Regarding Buffers

[nothing123]

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Hi guys,

If you have a buffer system set up (let's say a weak acid with its conjugate base) and you dilute it by adding H2O, what happens to the pH? My problem with this question is that if you use the HH equation, pH = pKa + [base-]/[acid], the volume of solution cancels out in the [base-]/[acid] portion of the equation and thus doesn't change the pH. However, if you use pH = -log[H+], which is the standard way of calculating it, adding more volume of solvent would obviously decrease the concentration of H+ and then increase the pH. Can someone explain to me why there is a discrepancy here?

Thanks.

 

[nothing123]

Whoops, sorry for posting it here; mods feel free to move it to Q&A.

 

[BerkReviewTeach]

Hi guys,

If you have a buffer system set up (let's say a weak acid with its conjugate base) and you dilute it by adding H2O, what happens to the pH? My problem with this question is that if you use the HH equation, pH = pKa + [base-]/[acid], the volume of solution cancels out in the [base-]/[acid] portion of the equation and thus doesn't change the pH. However, if you use pH = -log[H+], which is the standard way of calculating it, adding more volume of solvent would obviously decrease the concentration of H+ and then increase the pH. Can someone explain to me why there is a discrepancy here?

Thanks.

You have to use the HH equation here, because you have both an acid AND a base. By considering pH = -log[H+], you are only considering the acid in solution, and that's just half of the story. In a buffer, there is also a substantial amount of base, so you would also need to consider that pOH = -log[OH-]. Dilution lower [OH-] as well as [H3O+], so an approach considering one or the other won't work. You have to use the HH equation.

Consider the following equilibrium:

HA <=> H+ + A-.
Ka = ([A-]/[HA]) x [H+]
Given that Ka is a constant and the ratio of [A-]/[HA] is essentially constant when water is added, the [H+] cannot change, which means that pH also cannot change.

Does that help?

 

[nothing123]

Hi BerkReviewTeach,

The second part of your explanation certainly helps and I can fully relate to that but I was wondering if you could expand a little more on why you can't use the pH = -log[H+]. If I'm following correctly, I think you meant you can still use that equation but since [H+] doesn't change, the pH doesn't change. If you cared to explain why, that would be fantastic!

Thanks.