Max Buffering Capacity w/ 0.6 mol HAc & NaOH

[brake4country]

Homework Statement

0.6 moles of pure acetic acid (no salt) is added to water. How much NaOH would have to be added to get the solution to maximum buffering capacity?

Homework Equations

pH=pka + log [base]/[acid[

The Attempt at a Solution

HAc <---> H+ + Ac-
0.6 mol of H+
0.6 mol of Ac-

All that I know is that if NaOH is added, it will reduce the H+ concentration, making the solution less acidic. I am having problems with the calculations. Thanks in advance.

 

[Borek]

When does the buffer has its maximum capacity?

 

[brake4country]

When the pH = pKa

 

[Borek]

When the pH = pKa

Good.

And do you know when the buffer has pH=pKa?

 

[brake4country]

Yes. For acetic acid, that would be 4.76. I think my confusion is based on the question. I know that I need to find the ratio of [salt]/[acid] using the HH equation but I don't know what to do with the NaOH.

 

[Borek]

And what happens when you mix NaOH with acetic acid?

 

[brake4country]

It reduces the amount of H+ ions, increasing the pH.

 

[Borek]

It reduces the amount of H+ ions, increasing the pH.

While it is not false, it is - at best - only a half true.

How would you make sodium acetate?

 

[brake4country]

By adding a base like NaOH.

 

[Borek]

And you still don't know what NaOH does?

 

[epenguin]

Yes. For acetic acid, that would be 4.76. I think my confusion is based on the question. I know that I need to find the ratio of [salt]/[acid] using the HH equation but I don't know what to do with the NaOH.

Well if you substituted pH = pH in the HH equation (2. in your #1) that would tell you (or remind you of) something.

 

[brake4country]

And you still don't know what NaOH does?

Not sure if I was clear. I am new to the HH equation, however, I do know that 4.76 = 4.76 + log [salt]/[acid], which would mean that 0 = log [salt]/[acid]. To reverse the log, raise each side to base 10, which would give me:

10^0 = [salt]/[acid]
1 = [salt]/[acid]

This is the point where I don't know how to incorporate the NaOH. The correct answer is 0.2 moles but I don't know how to get that. Am I on the right track?

 

[brake4country]

Well if you substituted pH = pH in the HH equation (2. in your #1) that would tell you (or remind you of) something.

Yes, I responded to Borek: 0 = log + [salt]/[acid]
1 = [salt]/[acid], which would mean that we would have equal concentrations of salt and acid at maximum buffering capacity. I just don't know how to incorporate NaOH, for which the correct answer is 0.2 moles.

 

[epenguin]

0.2 moles is not the correct answer so don't try to make the answer come out that, hope that helps.

Your ratio 1 is right, key point.

Also in your HH eq you had base and acid, not salt and acid (though you could say... in a manner of speaking.)

 

[brake4country]

I am terribly sorry, the answer from the homework was 0.3 moles. That is why I was confused, that 0.2 moles was a typo. So, is the problem that simple? I thought that if we started with 0.6 moles of acid, then the acid would dissociate (not completely) to 0.6 moles of salt and 0.6 moles of H+. How is the correct answer 0.3 moles?

 

[Borek]

I tried to point you in the right direction, but apparently it didn't work. You are completely ignoring the fact NaOH neutralizes the acid producing the salt.

 

[epenguin]

Well within the problem there is a mechanical part of the calculation that happens to be that simple, I wouldn't say the overall problem is simple as that. It will be far less simple though than it is If you have misconceptions like 'acid dissociates into salt'. Concentrated weak acid like acetic into dissociates to a small extent into H⁺ and equal number of (negatively charged) acetate ions.

Then when you put into that some strong base, NaOH (Na⁺ and OH^-) what is going to happen?

For your question why Is the buffering capacity greatest when pH = pK I would say, first get clear the definition of buffering capacity d(base added)/d(pH), Secondly look in your book at a titration curve of a weak acid to convince yourself that is, then thirdly you can calculate uffering capacity from that definition and the HH equation if you're up to that kind of calculation (if not you can find it in plenty of places if you google "buffering capacity").

Now that you have seen that some questions are simpler than you thought the next thing you should do is to go back to your text and go through the subject again, because it is not profitable to discuss and explain when you have misconceptions and confusions like these.

 

[brake4country]

I tried to point you in the right direction, but apparently it didn't work. You are completely ignoring the fact NaOH neutralizes the acid producing the salt.

Conceptually I understand it but I am having difficulty with the calculations.

 

[Borek]

Conceptually I understand it but I am having difficulty with the calculations.

Just follow the most obvious stoichiometry, that's enough here. Start by writing the neutralization reaction.

 

[brake4country]

I was confused about this problem because my professor did not do a problem with us like this, but I used an ICE table to figure it out.

HOAc + OH- ------> H₂O + OAc-
I 0.6 x 0
C -x -x +x
E 0.6-x 0 x

pH = pKa + log [OAc-]/[HOAc]
4.76 = 4.76 + log x/0.6-x
0 = log x/0.6-x
1=x/0.6-x
0.6-x = x
2x=0.6
x=0.3 mol NaOH

I got the right answer I just need to confirm that my calculations are sound. Thank you.

 

[epenguin]

It looks to be OK. Maximum buffering power is when pH = pK. You could make a bufferwith pH = pK mixing an equal number of moles of acetic acid and of sodium acetate, according to your. You could say here that you are making this sodium acetate by adding NaOH to acetic acid. pH = pK when [salt] = [acid], or when when [added base] = [remaining Acid form] i.e. [Na⁺] = [CH₃CO₂^-] As you already know ufound in #12. So it is immediate that the added base is just half the total initial acid molarity. PS I believe you would find things easier to understand and give the impression of understanding if you framed more in terms of actual chemical species present and their concentrations (rather than 'acid', 'salt'), here HAc, Ac^-, Na⁺ - the major species, H⁺ a minor but key species and object of the excercise, and OH^- which you can justify leaving out of equations here but you have to be clear about its role.

 

[Borek]

ICE table is an overkill here. As epenguin wrote above, all you need is to know you have to neutralize exactly half of the acid, then it is trivial stoichiometry.

See some more detailed discussion of buffer calculations here:

http://www.chembuddy.com/?left=buffers&right=toc

(together with an example of how ICE table doesn't produce better results that will justify its use).

 

[brake4country]

Right, because of 1 to 1 ratio. Thank you.

 

[Borek]

For other ratios it would be still a trivial stoichiometry.

There are limitations to this approach, but they start to be important outside of the typical/practical application of buffers.