ps1265A
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Say if we have this buffer:

HA = H+ + A-

If we add H+, we will remove A-. Now the equilibrium shift to the left.

But shouldn't it be shifting to the right to restore A-?
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charco
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(Original post by ps1265A)
Say if we have this buffer:

HA = H+ + A-

If we add H+, we will remove A-. Now the equilibrium shift to the left.

But shouldn't it be shifting to the right to restore A-?
This is not a buffer system ..
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ps1265A
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(Original post by charco)
This is not a buffer system ..
Hmmm, okay, say if we have ethanoic acid and sodium ethanoate, the following buffer solution is established.


Original solution: CH3COOH = H+ + CH3COO-

Adding sodium ethanoate: CH3COOH + CH3COONa = ...

Adding H+, the position of equilibrium will shift so as to appose this addition, and so react with CH3COO-, the position of equilibrium will shift to the left.

I think my confusion is in understanding whether the second equation is in equilibrium. And also, when it says "the position of equilibrium sifts to the left" is that now referring to the first equation?
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charco
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(Original post by ps1265A)
Hmmm, okay, say if we have ethanoic acid and sodium ethanoate, the following buffer solution is established.


Original solution: CH3COOH = H+ + CH3COO-

Adding sodium ethanoate: CH3COOH + CH3COONa = ...

Adding H+, the position of equilibrium will shift so as to appose this addition, and so react with CH3COO-, the position of equilibrium will shift to the left.

I think my confusion is in understanding whether the second equation is in equilibrium. And also, when it says "the position of equilibrium sifts to the left" is that now referring to the first equation?
The first equation is an equilibrium, the second is just a reaction.

Check out this interactive on buffers, it may help.
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ps1265A
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(Original post by charco)
The first equation is an equilibrium, the second is just a reaction.

Check out this interactive on buffers, it may help.
Thanks! I understand it now. A buffer solution is 2 equations; one at equilibrium and one that goes into completion. I though it was a reaction between ethanoic acid and sodium ethanoate
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ps1265A
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(Original post by charco)
The first equation is an equilibrium, the second is just a reaction.

Check out this interactive on buffers, it may help.
The website states it OH- reacts with H+, I thought it reacts with HA to form water? This is what my book states
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charco
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(Original post by ps1265A)
The website states it OH- reacts with H+, I thought it reacts with HA to form water? This is what my book states
You mean on addition of OH-?

It would have to react with the hydrogen ions present. It has no reason to react with HA...
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langlitz
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(Original post by ps1265A)
The website states it OH- reacts with H+, I thought it reacts with HA to form water? This is what my book states
That's the same thing.

Remove H+ ions from the system by reacting with OH- to form H2O. By Le Chatelier's principle the system will shift to restore the H+ ions by dissociating more HA thus keeping the pH constant

OR

React OH- with HA to form H2O and A-. Thus no increase in H+ or OH- in the solution. Thus constant pH
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ps1265A
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(Original post by langlitz)
That's the same thing.

Remove H+ ions from the system by reacting with OH- to form H2O. By Le Chatelier's principle the system will shift to restore the H+ ions by dissociating more HA thus keeping the pH constant

OR

React OH- with HA to form H2O and A-. Thus no increase in H+ or OH- in the solution. Thus constant pH
Cheers
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ps1265A
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(Original post by charco)
You mean on addition of OH-?

It would have to react with the hydrogen ions present. It has no reason to react with HA...
As the user below said, it could? I think it makes more sense with it reacting HA as most questions tend to ask, e.g. what's the remaining moles of ethanoic acid after the addition of OH-, and so you could just do original HA - OH-
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charco
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(Original post by ps1265A)
As the user below said, it could? I think it makes more sense with it reacting HA as most questions tend to ask, e.g. what's the remaining moles of ethanoic acid after the addition of OH-, and so you could just do original HA - OH-
makes more sense for a negative charged species to react with a positive one...

... but suit yourself!
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ps1265A
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(Original post by charco)
makes more sense for a negative charged species to react with a positive one...

... but suit yourself!
Thanks for clarifying!
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ps1265A
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(Original post by charco)
makes more sense for a negative charged species to react with a positive one...

... but suit yourself!
Why can we use the equation for Ka when working out the pH of a buffer? Because a buffer comprises of 2 equations, we cannot just combine the two equations to use the equation for Ka can we?
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ps1265A
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(Original post by charco)
makes more sense for a negative charged species to react with a positive one...

... but suit yourself!
Q) What could you add to ammonia to form a buffer
A) NH4Cl

I thought the answer was a weak acid. Ammonia is a strong base and adding a weak acid would form a buffer?
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charco
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(Original post by ps1265A)
Q) What could you add to ammonia to form a buffer
A) NH4Cl

I thought the answer was a weak acid. Ammonia is a strong base and adding a weak acid would form a buffer?
ammonia is a weak base ...
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ps1265A
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Ah, thanks! The answer could be HCl as well surely
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Pigster
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(Original post by langlitz)
That's the same thing.

Remove H+ ions from the system by reacting with OH- to form H2O. By Le Chatelier's principle the system will shift to restore the H+ ions by dissociating more HA thus keeping the pH constant

OR

React OH- with HA to form H2O and A-. Thus no increase in H+ or OH- in the solution. Thus constant pH
Easy peasy way of losing a mark.

Buffers do not keep pH constant, they minimise pH changes on addition of small amounts of an acid or a base.
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charco
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(Original post by ps1265A)
Ah, thanks! The answer could be HCl as well surely
No, because the acid is in excess. With HCl you will have a strong acid and its salt = not a buffer.
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ps1265A
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(Original post by charco)
ammonia is a weak base ...
I also have another question: the dissociation of an acid HA is endothermic, what effect will this have on the Ka if the concentration of the acid increased. I thought that because it's more concentrated, more H+ dissociated, so the conc of products becomes greater, therefore Ka because bigger. But the answer is no change.
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ps1265A
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(Original post by charco)
No, because the acid is in excess. With HCl you will have a strong acid and its salt = not a buffer.
Sorry, I mean NaOH
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