Buffer

[Temperature101]

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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

The answer is D...Can someone explain the reasoning behind each answer choice? Why 2 equivalents in answer choices A and B?

 

[chiddler]

A forms a buffer because there are roughly equal equivalents of HA and A-.

B forms a buffer for the same reason.

C forms a buffer because 1 equivalent removes one proton from the H2CO3, leaving HCO3-. The last half equivalent leaves the solution with 0.5 CO3(2-) and 0.5 HCO3(-) which is a good buffer, yes? HCO3(-) is like HA, and CO3(2-) is like A-. So this works well.

D. This is a monoprotic weak base, so after 1 equivalent of HCl, all the base is protonated (H3CNH3+ forms). Another half equivalent of HCl is overkill; there is no more base to protonate! This is not a good buffer.

 

[Temperature101]

A forms a buffer because there are roughly equal equivalents of HA and A-.

B forms a buffer for the same reason.

C forms a buffer because 1 equivalent removes one proton from the H2CO3, leaving HCO3-. The last half equivalent leaves the solution with 0.5 CO3(2-) and 0.5 HCO3(-) which is a good buffer, yes? HCO3(-) is like HA, and CO3(2-) is like A-. So this works well.

D. This is a monoprotic weak base, so after 1 equivalent of HCl, all the base is protonated (H3CNH3+ forms). Another half equivalent of HCl is overkill; there is no more base to protonate! This is not a good buffer.

I understand why C is correct and D is wrong. I happen to get the right answer for that question. But I dont understand why in A and B they use 2 equivalents of H3CCO2K and NH4CL respectively. Why not using one (1) equivalent of them since they are the salts of the conjugate base (answer A) and conjugate acid (answer B).

 

[chiddler]

Lets pretend the pKa is 5. So lets plug in the numbers to the H.H equation:

pH = pKa + log (conjB/Acid)
pH = 5 + log (H3CCO2K / H3CCO2H)
pH = 5 + log (2/1) = 5 + log (2) = 5.3

This works. Why do these calculations show that this works? Because, according to the book (I assume this is empirically confirmed), buffers are effective when pH = pKa + or - 1. In other words, the log(conjB/Acid) must be between -1 and +1.

This means that it is ineffective if the HA/A- ratio or the A-/HA ratio exceed 10:1. This ratio is 2:1 so therefore it is a workable buffer system.

 

[Temperature101]

Lets pretend the pKa is 5. So lets plug in the numbers to the H.H equation:

pH = pKa + log (conjB/Acid)
pH = 5 + log (H3CCO2K / H3CCO2H)
pH = 5 + log (2/1) = 5 + log (2) = 5.3

This works. Why do these calculations show that this works? Because, according to the book (I assume this is empirically confirmed), buffers are effective when pH = pKa + or - 1. In other words, the log(conjB/Acid) must be between -1 and +1.

This means that it is ineffective if the HA/A- ratio or the A-/HA ratio exceed 10:1. This ratio is 2:1 so therefore it is a workable buffer system.

Thanks..Got it now. Would never thought of using the H.H equation on that one.

 

[SaintJude]

Also, since time is of the essence on the MCAT, Kaplan gives you a quick rule on eliminating buffers. They can never include a strong acid, like HCl in choice D. That takes you like 5 seconds..

So, I guess you could do all the math to understand it in practice, but you may never have to on the exam.

 

[theseeker4]

Also, since time is of the essence on the MCAT, Kaplan gives you a quick rule on eliminating buffers. They can never include a strong acid, like HCl in choice D. That takes you like 5 seconds..

So, I guess you could do all the math to understand it in practice, but you may never have to on the exam.

Hmm. While this may be true in the questions they usually ask on the MCAT, buffers CAN include a strong acid, as long as there is a weak base. As long as the acid is added to such an extent that ~1/2 of the base is protonated, you have a buffer. Many buffers are made by adding a strong acid to a weak base, or a strong base to a weak acid.

Kaplan may be right in that they usually don't ask questions with a strong base being in a buffer, but in the real world, buffers can indeed be made with a strong acid (or base).

 

[MT Headed]

Also, since time is of the essence on the MCAT, Kaplan gives you a quick rule on eliminating buffers. They can never include a strong acid, like HCl in choice D. That takes you like 5 seconds..

So, I guess you could do all the math to understand it in practice, but you may never have to on the exam.

I don't like it. Answer C has a strong base, and yet it does create a buffer.

Answer D, if it only had 0.5 equivalents of HCl, would have created a buffer.

A better approach would be that a buffer can never have leftover strong acid or strong base.