Bertybassett
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For Q1)A.) on this I don't understand why you cant just say that 1.74 x 10^-5 = H+ concentration squared divided by ethanoic acid conc? And then you get an answer of 0.186. I'm really confused on all of these buffer questions. My textbook says you can use H+ squared as the numerator of the calculation if it's a weak acid? https://pmt.physicsandmathstutor.com...n%201%20QP.pdf
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ChemistryWebsite
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It's because you are making a calculation for a buffer solution.

In a buffer [H+] isn't equal to [A-] as there is anion present from the salt too.
In a buffer use the assumption that [A-] is the conc of the salt present, and put this value into your Ka expression.
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Bertybassett
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(Original post by ChemistryWebsite)
It's because you are making a calculation for a buffer solution.

In a buffer [H+] isn't equal to [A-] as there is anion present from the salt too.
In a buffer use the assumption that [A-] is the conc of the salt present, and put this value into your Ka expression.
Thanks for the response. In regards to the assumptions of assuming the A- conc is equal to the conc of salt, what other assumptions do we make, and how do we make it? E.g. i thought we assume all the salt has all dissociated, so at equilibrium, would the conc of the salt itself be zero? What about for acids? Many thanks
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Pigster
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The two assumptions you make (and this is why we do calcs involving acids like ethanoics at A level) are:

1. Since the acid is rather weak, not much will dissociation, i.e. [dissociated acid] = [acid] i.e. effectively you ignore the degree of dissociation when considering the bottom term in the Ka calc.

2. Since the acid isn't very weak, the only source of H+ ions is from the dissociation of the acid, i.e. [H+] = [A-]. For very weak acids, H+ ions can come from the selfionisation of water (think Kw) and hence [H+] =/= [A-], so you can't use the [H+]2 method.
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Bertybassett
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(Original post by Pigster)
The two assumptions you make (and this is why we do calcs involving acids like ethanoics at A level) are:

1. Since the acid is rather weak, not much will dissociation, i.e. [dissociated acid] = [acid] i.e. effectively you ignore the degree of dissociation when considering the bottom term in the Ka calc.

2. Since the acid isn't very weak, the only source of H+ ions is from the dissociation of the acid, i.e. [H+] = [A-]. For very weak acids, H+ ions can come from the selfionisation of water (think Kw) and hence [H+] =/= [A-], so you can't use the [H+]2 method.
Hi, thanks for the response. I am a bit confused by what you mean in 1 by the acid is weak, but it 2 you say the acid isn't very weak? Also, what do we assume about the salt? Has this fully dissocaited?
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Pigster
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(Original post by Bertybassett)
Hi, thanks for the response. I am a bit confused by what you mean in 1 by the acid is weak, but it 2 you say the acid isn't very weak? Also, what do we assume about the salt? Has this fully dissocaited?
Acids are either strong (we assume they fully dissociate, but actually they just dissociate to such an extent that we just ignore the tiny fraction that remain undissociated) or they are weak (they only partially dissociate).

Weak acids partially dissociate to different degrees. Some (the stronger weak acids) dissociate a lot, in fact could dissociate just a little short of the threshold needed to be called a strong acid. Others dissociate barely at all, in fact just short of the amount before you’d no longer call them acids (as they would dissociate in solution less than the water they are dissolved in).

Ethanoic is a weak acid. It is not too weak and not too strong, but just right for A level students to do calculations on without having to worry about using the quadratic equation or to worry about the effects of Kw.

Ionic salts of weak acids (such as sodium ethanoate) fully dissociate.
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