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Offline laze80

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Confusing Henderson-Hasselbalch Question
« on: November 12, 2013, 12:29:01 AM »
I am having the most difficult time finding out how to answer this question and my take home test is due tomorrow. Any explanation would save my life. Thank you.


9. You have some different unknown acids, which you must rank in terms of weakest to strongest.
All that you know is the concentrations of each acid and conjugate base at equilibrium.  Given the acid/base equilibrium reaction and the concentrations of each, rank the acids from weakest to strongest.  REMEMBER YOUR MOLARITY!!! (4 points)




Acid #   [AH]      [A-]   
   
 A      0.1M      5M      
 B      1M      1M      
 C      5M      10M      
 D      100mM      1M      
 E      100M      300M         
 F      1M      100mM      


weakest   ____   ____   ____   ____   ____   ____  strongest

Offline gritch

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Re: Confusing Henderson-Hasselbalch Question
« Reply #1 on: November 12, 2013, 01:34:49 AM »
I am having the most difficult time finding out how to answer this question and my take home test is due tomorrow. Any explanation would save my life. Thank you.

I'll not do your test for you but maybe a little nudge might help. For a strong acid would you expect a higher concentration of the acid or conjugate base form? Remember your definition of a strong acid and you should be able to piece this one together without even touching Henderson-Hasselbalch equation.

Offline kriggy

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Re: Confusing Henderson-Hasselbalch Question
« Reply #2 on: November 12, 2013, 01:38:39 AM »
Do you have any ideas? What about the Henderson-Hasselbalch eq?
What about guilberg-waage law?

Offline laze80

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Re: Confusing Henderson-Hasselbalch Question
« Reply #3 on: November 12, 2013, 01:55:24 AM »
Ok. Thanks for the tips.

Well I know a strong acid is one which is virtually 100% ionized in water. Which would mean that at equilibrium, [A-]/[HA] would have to be largest for the strongest acids and lowest for the weakest acids?

Using that logic I would guess the correct answer, weakest to strongest, to be;
F,B,C,E,D,A

Any feedback would be appreciated.

Offline Borek

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Re: Confusing Henderson-Hasselbalch Question
« Reply #4 on: November 12, 2013, 03:02:04 AM »
I don't think it is about Henderson-Hasselbalch equation, I believe you are expected to calculate Ka for each acid from the data given (which requires one assumption). Than just order them by Ka values.

Comparing [A-]/[HA] can be misleading, as it can be high for a diluted weak acid, and lower for a concentrated much stronger acid.

At first sight looks like you are right about F being the weakest, but it is E that is the strongest (I have not checked those in between). However, I wonder if you have not made a mistake entering the data here, as 300M solution looks like a typo - shouldn't it be 300 mM? If so, it is A that is the strongest.
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Offline laze80

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Re: Confusing Henderson-Hasselbalch Question
« Reply #5 on: November 12, 2013, 03:05:24 AM »
I was looking online as to how to calculate Ka when only given the molarity and couldn't find the answer. Could you please advise me the correct way to do this? Thanks.

Offline Borek

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Re: Confusing Henderson-Hasselbalch Question
« Reply #6 on: November 12, 2013, 03:50:05 AM »
Write Ka definition. What stops you from calculating the value?
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Offline laze80

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Re: Confusing Henderson-Hasselbalch Question
« Reply #7 on: November 12, 2013, 11:30:21 AM »
In my notes it says the equation for Ka is the concentration of the reactants over the concentration of the products, or [A-][H+]/[HA]. But in the question I am not given the concentration of H+...I feel like I am missing an obvious point but cannot figure it out.

Offline Borek

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Re: Confusing Henderson-Hasselbalch Question
« Reply #8 on: November 12, 2013, 11:40:11 AM »
You are on the right track, and yes, you are missing obvious ;)

Take a look at the dissociation reaction:

HA ::equil:: H+ + A-

What does it tell you about how much H+ was produced? Think simple stoichiometry.
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Offline laze80

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Re: Confusing Henderson-Hasselbalch Question
« Reply #9 on: November 12, 2013, 01:27:21 PM »
I'm so sorry, but I am lost. I have racked my brain over this for over a day. I would be very grateful if you could tell me what concept I am missing. Thanks for all of your help.

Offline Borek

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Re: Confusing Henderson-Hasselbalch Question
« Reply #10 on: November 12, 2013, 01:42:35 PM »
1 mole of HA when dissociated produces 1 mole of A- and 1 mole of H+.

2 moles of HA when dissociated produces 2 moles of A- and 2 moles of H+.

n moles of HA when dissociated produces n moles of A- and n moles of H+.

1=1
2=2
n=n

Do you see it now?
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Offline laze80

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Re: Confusing Henderson-Hasselbalch Question
« Reply #11 on: November 12, 2013, 01:44:50 PM »
So essentially A- is H+?

The part that boggles my mind is how I didn't find this during my hours of research. I feel slightly moronic. Thanks again.


Offline laze80

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Re: Confusing Henderson-Hasselbalch Question
« Reply #12 on: November 12, 2013, 01:45:36 PM »
*The molar concentration of A- is equal to the molar concentration of H+

Offline Borek

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Re: Confusing Henderson-Hasselbalch Question
« Reply #13 on: November 12, 2013, 01:59:58 PM »
*The molar concentration of A- is equal to the molar concentration of H+

Yes.

Note: it is not universal. It holds as long as the acid in question is the only important source of H+ (so no other acids present and concentration and dissociation degree high enough so that the water autodissociation can be safely ignored). That's the case here, all concentrations are large enough.

The part that boggles my mind is how I didn't find this during my hours of research.

Happens to everyone.
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Offline zsinger

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Re: Confusing Henderson-Hasselbalch Question
« Reply #14 on: November 12, 2013, 02:24:25 PM »
Gritch had it CORRECT from the start.
"The answer is of zero significance if one cannot distinctly arrive at said place with an explanation"

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