Proving inflection points

[apchemstudent]

I know this is a chem question in general, but the steps needed to solve this are more mathematical.

This is not a homework question, but my prof came up with the concept of determining the inflection point on a titration graph with 2 pKas. He simply told us that it was the average of the 2 pKa values. However, he also just mentioned that it can also be proven through calculations, but he never showed us how.

I'm just curious as to how this can be proven, and remember, this is not a homework question. Can some one show this to me? Thanks.

The example we were talking about in class was glycine :P, again. so the COOH group's pKa value was 2.4, and the NH3(+) group's pKa value was 9.6.

I just don't know how to prove it...

here's the henderson hasselbach equation that we use

pH = pKa + log(conjugate base/conjugate acid) to determine the pH at different conjugate base to acid ratios.

the dissociation of COOH(conjugate acid) <-> COO(-) (conjugate base) + H(+)
dissociation of NH3(+)(conjugate acid) <-> NH2(conjugate base) + H(+)

I hope this helps as a background. Please help, thanks.

 

[shaner-baner]

you are talking about inflection points, so I assume you know calculus. If you don't I'm not sure you'll understand the answer. First, you need a function that describes the titration curve, then you find possible points of inflections by setting the second derivative equal to 0. Finally you check to see if what you found were really p.i.'s. I probably don't have enough background to come up with the specific equation.

 

[tacman]

To prove the inflection point on a titration graph with two pKa values, we can use the Henderson-Hasselbalch equation and the concept of buffering.

First, let's define what an inflection point is in this context. An inflection point on a titration graph occurs when the pH changes rapidly with small additions of titrant. This happens when the solution is at the buffering capacity, meaning that the concentration of the conjugate base and conjugate acid are equal.

Now, let's consider the titration of glycine. At the start of the titration, the solution contains only the conjugate acid (COOH) with a pKa of 2.4. As we add base, the pH increases and the concentration of the conjugate base (COO-) increases. At a certain point, the concentration of the conjugate base is equal to the concentration of the conjugate acid, and the solution is at the buffering capacity. This occurs at the first inflection point, which can be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log([COO-]/[COOH])

At this point, the pH equals the pKa, which is 2.4. This is the first inflection point on the titration graph.

As we continue to add base, the pH increases until we reach the second inflection point. At this point, the concentration of the conjugate base is equal to the concentration of the conjugate acid for the second dissociation (NH2- and NH3+), which has a pKa of 9.6. This can also be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log([NH2-]/[NH3+])

Again, the pH equals the pKa, which is 9.6. This is the second inflection point on the titration graph.

Now, to prove that the inflection point is the average of the two pKa values, we can use the concept of buffering. At the buffering capacity, the concentration of the conjugate base and conjugate acid are equal, meaning that the ratio of [conjugate base]/[conjugate acid] is 1. This allows us to simplify the Henderson-Hasselbalch equation to:

pH = pKa + log(1)

Since the logarithm of 1 is 0, the pH at the inflection point is simply