Why is one a buffer solution but not the other?

[A1s2s2p]

Homework Statement

A student prepares two solutions:
Solution A is prepared by mixing 50cm³ of 0.100mol dm^-3 CH₃COOH_(aq) with 25cm³ of 0.100mol dm^-3 NaOH_(aq)
Solution B is prepared by mixing 25cm³ of 0.200mol dm^-3 CH₃COOH_(aq) with 50cm³ of 0.100mol dm^-3 NaOH_(aq)

Explain why solution A is a buffer solution, but solution B is not.

Homework Equations

None

The Attempt at a Solution

I believe it is because of the reaction
CH₃COOH_(aq) + NaOH_(aq) ↔ CH₃COONa_(aq) + H₂O_(aq)

In solution A, the reacants are in the same proportion (concentration wise) as the stoichiometric constants ij the equation, but in solution B the concentrations are in the ratuo 2:1.

 

[Borek]

What do you know about buffers? What is the definition of the buffer solution?

 

[A1s2s2p]

A buffer solution is a solution that is made up of a weak acid and conjugate base, which minimises a change in pH when lsmall amounts of acid or base are added. In the question, both solutions are made up of the same weak acid and base, the only differences are volumes used and concentratioj.

 

[Borek]

a solution that is made up of a weak acid and conjugate base

Have you calculated amounts of acid and conjugate base in each case? Just by following stoichiometry?

 

[A1s2s2p]

You mean by n=C*v?
n - number of moles (mol)
C - concentration (mol dm^-3)
v - volume (dm³)

 

[Borek]

There were reactions taking place - what is present in the solutions afterwards?

 

[A1s2s2p]

H₂O, NaOH, CH₃COOH and CH₃COONa?

 

[Borek]

Check their concentrations. Use simple stoichiometry (hint: check what is the limiting reagent).

 

[A1s2s2p]

Well, the concentrations in solution A in the ratio of CH₃COOH:NaOH is 1:1, however in solution B it is 2:1, the stoichiometry ratio is 1:1.

Also, what is the limiting reagent? I don't remember doing it, I don't think it's part of our course.

 

[Borek]

If you are expected to do buffers, you are expected to know how to do stoichiometry calculations, limiting reagents are part of that.

Well, the concentrations in solution A in the ratio of CH3COOH:NaOH is 1:1, however in solution B it is 2:1, the stoichiometry ratio is 1:1.

No, you are looking at concentrations of reactants, not at concentrations of products of the reaction.

Write reaction equation first.

 

[epenguin]

I am not seeing that this: "solution A is a buffer solution, but solution B is not." is correct.

 

[A1s2s2p]

With buffers I know we only need to know what a buffer is, how to calculate its pH and find the ratio of the two compounds to get the pH, amongst a few other things.

I'm OK with the definition and a few other things, but not good with the maths and equations for buffers.

The question I asked in the OP was an example question

CH₃COOH_(aq)+NaOH_(aq)↔CH₃COONa_(aq)+H_2(l)O

Products are still 1:1.

In our second year textbook I don't find limiting reagent in the index, maybe it could be in the first year textbook.

 

[A1s2s2p]

I am not seeing that this: "solution A is a buffer solution, but solution B is not." is correct.

Let me post the image.

 

[A1s2s2p]

I am not seeing that this: "solution A is a buffer solution, but solution B is not." is correct.

Not the best of quality. NaOH is partially cut off from the left hand side.

 

[A1s2s2p]

All I know is that the mole ratio of CH₃COOH_(aq):NaOH_(aq) in solution A is 2:1 and in solution B it is 1:1, while the stoichiometric ratio is 1:1.

 

[epenguin]

Well maybe a speck of dust has surrounded my brain, that is one hypothesis.

However I am still seeing the statement you are required to explain as wrong. It requires a lot of ingenuity the student to give the justification for a wrong statement. If I give you a counter authority to not have to justify this statement, but just to work out what would be the right answer, e.g. which of these solutions is a better buffer, then maybe that is easier for you to do.

I think your conclusions in #9 are wrong but maybe you were trying to force them to conform to what you’re being told,So have another look. What are the ratio of moles in the two cases?

Edit: I see you have corrected in #15

 

[A1s2s2p]

Well maybe a speck of dust has surrounded my brain, that is one hypothesis.

However I am still seeing the statement you are required to explain as wrong. It requires a lot of ingenuity the student to give the justification for a wrong statement. If I give you a counter authority to not have to justify this statement, but just to work out what would be the right answer, e.g. which of these solutions is a better buffer, then maybe that is easier for you to do.

I think your conclusions in #9 are wrong but maybe you were trying to force them to conform to what you’re being told,So have another look. What are the ratio of moles in the two cases?

Edit: I see you have corrected in #15

Yeah, they have made mistakes before. Once, in the first year physics textbook, they quoted the density of seawater as being 10 times less than what it is meant to be. It was for an exam style question.

So, what do I do with the tree moe ratios?

 

[epenguin]

Okay to get to the answer, at a given acid-base ratio do you understand whether increasing the total concentration makes the Buffer better or worse.?

And at given concentration What is the acid base ratio that makes the best buffering?

 

[Borek]

CH₃COOH_(aq)+NaOH_(aq)↔CH₃COONa_(aq)+H_2(l)O

OK, for both cases calculate how much acid and how much conjugate base is present after the reaction.

 

[A1s2s2p]

Okay to get to the answer, at a given acid-base ratio do you understand whether increasing the total concentration makes the Buffer better or worse.?

And at given concentration What is the acid base ratio that makes the best buffering?

Based on what I remember reading from the textbook, changing the concentrations of the acid and/or base changes the pH of the buffer solution.

 

[epenguin]

Oh dear, the point of textbooks is to supplement memory, you can go back to them.

Evidently you need to know about strong and weak acids, strong and weak bases. About buffering power. All of which follow from the simple equilibrium equations for, in this case, acid dissociation. So far in this thread we have not seen a chemical equation for the ionic equilibrium involving acetic (ethanoic) acid...

Everything follows from those things.

 

[A1s2s2p]

But I was just saying that the textbook doesn't mention anything to do woth how good of a buffer something is, only what a buffernsolution is, how to work out its pH, how to wrk out the ratio of salt to base/acid for a given pH, stuff to do with buffers and pH curves, and controlling pH in blood, there's nothing to do with how good or bad a buffer solution is.

 

[A1s2s2p]

OK, for both cases calculate how much acid and how much conjugate base is present after the reaction.

How would I do that, especially for after? Look, I said I'm not good with the maths and equations involved with buffers, I'm trying to make sure I know what I need to know for the exam. I don't want to write something down which is completely wrong and useless and look stupid, I just want make sure I know what I'm doing.

The questions below on the image, ones with benzoic acid, and ammonia, I can do

 

[Borek]

How would I do that, especially for after?

Follow the stoichiometry - what reacted, what was left. It doesn't require any equations related to buffers, it does require things that you were taught in the past.

For solution A - how many moles of NaOH were used? How many moles of acetic acid were used? They reacted according to the equation you have already listed. Assuming reaction went to completion - what was left? What was produced? How many moles of each? (Ignore water, we are interested only in the acid and its conjugate base).

 

[epenguin]

How would I do that, especially for after? Look, I said I'm not good with the maths and equations involved with buffers, I'm trying to make sure I know what I need to know for the exam. I don't want to write something down which is completely wrong and useless and look stupid, I just want make sure I know what I'm doing.

Okay, if you don’t want to look stupid you need to be able to answer not just those questions but any question like that.
Look in your book about titration curves. At what point in the curve is is the rate of change of pH with added strong base or acid the least? Which is to say, buffering is best. Okay, ‘good’ or less good buffering is vague language, instead we can define “buffering capacity“ as d[NaOH]/d(pH). The higher that is the better buffering.

If you want to take this further here, you need to write out the equilibrium equations for the ionic equilibrium in play here, and also the relevant ‘electroneutrality’ equation or condition which is no doubt in your book.

 

[A1s2s2p]

Follow the stoichiometry - what reacted, what was left. It doesn't require any equations related to buffers, it does require things that you were taught in the past.

For solution A - how many moles of NaOH were used? How many moles of acetic acid were used? They reacted according to the equation you have already listed. Assuming reaction went to completion - what was left? What was produced? How many moles of each? (Ignore water, we are interested only in the acid and its conjugate base).

Not sure how much of this, or what bits, was needed but I wrote as much as I thought would be.

1. In the reaction, for stoichiometry it was 1, in terms of actual moles, 1.
2. In the reaction, for stoichoimetry it was 1, in terms of actual moles, 2.

Well, for the forward reaction (assuming completion): CH₃COOH_(aq)+NaOH_(aq)→CH₃COONa_(aq)(+H₂O_(l)), 1 mole of sodium ethanoate would be produced (along with water).

The solution would be a mix of sodium ethanoate, ethanoic acid, and sodium hydroxide (since in reality it is reversible, and ethanoic acid partially dissociates and equilibrium lies far to the left in the reaction: CH₃COOH ↔H⁺+CH₃COO^-)

Solution A would have 1 mole of sodium hydroxide dissociate into Na⁺, and ethanoic acid would partially dissociate into H⁺ and CH₃COO^-, with 1 mole excess of ethanoic acid.

The 1 mole of excess would dissociate as well into its respective ions, but those ions will react with the existing ethanoate and hydrogen ions in the reaction:
CH₃COO^-+H⁺↔CH₃COOH

 

[A1s2s2p]

Okay, if you don’t want to look stupid you need to be able to answer not just those questions but any question like that.
Look in your book about titration curves. At what point in the curve is is the rate of change of pH with added strong base or acid the least? Which is to say, buffering is best. Okay, ‘good’ or less good buffering is vague language, instead we can define “buffering capacity“ as d[NaOH]/d(pH). The higher that is the better buffering.

If you want to take this further here, you need to write out the equilibrium equations for the ionic equilibrium in play here, and also the relevant ‘electroneutrality’ equation or condition which is no doubt in your book.

I know that, as I said, I'm fine with the other questions in the image except for that one.

The questions in the image come under 'Buffer solutions' which covers the following:

• What a buffer solution is
• Calculating the pH of a buffer solution
• A buffer solution made from a weak acid and conjugate base.
• How a buffer works
• The Henderson-Hasselbalch equation
• A buffer solution made from a weak base and conjugate acid.
  
• How to calculate the salt:(acid/base) ratio to get a given pH.
• How pH is controlled in the blood.

Covered in 4 pages.

The next section is on buffer solutions and pH curves, but it only talks about the buffer range and how to find the K_a using the half-equivalence point.

 

[Borek]

1. In the reaction, for stoichiometry it was 1, in terms of actual moles, 1.
2. In the reaction, for stoichoimetry it was 1, in terms of actual moles, 2.

Sorry, what you wrote doesn't mean anything. 1 of what? Actual moles of what?

Well, for the forward reaction (assuming completion): CH₃COOH_(aq)+NaOH_(aq)→CH₃COONa_(aq)(+H₂O_(l)), 1 mole of sodium ethanoate would be produced (along with water).

No. Just because the reaction equation states "1 mole of NaOH reacts with 1 mole of acetic acid" doesn't mean 1 mole of NaOH reacts every time.I asked you how much acid and how much NaOH were initially put in the solution, have you calculated these numbers? Was there 1 mole of any of these substances present at any moment?

The solution would be a mix of sodium ethanoate, ethanoic acid, and sodium hydroxide

And it is their relative amounts that will tell you which solution is a buffer and which is not. That's what I am asking you to calculate, that's what epenguin hints at in his posts.

 

[A1s2s2p]

And it is their relative amounts that will tell you which solution is a buffer and which is not. That's what I am asking you to calculate, that's what epenguin hints at in his posts.

Not sure if its right or not, but after using K_a and an ICE table, I got a concentration ratio of CH₃COOH:NaOH:CH₃COONa as:
Solution A - 73:74:1
Solution B - 105:52:1

It seems like solution A is the buffer contains around equal concentrations of ethanoic acid and sodium ethanoate at equilibrium, while solution B contains almost twice the concentration of ethanoic acid to sodium ethanoate.
Results:

Some information I found (from page 2 of http://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Notes/Edexcel/12-Acid-Base-Equilibria/Set-A/Acids%20and%20Bases%20-%20Buffer%20Solutions%20and%20Uses.pdf):

 

[A1s2s2p]

Hang on.

 

[Borek]

but after using Ka and an ICE table

I told you to follow simple stoichiometry and assume reaction went to completion, not to attempt to calculate equilibrium.

(Besides, numbers you posted are definitely wrong.)

 

[A1s2s2p]

I said 'hang on', because I had to check something. I wasn't sure about what I did.

 

[A1s2s2p]

Whenever I try to find similar questions, I only get unrelated questions involving a weak acid and a (usually sodium) salt based off of the conjugate base of the acid.

 

[epenguin]

CH₃COO^-+H⁺↔CH₃COOH

Yes that is the chemical equation for disassociation of ethanoic (acetic) 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.)

 

[epenguin]

Whenever I try to find similar questions, I only get unrelated questions involving a weak acid and a (usually sodium) salt based off of the conjugate base of the acid.

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.

 

[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.