Henderson-Hasselbalch equation

[pisluca99]

Is it possible to calculate the degree of dissociation of an ionizable substance using the Henderson-Hasselbalch equation, in the context of an unbuffered system?
For example: you put acetic acid in a NaOH solution. At the end of the reaction (at equilibrium) is it possible to exploit the H-H equation to evaluate the degree of dissociation of the acetic acid, even if we do not have a buffer, using the final pH?

 

[Borek]

You do know that HH equation is nothing else but just rearranged equation for dissociation equilibrium?

 

[pisluca99]

You do know that HH equation is nothing else but just rearranged equation for dissociation equilibrium?

Sure, we can reach the H-H equation starting from Ka. Usually It Is used for buffered solution, for example to evaluate the pH of a buffer, but also to evaluate the degree of dissociation of an ionisable compound.
However, I don't understand if the degree of dissociation can be evaluated only in buffered solutions or also for non-buffered solutions, as in the previous example, and therefore by exploiting the final pH (different from the initial one) obtained following the addition of acetic acid. In theory, I think it can be done, but I'm not sure, as H-H generally Is introduced when talking about buffers.

 

[Borek]

Can you solve this problem:

"Calculate the degree of dissociation of acetic acid in 0.1M sodium acetate solution if pKa=4.75."

 

[pisluca99]

Can you solve this problem:

"Calculate the degree of dissociation of acetic acid in 0.1M sodium acetate solution if pKa=4.75."

From the concentration we can find the starting pH of the solution (basic), then, with the H-H we can find relative quantities of CH3COO- and CH3COOH:
log ([CH3COO-]/[CH3COOH]) = pH - pKa , so [CH3COO-]/[CH3COOH] = 10^(pH-pKa) = ā.
Then the degree of dissociation DD = ā/(1+ā).

The problem is that we don’t consider the final pH after we add acetic acid to the solution: pH must drop. So do we consider the final pH? is this reasoning (H-H equation) valid also for unbuffered solutions?

 

[Borek]

Sorry, I misread you first post, I was under impression you were talking about a stoichiometric mixture of acid and base.

Not that it changes anything. I feel like you are missing the point - HH equation doesn't add anything to the situation, as it is not in any way different from just the dissociation equilibrium equation. Whenever you feel "I can calculate something from the HH equation" it means you can calculate exactly the same thing with the Ka formula.

Typical application of HH equation is based on the simplifying assumption that the reaction went to the end and amounts of acid and its conjugate base are given just by the neutralization stoichiometry. You don't need HH equation to use this assumption (and it not always holds).

For general case of calculation of pH of acid/base mixture see https://www.chembuddy.com/calculation-of-pH-of-salt-solutions

 

[pisluca99]

From the concentration we can find the starting pH of the solution (basic), then, with the H-H we can find relative quantities of CH3COO- and CH3COOH:
log ([CH3COO-]/[CH3COOH]) = pH - pKa , so [CH3COO-]/[CH3COOH] = 10^(pH-pKa) = ā.
Then the degree of dissociation DD = ā/(1+ā).

The problem is that we don’t consider the final pH after we add acetic acid to the solution: pH must drop.

Not that it changes anything. I feel like you are missing the point - HH equation doesn't add anything to the situation, as it is not in any way different from just the dissociation equilibrium equation. Whenever you feel "I can calculate something from the HH equation" it means you can calculate exactly the same thing with the Ka formula.

Therefore, since the H-H is a simple rearrangement of the Ka, everything that can be evaluated with the ICE method will have to coincide with the H-H itself. Therefore, the concentrations [CH3COO-] and [CH3COOH] which are found with the ICE method, at the end of the reaction with NaOH, can also be found with the H-H by exploiting the final pH. Right?

So, in theory, we can Always find the degree of dissociation in every situation, simply knowing equilibrium pH and pKa, with H-H.
For the same reason, we can always find the pH of a solution, with H-H, by using equilibrium concentrations (for example pH of a simple acetic acid solution 0,2 M).
If acetic acid reacts with NaOH, we can always find the degree of dissociation with H-H, but not the pH, because we must consider the effect of remaining (eventually) NaOH.

 

[Borek]

Therefore, since the H-H is a simple rearrangement of the Ka, everything that can be evaluated with the ICE method will have to coincide with the H-H itself. Therefore, the concentrations [CH3COO-] and [CH3COOH] which are found with the ICE method, at the end of the reaction with NaOH, can also be found with the H-H by exploiting the final pH. Right?

Yes, but in typical situation once you use ICE to find equilibrium you already have H⁺ (or OH^-) concentration, so there is really no need to plug them into HH if all you need is a simple -log.

So, in theory, we can Always find the degree of dissociation in every situation, simply knowing equilibrium pH and pKa, with H-H.

There are simpler ways of doing the same, no idea why you are sticking to detour when there is a straight, well known route ahead.

For the same reason, we can always find the pH of a solution, with H-H, by using equilibrium concentrations (for example pH of a simple acetic acid solution 0,2 M).

But where are you going to take the equilibrium concentrations from?

If acetic acid reacts with NaOH, we can always find the degree of dissociation with H-H, but not the pH, because we must consider the effect of remaining (eventually) NaOH.

Nope, there is definitely something wrong with your understanding of the equilibrium processes. You can't calculate degree of dissociation (no matter what equation you use) without "considering effect of remaining (eventually) NaOH".

 

[pisluca99]

Yes, but in typical situation once you use ICE to find equilibrium you already have H⁺ (or OH^-) concentration, so there is really no need to plug them into HH if all you need is a simple -log.
There are simpler ways of doing the same, no idea why you are sticking to detour when there is a straight, well known route ahead.
But where are you going to take the equilibrium concentrations from?

We can find concentrations with ICE method. So you mean that It Is useless to exploit H-H if we can find all with ICE. It makes sense.

Nope, there is definitely something wrong with your understanding of the equilibrium processes. You can't calculate degree of dissociation (no matter what equation you use) without "considering effect of remaining (eventually) NaOH".

What do you mean?
What i mean Is that, when the reaction between acetic acid and NaOH happens, the solution will have a certain final pH. Using this pH we can find the CH3COO-/CH3COOH ratio (i suppose), with H-H. But as you said, we can find all with ICE already, so i don't know if this makes really sense.

So H-H is really useful only when we have a buffer.