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Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: laze80 on November 12, 2013, 12:29:01 AM

Title: Confusing Henderson-Hasselbalch Question
Post by: laze80 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
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: gritch 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.
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: kriggy on November 12, 2013, 01:38:39 AM
Do you have any ideas? What about the Henderson-Hasselbalch eq?
What about guilberg-waage law?
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: laze80 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.
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: Borek 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.
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: laze80 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.
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: Borek on November 12, 2013, 03:50:05 AM
Write Ka definition. What stops you from calculating the value?
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: laze80 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.
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: Borek 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.
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: laze80 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.
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: Borek 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?
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: laze80 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.

Title: Re: Confusing Henderson-Hasselbalch Question
Post by: laze80 on November 12, 2013, 01:45:36 PM
*The molar concentration of A- is equal to the molar concentration of H+
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: Borek 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.
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: zsinger on November 12, 2013, 02:24:25 PM
Gritch had it CORRECT from the start.
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: Borek on November 12, 2013, 03:04:43 PM
Gritch had it CORRECT from the start.

No, he got it wrong from the start, which was explained earlier in the thread. Easy to check that comparing just ratios of acid and conjugate base gives different order than the one given by calculation of Ka.
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: zsinger on November 12, 2013, 03:18:48 PM
Hah….I asked this same question to my kids on a Chem 2 A/B test, but I asked for Normalities, and gave the Acids.
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: gritch on November 13, 2013, 01:02:14 AM
Gritch had it CORRECT from the start.

No, he got it wrong from the start, which was explained earlier in the thread. Easy to check that comparing just ratios of acid and conjugate base gives different order than the one given by calculation of Ka.

I don't believe I suggested anything of the sort actually. I would propose rather we look at the [A-] relative to the overall concentration of every species( [A-]+[HA]), which I find mathematically simpler and more intuitive. Basically calculate the relative amount of depronotated form given the initial concentration. The higher the relative amount of deprotonated form the more acidic the acid should be.

For example:
A = 5/(0.1+5)=0.98
B=1/(1+1)=0.5
C=5/(10+5)=0.33
D=1/(1+0.1)=0.91
E=300/(100+300)=0.75
F=0.1/(1+0.1)=0.091

I could have been more precise but I thought given this was a test it would've been better to nudge in the correct direction rather than give the answer outright but since it's out there now
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: Borek on November 13, 2013, 03:21:25 AM
I don't believe I suggested anything of the sort actually.

That's how I read your proposal, it was a little bit ambiguous.

Quote
I would propose rather we look at the [A-] relative to the overall concentration of every species( [A-]+[HA]), which I find mathematically simpler and more intuitive. Basically calculate the relative amount of depronotated form given the initial concentration. The higher the relative amount of deprotonated form the more acidic the acid should be.

That's still wrong. You suggest to use dissociation degree to judge acid strength. That would work only for solutions of identical total acid concentration. As the totals are different, comparison of dissociation degree doesn't say anything.

For example solution C has 5 M of HA and 10 M of A-, so the dissociation degree is 2/3 or 0.66 (you calculated it as 1/3 by accidentally switching A with HA). Solution E has dissociation degree of 0.75 - so the substance from solution E is dissociated more. Does it mean it is a stronger acid? No! Ka for E is 0.9, while Ka for C is 20, so it is C that is a much stronger acid.

That's all because solution of C has a total concentration almost 40 times higher.
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: gritch on November 13, 2013, 02:26:01 PM
For example solution C has 5 M of HA and 10 M of A-, so the dissociation degree is 2/3 or 0.66 (you calculated it as 1/3 by accidentally switching A with HA). Solution E has dissociation degree of 0.75 - so the substance from solution E is dissociated more. Does it mean it is a stronger acid? No! Ka for E is 0.9, while Ka for C is 20, so it is C that is a much stronger acid.

I still don't agree with you. I admit I'm a bit rusty on the subject but something still feels off. You state at the Ka of E is 0.9 which I assume you got by assuming it was initially a typo and that [HA]=100mM and [A-]=300mM. If we assume it is not a typo (as ridiculous as that is) we get the value 0.9 for the Ka. Furthermore if we assume [HA]=100μM and [A-]=300μM we get the value 9×10-6.

Unless I did my calculations wrong (which I won't rule out) something seems off.  There seems to be some sort of of bias in the system due to ionic strength considerations. When comparing the inherit acidicities of two compounds I would argue it is inappropriate to account for ionic strength.

I honestly can't argue with your mathematics - that's how we define Ka and that's generally the metric used to determine acidity of a substance is just... doesn't feel right.
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: Borek on November 13, 2013, 04:47:04 PM
It is rather easy to find an example where your approach will fail for solutions where ionic strength is irrelevant.

Acid A, Ka=10-2, C=10-2 M, [HA]=3.8e-3 M, [H+]=6.2e-3 M, α=62%

Acid B, Ka=10-3, C=10-4 M, [HA]=8.4e-4 M, [H+]=9.1e-3 M, α=91%

Using your approach acid B looks stronger, when comparison of Ka clearly shows it is weaker.

You won't defend it, don't waste your time  :P
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: DrCMS on November 14, 2013, 05:37:24 AM
Acid B, Ka=10-3, C=10-4 M, [HA]=8.4e-4 M, [H+]=9.1e-3 M, α=91%

is [H+] not ~9.2e-4 M to get a Ka of ~10-3?

Anyway if we go back to the initial question then a calculation of Ka using the concentrations given (including the unlikely 100M and 300M) is F < B < D < C < A < E do we agree on that? 

If we assume the concentrations are 100mM and 300mM we get the Ka order F < E < B < D < C < A agreed?


However if you just look at the at relative amounts of A- to HA you get the wrong sequence given previously agreed? 

And that I assume is the point of this homework exercise to make the correct answer counter-intuitive but easy if you calculate the Ka rather than look and guess.
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: Borek on November 14, 2013, 06:17:59 AM
is [H+] not ~9.2e-4 M to get a Ka of ~10-3?

Actually 9.2e-5 M (9.161e-5), sorry about that. But the conclusion still holds.

Quote
Anyway if we go back to the initial question then a calculation of Ka using the concentrations given (including the unlikely 100M and 300M) is F < B < D < C < A < E do we agree on that?

If we assume the concentrations are 100mM and 300mM we get the Ka order F < E < B < D < C < A agreed?

Yes.

Quote
However if you just look at the at relative amounts of A- to HA you get the wrong sequence given previously agreed?

And that was my point all the time :)
Title: Re: Confusing Henderson-Hasselbalch Question
Post by: DrCMS on November 14, 2013, 09:26:10 AM
And that was my point all the time :)

I know which is why I usually defer to you on these pH problems but the thread seemed to be going off at a tangent so I thought I'd jump in. 

I still don't agree with you. I admit I'm a bit rusty on the subject but something still feels off. You state at the Ka of E is 0.9 which I assume you got by assuming it was initially a typo and that [HA]=100mM and [A-]=300mM. If we assume it is not a typo (as ridiculous as that is) we get the value 0.9 for the Ka. Furthermore if we assume [HA]=100μM and [A-]=300μM we get the value 9×10-6.


Unless I did my calculations wrong (which I won't rule out) something seems off

Yes it is because you did get your maths wrong,  if it is 100M and 300M Ka = 900, if it is 100mM and 300mM Ka = 0.9 and if it is 100μM and 300μM Ka = 9x10-4

However do you not think that would have been written 0.1M and 0.3M or 0.1mM and 0.3mM if that was what was meant.
100mM is just the wrong way to quote it.


I would propose rather we look at the [A-] relative to the overall concentration of every species( [A-]+[HA]), which I find mathematically simpler and more intuitive.

That may be so but it is the wrong approach.

There seems to be some sort of of bias in the system due to ionic strength considerations. When comparing the inherit acidicities of two compounds I would argue it is inappropriate to account for ionic strength.

I honestly can't argue with your mathematics - that's how we define Ka and that's generally the metric used to determine acidity of a substance is just... doesn't feel right.

Well it is right and the rest of the chemical world does use it so unless you want to be the lone voice start using it as well.