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Topic: Buffer with an Ester in it?  (Read 5076 times)

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Offline Halogen876

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Buffer with an Ester in it?
« on: June 07, 2014, 01:51:37 PM »

I have this question about a buffer that I'm not sure how to approach. (It's not homework.)

Here is the question:

You wish to study the saponification of an ester at pH 10.00:
RCOOR' + OH-  :rarrow: RCOO- + HOR'
You choose a buffer containing ammonia and ammonium chloride. What must the formalities of these substances be to hold the pH constant to within 0.10 unit if the formality of the ester (RCOOR') at the start of the reaction is 0.0500 F? You may choose a specific volume for the reaction if you wish.

I'm familiar with the concept of keeping pH within a certain range. It is the part about the ester that is throwing me off. I've successfully completed several similar questions regarding holding pH constant, but none of them had esters in them so I am completely at a loss as to how to start this problem. :-\

Any suggestions would be greatly appreciated!
Thanks!

Offline Borek

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Re: Buffer with an Ester in it?
« Reply #1 on: June 07, 2014, 02:24:31 PM »
Hydrolysis of the ester means you add an acid to the solution. Question asks what must be concentration of the buffer so that the pH change after acid addiction won't be larger than 0.1 unit.
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Offline Halogen876

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Re: Buffer with an Ester in it?
« Reply #2 on: June 08, 2014, 12:43:22 PM »

Thanks for the reply.

I know that hydrolysis of an ester means to add acid but in the equation given in this question, it shows that base is being added so I'm not sure how that figures into the calculations. Any more insight would be appreciated!


Offline Borek

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Re: Buffer with an Ester in it?
« Reply #3 on: June 08, 2014, 12:54:53 PM »
I know that hydrolysis of an ester means to add acid but in the equation given in this question, it shows that base is being added so I'm not sure how that figures into the calculations. Any more insight would be appreciated!

I think you misunderstood what I tried to wrote.

What are PRODUCTS of the hydrolysis?
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Offline Halogen876

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Re: Buffer with an Ester in it?
« Reply #4 on: June 08, 2014, 03:08:41 PM »
Ok, so if you add acid to the ester it will hydrolize it and form a base? and that's why OH- is given in the equation in the question?

Also, although I think I understand the general concept of hydrolyzing an ester, I am a little confused as to why you mentioned hydrolysis since it is not mentioned in the question. Is there something in the question that I should have picked up on that would tell me I should be thinking about hydrolysis?

Offline Borek

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Re: Buffer with an Ester in it?
« Reply #5 on: June 08, 2014, 05:20:20 PM »
Saponification and hydrolysis mean the same thing.

I think you are still missing the point of the question.

What are products of the saponification (hydrolysis)?
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Offline Halogen876

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Re: Buffer with an Ester in it?
« Reply #6 on: June 09, 2014, 02:22:34 PM »
Ok thanks. I didn't realize that saponification and hydrolysis meant the same thing.

So if you add acid to an ester you would get a carboxylic acid and an alcohol.

But in this question, they're doing a hydrolysis with base, in which case your produsts are the conjugate base of the carboxylic acid and the same alcohol.

So I would have a weak base in solution with the NH3/NH4+ buffer system. If that's correct though, I'm not sure how I would deal with these 2 separate weak base/conjugate acid pairs in a solution together...

Offline Borek

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Re: Buffer with an Ester in it?
« Reply #7 on: June 09, 2014, 02:49:30 PM »
produsts are the conjugate base of the carboxylic acid

Good. It is equivalent to adding an acid to the solution. That's what I told you in my first post in this thread.

You can write the hydrolysis as

RCOOR' + H2O :rarrow: RCOOH + HOR'

followed by the neutralization. Do you see now why the reaction is equivalent to adding an acid to the solution? It is not an acid to catalyze the hydrolysis, it is an acid that is produced by the hydrolysis.
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Offline Halogen876

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Re: Buffer with an Ester in it?
« Reply #8 on: June 10, 2014, 09:17:19 AM »
I do see now that an acid (or the salt of the acid) is produced in this reaction but if OH- is added you would first make the salt of the carboxylic acid (the conjugate base - as the equation shows in the question in my original post) and then this would react with water to make the acid? But since it's not a strong acid/base pair that reaction wouldn't go to completion would it?

Offline Borek

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Re: Buffer with an Ester in it?
« Reply #9 on: June 10, 2014, 02:33:47 PM »
I do see now that an acid (or the salt of the acid) is produced in this reaction but if OH- is added

OH- is not added.

You have a solution, buffered at 10.00 (which means it is 10-4 in OH-, but it doesn't matter much and you can safely ignore this information, it is not necessary to solve the problem).

You add the ester so that its concentration is 0.05F. Ester starts to hydrolyse. When the hydrolysis goes to completion, what you have is a solution that is identical to the one prepared by using original buffer solution at pH 10.00 and adding 0.05 moles of the acid and 0.05 moles of alcohol per liter.

This acid will protonate NH3, shifting buffer pH. In the worst case acid is strong enough for the protonation reaction to go to completion. You want buffer to be strong enough so that even if the protonation goes to completion pH change is below 0.1.
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Offline Halogen876

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Re: Buffer with an Ester in it?
« Reply #10 on: June 11, 2014, 03:12:03 PM »
Ok, I can see now that the OH- is not added. I think what that equation in my original post is showing is that the ester just reacts with OH- that is already present in the buffer (since there is lots of it due to the high pH of the buffer).

The question says you can assume a volume so if I assume a volume of 1L then that means that originally I have 0.0500 mol of ester. The reaction goes to completion so at the end of the reaction, you'd have 0.0500mol of alcohol and 0.0500mol of RCOO-, according to the equation given in the problem. I can see that after this happens, the solution would be identical to one prepared by adding 0.05 moles of alcohol and 0.05 moles of acid - my question now with that part is why is it acid and not the conjugate base like it shows in the problem?

Thanks for your continuing help and patience :)


Offline Borek

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Re: Buffer with an Ester in it?
« Reply #11 on: June 11, 2014, 03:53:58 PM »
my question now with that part is why is it acid and not the conjugate base like it shows in the problem?

Actually it doesn't matter, as both approaches are equivalent. It is just that the one with acid is easier to grasp.
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Offline Halogen876

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Re: Buffer with an Ester in it?
« Reply #12 on: June 12, 2014, 01:32:23 PM »
Ok, I think that clears up one big issue!

So now I think I might know basically how to start. The reaction goes to completion so the ester is completely gone so we don't need to worry about the ester at all. The only reason it was mentioned was because it tells you how many moles of acid (RCOOH) you're going to form (0.05moles) (assuming a volume of 1L). I think also the alcohol that is formed is irrelevant?

So then I need to assume the acid (RCOOH) reacts completely with NH3 to make NH4+.

Since I don't know exactly what acid RCOOH is, then the pKa value that I'll need to use to carry out the calculations, would be the pKa for the ammonia/ammonium system, since that is what wil be left after th ereaction goes to completion?

Offline Halogen876

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Re: Buffer with an Ester in it?
« Reply #13 on: June 14, 2014, 03:23:07 PM »

I got this question all figured out!

Thanks again for all your insight into it - it was very helpful and very much appreciated!  :)

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