Who do you side with for this one?

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Spiker

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HX is generally used as a buffer. If pH 8.0 is a good buffering region for X, then:

I. the pKa of HX must be near pH 8.0
II. if HX is titrated with acid, the titration curve will possess a steep region near pH 8.0.
III. a great deal of NaOH would have to be added to pH 8.0 HX in order to significantly affect the pH.

A. I only
B. III only
C. I and II only
D. I and III only

I picked A because I is ture. III could be true but that depend on how concentrated the X and HX are to started with.

However the book say it is D.

What do you think?
 
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I would have said A too...but i guess I can see how they pick D? But III is kind of a vague statement IMO..

makes sense...but i would have gotten it wrong too..sorry that was not helpful haha
 
D for sure... and III is essentially the opposite of II... so its either C or D...

its a BUFFER!... buffers stop changes in pH, or slow them rather, so you need more of an acid or base to affect Ph. so thats what III says.
 
D for sure... and III is essentially the opposite of II... so its either C or D...

its a BUFFER!... buffers stop changes in pH, or slow them rather, so you need more of an acid or base to affect Ph. so thats what III says.

But if the concentration of buffer is like 0.0001/0.0001 it isnt going to help much.
 
well the question is assuming you're using the HX buffer in a solution... and you wouldn't buffer something with that little.
 
well the question is assuming you're using the HX buffer in a solution... and you wouldn't buffer something with that little.


I think the point is that the question is vague. I would have also picked A.
 
If you work it out you would have to add 5.4 x 10^19 molecules of NaOH to increase the PH by 1 to PH = 9. I would consider that a great deal of NaOH.

Math

PH + POH = 14

POH = 6
POH = -log[OH]

[OH] = .00001
we want to increase PH by 1 or decrease POH by 1 which would mean we want [OH] = .0001

We would have to add .00009.

.00009 x (6.022x10^23) = 5.4 x 10^19 molecules of OH
 
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I could definitely see why you would pick A. The question never specifies anything about the concentration of the HX.


However, if the question is telling you that HX is generally used as a buffer, I would just assume that there is a significant amount of HX in the solution. Because, like "Doodl3s" said, you wouldn't use such a small quantiy of something if you were using it as a buffer.
 
If you work it out you would have to add 5.4 x 10^19 molecules of NaOH to increase the PH by 1 to PH = 9. I would consider that a great deal of NaOH.

Math

PH + POH = 14

POH = 6
POH = -log[OH]

[OH] = .00001
we want to increase PH by 1 or decrease POH by 1 which would mean we want [OH] = .0001

We would have to add .00009.

.00009 x (6.022x10^23) = 5.4 x 10^19 molecules of OH

That's assuming this solution is one liter though right?

The problem is that this question never said anything about volumes or concentrations.
 
If you work it out you would have to add 5.4 x 10^19 molecules of NaOH to increase the PH by 1 to PH = 9. I would consider that a great deal of NaOH.

Math

PH + POH = 14

POH = 6
POH = -log[OH]

[OH] = .00001
we want to increase PH by 1 or decrease POH by 1 which would mean we want [OH] = .0001

We would have to add .00009.

.00009 x (6.022x10^23) = 5.4 x 10^19 molecules of OH


Dude, I see what you're saying, but you're talking molecules... That's ridiculous for a problem where we're discussing concentration without values.

The real value is the mole value, which is 9.0x10^-5. That's not much if you calculate the grams of OH then you have:

9.0x10^-5/1.7x10^1 = ~ 5.3x10^-4 grams I wouldn't call that a large amount, so that's where the ambiguity comes from.
 
HX is generally used as a buffer. If pH 8.0 is a good buffering region for X, then:

I. the pKa of HX must be near pH 8.0
II. if HX is titrated with acid, the titration curve will possess a steep region near pH 8.0.
III. a great deal of NaOH would have to be added to pH 8.0 Tris in order to significantly affect the pH.

A. I only
B. III only
C. I and II only
D. I and III only

I picked A because I is ture. III could be true but that depend on how concentrated the X and HX are to started with.

However the book say it is D.

What do you think?

D for sure. It's a buffer-you have to add a lot of NaOH to cause a pH change.
 
Dude, I see what you're saying, but you're talking molecules... That's ridiculous for a problem where we're discussing concentration without values.

The real value is the mole value, which is 9.0x10^-5. That's not much if you calculate the grams of OH then you have:

9.0x10^-5/1.7x10^1 = ~ 5.3x10^-4 grams I wouldn't call that a large amount, so that's where the ambiguity comes from.

III. a great deal of NaOH would have to be added to pH 8.0 Tris in order to significantly affect the pH.

its talking about the molecules not grams, i just showed the math to prove it would be alot. It should be known that it would take alot to increase the basicity of a already basic solution.
 
That's assuming this solution is one liter though right?

The problem is that this question never said anything about volumes or concentrations.

The definition of ph is:

ph = -log[H+]

If you know the PH you know the H+ or OH- concentrations. You do not have to be given the volume since ph inherently contains this value.
 
If you work it out you would have to add 5.4 x 10^19 molecules of NaOH to increase the PH by 1 to PH = 9. I would consider that a great deal of NaOH.

Math

PH + POH = 14

POH = 6
POH = -log[OH]

[OH] = .00001
we want to increase PH by 1 or decrease POH by 1 which would mean we want [OH] = .0001

We would have to add .00009.

.00009 x (6.022x10^23) = 5.4 x 10^19 molecules of OH

That's still less than 1 mol of OH-, no?
 
Yup its less then 1mol, 1 mol is 6.022x10^23 molecules.

I know, I was just trying to point out that since it's less than 1 mol, "a great deal" may be a bit vague, even though I still stand by D since it's a buffer!
 
I know, I was just trying to point out that since it's less than 1 mol, "a great deal" may be a bit vague, even though I still stand by D since it's a buffer!

Its saying a great deal of "NaOH", to me that implies the "molecule" as the unit. Not moles, not grams, just molecules, but yes the term "a great deal" is very a very relative term and kind of a ****ty answer choice because you have nothing to compare "a great deal" to.
 
Its saying a great deal of "NaOH", to me that implies the "molecule" as the unit. Not moles, not grams, just molecules, but yes the term "a great deal" is very a very relative term and kind of a ****ty answer choice because you have nothing to compare "a great deal" to.


DING!!!


Exactly! I appreciate your earlier explanation, but I still think the questions is a bit vague. I see why it should be D, I would have selected A though prior to this thread.

On another note, I've done 6 of the AAMC exams and I've yet to see a question that ambiguous, so while it is still possible to get one, it is unlikely.
 
DING!!!


Exactly! I appreciate your earlier explanation, but I still think the questions is a bit vague. I see why it should be D, I would have selected A though prior to this thread.

On another note, I've done 6 of the AAMC exams and I've yet to see a question that ambiguous, so while it is still possible to get one, it is unlikely.

You will notice that AAMC exam questions wont have such a grey area because they need to be able to defend their answers choices to a T. On the other hand prep companies could care less if you agree or disagree with their answers lol.

I take my MCAT the 14th, im going to start the AAMC exams in about 4 days. Haven't taken a diagnostic test yet, just been preping for close to 2 months now.
 
You will notice that AAMC exam questions wont have such a grey area because they need to be able to defend their answers choices to a T. On the other hand prep companies could care less if you agree or disagree with their answers lol.

I take my MCAT the 14th, im going to start the AAMC exams in about 4 days. Haven't taken a diagnostic test yet, just been preping for close to 2 months now.


Yeah, I would have to agree with you.


By the sounds of it, I'm sure you'll do great on your practice exams. Good luck!
 
Its saying a great deal of "NaOH", to me that implies the "molecule" as the unit. Not moles, not grams, just molecules, but yes the term "a great deal" is very a very relative term and kind of a ****ty answer choice because you have nothing to compare "a great deal" to.


NaOH doesn't occur as a molecule in the laboratory. If you ask for some NaOH, you're going to get a solution. The question is basically (no pun intended) testing to see if you know the difference between a buffered solution and a non-buffered solution. When comparing the two together, the buffered solution is going to require a GREAT DEAL of NaOH.
 
The definition of ph is:

ph = -log[H+]

If you know the PH you know the H+ or OH- concentrations. You do not have to be given the volume since ph inherently contains this value.


Yeah, but you used that calculation to obtain an ABSOLUTE value of molecules of NaOH needed (you said it was xxx number of molecules)

If this solution was contained in a 100,000 gallon fish tank, you would obviously need a lot more NaOH than if this solution was contained in a shot glass.

If it was a 100,000 gallon fish tank, a "great deal" of NaOH would be needed. If it was in a shot glass, then not much NaOH would be needed.
 
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