Linked equilibria: Microscopic view vs. Macroscopic view

In the last two lectures we looked at a very simple binding equilibrium and determinedthat such equilibria can be described by the Langmuir formula. We then went on andlinked conformational equilibria to this simple binding equilibrium. The result we gotwere modified forms of the Langmuir.

Our goal now is to go on and look at more complicated schemes of equilibria and to findways to describe their properties.

Before we go on, we have to think about two fundamental issues that will come up in theprocess. The first issue is the issue of microscopic equilibrium constants vs. macroscopicequilibrium constants. The second one is the question of statistical prefactors.

Macroscopic vs. Microscopic equilibrium constants

Lets consider a relatively simple binding equilibrium: The pH titration of glycine. If wewanted to understand the equilibrium constant for proton binding, we would probably dothe following macroscopic binding experiment.

We would start with a very acidic solution of glycine and we would slowly dribble in aconcentrated solution of a base. We can then record the change in the pH of the solutionas a function of this added base.

If the glycine molecules in solution take up protons, then the rise in pH per volume ofadded base is decreased. Here is what this curve would look like.

From this curve we can immediately read off the characteristic pKa values of Glycine.These values are pK1=2.35 and pK2=9.78. The fact that we observe two pKas makessense, because Glycine has two ionizable groups.

This gives us the following reaction scheme:

with two reaction steps

GH2+ ‡ GH + H+

GH ‡ G- + H+And the corresponding equilbria

†

K1 =[GH][H +]

[GH2+]

=10-2.35 Æ pK1 = 2.35

K2 =[G-][H +]

[GH]=10-9.78 Æ pK1 = 9.78

But if we look at the binding reaction microscopically, we will see that the situation isactually more complicated.

There are two forms of GHThe complication comes from the fact that there are two microscopic forms of GH.The first one is the neutral form, which I will call HG

The second one is the zwitter ionic form, which I call GH

The difference between them could be very important for the molecular function, Forexample the neutral form would bind to a rather different binding site than the chargedform.

So our microscopic picture is more complicated than we first thought. Instead of havingthree species and two equilibrium constants and we actually have four species and fourequilibrium constants.

With the following definitions for the microscopic binding constants. (I refer to themicroscopic constants by their lower case form.)

†

k1 =[GH][H][HGH]

k2 =[HG][H][HGH]

k3 =[G][H][GH]

fi k3-1 =

[GH][G][H]

k4 =[G][H][HG]

fi k4-1 =

[HG][G][H]

Relating microscopic and macroscopic dissociation constantsHow are these microscopic constants related to our macroscopic constants? We havefour unknowns so we need four relationships in order to work this problem out.Two relationships are pretty obvious.

Since

†

[G1H1] = [GH]+ [HG]

K1 =([GH]+ [HG])[H]

[HGH]= k1 + k2

and

K2 =[G][H]

[GH]+ [HG]=

1k3

-1 + k4-1

We also know that

†

k1 ⋅ k3 = k2 ⋅ k4

This we know because the product of k1 and k3 determines the equilibrium constantbetween the doubly protonated form and the fully unprotonated form. Since theequilibrium constant is related to the energy of converting HGH into G+2H and theenergy for this conversion is a state function (it is independent of the path) then productof k2 and k4 which link the same beginning and end points must equal the product of k1and k3.

Now we have three equations and four unknowns. So to relate the macroscopic bindingconstants to the microscopic ones we need one additional piece of information.In our case this piece of additional information comes from a different experiment.We can experimentally measure the equivalent of k2 by measuring the pKa of the NH2group of a glycine ester in which we block the carboxyl group from deprotonating. Inother words we isolate (HGH ‡ HG+H) from the rest of the equilibria.

The “measured” value of pk2 = 7,7

Now we have three unknowns and three equations to solve them.Lets get to it!

First

†

k1 = K1 - k2 fi 10-2.35 -10-7.7 =10-2.35

k2 is so small that its subtraction does not make much of a difference.

Next:

†

k1k3 = k2k4

k3 =k2k4k1

substituting int o equation for K2

K2 =1

1k4

+k1

k2k4

K2 =1

k2k2k4

+k1

k2k4

K2 =1

k2 + k1k2k4

substituting K1 = k1 + k2

K2 =1K1

k2k41

K2=

K1k2k4

k4 =K1K2

k2=

10-2.35 ⋅10-9.78

10-7.7=10-4.43 fi pk4 = 4.43

also have the same If the Still our macroscopic binding constant does not give us anyindication about which of the molecular forms is formed during the titration experiment.

Substituting back into equation for k3 we get

†

k3 =k2k4k1

=10-7.7 ⋅10-4.43

10-2.35=10-9.78 fi pk3 = 9.78

So our microscopic equilibrium constants are:

k1=10-2.35k2=10-7.7k3=10-9.78k4=10-4.43

So our reaction scheme looks like this:

If we compare the size of k1 to the size of k2, we see that the formation of the zwitterionic form predominates (large k1 means relatively large amount of product forming at agiven pH). The same is true for the k3 and k4: k4 is much larger than k3, so if we have alot of G-, very little of it will go into HG and a lot will go into GH.

So while there will inevitably be some very small fraction of HG it is all but negligibleand for this particular case we can treat the reaction like the simple macroscopic model.So if we know which route the reaction takes, our analysis all of a sudden has becomevery simple, because our macroscopic binding constants are equivalent to ourmicroscopic binding constants.

We will later see that such elimination of whole areas of our binding scheme will convertincredibly complicated binding schemes into manageable ones.

Statistical factorsLets now consider a second scenario. We have a polyethylene chain that carries twocarboxylic groups at either end of the chain. Lets further assume that the chain is so longthat the two ionization events do not influence one another. (Side question, how far dothese two groups have to be apart for this to be true? => Brejum Length => 7.7 Angstromin Water. So we need a chain that is substantially longer than 7.7 Angstrom so that onaverage the random walk of the chain takes the ends more than 7.7 Angstroms apart)

Equal binding constants in the microscopic worldSince the two ends are chemically identical and since their environments are identicalaveraged over the time of the experiment, the microscopic energy and therefore themicroscopic equilibrium constant for each of the two groups is identical.

If we see the molecule with its two ends before us we can formulate for reactions.

A+H ‡ HAA+H ‡ AHHA+H ‡ HAH

AH+H ‡ HAH

The equilibrium constant for all four reactions is the same

†

k = [HA][A][H]

=[AH]

[A][H]=

[HAH][HA][H]

=[HAH]

[AH][H]

The Macroscopic viewIf we now look at the macroscopic level, we are not able to distinguish between the twoends of the molecules. After all we are making bulk measurements!So we have two reactions:

A+H ‡ AH1AH1+H ‡ AH2

With the following equilibrium constants:

†

K1 =[AH1][A][H]

and

K2 =[AH2]

[AH1][H]with[AH1] = [AH]+ [HA]

K1 =[AH1][A][H]

=[AH]+ [HA]

[A][H]=

[AH][A][H]

+[HA]

[A][H]= 2k

and

K2 =[AH2]

[AH1][H]=

[AH2]([AH]+ [HA])[H]

=[AH2]

[AH][H]+ [HA][H]

K2-1 =

[AH][H]+ [HA][H][AH2]

=[AH][H]

[AH2]+

[HA][H][AH2]

= 2k-1

K2 = k /2

giving us a K1/K2 ratio of 4.

So if we do independent experiments to measure the concentration of AH1 and AH2 wewill actually see two distinct pKa values even though the microscopic binding energieswere absolutely identical. The reason is simply that there are two times as many ways wecan be singly protonated than we can be doubly protonated. This simple statistical factgives us a very noticeable macroscopic effect even though the microscopic bindingevents were absolutely identical.

This effect should be reminiscent of the explanation I gave for the apparent change inbinding affinity we observe at high concentrations of free ligand in the Langmuir case.

Here is the confusing partNow if we would do a titration of this acid the way we did a titration of glycine.I.e. we would slowly add base and monitor the pH, what would we see?Just one pKa!!!!. What ?????Yes that is right we would see only one pKa, because what we would be looking at in thistitration is the concentration of free protons. (Or the other way around the totalconcentration of occupied protonation sites on the acid). In this titration experiment wedo not distinguish if the sites we are titrating is on a molecule, which has its other endprotonated or not. All we detect is if a particular site is available.

This situation is different if we follow the concentration of the singly protonated form ofthe acid. In this case the concentration of the singly protonated species is controlled bythe two equilibrium constants (K1 and K2) we derived. Plotting the fraction of acid in thesingly protonated form as a function of free proton concentration, we get a plot with twocharacteristic pKa values. Where pK1 describes the buildup of this singly charged species(as we protonate the unprotonated form) and pK2 describing the consumption of singlyprotonated species as the doubly protonated species is formed. The curve will look sort oflike bell curve with the pH of maximum activity located between pK1 and pK2.

A different way to think about statistical factorsThe fact that we get qualitatively very different curves depending on the exact species wetrack in an equilibrium reaction is confusing to many people.Thinking about the equilibrium constants (k, K1 and K2) in terms of a probability (at agiven pH) of observing a protonation event may make it easier to understand suchstatistical effects.

In the case of the di-basic acid.

k is related to the the probability of adding a proton to a single site (each site has onechance).

therefore:

K1 is related to the probability that either site becomes protonated (two chances permolecule)

K2: is related to the probability that the second site becomes protonated as well (onechance per two sites)

The moral of the storyThe equilibrium constants we observe are strongly dependent on what molecular specieswe are tracking during our equilibration experiments. So if we perform experiments on

systems of linked equilibria, we have to consider very carefully for which equilibrium wewant to monitor our constants and how the macroscopic constants we are observing arerelated to microscopic events.