Chem Buffer Question (TBR)

[moto_za]

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Would greatly appreciate some help with understanding this question:

Which mixture does NOT produce a buffer?

A . H3CCO2H with 2 equivalents of H3CCO2K
B . NH3 with 2 equivalents of NH4CI
C . H2CO3 1.5 equivalents of KOH
D . H3CNH2 with 1.5 equivalents of HCI

 

[BerkReviewTeach]

Would greatly appreciate some help with understanding this question:

Which mixture does NOT produce a buffer?

A . H3CCO2H with 2 equivalents of H3CCO2K
B . NH3 with 2 equivalents of NH4Cl
C . H2CO3 1.5 equivalents of KOH
D . H3CNH2 with 1.5 equivalents of HCl

This question starts with identifying what forms a buffer. A buffer is formed when the solution contains roughly equal mole portions of a weak acid and its conjugate base. So for this question, you need to look at each answer choice one-by-one and decide if that is what forms after mixture (and in some cases reaction).

Choice A is a weak acid (H3CCO2H) with its conjugate base (H3CCO2-) in a 1 : 2 ratio, which as far as buffers go is a roughly equal mole quantity. Choice A makes a buffer with the pH about 0.3 greater than the pKa of H3CCO2H.

Choice B is a weak base (NH3) with its conjugate acid (NH4+) in a 1 : 2 ratio, which as far as buffers go is a roughly equal mole quantity. Choice B makes a buffer with the pH about 0.3 less than the pKa of NH4+.

Choice C involves the mixture of a weak diprotic acid with a strong base, so there will be a reaction to consider. There are 1.5 equivalents of strong base, so the first proton will be completely removed and the second proton will be half removed. This will result in a solution of a weak acid (HCO3-) in equal mole quantity with its conjugate base (CO32-). Choice C makes a buffer with the pH equal to the pKa of HCO3- (which happens to be pKa2 for H2CO3).

Choice D involves the mixture of a weak base with a strong acid, so there is a reaction to consider again. There are 1.5 equivalents of strong acid, but the base only has one lone pair to donate. This will result in a solution of one full equivalent of weak acid (H3CNH3+) plus a half-equivalent of strong acid (HCl) leftover. Choice D results in a mixture of a weak acid and a strong acid, which does not result in a buffer.

This was a great deal of work for a single question, but if you thoroughly work through each answer choice, you learn a great deal about buffers.

 

[loveoforganic]

A buffer is a mixture of a weak acid and its weak conjugate base within the range of a 10:1 or 1:10 ratio.

So let's start by identifying those answers that obviously meet that condition.

A. A mixture of acetic acid and its conjugate base, acetate ion, in a 1:2 ratio. Buffer.

B. A mixture of ammonia and its conjugate acid, ammonium ion, in a 1:2 ratio. Buffer.

C. Not obviously a buffer.

D. Not obviously a buffer.

So for C and D, it looks like we have a weak acid (carbonic acid) reacting with a strong base (potassium hydroxide) and a weak base (methylamine) reacting with a strong acid (hydrochloric acid), respectively.

Let's start by looking at how many protons the acids can donate and the bases can accept.

Carbonic acid has two acidic protons (it's a dicarboxylic acid), HCl has one acidic proton.

Methylamine can accept one proton, KOH can accept one proton.

Now let's write balanced equations for the reactions of each.

H2CO3 + 2KOH --> K2CO3 + 2H2O

H3CNH2 + HCl --> H3CNH3Cl

Now let's determine which makes a buffer solution.

C. H2CO3 + 2KOH --> K2CO3 + 2H2O

It takes 2 equivalents of KOH to fully neutralize (take all the protons from) H2CO3. We only used 1.5 equivalents however. So what're we left with? The first equivalent pulls off one proton in the reaction

H2CO3 + KOH --> KHCO3 + H2O

The half equivalent then reacts with half of the remaining KHCO3 in the reaction

KHCO3 + KOH --> K2CO3 + H2O

You're left with a mixture of KHCO3 and K2CO3, a weak acid and its conjugate base in a 1:1 ratio. Buffer.

So we know the answer is D, but we'll look at why.

D. H3CNH2 + HCl --> H3CNH3Cl

The first equivalent of HCl reacts with all of the methylamine, as written in the above reaction. So what happens to the remaining half equivalent of HCl? Nothing! It has nothing to react with, so it dissociates fully in solution, as a strong acid, lowering the pH significant. Not a buffer.


Edit: Beat to it!!

 

[PltEnthusiast]

Deleted

 

[BerkReviewTeach]

I'm really sorry LoOrg, because you spent so much time putting together a great answer. Sorry to steal your thunder by seconds.

To figure out the equivalents, you look at the number of Hs and there are 3 in this case (H3CNH2) so, half equivalents will be 1.5. Thus, 1/5 equivalents of the strong acid (HCl).

Hey Unforg, you have a great explanation up until this point. There are not three Hs to lose on that compound, as Hs on carbon don't get deprotonated very easily, and if it were to lose a proton, it be lost from N. At most it would lose only one proton, and it would take an extremely strong base such as BuLi to do it. In this question, under aqueous conditions, H3CNH2 is a base with one lone pair to donate. That is the problem with this answer choice, because 1.5 eq of strong acid will completely protonate it and not result in any conjugate base being left.

 

[loveoforganic]

I was kidding. No hard feelings.

 

[PltEnthusiast]

Oh great! I don't know what I wrote. You are absolutely right. When I thought about the answer, I used the right strategy and got it all mixed up when I started typing.

I guess I should sleep instead of giving wrong explanations 🙁 Sorry..

 

[BerkReviewTeach]

Oh great! I don't know what I wrote. You are absolutely right. When I thought about the answer, I used the right strategy and got it all mixed up when I started typing.

I guess I should sleep instead of giving wrong explanations 🙁 Sorry..

Yeah, I figured that must have happened, because you have such a systematic and logical answer going except for one small part. At least it's just a silly online forum, because I tend to screw up and type my half-awake messages in emails to my boss or lab director. Damn I wish there was an "unsend" button on email. 😳

 

[loveoforganic]

There's always a solution...