Why is one a buffer solution but not the other?

[A1s2s2p]

Yes that is the chemical equation for disassociation of an acid.

Now what is the equation that relates the concentrations of those three things? This normally is what would be called for in section 2 of the first post.

Plus I'm sure you will find (if just the words themselves are not enough to tell you what it is) something about electroneutrality. So give that (in terms of concentrations of the ions that you have in these solutions.)

The only equation I know of that would work here would be $K_a=\frac{[H^+]CH_{3}COO^-]}{[CH_{3}COOH]}$ but doesn't count in $[NaOH]$

Unless there is another word for electroneutrality, then I've probably never heard of it before. Maybe I have by definition but not by name.

 

[A1s2s2p]

Well actually you have exactly that here. It is not 'unrelated'. You could say that you have the weak acid certainly, and that by adding alkali to if you have made some salt. That's actually it can be confusing to think in these terms, I think the best way is to think of the concentrations of the species in the solution, some of which are ions.

One exanple I found was to find the pH of a buffer solution containg $acm^3# of$bmol## $dm^{-3}} ethanoic acid with$xcm^3# of $ymol$ ##dm^{-3}} sodium ethanoate. None involving an acid with a base, only acid with salt.

 

[Borek]

None involving an acid with a base, only acid with salt.

Then you don't understand what is base in the buffer context. Buffer is a solution containing an acid and its conjugate base. What is conjugate base of the acetic acid?

 

[epenguin]

One exanple I found was to find the pH of a buffer solution containg $acm^3# of$bmol## $dm^{-3}} ethanoic acid with$xcm^3# of $ymol$ ##dm^{-3}} sodium ethanoate. None involving an acid with a base, only acid with salt.

I have explained that point in #35

 

[A1s2s2p]

Then you don't understand what is base in the buffer context. Buffer is a solution containing an acid and its conjugate base. What is conjugate base of the acetic acid?

CH₃COOH^-

 

[A1s2s2p]

Well actually you have exactly that here. It is not 'unrelated'. You could say that you have the weak acid certainly, and that by adding alkali to if you have made some salt. That's actually it can be confusing to think in these terms, I think the best way is to think of the concentrations of the species in the solution, some of which are ions.

But how would $[CH_{3}COONa]$ be found using the reaction equation, and the volumes and concentrations of the acid and alkali used?

 

[Borek]

So if conjugate base of acetic acid is

CH₃COOH^-

why do you state about question you have found:

None involving an acid with a base, only acid with salt.

What anion does the salt contain?

 

[Borek]

But how would $[CH_{3}COONa]$ be found using the reaction equation, and the volumes and concentrations of the acid and alkali used?

Using simple stoichiometry, which is what I am asking you to do from the very beginning.

Have you calculated initial number of moles of acetic acid and NaOH for each solution?

 

[epenguin]

The only equation I know of that would work here would be $K_a=\frac{[H^+]CH_{3}COO^-]}{[CH_{3}COOH]}$ but doesn't count in $[NaOH]$

Unless there is another word for electroneutrality, then I've probably never heard of it before. Maybe I have by definition but not by name.

CH₃COOH^-

But how would $[CH_{3}COONa]$ be found using the reaction equation, and the volumes and concentrations of the acid and alkali used?

Basically in solution there is no such thing as CH₃COONa. NaOH is a strong base.

 

[A1s2s2p]

Using simple stoichiometry, which is what I am asking you to do from the very beginning.

Have you calculated initial number of moles of acetic acid and NaOH for each solution?

OK, using the reaction, there is 1 mooe of ethanoic acid with 1 mooenof sodium hydroxide.

Solution A:
2 moles of ethanoic acid with 1 mole of sodium hydroxide.

Solution B:
1 mole of ethanoic acid with 1 mole of sodium hydroxide.

 

[A1s2s2p]

What anion does the salt contain?

OH^-

 

[Borek]

Solution A:
2 moles of ethanoic acid with 1 mole of sodium hydroxide.

Solution B:
1 mole of ethanoic acid with 1 mole of sodium hydroxide.

None of these is correct. To begin with: 50cm³ of 0.100mol dm-3 CH₃COOH(aq) doesn't contain one mole of the acid.

OH^-

No, that's an anion in the hydroxide, not an anion in the salt.

 

[A1s2s2p]

No, that's an anion in the hydroxide, not an anion in the salt.

Sorry ^^; Got it mixed up somehow.

CH₃COO^-

 

[A1s2s2p]

None of these is correct. To begin with: 50cm3 of 0.100mol dm-3 CH3COOH(aq) doesn't contain one mole of the acid

Sorry, got confused with the mole ratio, it has $\frac 1 {200}$ moles

 

[epenguin]

And my own posts are getting mixed up too because I have been several times interrupted where I am.

I don’t think we can make much progress by Socratic dialogue here as you have not sufficiently studied or grasped the basics.

I will give you first my answers to those questions, later maybe a summary of the subject you have to study

Look at a titration curve in your book and you will find that when the moles of added strong base are half those of the weak acid, which is what you have in case B, you are at an inflection point in the titration curve, the rate of increase of pH per mole of base is added is at a minimum. In other words you have the maximum buffering you can have for that concentration of the acid. The more concentrated the acid, the more NaOH you will have to add for any given change of pH. In other words the more concentrated the acid (in all its forms) the more it is buffered. Your solutions are quite concentrated, so in case B for these two reasons what you have is what would be called a good buffer, contrary to the statement of the original question. In case A the moles of base are different from a half of total acid, and it is less concentrated than B; it is still a buffer, but does not buffer so much as B.

 

[Borek]

Sorry, got confused with the mole ratio, it has $\frac 1 {200}$ moles

OK, now follow with the calculations I asked you to do several times earlier in the thread, repeating myself for n^th time will be just waste of time.