[biobufferguy]

Freshening up on acid/base chemistry and came across this problem.
I can't remember how to work this. Help?

One liter of a 0.1M buffer (pK=8.3) is prepared and adjusted to a pH of 2.0. (yes, you read that correctly)

a)What are the concentrations of the conjugate base and weak acid at this pH?
b)What is the pH when 1.5mL of 3.0M HCl is added to 1.0L of the buffer?
c) What is the pH when 1.5mL of 3.0M NaOH is added to 1.0L of the buffer?

a) I calculate [HA] = 0.1M and [A-] = 5 x 10^-8M (so essentially no weak base available to buffer any acid)
b) I calculate pH = 1.83
c) Using HH equation the pH would be about 7, but I can't make sense of that. It seems like it would still be acidic. I think I am missing something.

Thank you!

[Borek]

c) Using HH equation the pH would be about 7, but I can't make sense of that. It seems like it would still be acidic. I think I am missing something.

Your intuition seems to be right. Show what you did.

[biobufferguy]

c) Using HH equation the pH would be about 7, but I can't make sense of that. It seems like it would still be acidic. I think I am missing something.

Your intuition seems to be right. Show what you did.

Assumed [HA] = 0.1M therefore there is 0.1 mol of HA.  0.0045 mol of OH- is added.  Assuming the OH- reacts with the HA, now [HA] = 0.0955M and [A-] = 0.0045. Plug that into HH and you get pH = 7.

However that doesn't take into account that you start with an [H+] of 0.01M. I can't figure out how that comes into play, if at all. Thanks for your help...

[biobufferguy]

Full disclosure, this is not a homework problem. This is a review problem for a biochem class and the answer given by the instructor is pH = 7. I am convinced that's incorrect and I have not been able to convince my teacher and another classmate, so I am desperately wondering if I'm crazy. Thanks.

If I assume all of the -OH is condensed to H2O, then there is a drop in [H+] from 0.1 to 0.055 but that only brings pH up to ~2.26.

[Borek]

You are not crazy, 2.26 is the correct answer, and your argument (about using all OH^- and neutralizing H⁺ only partially) is perfectly valid.

To get to pH 7 they would need to

1. Neutralize 0.01 M of H⁺ (actually 0.01-10^-7)
2. Neutralize part of the HA (at pH 7 concentration of A^- is around 0.005M)

At 1L that means 0.01 moles of NaOH + 0.05 moles of NaOH = 0.015 moles of NaOH - 5 mL of 3M solution, not 1.5 mL.

[DanChen2014]

Think about the reaction happening here, the NaOH is strong base, so it will react with HA quickly and lead to the [A^-] raised, a new equilibrium will be built up.Hence, we have:

 pH=pK+log[A^-]/[HA]=8.3+log(0.0045/(0.1-0.0045))=6.97

[DanChen2014]

You are not crazy, 2.26 is the correct answer, and your argument (about using all OH^- and neutralizing H⁺ only partially) is perfectly valid.

To get to pH 7 they would need to

1. Neutralize 0.01 M of H⁺ (actually 0.01-10^-7)
2. Neutralize part of the HA (at pH 7 concentration of A^- is around 0.005M)

At 1L that means 0.01 moles of NaOH + 0.05 moles of NaOH = 0.015 moles of NaOH - 5 mL of 3M solution, not 1.5 mL.

This is not right. It's dynamic equilibrium. OH^- will change into water and cause more A^- in the solution, as a result it will change the equilibrium of the reaction happened in the buffer.
PH is a log value, 0.0045M is big enough to change it dramatically.
Think about why buffer is buffer. After reaching the lower limit, the pH will lower dramatically with more acid adding in. You can try to find the range of the buffer. A small amount of NaOH is enough to bring it back.

[Borek]

This is not right. It's dynamic equilibrium. OH^- will change into water and cause more A^- in the solution, as a result it will change the equilibrium of the reaction happened in the buffer.

Calculate first, comment later.

Yes, this is kind of an approximation, but it works perfectly well here.

[Borek]

Think about the reaction happening here, the NaOH is strong base, so it will react with HA quickly and lead to the [A^-] raised, a new equilibrium will be built up.

And what about 0.01M H⁺ that was present at pH 2? Did it miraculously disappear?  No, it has to be neutralized as well. Actually it will be neutralized first.

[DanChen2014]

And what about 0.01M H⁺ that was present at pH 2? Did it miraculously disappear?  No, it has to be neutralized as well. Actually it will be neutralized first.

If it's neutralized what happen to the reaction? It will shift to the right and more [A^-] present. It doesn't matter where the [H⁺] from.

[Borek]

It doesn't matter where it comes from, but it has to be neutralized nonetheless.

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]

Actually, the reaction cause the dramatically raise of PH is

A^-+H₂O HA+OH^-

[Borek]

Actually, the reaction cause the dramatically raise of PH is

A^-+H₂O HA+OH^-

No, the only reaction that changes pH is the one that changes concentration of H⁺. That's by definition of pH.

You have ignored my 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]

Actually, the reaction cause the dramatically raise of PH is

A^-+H₂O HA+OH^-

No, the only reaction that changes pH is the one that changes concentration of H⁺. That's by definition of pH.

You have ignored my 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 need to check what makes buffer have the ability to stable the pH and what really happen in there.

[Borek]

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.