[Borek]

Do you agree that it's dynamic equilibrium and the two reactions happening inside the solution?

I can list even a few more. It doesn't chaneg anything.

Yes, pH=2 means 0.01 M H⁺.

Good, finally we are coming somewhere. Do you agree that to get the pH to 7 you need to neutralize almost all 0.01 M of H⁺?

It doesn't matter where it is from.

Yes and no. It doesn't matter from which acid it comes, but it has to come from somewhere. See below.

Do you agree that it could be changed dramatically when at the edge of the buffer limit?

What "it" could change? At the buffer limit buffer capacity is low, but at low and high pH it grows back. See http://www.chembuddy.com/?left=pH-calculation&right=pH-buffer-capacity

You are still ignoring the crucial part: where does the 0.01 M of H⁺ come from? We agreed it IS there, didn't we?

See if you can use pKa 8.3 acid to design a pH 2.0 buffer. 1M solution of such an acid has pH of 4.15. What are you going to do to lower pH to 2.0?

[DanChen2014]

Of course, in order to make pH=7, we need to make [H⁺]=10^-7.
But "the crucial part" is what neutralize H⁺ ?  Should all the OH^- needed  come from NaOH ?

[DanChen2014]

See if you can use pKa 8.3 acid to design a pH 2.0 buffer. 1M solution of such an acid has pH of 4.15. What are you going to do to lower pH to 2.0?

I will ignore this kind of question, you have the answer already. Show your answer or show your process.

[Borek]

Should all the OH^- needed  come from NaOH ?

Yes.

I will ignore this kind of question, you have the answer already. Show your answer or show your process.

End of thread for me. You don't want to learn, you are here just to troll.

Despite of what you are trying to prove you don't understand the acid-base equilibrium, and with this attitude you won't learn it ever. So be it.

[DanChen2014]

Should all the OH^- needed  come from NaOH ?

Yes.

The answer is no. This is "the crucial part".

Anyway, I don't need to prove anything.
Keep open mind.
You can be wrong even you think you are an expert. At least, I can't learn anything from you in this topic except English.

[Durlag]

I meant the pH of the 0,1M buffer (pKa = 8,3). Not adjusted.

No such thing as "0.1 M buffer, pKa=8.3". Please elaborate.

I meant the pH of the 1L solution that contains 0,1M of the species that has pK=8,3. It's said exactly like that in the wording of the problem, "one liter of a 0,1M buffer (pKa = 8,3)".
What pH would you find for this solution? Understanding this could solve a lot of my confusion. Because I find different solutions (with my intuition telling me that the correct answer is pH 10,65).

EDIT: actually now I think more that it's pH 4,65.

By the way, with a pKa like this, shouldn't we call the couple HB⁺/B instead of HA/A^-?

H₂PO₄^- has pKa of 7.2, would you name it HA^- or HB⁺?

Got it.

I also did the question c).
I found pH = 2,26 and at this pH, 9,12.10^-5 % of HA is dissociated.
I have the intuition that the OH^- first reacts with H⁺ to produce H₂O, but I don't really understand why.

Order doesn't matter. H⁺ is definitely the stronger acid of the two present, so it is easier to think in terms of it reacting first. But even if, it is immediately followed by the HA dissociation and we are back to the H⁺ present (just in smaller amounts).

Technically it is equivalent to question whether neutralization of a weak acid is a one step reaction:

HA + OH^-  A^- + H₂O

or two step reaction:

HA  H⁺ + OH^-

H⁺ + OH^-  H₂O

I don't know the answer, and the exact mechanism doesn't matter for the final equilibrium.

I think I got it. The H⁺ from the strong acid reacts with OH^- to form water, and the HA dissociation happens simultaneously. When all the H⁺ has reacted though, at pH 2,26, the HA dissociation is very small.

Did you mean

HA  H⁺ + A^-

H⁺ + OH^-  H₂O

?

Also, maybe this is a stupid question, but could this happen:

HA + H₂O  A^- + H₃O⁺

?

Or maybe it doesn't happen in our case because there is OH- in solution, and pKa H₂O/OH^- is superior to pKa H₃O⁺/H₂O, so OH^- wins over the water?

[Borek]

I meant the pH of the 1L solution that contains 0,1M of the species that has pK=8,3. It's said exactly like that in the wording of the problem, "one liter of a 0,1M buffer (pKa = 8,3)".
What pH would you find for this solution? Understanding this could solve a lot of my confusion. Because I find different solutions (with my intuition telling me that the correct answer is pH 10,65).

You can't tell anything about the buffer pH without further information. All you know is that [HA]+[A^-]=0.1 M and what is the pKa value, but it is not enough information to tell what is the solution pH. All you know is that the pH was adjusted later to 2.0.

I also did the question c).
I found pH = 2,26 and at this pH, 9,12.10^-5 % of HA is dissociated.
I have the intuition that the OH^- first reacts with H⁺ to produce H₂O, but I don't really understand why.

Order doesn't matter. H⁺ is definitely the stronger acid of the two present, so it is easier to think in terms of it reacting first. But even if, it is immediately followed by the HA dissociation and we are back to the H⁺ present (just in smaller amounts).

Technically it is equivalent to question whether neutralization of a weak acid is a one step reaction:

HA + OH^-  A^- + H₂O

or two step reaction:

HA  H⁺ + OH^-

H⁺ + OH^-  H₂O

I don't know the answer, and the exact mechanism doesn't matter for the final equilibrium.

I think I got it. The H⁺ from the strong acid reacts with OH^- to form water, and the HA dissociation happens simultaneously. When all the H⁺ has reacted though, at pH 2,26, the HA dissociation is very small.

Note: once HA dissociated, H⁺ produced from HA dissociation becomes part of the general H⁺ pool in the solution. You can't tell "this H⁺ comes from the strong acid, and this H⁺ comes from the weak acid". Besides, at equilibrium both dissociation and protonation processes still occur, so even if you were able to mark H⁺ somehow, you will see they exchange their positions all the time.

Did you mean

HA  H⁺ + A^-

H⁺ + OH^-  H₂O

Yes, sorry about that, a careless mistake. Corrected.

Also, maybe this is a stupid question, but could this happen:

HA + H₂O  A^- + H₃O⁺

Or maybe it doesn't happen in our case because there is OH- in solution, and pKa H₂O/OH^- is superior to pKa H₃O⁺/H₂O, so OH^- wins over the water?

This is a matter of convention, I use H⁺ as a shortcut to H₃O⁺.

Note that in reality what is present in the solution is neither H⁺ nor H₃O⁺, there is an ongoing equilibrium with several even larger cations present (of the (H₂O)_nH⁺ form - so H₃O⁺, H₅O₂⁺, H₇O₃⁺ and so on, think about them as solvated proton).

[Durlag]

I meant the pH of the 1L solution that contains 0,1M of the species that has pK=8,3. It's said exactly like that in the wording of the problem, "one liter of a 0,1M buffer (pKa = 8,3)".
What pH would you find for this solution? Understanding this could solve a lot of my confusion. Because I find different solutions (with my intuition telling me that the correct answer is pH 10,65).

You can't tell anything about the buffer pH without further information. All you know is that [HA]+[A^-]=0.1 M and what is the pKa value, but it is not enough information to tell what is the solution pH. All you know is that the pH was adjusted later to 2.0.

But why? What is missing? Couldn't you do an ICE table and calculate [H⁺] at equilibrium from the Ka?

Note: once HA dissociated, H+ produced from HA dissociation becomes part of the general H+ pool in the solution. You can't tell "this H+ comes from the strong acid, and this H+ comes from the weak acid". Besides, at equilibrium both dissociation and protonation processes still occur, so even if you were able to mark H+ somehow, you will see they exchange their positions all the time.

Got it.

Also, maybe this is a stupid question, but could this happen:

HA + H2O  A- + H3O+

Or maybe it doesn't happen in our case because there is OH- in solution, and pKa H2O/OH- is superior to pKa H3O+/H2O, so OH- wins over the water?

This is a matter of convention, I use H+ as a shortcut to H3O+.

Note that in reality what is present in the solution is neither H+ nor H3O+, there is an ongoing equilibrium with several even larger cations present (of the (H2O)nH+ form - so H3O+, H5O2+, H7O3+ and so on, think about them as solvated proton).

Ok. I understand that writing HA + H2O  A- + H3O+
is equivalent to writing HA  H+ + A-

However, in the case of this problem, it seems that the free H⁺ goes preferentially to the OH^- brought by NaOH than to H₂O (or H₄O₂ etc.). Is this right? Is it pointless, since pH=-log[H⁺]=-log[H₃O⁺]?

[Borek]

But why? What is missing? Couldn't you do an ICE table and calculate [H⁺] at equilibrium from the Ka?

You can do it for an acid solution, but you are told "buffer". That means part of the acid is already neutralized - but you are not told what part.

Is it pointless, since pH=-log[H⁺]=-log[H₃O⁺]?

Yes, its a moot.

[Durlag]

But why? What is missing? Couldn't you do an ICE table and calculate [H⁺] at equilibrium from the Ka?

You can do it for an acid solution, but you are told "buffer". That means part of the acid is already neutralized - but you are not told what part.

Thanks. That helps.

Yes, its a moot.

Ok.

To summarize, what happens when you add the NaOH, there is a pool of free H⁺, that is caused by the dissociation of the strong acid and of HA. When all the OH^- has reacted, the pH stabilizes.

You could write what is happening for HA as (one step):
HA + OH^-  A^- + H₂O
Which is equivalent to writing (2 steps):
HA  H⁺ + A^-
H⁺ + OH^-  H₂O (the H+ can also come from the strong acid, it's the same pool of H⁺).

You could also consider that the following reaction is happenning:
HA + H₂O  A^- + H₃O⁺
But that's the same as
HA  H⁺ + A^-
Because you could consider that the H⁺ is taken from H₃O⁺ to react like this:
H⁺ + OH^-  H₂O
Or that H₃O⁺ reacts directly with water:
H₃O⁺ + OH^-  2H₂O

Does that make sense?

I know i have a big tendency to complicate stuff and that's why I get confused. In the begenning I was confused because for me, consuming HO- to create H2O and consuming H2O to create H3O+ had different effects on pH. Now I feel that it has the same effect on pH.
Kw=[H+][HO-] and Kw is a constant, so if one changes, the other changes in the "opposite direction".

[Borek]

Does that make sense?

Nothing cries out loud "I am wrong!"

I know i have a big tendency to complicate stuff

Hard to deny. But learning how to simplify things without ignoring something important is doable.

[Durlag]

Does that make sense?

Nothing cries out loud "I am wrong!"

Haha. Does that mean it's wrong?

Hard to deny. But learning how to simplify things without ignoring something important is doable.

I hope I can learn to do that with chemistry. It's just that if I don't understand everything I will have trouble remembering. I can learn by heart without understanding but it will get out of my brain very quicly.

[Borek]

Haha. Does that mean it's wrong?

No. It is just a little bit convoluted which means you are probably still able to confuse yourself with all these equations.

[Durlag]

True. Thanks for your answer.

[biobufferguy]

I didn't expect to create such a dialogue when I posted the question.

My professor responded to my email by reiterating the textbook's approach and their solution of pH = 7. At the next class she asked if I was clear on the answer and I said that I still didn't understand how that could be correct. I asked if I could write my approach on the board so I stood up in front of the class and explained how you need to find the new equilibrium and that it would lead to a pH much lower than 7. Unfortunately no one in my class backed me up and the professor continued to push the HH approach. I started to think she was right. Another student had a newer edition of the textbook so we looked up the problem to see if they had a different solution in the new edition and it STILL said pH = 7, so then I really started to think I was crazy.

I put the approach in an Excel file and tried to explain it a couple different ways and emailed it to my professor, who was probably super annoyed with my by this point. But she emailed back and said that the calculation looks correct and pH should be around 2.26.