TBR Chemistry Buffers and Titrations

[fluff]

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8. Which mixture does NOT produce a buffer?

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

I understand the explanations for C and D, but why would A and B both make good buffers? I thought that, in order to make a good buffer, you want equimolar or equivalent amounts of the weak acid and its conjugate base?? Someone please help?

 

[joshto]

8. Which mixture does NOT produce a buffer?

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

I understand the explanations for C and D, but why would A and B both make good buffers? I thought that, in order to make a good buffer, you want equimolar or equivalent amounts of the weak acid and its conjugate base?? Someone please help?

to make a good buffer, you need to have weak acid (or a weak base), with its conjugate, present in a 10:1 or 1:10 molar ratio. You don't need exactly 1:1 ratios. this is why for buffers, the pH can range to pKa +/- 1

 

[puffylover]

what's the answer for this? i know a and b are buffers... but c and d both don't qualify as buffers, no?

 

[09grad]

I'll take it choice by choice:

D: wrong. 1 equiv of HCl reacts with the base to form H3CNH3. The other half equiv of HCl reacts with H3CNH3 to form H3CNH4+. Thus, we have H3CNH4+ and H3CNH3 in equilibrium.

C: wrong. 1 equiv of KOh reacts with H2CO3 to form HCO3-. The remaining half equiv of KOh reacts with HCO3- to form CO3(2-). Thus, HCO3- is in equilib with its conjugate base.

B: wrong. NH3 reacts with 1 equivalent of NH4Cl (an acid) to form NH4+ and NH3. However, the original NH4+ became NH3 after this reaction. Thus, we have some equilibrium of NH4+ and NH3. Whenever they react, the conjugate acid (NH4+) becomes the conjugate base (NH3), so theres always some combination of these.

A: correct. 1 equiv of H3CCO2K reacts with the H3CCO2H to form H3CCO2-. The remaining equiv of H3CCO2K does not react, and so all we have left is base.

 

[puffylover]

C. understand

D. H3CNH4+ or H3CNH3+? aren't there too many H's? don't really understand this.

my thought process. 1 eq of HCl reacts with H3CNH2 to form H3CNH3+. what happens the leftover 0.5 equivalent of HCl? There's nothing else to protonate, right?

B. kinda get... don't really get the significance of the 2M though. could it be anything? like 1M and it wouldn't matter?

A. why would 1 equiv of H3CCO2K reacts with the H3CCO2H to form H3CCO2-? Is this complete (H3CCO2K isn't a strong base is it?)

 

[loveoforganic]

D: H3CN3+ is correct

B: Where is 2M present? It says 2 equivalents. They aren't the same thing. An equivalent just means one substance is reacted with an equal amount of moles of another substance.

A: There is no reaction here - this is just a formed buffer solution. A weak base (H3CCO2K) and its conjugate weak acid (H3CCO2H)

 

[09grad]

Yeah my bad. I counted one too many hydrogen's on the N

Choice A, a combination of weak acid and weak base directly forms a buffer, even with no reaction.

Choice B is one equivalent conjugate base with 2 of conjugate acid. Because they are in less than 10:1 (or 1:10) ratio, it's a buffer.

In D, all of the conjugate base is protonated to form conjugate acid, therefore it does not form a buffer