The Student Room Group

Buffer solutions

Hi,
I was thinking if you have a buffer solution say ethanoic acid and sodium ethanoate to produce:

CH3COOH <--> H+ + CH3COO-

and you increase the concentration of H+, i know equilibrium will shift to the left and so the CH3COO- will react with the H+ using up most of it. What i dont understand is why its only most, is it because the acid dissociation constant needs to be restored and both H+ and CH3COO- will decrease when they are reacting so H+ does not need to decrease to its intitial level?

Also, because H+ does not decrease to its intial level surely that implies that the concentration of OH- ions has reduced to keep the constant value of kw, but how does this happen if it does?

Thanks for your help :smile:
Original post by 111davey1
Hi,
I was thinking if you have a buffer solution say ethanoic acid and sodium ethanoate to produce:

CH3COOH <--> H+ + CH3COO-

and you increase the concentration of H+, i know equilibrium will shift to the left and so the CH3COO- will react with the H+ using up most of it. What i dont understand is why its only most, is it because the acid dissociation constant needs to be restored and both H+ and CH3COO- will decrease when they are reacting so H+ does not need to decrease to its intitial level?

Also, because H+ does not decrease to its intial level surely that implies that the concentration of OH- ions has reduced to keep the constant value of kw, but how does this happen if it does?

Thanks for your help :smile:


It is only "most" because the initial value of ka must be re-established.

Your last paragraph:

If you increase the concentration of hydrogen ions you must reduce the concentration of hydroxide ions due to the kw equilibrium:

H+ + OH- <==> H2O

Here's a little interactive on buffers that I made a while ago ...
... it may help.
(edited 7 years ago)
Reply 2
Original post by charco
It is only "most" because the initial value of ka must be re-established.

Your last paragraph:

If you increase the concentration of hydrogen ions you must reduce the concentration of hydroxide ions due to the kw equilibrium:

H+ + OH- <==> H2O

Here's a little interactive on buffers that I made a while ago ...
... it may help.


Ah yes that explains it thank you.
Reply 3
Original post by charco
It is only "most" because the initial value of ka must be re-established.

Your last paragraph:

If you increase the concentration of hydrogen ions you must reduce the concentration of hydroxide ions due to the kw equilibrium:

H+ + OH- <==> H2O

Here's a little interactive on buffers that I made a while ago ...
... it may help.


Hi,
And by reducing the concentration of OH- ions would you also be further decreasing the concentration of H+ ions?
Thanks
Original post by 111davey1
Hi,
And by reducing the concentration of OH- ions would you also be further decreasing the concentration of H+ ions?
Thanks


No, remember that kw is a constant and equal to [H+][OH-] = 1 x 10-14 at 298K

If you increase one you must decrease the other and vice versa.
Reply 5
Original post by charco
No, remember that kw is a constant and equal to [H+][OH-] = 1 x 10-14 at 298K

If you increase one you must decrease the other and vice versa.


Thanks
but surely you cant just decrease one. I thought the way to reduce the OH- would be for it to react with the H+ meaning that reduces as well with the H+ still being higher than it intially was in the buffer solution so the kw constant is intact.
(edited 7 years ago)
Original post by 111davey1
Thanks
but surely you cant just decrease one. I thought the way to reduce the OH- would be for it to react with the H+ meaning that reduces as well with the H+ still being higher than it intially was in the buffer solution so the kw constant is intact.


You don't seem to understand that kw is constant, i.e. its value cannot change providing the temperature is constant.

Logically if

y x z = n

then decreasing y must increase z
Reply 7
Original post by charco
You don't seem to understand that kw is constant, i.e. its value cannot change providing the temperature is constant.

Logically if

y x z = n

then decreasing y must increase z



i know but what i mean is can it change temporarily before the equilibrium sorts it out. (like the equilibrium constant can change temporarily)

In the buffer solution you increase H+ equilibrium will shift to the left using up most but leaving some.

As some is left if the solution was to stay like this kw would not be constant it would be higher.

so the equilibrium H+ + OH- <--> H20 shifts to the right to decrease H+ and OH- until kw is re-established. In this process both of the concentrations decrease. (This is the problem as i can't see how OH- decreases without H+ decreasing also)

I might be talking nonsense but if i am could you tell me how OH- decreases, after the buffer has left some extra H+. Thanks.
(edited 7 years ago)
Original post by 111davey1
i know but what i mean is can it change temporarily before the equilibrium sorts it out. (like the equilibrium constant can change temporarily)

In the buffer solution you increase H+ equilibrium will shift to the left using up most but leaving some.

As some is left if the solution was to stay like this kw would not be constant it would be higher.

so the equilibrium H+ + OH- <--> H20 shifts to the right to decrease H+ and OH- until kw is re-established. In this process both of the concentrations decrease. (This is the problem as i can't see how OH- decreases without H+ decreasing also)

I might be talking nonsense but if i am could you tell me how OH- decreases, after the buffer has left some extra H+. Thanks.


The equilibrium constant cannot "change temporarily", what changes is the quotient. The system responds until the quotient is once again equal to the equilibrium constant.

Once you initially increase [H+] none of the quotients equal the equilibrium constants so there is net movement in all systems.
Reply 9
Original post by charco
The equilibrium constant cannot "change temporarily", what changes is the quotient. The system responds until the quotient is once again equal to the equilibrium constant.

Once you initially increase [H+] none of the quotients equal the equilibrium constants so there is net movement in all systems.


Oh thanks i didn't know about that, so you add the H+ and both the kw quotient and the equilibrium quotient change. in the case of kw, there needs to be a shift in the equilibrium to the right to restore the constant kw. This decreases both H+ and OH- in the process??

Is this true?
Original post by 111davey1
Oh thanks i didn't know about that, so you add the H+ and both the kw quotient and the equilibrium quotient change. in the case of kw, there needs to be a shift in the equilibrium to the right to restore the constant kw. This decreases both H+ and OH- in the process??

Is this true?


yes

Quick Reply

Latest