic equilibrium constants. These are connected to the free energy difference of the relevant microstates between the protonation state vector of the product state and reactant state:
\[\large G_p^0-G^0_r=-\beta ^{-1} \ln(10) pK_r^p\]

Some thermodynamics you know: (:D)
Because the energy of the states must be independent from the way to reach the state, we can setup a thermodynamic cycle:
\[\large pK_{11}^{01}+pK_{01}^{00}=pK_{11}^{10}+pK_{10}^{00}\]
We therefore design an interaction energy \(W\) between the sites as:
\[\large W=pK_{10}^{00}-pK_{11}^{01}=pK_{01}^{00}-pK_{11}^{10}\]

Answer 13

The "true" energy between the state is then calculated using:
\[\large W=k_bT \ln(10) \Delta pK_a\]

Answer 14

Ahhh ok now I see what W you are using, I was thinking work but this is the thingy that puts things into states.

Ok so small question on notation, up above you're using (01) and (10) to denote unprotonated/protonated but here in this last equation, that (10) doesn't mean that right? That just means you're using the decimal logarithm is that right? In other words just to clarify:

\[\large W=\beta \ln_{10}( \Delta pK_a)\]

Answer 15

100 correct :P

Answer 16

Why I also in my thesis choose to write W = 1.38 d pKa kcal/mol at 25 degrees C

Answer 17

Poor souls and my notation.

Answer 18

Then you can do the partition function, first as a function of pH:
\[\large Z=1+10^{pK_{01}^{00}-pH}+10^{pK_{10}^{00}-pH}+10^{pK_{01}^{00}+pK_{10}^{00}-W-pH}\]

Answer 19

and you can do the same with energies just do substitution and use the chemical potential for pH.

Answer 20

Aha haha. I am really appreciating this trick with the state functions thingy on the pKas, that's really cool. I have to think about this more to figure out what I don't understand about it cause I'm not too sure.

I see how you've defined the interaction energy but I need to think about that more to understand the implication of that, that's interesting.

Ok, now I have a question, I've only ever seen partition functions for energies not for pH. I guess I'm slowly thinking about this, I'll have to look this up and read a little more about chemical potentials which is good for me. I guess you're using Gibbs energy in the partition function which I guess ends up being something like... Is this right?

\[\frac{\partial G}{\partial n_i} = \mu_i \]

Answer 21

Actually it sorta looks like you already answered my question I just missed that whoops lol

Answer 22

Yup. since it is protons only it is the chemical potential. Your thinking is correct.

Answer 23

And then I can give you a challenge NOBODY have answered so far (at least to such a convincing amount they dared to publish it): What does negative interaction energies mean? ;)

Answer 24

I'm missing some details here, I think specifically I am not too sure about where the formulas are coming from so that's just something I guess I gotta look up on my own to see how they're derived. I'm not super clear on how, but I know where to find the derivation of the famous
\[\Delta G^o = RT \ln K\]
so I'll just have to find that.

Actually slightly unrelated question but I think I accidentally stumbled onto the concept of entropy maybe you can confirm or help me a little. One sec.

Answer 25

Haha maybe negative interaction energies means you need to shift your ground state energy up? :P

Answer 26

I think only energy differences are observable!

Answer 27

Ok here's how I think I accidentally stumbled onto entropy, first I was looking at the partition function with degeneracies \(g_i\):

\[q = \sum_i g_i e^{-\beta E_i}\]

Then I thought, hey, I can equivalently look at a state with degeneracies as having NO degeneracies but instead having a lowered energy that's proportional to the temperature, simple algebra!

\[\large q = \sum_i g_i e^{-\beta E_i}\]\[\large q=\sum_i e^{\ln g_i}e^{-\beta E_i}\]\[\large q=\sum_i e^{-\beta E_i+\ln g_i}\]\[\large q=\sum_i e^{-\beta( E_i-\frac{\ln g_i}{\beta})}\]

So it looks like we can call the reduced energies:

\[\varepsilon_i = E_i-k_bT\ln g_i\]
Then I realized we could look at energy differences and that'd correspond to multiplication in the logarithm so, we could end up looking ta stuff like

\[-k_bT \ln W\]

which I think is Boltzmann's entropy formula? :O

Answer 28

Ahhhh ok right you're talking about Gibbs energies, then I'm totally off. Negative interaction energies must mean some kind of spontaneous thing then eh? Beats me, I think I barely know what a positive interaction energy is haha xD

Answer 29

Just remove the minus sign. Cause you look at the difference?

Answer 30

But that is actually very smart.... never thought of that.

Answer 31

Oh right, it shouldn't be there cause it's treated as the amount that it's lowered by and entropy is always positive. Yeah I think I just got lucky on this one haha xD

Answer 32

Even though this is some pretty awesome insight I think, I still don't fully understand entropy I feel but I think this is a good step so I'm pretty excited about this. I just found this like 3 days ago or something xD

Answer 33

Oooh ok what are interaction energies typically? 3.58 kcal/mol seems like a lot but then again I have no reference. How'd you come across this mutant protein anyways, like did you come up with the idea or was it just an accident?

Answer 34

It also helps one get an better understanding of the weight \(W\) in his formula or Wahrscheinlichkeit as he called it.

Answer 35

As I like to think degeneracy is easier to understand.

Answer 36

like 0.3 kcal/mol I would guess.

Answer 37

3.58 kcal/mol is the stability of the protein it self :P

Answer 38

Maybe I should change unit since it is so small numbers.... to kJ or would it make the americans sad?

Answer 39

I understand W to be the generalization of the binomial, also known as the multinomial coefficients, which are defined as:

\[\large W = \binom{N_1+N_2+\cdots +N_k}{N_1,N_2,\cdots, N_k} = \frac{(N_1+N_2+\cdots +N_k)!}{N_1!N_2!\cdots N_k!}\]

I don't know if this is the most general form of W or if this is just a special case for like 'nice stuff' whatever that means. Sometimes when I learn this I never know if it's only for ideal and distinguishable stuff or if it's more general or less general. :X

Answer 40

Hahaha actually I prefer SI units, I don't use cal. I do all chemistry in Celsius and Kelvin :P

Answer 41

I think everyone who does chemistry and physics in American universities do it in SI units, it's just the uneducated americans who still use feet and yards and stuff haha :P

Answer 42

Yeah but still I just hate to try explain physical chemistry students that \(W\} is suppose to be the "most probable distribution of microstates" they don't even have statistics and probability courses before physical chemistry so I don't trouble my self :P

Answer 43

Then I might as well change that for my thesis.

Answer 44

Haha, yeah in QMC Atkins he gives the derivation of the partition function using Lagrange multipliers at the very end which I wish he had explained a little better since it seems pretty important.

It's cool cause I actually can remember (well enough to figure out and rederive it on the fly if I need to now) how to derive the partition function from \(d \ln W = 0\) :P

Answer 45

You made a comment earlier about titration curves, I have never known, mathematically speaking, how to derive those curves. Are these: https://figures.boundless-cdn.com/13986/large/di.png

literally what you get from the partition function, \(\langle x_i \rangle\)? :O

Answer 46

Exactly the fraction of protons bound. Thermodynamics is much more fun when applied for anything BUT the things they mention in the textbooks :P

Answer 47

Hahaha I wish I was taught about the partition function in undergrad. It's that one simple, beautiful thing that allows you to tie the quantum energies to the thermodynamic properties so damn good ahhhh lol.

Answer 48

The circles are the actual data, the line is the fit to the hill-equation. (from the wiki page)

Answer 49

The hill equation is in principle 100% identical to the HH equation. So it is very evident that the titration curves does not follow the HH-equation :P

Answer 50

haha yeah I can see it's definitely off. So what exactly is the HH assuming that makes it break?

Answer 51

That the sites titrate independently i.e. does not interact with each other.

Answer 52

Oh ok, just like we've been talking about this entire time, ok makes sense.

Answer 53

But I mean, that seems like a really bad approximation when you add 2 acids groups right next to each other. In this case, I showed data for, it is an amide and a carboxyl acid

Answer 54

Do you have any good intro/mid level books on this sorta stuff you would suggest for physical biochem like this? I have never really dealt with pchem applied to stuff larger than organic molecules so I'm kinda curious. I have a vague idea of the basics like Lineweaver Burke plots and I think the more math the better for me. I don't know how serious I'll get right now into this but this is fun right now... :P

Answer 55

Hmmm... How about I just give you the book of the man him self whos work on protein folding is legendary among biochemists: Alan Fersht
http://www.fersht.com/Structure_And_Mechanism_in_Protein_Science.pdf

Answer 56

Ok, here's my solution. First, we pick one protonation site of interest, since there are 3 to pick from I'll choose the 2nd one for fun. Let's call this:

\[pK_2= pK_{-1-}^{-0-}\]

This is poorly defined cause there are 4 ways the other two sites can be protonated, so let's go ahead and let's define the 'interaction difference' as deprotonated minus protonated in the 1st slot:

\[\Delta_1 pK_2 = pK_{01-}^{00-}-pK_{11-}^{10-}\]

Alternatively we could have used the 3rd slot, it doesn't matter yet.

Now if this was just a diprotic acid, then this would be all we need, and we could define this to be the interaction energy as we've already seen!

\[W = pK_{01}^{00} - pK_{11}^{10}\]

However we're now free to continue from here and do:

\[\Delta_3 (\Delta_1 pK_2 )= [pK_{010}^{000}-pK_{110}^{100}]-[pK_{011}^{001}-pK_{111}^{101}]\]
\[=pK_{010}^{000}-pK_{110}^{100}-pK_{011}^{001}+pK_{111}^{101}\]

And this is how I think you should define your interaction of interaction energies.

Notice a couple things here, it really doesn't matter what order you do this in, so you can end up just calling this \(W_2\) since that's the real focus of this term:

\[W_2 = \Delta_3 \Delta_1 pK_2 = \Delta_1 \Delta_3 pK_2\]

It's sorta confusing notation maybe, but I don't know how better to organize this. This allows you to look at any polyprotic acids and look at interaction of interaction of interaction... energies hahahaha.

Hopefully this makes sense; it's more complicated than it sounds I hope, maybe we can talk in a call if that would be better.

Answer 57

n1g@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$