Question

http://openstudy.com/study#/updates/4f5a784ae4b0636d89060452

Answer 1

That's pretty long... What don't you understand exactly?

Answer 2

Basically...

If you place the conjugate base of a weak acid into a solution, why will the conjugate base not just grab a bunch of protons and make more acid? Why does it produce a buffer with similar concentrations of weak acid and conjugate base?

Answer 3

Well, the weak acids and weak bases are constantly reacting and forming each other, but because they are both "weak" an equilibrium is formed that maintains their concentrations.

Answer 4

But since the equilibrium concentration of weak acid HA and the product of the equilibrium concentrations of H3O and A- must remain the same, this means that if you stuff a bunch of conjugate base into the solution, if it's not converting mostly back into HA, then the H3O concentration must decrease significantly, yes? How is that occurring? What is grabbing so many protons out of the water?

Answer 5

I don't really get what your saying.

Lets say you have [HA] and [A-]. If you add more A-, the reaction will shift in the reverse direction until a new equilibrium, with new concentrations, but with the same ka/kb, is created.

Answer 6

Ok, let me type up an example.

Answer 7

You have weak acid, HA, dissociating in water:
\[HA(aq)+H_2O(l) \rightarrow H_3O^+(aq)+A^-(aq)\]Now let's say that this weak acid, HA that...\[K_a(HA)=1.0 \times {10^{-5}}\]One example of these concentrations would be... \[1.0 \times {10^{-5}}=\frac{[H_3O][A^-]}{[HA]}=\frac{(5.0 \times {10^{-6}})(5.0 \times {10^{-6}})}{1}\]

Answer 8

Actually, they would be this. Sorry:\[1.0 \times {10^{-5}}=\frac{(0.00316M)(0.00316M)}{1M}\]

Answer 9

Now, if we add enough conjugate base to give equal concentrations of conjugate base and weak acid, then we have these example concentrations... \[1.0 \times 10^{-5} = \frac{(1M)(1.0 \times 10^{-5}M)}{1M}\]So doing this decreases the pH by about 2.

I apologize. I'm spending a considerable amount of time thinking. I'm just having trouble grasping this concept. Let me think for a moment, please.

Answer 10

Alright, my question is... HOW does the hydronium concentration decrease in the above case? The only way I know for it to decrease is for the conjugate base to react with it to make more weak acid.

Answer 11

Oh! I see. Only a small amount of the newly added conjugate base needs to react with the H+ to reduce it to that concentration, right?

Answer 12

If you think about it in terms of moles before an equilibrium is established...\[\frac{(1mol)(0.00316mol)}{1mol}\]That means that to reduce the hydronium concentration to 1.0 x 10^-5, we need to react about...\[0.00316mol-1.0 \times 10^{-5} mol =0.00315mol\ A^-\]Which is negligible compared to the 1 mole which has just been added. Obviously once it's reacted, there's going to be another slight equilibrium shift, but nothing significant.

Does this logic seem correct?

Answer 13

Alright, I don't think this is the correct way of thinking of buffers. A buffer is a solution consisting of a mixture of a weak acid and its conjugate weak base. If their concentrations are equal, the pH is simply going to be the pKa. Before you add anything to the solution, there is an equilibrium between the weak acid & base.
If some amount of protons are added, the conjugate weak base can neutralize it. If some amount of OH- is added, the conjugate weak acid can neutralize. Adding such things will not lead of a significant pH change until a certain extent.
Once you add something, a new equilibrium is going to be established.

For your example, if you add A-, there really isn't going to be that much of a change; the solution will become slightly more basic with a decrease in H+.

If you add HCl to a buffer of acetic acid and acetate, some of the acetate is going to react with the HCl to form acetic acid. If you add NaOH, some of the acetic acid will react to form acetate. By forming conjugate weak acids/bases that do not dissociate/associate very strongly, the pH isn't changed much.

Answer 14

Well, adding the A- to the solution actually changes the pH by about 2 in the above example, but this is only because the ratio of acid to base was fairly high.

I wasn't so concerned about the mechanism behind a buffer, but rather the preparation of one. I think just talking through this has helped me, though. I'd not looked at it quantitatively that in-depth before.

Thank you very much for your explanations -- they've helped me considerably.

Answer 15

Preparation of a buffer... Hmm, wouldn't mixing 2 solutions of conjugate acids and bases with equal concentrations work?

Answer 16

I just misunderstood exactly how the newly added conjugates base reacted with the current solution. But I understand now.