Why does a buffer require a weak acid and base?

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chiddler

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TBR has a good page explaining this mathematically but I don't understand very well.

They give an example and compare different conditions. This is the first:

Initially A- / HA = 1 / 999. After adding one part OH, then it becomes 2 / 998.

Because Ka = [H+] * [A-] / [HA], and Ka is constant, then [A-] / [HA] doubles, [H+] must be cut in half. pH changes and thus this is not a buffer.

First of all, i'm not understanding what is being said here. Second, "[H+] must be cut in half"; should it be equal to [A-]?

thank you!

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Ka is a constant. In this example, HA is also a constant (999 equals 998). So since A- doubled going from the first solution to the next, the H+ concentration must have been halved from the first solution to the next. It has to, because A- times H+ divided by HA is a constant.

So you added a tiny amount of base and the H+ concentration changed a lot, and therefore the pH also changed by a lot. such extreme sensitivity to a small amount of base is not behaving like a buffer.

If you ran the same scenario with 500/500 and 499/501 you'd see that adding the same amount of base caused the H+ concentration and therefore the pH to hardly budge. A buffer requires a lot of weak acid and it conjugate base.
 
oh so my title question is really irrelevant. they are showing why the concentrations of acid + conjugate should be about the same.

...well maybe not. the answer to my original question is that strong acids and bases cannot hold a solution in such conditions because they dissociate so easily. right?
 
Right. You need a mix of an acid and its base sitting in a happy equilibrium with lots of each, in order to create a buffer. And a strong acid would never have a happy equilibrium with lots of each, it would always shift completely to the A- state.
 
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thanks very much. A tangentially relevant question, please. So i'm looking at the henderson hasselbalch equation:

pH = pKa + log(A-/HA)

When pH = pKa, then A- = HA. So depending on what compound is used, the pH can vary because they all have different pKa's.

A- should also be equal to H+, right? If it is, then when pH = pKa, then A- = HA = H+.

Therefore, under these conditions, pH should be the same regardless of what compound is used because the amount of H+ in solution is fixed: it is equal to A- and equal to HA. But this can't be true because when log(A-/HA) = 0, then the pH depends on the pKa and this varies with the type of compound.
 
A- is only equal to H+ if you start with water and then add HA, and let it dissociate some to find its equilibrium.

Buffers are typically built by mixing a weak acid like acetic acid with its conjugate base, say sodium acetate (sodium ion being a spectator ion), so the H+ concenteation will be unrelated to the A- concentration, except that we know that A- times H+ divided by HA will be a constant.
 
thanks very much. A tangentially relevant question, please. So i'm looking at the henderson hasselbalch equation:

pH = pKa + log(A-/HA)

When pH = pKa, then A- = HA. So depending on what compound is used, the pH can vary because they all have different pKa's.

A- should also be equal to H+, right? If it is, then when pH = pKa, then A- = HA = H+.

Therefore, under these conditions, pH should be the same regardless of what compound is used because the amount of H+ in solution is fixed: it is equal to A- and equal to HA. But this can't be true because when log(A-/HA) = 0, then the pH depends on the pKa and this varies with the type of compound.

Why would this be the case? Recall that [H+] = [HA] for strong acids, which is precisely what we aren't using to create our buffer.

Just my 2 cents. MTHeaded has a much more solid grasp of these gen chem subtleties than I do.
 
Why would this be the case? Recall that [H+] = [HA] for strong acids, which is precisely what we aren't using to create our buffer.

Just my 2 cents. MTHeaded has a much more solid grasp of these gen chem subtleties than I do.

no problemo i appreciate your help regardless lol

The reason I would think this is the case is because when HA dissociates, it forms H+ and A- so they should be equi..SHOULD NOT BE EQUIVALENT because we're adding our own A- to solution!!

ohhh....ok!

thanks :D
 
Need more help :(. My previous conclusion was that there must roughly equal amounts of acid and base for a buffer to form because if use strongs, then pH changes too much with introduction of disturbance.

Look at this graph of HCl + NaOH

sasb2.gif


I'm looking at the beginning of the titration. pH doesn't change very much despite addition of strong base. So why is this not a good buffer system?

Comparing to weak/weak (ethanoic acid + ammonia):

wawb.gif


where this change is more continuous rather than bursty.
 
See if you can make the graph for acetic acid + its conjugate base and then start adding ammonia to that. It will be a good start, I can post more later when not on a phone.
 
See if you can make the graph for acetic acid + its conjugate base and then start adding ammonia to that. It will be a good start, I can post more later when not on a phone.

I can imagine the graph will start out at some pH that is +-1 from pKa of acetic acid. Adding ammonia will not cause it to change much, so line will be straight more or less.
 
You need a mix of an acid and its base sitting in a happy equilibrium with lots of each, in order to create a buffer.
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I can imagine the graph will start out at some pH that is +-1 from pKa of acetic acid. Adding ammonia will not cause it to change much, so line will be straight more or less.

Yes, it's very much like the graph for the titration of the strong acid when you look at it sideways - steep ends and a fairly flat middle part. Buffers rely on having the [HA]<->[H]+[A] established at fairly high (and similar) concentrations of [HA] and [A]. When you add strong acid or base, the equilibrium gets pushed to the left or the right. Since the amount of added base/acid is much smaller than the amount of the buffer, you end up with a fairly small change in [A] and [HA], which makes your new [H] = Ka*([HA]+&#916;)/([A]-&#916;) very similar to the old [H].

This does not work well when trying a strong acid as a buffer, because the acid dissociates completely and when you add more [OH] there is nothing to push the equilibrium back. The lower the concentration of the strong acid is, the worse that behavior becomes. You can see that in the middle of your titration graph for the strong acid - minimal additions of base lead to huge changes of pH.

And if you want to be really, really strict, high concentrations of strong acid can be somewhat close in resistance to pH changes to a buffer. But that's true only for very low pH which makes it less interesting than the buffers of weak bases/acids.
 
It doesn't necessarily have to be concentrated acid, does it? The graph I linked has some buffer activity when NaOH is added but only in small amounts. Though what you said is true; it's not very useful!
 
It doesn't necessarily have to be concentrated acid, does it? The graph I linked has some buffer activity when NaOH is added but only in small amounts. Though what you said is true; it's not very useful!

If you're adding something different from strong acid, you'll be affecting the pH even slower. :confused:

Adding strong base works in the same way but pushes the equilibrium the other way.

Sorry if that's not helpful. Could you re-phrase your question somehow? I'm not sure what else to would be helpful right now.
 
I'm sorry I wasn't clear.

You said that you can create a buffer system using concentrated acid. I'm questioning if "concentrated" is a required condition. The reason i'm questioning it is because if you look at the titration graph of HCl + NaOH that I linked, adding base does not immediately change the pH very much which looks like buffering activity!

A buffer doesn't change pH very much when disturbance is added, right? Well that is characteristic of this graph. So why isn't this considered a buffer system?
 
I'm sorry I wasn't clear.

You said that you can create a buffer system using concentrated acid. I'm questioning if "concentrated" is a required condition. The reason i'm questioning it is because if you look at the titration graph of HCl + NaOH that I linked, adding base does not immediately change the pH very much which looks like buffering activity!

I should not have called it a buffer, it only made things more confusing. Yes, concentrated is required in this buffer-like case. What you're doing here is dumping a huge amount of H+ and saying that you have to add a lot of OH-/H+ to make any noticeable difference in pH.

Using a low concentration of strong acid will make it easier to change the pH since you'll need less OH- to do that. In your graph, that would be like pulling the graph up and moving it to the left, with the area of somewhat stable pH becoming even smaller.

What you are doing with a real buffer is providing a mechanism which can store and release H+ (within some limits) to compensate for the changes that you make to the solution.

A buffer doesn't change pH very much when disturbance is added, right? Well that is characteristic of this graph. So why isn't this considered a buffer system?
Because the strong acid solution will change pH faster (with smaller disturbance) than a traditional buffer.

For each OH- that you add to the strong acid "buffer" you are going to lose one H+.

If you add 10 OH- to your traditional buffer, the equilibrium of [HA]<->[H]+[A] can easily lead to 8 or 9 or even 10 H+ being dissociated and the solution returning to your original pH. In this case you have only 1:10 "loss" of H+ vs OH-. (numbers are made up, real buffers may perform better or worse)
 
Why is this buffering region there to begin with? Why doesn't the pH drop immediately when base is added to strong acid? I think that it's because there are undissociated HCl's, as few as they are, in solution, yes?
 
Why is this buffering region there to begin with? Why doesn't the pH drop immediately when base is added to strong acid? I think that it's because there are undissociated HCl's, as few as they are, in solution, yes?

It's not a buffering region. The curve there is still steeper than what you would get in the buffering region of a real buffer. The pH starts changing immediately after you start adding NaOH.
 
Because pH is logarithmic, so the first part of the graph is exponential.

By the time you have added half as much strong base as original strong acid, your pH will have budged by only log 2 = 0.3
 
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