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Topic: Problem of the week - 20/08/2012  (Read 8967 times)

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Offline Borek

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Problem of the week - 20/08/2012
« on: August 20, 2012, 09:24:36 AM »
Concentration of A2- in the solution of a weak, diprotic acid H2A, was determined to be 5.1×10-10 M. Drop of a methyl orange added to the same solution colors it red. Can you calculate pKa2 of the acid?
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Offline Rutherford

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Re: Problem of the week - 20/08/2012
« Reply #1 on: August 20, 2012, 09:47:47 AM »
Not sure, but after a lot of approximations I got pKa2≈9.3.

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Re: Problem of the week - 20/08/2012
« Reply #2 on: August 20, 2012, 02:20:07 PM »
after a lot of approximations

Care to list them?
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Offline Rutherford

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Re: Problem of the week - 20/08/2012
« Reply #3 on: August 20, 2012, 03:30:42 PM »
Methyl orange became red=very acidic solution. The concentration of A2- is very low, so it means that Ka1 is enough big (therefore the solution is acidic).
H2A :rarrow: HA-+H+ Ka1 reasonable big
HA- ::equil:: A2-+H+ Ka2=?
If I mark the concentration of H2A by x, then the concentration of HA- and H+ are both x, too.
Ka2=[A2-]*[H+]/[HA-]
The concentration of A2- is very low, meaning that Ka2 is very low, too, so the amount of H+ produced can be neglected to the amount of H+ produced in the first step. The amount of HA- that dissociated can also be neglected, so:
Ka2=5.1×10-10*x/x=5.1×10-10
pKa2≈9.3.

Offline Borek

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Re: Problem of the week - 20/08/2012
« Reply #4 on: August 20, 2012, 03:37:33 PM »
If I mark the concentration of H2A by x, then the concentration of HA- and H+ are both x, too.

This part is wrong. You are right Ka1 is much higher than Ka2, but you don't know if it is high enough for H2A to be completely dissociated. Besides, what you wrote suggests [H2A]=[HA-]=[A2-] - are you sure that's what you mean?
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Offline Rutherford

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Re: Problem of the week - 20/08/2012
« Reply #5 on: August 20, 2012, 04:10:06 PM »
If I mark the concentration of H2A by x, then the concentration of HA- and H+ are both x, too.

This part is wrong. You are right Ka1 is much higher than Ka2, but you don't know if it is high enough for H2A to be completely dissociated. Besides, what you wrote suggests [H2A]=[HA-]=[A2-] - are you sure that's what you mean?
I am not sure how did you mean that the thing I wrote suggests that. If Ka2 is very low, that can't be true.

Offline Borek

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Re: Problem of the week - 20/08/2012
« Reply #6 on: August 20, 2012, 04:13:10 PM »
Sorry, a typo.

[H2A]=[HA-]=[H+]
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Offline Rutherford

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Re: Problem of the week - 20/08/2012
« Reply #7 on: August 20, 2012, 04:18:59 PM »
Okay, using now a different approach:
H2A ::equil:: HA-+H+ Ka1
y-x        x     x
Ka1=x2/(y-x), from this y-x=x2/Ka1

Ka1*Ka2=[H+]2[A2-]/[H2A]
Ka1*Ka2=x2*5.1*10-10/(y-x)
When I change x2/Ka1, derived above, for y-x, Ka1 and Ka1 are negating each other, and x2 and x2 do the same thing so again I get: Ka2=5.1*10-10, pKa2≈9.3.
Is this way maybe correct?

Offline Borek

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Re: Problem of the week - 20/08/2012
« Reply #8 on: August 20, 2012, 04:35:57 PM »
Stop worrying about H2A, you were much closer earlier. There is a single assumption needed - and you already did it, you are just unnecessarily making thing more difficult than they are.
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Re: Problem of the week - 20/08/2012
« Reply #9 on: August 21, 2012, 05:08:41 AM »
I am not sure what's wrong in this way, but I will re-write the first way from beginning with the correction (without introducing Ka1):
H2A ::equil:: HA-+H+
y-x        x    x
HA-       ::equil::      A2-     +     H+
x-5.1*10-10     5.1*10-10     x+5.1*10-10
Because 5.1*10-10 is probably much less that x, it can be neglected for [H+](without this assumption it couldn't be solved, but that is way the methyl orange is introduced I think) and for [HA-].
Ka2=5.1*10-10*x/x=5.1*10-10
pKa2=9.29

Offline Borek

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Re: Problem of the week - 20/08/2012
« Reply #10 on: August 21, 2012, 05:19:56 AM »
Let's say it is OK now.

Basically it is all about rewriting Ka2:

[tex]K_{a2} = [A^{2-}] \frac {[H^+]}{[HA^-]}[/tex]

If the second dissociation step is much weaker than the first, we can safely assume [H+] = [HA-] - and they cancel out. This is the only required assumption, and it holds reasonably good for a surprisingly wide range of Ka1 and Ka2 values. Methyl orange was there just to make sure Ka1 is reasonably large and solution is not too diluted.
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