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Topic: Acid/base equilibrium question  (Read 9142 times)

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Offline daylight

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Acid/base equilibrium question
« on: March 16, 2009, 12:24:35 PM »
I have a problem that I need help with understanding why it is set up this way.
Calculate pH of 0.100M solution of Na2C2O4. for H2C2O4 Ka1=5.9x10-2; Ka2= 6.4x10-5.
the solutoins has:
Cr2O42-, Kb1+Kw/Ka2, and Kb2=Kw/Ka1.
so why is b1 not kw/ka1 and b2 not kw/ka2? this is confusing if Kb=Kw/Ka *delete me*?? ???

Offline Borek

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Re: Acid/base equilibrium question
« Reply #1 on: March 16, 2009, 12:39:53 PM »
http://www.chembuddy.com/?left=pH-calculation&right=polyprotic-dissociation-constants

Scroll down the page.

Although I am not sure what you are really asking about, as there are several typos in your post.
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Offline daylight

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Re: Acid/base equilibrium question
« Reply #2 on: March 16, 2009, 01:26:35 PM »

OK< I don't know how the 'delete me' got in there, something I clicked I guess, but I am asking why is Kb1= to Kw/Ka2 and not Kb1=Kw/Ka1...

Offline enahs

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Re: Acid/base equilibrium question
« Reply #3 on: March 16, 2009, 02:18:53 PM »
Write the reactions of Ka1 and Ka2.
Write the reactions of Kb1 and Kb2.

Which one are compliments of each other?

Offline Borek

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Re: Acid/base equilibrium question
« Reply #4 on: March 16, 2009, 02:33:20 PM »
but I am asking why is Kb1= to Kw/Ka2 and not Kb1=Kw/Ka1...

Have you visited the page at link posted? Have you read it?
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Offline daylight

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Re: Acid/base equilibrium question
« Reply #5 on: March 16, 2009, 02:34:42 PM »
H2C2O4 + H20  ::equil:: H3O + HC2O4
HC2O4 + H20  ::equil:: H30 + C2O4

C2O4 + H2O  ::equil:: HC2O4- + OH-
I'm not sure if I should then have H2CO4 or what?

Offline daylight

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Re: Acid/base equilibrium question
« Reply #6 on: March 16, 2009, 02:37:29 PM »
Yes, I've read it, but it's still confusing. It would help if it were a real problem...

Offline Borek

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Re: Acid/base equilibrium question
« Reply #7 on: March 16, 2009, 03:20:26 PM »
OK, your oxalic acid is almost OK, just please DON'T ignore charges. You may wrote H+ between products, while we all know it is H3O+ you will have to ignore water in dissociation constants, that's additional step tham may lead to confusion.

You need second hydrolysis reaction, and you are right - undissociated oxalic acid will be between products.

Now, write definitions of Ka and Kb for each reaction. Assign numbers (a1, a2, b1, b2) to them.
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Offline daylight

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Re: Acid/base equilibrium question
« Reply #8 on: March 16, 2009, 03:26:18 PM »
ka1= H2C2O4 + H20  ::equil:: H3O+ + HC2O4
ka2= HC2O4 + H20  ::equil:: H30+ + C2O4-

Kb1= C2O4- + H2O  ::equil:: HC2O4 + OH-
Kb2= HC2O4- + H2O  ::equil:: H2C2O4
I'm not sure about b1 b2, I haven't done one like that before... so then Kb2 would be the reverse of ka1 and kb1 the reverse of ka2?

Offline Borek

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Re: Acid/base equilibrium question
« Reply #9 on: March 16, 2009, 04:08:52 PM »
You are still missing some charges, and I would not wrote

ka1= H2C2O4 + H20  ::equil:: H3O+ + HC2O4

as

Ka1 = [H+][HC2O4-]/[H2C2O4]

but you are gettng it - Kax and Kbx are numbered in the opposite directions, so Kafirst*Kblast=Kw and so on.
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