[DanChen2014]

Please show your calculations: how much NaOH has to be added to the original solution to change pH to 7.0.

I have shown my answer, and I explained how I got it. So far you are arguing I am wrong, but apart from waving your hands and suggesting I am wrong you have not presented a single numerical argument. Do the calculations and we will start from there.

To make the original solution to pH=7.0, we need 1.67ml 3M NaOH. To make it clear, I do it in this way.
Kw=K_aK_b, here I assume Kw=14 and pK_a was given as 8.3. Assume we need x ml 3M NaOH

     A^-+ H₂O HA + OH^- K_b=10^-5.7

3x/1000         0.1+10^-7    10^-7   

 (0.1+10^-7)*10^-7/(3x/1000)= 10^-5.7
x=1.67 ml

You can get this through the other reaction:
  HA+ H₂O A^- + H₃O⁺        pK_a=8.3
You will have the same result. you know why? Both these reactions happened in the solution, when pH changed, one dominates over the other. It's all about the equilibrium among [HA], [A^-], [OH^-] AND [H₃O⁺]. You can calculate through different ways but only have one result.

I hope this will be helpful and not against the forum's rule. It's not an argue.

[Borek]

To make the original solution to pH=7.0, we need 1.67ml 3M NaOH

OK, now answer my earlier question: Imagine you have 1 L of the pH 2 solution without HA - just a strong acid. How many mL of the 3M NaOH do you have to add to change pH to 7?

[DanChen2014]

To make the original solution to pH=7.0, we need 1.67ml 3M NaOH

OK, now answer my earlier question: Imagine you have 1 L of the pH 2 solution without HA - just a strong acid. How many mL of the 3M NaOH do you have to add to change pH to 7?

You already have the answer, I will not answer this kind of question. If you can not have it, show your process.

[DanChen2014]

For this thread, I answered all I thought it should be. It's clear enough, about buffer, I don't know anything more than this.

[Borek]

Nothing is clear, you are still missing the big picture, unfortunately, you are refusing to try to follow the path that would lead you to the correct conclusion.

pH 2.0 solution is 0.01 M in H⁺. Let's assume for a moment solution doesn't contain the HA/A^- buffer. That means to bring the solution to pH 7 we need to neutralize 0.01 moles of H⁺. As the reaction is

H⁺ + OH^- H₂O

we need 0.01 moles of NaOH, or 3.3 mL of 3M NaOH.

Oops, there is a problem - you calculated 1.67 mL of the NaOH solution is enough, yet the solution WITHOUT buffer needs more NaOH than that to be neutralized. Something is definitely wrong.

At pH 2 weak acid that is part of the buffer is fully protonated, so when bringing the buffer to pH 7 you will need even more NaOH than just 3.3 mL required for the solution without buffer, after all, it contains more acid than the solution without buffer, doesn't it?

Why the difference? Because you are blindly looking at the buffer presence, ignoring the fact that to bring the buffer to pH 2.0 you had to add ANOTHER acid. What you have at pH 2 is not just a buffer solution, it is a solution containing both buffer and another acid that was used to bring pH down to 2.0.

[Durlag]

b) I calculate pH = 1.83

How did you get this answer? I did the calculations and I don't find that, so i'd like to know how you did it. 

[Borek]

I don't know how he did it, but I got 1.84 - so the result looks OK.

Show what you got and how and we will see where the problem is.

[Durlag]

Well at first I misunderstood the wording for question b). I thought that the 1,5 mL of 3M HCl was added to 1L of the buffer, but the original buffer, not the one that had been adjusted to pH 2.
By the way, for this I found the original pH of the buffer to be 10,65 (assuming that the pK given was a pKa and thus that it was the pKa of a weak base, and that this weak base would take a proton from water and generate OH^-). After adding the 1,5 mL 3M HCl I found the new pH to be 9,63. I don't know if this is right, and if you're interested I can show you how I did this.

So now to the real question b): after correction I find 1,84 as well!
Indeed, in solution we have the H⁺ brought by the strong acid that was used to bring the pH of the buffer to pH 2. That's 0.01 mol of H⁺. Then you add the HCl, and that brings 4,5.10^-3 mol of H⁺. That's a total of 0.0145 mol of H⁺, that you divide by 1,0015 L, you take the log of this, and that gives pH 1,84.

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.
I also got the intuition that it's like the tug-of-war bewteen two bases for a proton, except that this time it's a tug-of-war between two acids for an OH-. In the case of the tug-of-war between the 2 bases, the base with the higher pKa wins. In this case, the acid with the lower pKa wins. And that's the strong acid that was used to bring pH down to 2. So it "takes" the OH^-, or more accurately gives its H⁺ to OH^-. Is this right?
Also, when first thinking about the reaction between HA and NaOH, I was confused wether:
HA + OH^-  A^- + H₂O or
HA + H₂O  A^- + H3O⁺

[DanChen2014]

Nothing is clear, you are still missing the big picture, unfortunately, you are refusing to try to follow the path that would lead you to the correct conclusion.

pH 2.0 solution is 0.01 M in H⁺. Let's assume for a moment solution doesn't contain the HA/A^- buffer. That means to bring the solution to pH 7 we need to neutralize 0.01 moles of H⁺. As the reaction is

H⁺ + OH^- H₂O

we need 0.01 moles of NaOH, or 3.3 mL of 3M NaOH.

Oops, there is a problem - you calculated 1.67 mL of the NaOH solution is enough, yet the solution WITHOUT buffer needs more NaOH than that to be neutralized. Something is definitely wrong.

At pH 2 weak acid that is part of the buffer is fully protonated, so when bringing the buffer to pH 7 you will need even more NaOH than just 3.3 mL required for the solution without buffer, after all, it contains more acid than the solution without buffer, doesn't it?

Weak acid() but need less NaOH? It's may not be weak acid.
Do the experiment and give an answer.

[Borek]

By the way, for this I found the original pH of the buffer to be 10,65 (assuming that the pK given was a pKa and thus that it was the pKa of a weak base, and that this weak base would take a proton from water and generate OH^-). After adding the 1,5 mL 3M HCl I found the new pH to be 9,63. I don't know if this is right, and if you're interested I can show you how I did this.

No idea what you are talking about. No such thing as "original buffer". Unless you mean just a salt (conjugate base) solution.

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⁺ + A^- // edit: corrected a careless mistake

H⁺ + OH^-  H₂O

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

[Durlag]

No idea what you are talking about. No such thing as "original buffer". Unless you mean just a salt (conjugate base) solution.

I meant the pH of the 0,1M buffer (pKa = 8,3). Not adjusted.
By the way, with a pKa like this, shouldn't we call the couple HB⁺/B instead of HA/A^-?

[Borek]

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.

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⁺?

[DanChen2014]

we need 0.01 moles of NaOH, or 3.3 mL of 3M NaOH.

Oops, there is a problem - you calculated 1.67 mL of the NaOH solution is enough, yet the solution WITHOUT buffer needs more NaOH than that to be neutralized. Something is definitely wrong.

At pH 2 weak acid that is part of the buffer is fully protonated, so when bringing the buffer to pH 7 you will need even more NaOH than just 3.3 mL required for the solution without buffer, after all, it contains more acid than the solution without buffer, doesn't it?

Why the difference? Because you are blindly looking at the buffer presence, ignoring the fact that to bring the buffer to pH 2.0 you had to add ANOTHER acid. What you have at pH 2 is not just a buffer solution, it is a solution containing both buffer and another acid that was used to bring pH down to 2.0.

Yes, for strong acid you need 3.3ml 3M NaOH. But here is not strong acid. Why is it different? Because, when pH raised, A^- start to dominate. why do we call A^- conjugate base? Because it will produce OH^- from water, and act as base.

Please reconsider what I said before. Think about the reaction happening in the solution
   
 A^-+ H₂O  HA + OH^- Kb=10^-5.7
 HA+ H₂O  A^- + H₃O⁺        pKa=8.3

"Both these reactions happened in the solution, when pH changed, one dominates over the other. It's all about the equilibrium among [HA], [A-], [OH-] AND [H3O+]. You can calculate through different ways but only have one result."

I didn't follow your question, because that is a kind of misleading, not to "correct conclusion".

What's inside the buffer solution?
What's titration curve look like for weak acid? For H₃PO₄ ?
How to make a buffer solution?

All these information is quite open and you can find it anywhere through the internet. MIT and UC Berkley have the open course for this one.

[Borek]

Yes, for strong acid you need 3.3ml 3M NaOH. But here is not strong acid.

So, how did the pH got down to 2.0?

Do you agree that pH 2 solution contains - by definition - 0.01 M H⁺?

Where did it come from?

[DanChen2014]

So, how did the pH got down to 2.0?

Do you agree that pH 2 solution contains - by definition - 0.01 M H⁺?

Where did it come from?

Do you agree that it's dynamic equilibrium and the two reactions happening inside the solution?
Yes, pH=2 means 0.01 M H⁺. It doesn't matter where it is from. Do you agree that it could be changed dramatically when at the edge of the buffer limit?