# Homework Help: Why is one a buffer solution but not the other?

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1. Dec 4, 2017

### A1s2s2p

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

2. Dec 4, 2017

### Staff: Mentor

So if conjugate base of acetic acid is

why do you state about question you have found:

What anion does the salt contain?

3. Dec 4, 2017

### Staff: Mentor

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?

4. Dec 4, 2017

### epenguin

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

Last edited: Dec 4, 2017
5. Dec 4, 2017

### A1s2s2p

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.

6. Dec 4, 2017

OH-

7. Dec 4, 2017

### Staff: Mentor

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

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

8. Dec 4, 2017

### A1s2s2p

Sorry ^^; Got it mixed up somehow.

CH3COO-

9. Dec 4, 2017

### A1s2s2p

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

10. Dec 4, 2017

### 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.

Last edited: Dec 5, 2017
11. Dec 4, 2017

### Staff: Mentor

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