Buffer solution question

[bidiboom]

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Hi guys,

I am lost in a question and I think there is a problem with the question per se 🙁 its from Kaplan SAT-Chem:

For each of chemicals below, the amounts shown are dissolved, together, in water to make a 1.0 L of solution. Which mixture will form a buffer solution?

A) 0.2 mol of HCl and 0.1 mol of K2SO3
B) 0.2 mol of NaOH and 0.4 mol of HF
C) 0.1 mol of HBr and 0.1 mol of Ba(OH)2
D) 0.4 mol of HCl and 0.2 mol of NH3
E) 0.2 mol of NaOH and 0.4 mol of HCl

To me, there is no correct answer among them, since to have a buffer sol. we need a weak acid and its conjugate base, vice versa.. the only weak acid here is HF, but NaOH is not its conjugate base, its an irrelevant strong base and consumes HF to the end..

If K2SO3 would be something like KHSO3, then I would take it as a weak conjugate base, but with 2 K+ ions, it is not conjugate of H2SO3 as well..

Am I wrong? Please correct me if I am, I have only a couple of weeks before the test and I feel distressed even now

Thanks a lot..

Oh, by the way, the answer of the book is B.

 

[Erythropoietin]

Hi guys,

I am lost in a question and I think there is a problem with the question per se 🙁 its from Kaplan SAT-Chem:

For each of chemicals below, the amounts shown are dissolved, together, in water to make a 1.0 L of solution. Which mixture will form a buffer solution?

A) 0.2 mol of HCl and 0.1 mol of K2SO3
B) 0.2 mol of NaOH and 0.4 mol of HF
C) 0.1 mol of HBr and 0.1 mol of Ba(OH)2
D) 0.4 mol of HCl and 0.2 mol of NH3
E) 0.2 mol of NaOH and 0.4 mol of HCl

To me, there is no correct answer among them, since to have a buffer sol. we need a weak acid and its conjugate base, vice versa.. the only weak acid here is HF, but NaOH is not its conjugate base, its an irrelevant strong base and consumes HF to the end..

If K2SO3 would be something like KHSO3, then I would take it as a weak conjugate base, but with 2 K+ ions, it is not conjugate of H2SO3 as well..

Am I wrong? Please correct me if I am, I have only a couple of weeks before the test and I feel distressed even now

Thanks a lot..

Oh, by the way, the answer of the book is B.

Choice E is out right away because you can't make a buffer with two strong reagents.
Choices A&D say exactly the same thing: For an equivalent of weak base, two equivalent of strong reagent is needed. Which is incorrect.
Choice C is incorrect because you can only have 1:1 equivalency with a weak acid/base and its conjugate.
Choice B is the right answer because you have 1 equivalent of weak acid and half equivalent of a strong base.

FYI:
-you an only have 1:1 equivalency for weak reagents and their conjugates
-If you have weak + strong (like HF & NaOH), the only way to make a buffer is to have X moles of weak + X/2 moles of strong.

 

[Erythropoietin]

Forgot to mention that C & E can be eliminated for the same reason because Group II hydroxides are strong bases... So strong + strong doesn't make a buffer.

 

[bidiboom]

Choice E is out right away because you can't make a buffer with two strong reagents.
Choices A&D say exactly the same thing: For an equivalent of weak base, two equivalent of strong reagent is needed. Which is incorrect.
Choice C is incorrect because you can only have 1:1 equivalency with a weak acid/base and its conjugate.
Choice B is the right answer because you have 1 equivalent of weak acid and half equivalent of a strong base.

FYI:
-you an only have 1:1 equivalency for weak reagents and their conjugates
-If you have weak + strong (like HF & NaOH), the only way to make a buffer is to have X moles of weak + X/2 moles of strong.

Omg.. I didnt know that.. thank you really.. especially the last point you mention, 1 weak:half strong ratio, I didnt see it anywhere and its good to have this tip.. actually of course it should be that way, right? the strong base will consume all of the H+ to H2O(l) and the conjugate base came out, F-, will create the common ion effect.. I have never looked at the case from this perspective, half amount of stong acid/base I mean.. thank you once again 🙂

 

[Erythropoietin]

Omg.. I didnt know that.. thank you really.. especially the last point you mention, 1 weak:half strong ratio, I didnt see it anywhere and its good to have this tip.. actually of course it should be that way, right? the strong base will consume all of the H+ to H2O(l) and the conjugate base came out, F-, will create the common ion effect.. I have never looked at the case from this perspective, half amount of stong acid/base I mean.. thank you once again 🙂

You want to use only the half equivalent of a strong reagent, that way you can be at the buffer region. If you use 1 equivalent of a strong reagent with something weak, you will reach the equivalence point.
Anyways if you have TBR, they explain this clear in chapter 5.

 

[bidiboom]

Ok ok, I got the point.. thank you 🙂

 

[IH8ColdWeath3r]

You want to use only the half equivalent of a strong reagent, that way you can be at the buffer region. If you use 1 equivalent of a strong reagent with something weak, you will reach the equivalence point.
Anyways if you have TBR, they explain this clear in chapter 5.

yeah TBR really hammers this point across in their chem book. For everyone who already doesn't have it, GO GET IT!!

 

[Erythropoietin]

yeah TBR really hammers this point across in their chem book. For everyone who already doesn't have it, GO GET IT!!

out of ~40 chapters on mcat, buffers&titrations was my worst topic..but after reading the TBR chapter on it made it my #1 best topic now lol