Henderson Hasselbalch Equation - Request for Clarification

[njh]

TL;DR Summary

Where am I making a mistake in my understanding of the HH equation?

I understand the HH equation as:

##pH = pKa + log\frac{[A^-]}{[HA]}##

• The Numerator of the fraction is the conjugate Base of the original acid.
• The denominator is the undissociated/ unprotonated Acid.
• I understand that when the pH equals the PkA, then the solution is 50% dissociated, so for example 1 Mol of conjugate Base and 1 Mol of undissociated Acid. Therefore, if pH = 2 and pKA = 2, then the solution is 50% dissociated, or ##log\frac{1}{1}####=0##.
• I know that if the pH fell from 2 to 1, and the pKa remains 2, then the lower pH means a higher level of protonation, and so more conjugate Base and less undissociated Acid. The ratio of ##\frac{[A^-]}{[HA]}## falls from ##\frac{1}{1}## to ##\frac{1}{10}##
• I know that if the pH of a solution falls, then it becomes a stronger protonator, thereby distributing more H ions.

I think that somewhere above I am making a mistake, but I do not know what:

1. If a stronger acid protonates more, then this results in less undissociated Acid and more conjugate Base (the Acid distributes Hydrogen ions and converts to its conjugate Base). I would expect that to result in an INCREASED ratio of ##\frac{[A^-]}{[HA]}## i.e. MORE CONJUGATE BASE.
2. In my example above, lowering the pH leads to the ratio of ##\frac{[A^-]}{[HA]}## falling from 1 to 0.1 (##\frac{1}{1}## to ##\frac{1}{10}##), which means the concentration of conjugate Base REDUCES (or the concentration of undissociated Acid INCREASES).

These two statements can not both be correct, but I can not see where I am making an error. Any guidance would be appreciated, thankyou.\pH=pKa+log([A−]/[HA])

 

[Borek]

I know that if the pH of a solution falls, then it becomes a stronger protonator, thereby distributing more H ions.

What becomes the stronger "protonator"? Solution itself?

 

[njh]

Thank you for the question. At a lower pH, the original acid donates more Hydrogen ions (which I think is due to a difference in electronegativity). So the acid is the protonator.

 

[Borek]

Thank you for the question. At a lower pH, the original acid donates more Hydrogen ions

No, you are misunderstanding the scenario. pH doesn't change out pf a blue.

There are two ways of lowering pH of an acid solution. First is to add more acid. Second is to add another acid, preferably a stronger one. In both cases dissociation fraction of the acid will actually go down. In neither case it will become "a stronger protonator", its strength, defined by the Ka, doesn't change at all.

 

[njh]

Thank you, I understand now. Once an acid has reached equilibrium in a solution, it can not donate more Hydrogen ions, even if the concentration of the acid is increased. This is because there is nothing to accept these Hydrogen ions. As a result, the concentration of undissociated acid will increase. The only statement that I can make is that a strong acid is a greater protonator than a weak acid, everything else being equal. Thank you for that help.

 

[Borek]

Once an acid has reached equilibrium in a solution, it can not donate more Hydrogen ions, even if the concentration of the acid is increased.

No, if you add acid, the concentration of H⁺ from the dissociation goes up.

This is because there is nothing to accept these Hydrogen ions.

It is water that accepts these protons (H₃O⁺ being the product).

As a result, the concentration of undissociated acid will increase.

Yes, but so will the concentration of H⁺ and the conjugate base. Two things governing that are mass balance (total analytical concentration of the acid and the conjugate base reflecting amount of acid/base that was put into the solution) and the equilibrium equation.

 

[njh]

Thank you again. I was confusing the concepts of Protonation and Dissociation.

I have looked again at protonation with reference to titration of an amino acid.

I understand that "Protonation" refers to the extent that a molecule has an [H⁺] available to donate. In the case of COOH ↔ COO^-, at a lower pH the COOH is more protonated. This in itsself does not tell me the extent to which the COOH has donated protons via dissociation.

Separately, I understand that adding an acid into a H₂O solution results in the acid's dissociation into its conjugate base, donating an [H⁺] proton to convert H₂O to H₃O⁺. Once the acid has reached equilibrium, it is possible to 'force' the dissociation by increasing the concentration of the original acid, thereby lowering the pH of the solution.

I believe that I now understand the two concepts. Thank you.

 

[Mayhem]

You are not "forcing" the dissociation. Acids dissociate in solution because they react with the solvent. It is that simple.

 

[njh]

Thank you again for the clarification. For what should have been a simple concept, I feel that I made a three course meal of it; for which I appreciate people's effort.