Amino Acids are Polyprotic acid
Polyprotic acids
Part 1: Reactions and pKa values of polyprotic acids
To write out the reactions of a compound that has more than one acidic proton, you need to know
which proton is more acidic, because it will be the first one donated. You should often be able to
determine this using principles of organic chemistry structure, but you should always be able to
determine them based on pKa values, if known. For example, consider alanine. From chemical
principles, you know that carboxylic acids are more acidic than protonated amines. The pKa values are
2.4 and 9.9 respectively. That means that we can draw two equilibrium acid/base reactions for alanine,
and we can determine that loss of the carboxylic acid proton occurs first.
Confusion comes in when we ask the question, “Is the zwitterionic structure a conjugate acid or a
conjugate base?” The answer is yes to both—the key is that, when discussing conjugate acids and
bases, you need to define to which structure they are conjugate. Notice that the product of reaction 1 is
the reactant of reaction 2. We can put both reactions together since they are all in equilibrium.
KEY POINT: Be sure you know what reaction you are talking about before deciding if a structure is a
conjugate acid or a conjugate base!
Part 2: Using the Henderson-Hasselbalch reaction to calculate pH
The Henderson-Hasselbalch reaction essential relates three variables: the pH of the solution, the
strength of the weak acid (the pKa) and the relative amounts of conjugate acid/conjugate base present
(the [A-]/[HA] ratio.) Like any equation, if you know two of them, you can calculate the third. To better
understand polyprotic acids, let’s do an exercise in which pKa and acid/base ratio are know, then pH is
calculated.
This exercise is a thought experiment to investigate the titration curve, what it means, and its
implications for buffering. To make this abstract concept more concrete, let’s imagine that we start with
a solution of very low pH that contains 100 molecules of alanine.
Using this thought experiment, we can
 Qualitatively understand the shape of a titration curve
 Understand buffering regions and non-buffering regions
The thought experiment will involve looking at the following points on the titration curve. We will look
at what is happening at points A-J on the titration table.
Point A: At point A, the pH is so acidic that alanine exists predominately in its totally protonated state.
As a good estimation, that means that all 100 molecules of alanine exist in this form:
Point B: Add 0.09 equivalents of hydroxide
Even though it doesn’t look like it, the x-axis tells us the ratio of conjugate base to conjugate acid,
although it is labeled as “equivalents of hydroxide.” This axis means that, to our solution of alanine, we
are adding 0.09 equivalents of strong base. What is an equivalent? It just means “the same.” Since we
are starting with 100 molecules of alanine, 1.00 equivalents of base would be 100 molecules of
hydroxide. At point B, 0.09 equivalents of hydroxide is 9 molecules of hydroxide in our thought
experiment. Since strong acids react completely with weak acids, the hydroxide acts as a limiting
reagent. Nine molecule of alanine react and 91 are left unchanged.
We now have the ratio of conjugate base to conjugate acid to be 9 molecules/91 molecules, or 1/10.
(Notice that the units drop out of the ratio. That is why our thought experiment works—we are thinking
about molecules, but it is the same if we treat this in terms of moles, or molar concentration. All that
matters is the numeric ratio.)
We know that the pKa of the proton we are dealing with is 2.4, and we have figured out the ratio of base
to acid. If we plug these into Henderson-Hasselbalch, we can determine the pH of the solution.
pH = pKa + log [A-]/[HA] = 2.4 + log (1/10) = 2.4 – 1.0 = pH 1.4 at Point B
Point C: Continue with the reaction by adding 41 more molecules of hydroxide
If we add 41 more molecules of hydroxide, we have added a total of 50 molecules of hydroxide. This
brings us to point C, which is 0.5 equivalents of hydroxide. Since we are moving from point B to Point C,
the products of the reaction above are the starting reagents of this reaction, plus the added 41
molecules of hydroxide:
The ratio of base to acid is one to one, so the pH of the solution at point C is:
pH = 2.4 + log (1) = pH 2.4 at point C
Take home message: From B C, adding 41 molecules of base raised the pH by one unit.
Point D: Continue the reaction by adding 41 more molecules of hydroxide. We have added a total of
0.91 equivalents of hydroxide at point D
The ratio of base to acid is now 10/1, so the pH at point D is:
pH = 2.4 + log (10) = pH 3.4 at point C
Take home message: From C D, adding 41 molecules of base raised the pH by one unit.
Point E: Continue the reaction by adding 8 more molecules of hydroxide. We have added a total of 0.99
equivalents of hydroxide at point E.
The ratio of base to acid is now 99/1, which can be rounded to 100. The pH at point D is:
pH = 2.4 + log (100) = pH 4.4 at point C
Take home message: From D E, adding only 8 molecules of base raised the pH by one unit. At this
point in the graph, addition of a smaller amount of strong base raises the pH more substantially. We
are outside of the buffer range of alanine.
Point F: Continue the reaction by adding 2 more molecules of hydroxide. This point of the chart
requires a little more thought about what is happening, because we are now starting to make new
compounds after we cross over the 1.0 equivalent mark. Consider what happens when the first
molecule of hydroxide is added:
There is no longer any substantial amount of the +1 charged molecule, which we have been calling the
conjugate acid. (This is a simplification—there is always at least a small amount of the conjugate acid,
but at these higher pHs—when pH is more than 2 units greater than pKa—the amount is insignificant.)
That means there is only the Zwitterion left to react with hydroxide. Up until this point, the zwitterion
has been the conjugate base being formed. Now, it is the conjugate acid reacting with hydroxide. This
is the concept that makes polyprotic acids hard. We are doing a whole separate acid/base problem.
This also means that we have a new pKa value—pKa2, which is 9.9 for alanine, and corresponds to the
protonated amine.
Overall, the reaction from point E to point F is:
And the pH is equal to pKa2 + log ([A-]/[HA])
pH = 9.9+ log 1/100 = pH 7.9 at point F.
Take home message: From E F, adding only 2 molecules of base raised the pH from 4.4 to 7.9, an
increase of 3.5 units.
Exercises:
1. Finish doing this thought experiment and filling in the table for points G-J.
Point
Added HO-
F
-------
G
8 molecules
H
41 molecule
I
41 molecule
J
8 molecule
Total Equiv of Molecules
HOof A1.01
1
Molecules
of HA
99
[A-]/[HA]
About 1/100
pH
7.9
2. Where are the buffer regions for alanine? Explain how the solution acts as a buffer in these pH
regions.
3. Why did we have to start using a different value of pKa half-way through the problem? How do you
know which pKa value to use?
4. Imagine we had a 1 M triprotic acid with pKa1 = 3.4, pKa2 = 8.1, and pKa3 = 12.9. The solution starts at
a high pH so that the acid was completely deprotonated. What is the pH of the solution if:
A. 0.09 eq. of HCl are added?
B. 0.5 eq. of HCl are added?
C. 1.0 eq, of HCl are added?
D. 2.0 eq. of HCl are added?
E. 2.5 eq. of HCl are added?