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### Topic: Confusing Henderson-Hasselbalch Question  (Read 7811 times)

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#### Borek ##### Re: Confusing Henderson-Hasselbalch Question
« Reply #15 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.
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#### zsinger

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« Reply #16 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.
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#### gritch ##### Re: Confusing Henderson-Hasselbalch Question
« Reply #17 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

#### Borek ##### Re: Confusing Henderson-Hasselbalch Question
« Reply #18 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.
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#### gritch ##### Re: Confusing Henderson-Hasselbalch Question
« Reply #19 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.

#### Borek ##### Re: Confusing Henderson-Hasselbalch Question
« Reply #20 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.

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#### DrCMS

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« Reply #21 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.

#### Borek ##### Re: Confusing Henderson-Hasselbalch Question
« Reply #22 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 ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### DrCMS

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« Reply #23 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.