[Erichsen]

Hello there. Im new in here and wanna learn some more chemistry. I have allready read some books about buffers but non of them explains how its possible to create a buffer.

Lets say I want to make a 1M CH₃COOH/CH₃CHOO^-.
I take 1M CH₃COOH and the equilibrum gives:

CH₃COOH + H₂O <-> CH₃CHOO^- + H₃O⁺ (mostly left)
Then I add 1M CH₃COONa. The equilibrum then goes even more to the left to give the acetic acids Ka. But then Kb for this equilibrum cant be satisfied

CH₃COO^- + H₂O <-> CH₃COOH + OH^-

Or can it? Am I doing something wrong in my thinking?

Plz hlp me out here

[Borek]

But then Ks for this equilibrum cant be satisfied

CH₃COO^- + H₂O <-> CH₃COOH + OH^-

It IS satisfied. Always. You are making some mistake in your thinking, but to be honest - I have no idea where. Please try to elaborate, preferably with numbers, it will be easier to pinpoint the problem.

[jacobcolbert]

To prove it is satisfied you can write the K^b for the second equilibrium and then substitute [OH^-]=K_w/[H₃O⁺]. This will yield Kb=Kw/Ka. It is important to note that this Kb is for the conjugate base CH₃COO^- not acetic acid. The Kb for acetic acid would be some ridiculously low thing. Hope this is what you were talking about.

[Erichsen]

Jacobcolbert thx for your advice. I translate it to this

Kb= [CH3COOH][OH-] / [CH3COO]

Then your advice gives

kb = [CH3COOH](kw/[H3O]) / [CH3COO]

kb = [CH3COOH] kw / [H3O][CH3COO]

kb * [H3O][CH3COO]/[CH3COOH] = kw

kb * ks = kw

Does this explains my problem? I mean I start with a solution of 1M acetic acid. It gets mixed with 1M acetate. I cant understand why these two equilibrums (for ks and kb) doesnt affect each other? Or at least I cant see it mathematical how a solution containing 1M acetic acid and 1M acetate is possible without breaking the laws

[Borek]

System containing CH₃COOH, CH₃COO^-, H⁺, OH^- and H₂O is described by two equilibria (water dissociation and acetic acid dissociation) and three mass balances (carbon, oxygen, hydrogen). Five equations with five unknown concentrations. "Just" solve. "Just" because system is not linear...

Note that adding equation for acetate hydrolysis you don't add any new information to the system - you still have the same five unknowns and five independent equations. 6^th equation is just a comibnation of two others so it doesn't add new information about the system.

Oh, and by the way - your K_s will be better understood here if you will name it K_a

[Erichsen]

Oh sorry for my use of Ks, i really mean Ka. Its because im from Denmark where acid its called "syre" which leads to Ks . And 5 equations, thats alot, maybe thats why i couldnt see the connection. Can u write an example of a massbalance with carbon in this situation. Then i think i can solve the rest, or at least try. Thx for all the replies.

[Borek]

I know acid is Säuren in German so I got it, not without problems at first

You don't have to put all mass balances into your system of equations. You may safely assume that water concentration is constant and the only mass balance required is that of acetate/acetic acid:

[CH₃COOH] + [CH₃COO^-] = const

(const must be sum of both initial concentrations of acetic acid and acetate). Note that it is identical to carbon mass balance. Compare with - say - hydrogen balance:

4[CH₃COOH] + 3[CH₃COO^-] + [H⁺] + [OH^-] + 2[H₂O] = const

In most cases you don't have to use mass balances for every element, just like you don't have to balance every element separately when balancing reaction equations - say, you just balance SO₄^2-, not S and then O.

However, I did something wrong, I haven't mentioned very important charge balance:

[CH₃COO^-] + [OH^-] = [H⁺] + [Me⁺]

where Me⁺ is a cation entered into solution with the acetate. Looks to me like 5 equations/5 unknowns mentioned earlier won't work - probably out of three (C, H, O) there are only two independent mass balances, and third is some combination of these two, sorry about that.

See introduction to acid/base equilibrium calculation. In your case set of equations is:

[CH₃COOH] + [CH₃COO^-] = const

[CH₃COO^-] + [OH^-] = [H⁺] + [Me⁺]

[OH^-][H⁺] = K_w

[CH₃COO^-][H⁺]/[CH₃COOH] = K_a

There are four unknowns and four equations; fifth unknown was water (it is now hidden in the K_w; see water ion product). Solving it will be hard, browse my lectures to see ideas about how to deal with such problems.

[Erichsen]

Okay thank for your help. One last question. How can it be that when u create a buffersolution u add like 1mol acetic acid and 1mol natriumacetate, why not JUST 2 mol acetic acid? It gives the same ions/molekyles in the solution or (except no Na+)?

I bet it has something to do with the charge balance you talk about? (yes i know im a little slow on this bufferthing, but i really never totally understood it , except for its uses )