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Topic: dominant equilibrium of KI  (Read 8397 times)

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Offline slayer

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dominant equilibrium of KI
« on: December 11, 2006, 04:45:36 PM »
I was way off on this problem. But im not sure what is going on here.

Dom Eq of KI


KI (aq) <---> K+ (aq) + I- (aq)

how could it be anything else?

Offline Borek

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Re: dominant equilibrium of KI
« Reply #1 on: December 11, 2006, 05:15:40 PM »
In water solution? Four equilibria I can think of:

K+ + I- <-> KI
K+ + OH- <-> KOH
H+ + I- <-> HI
H2O <-> H+ + OH-

How do you define 'dominating' equilibrium?
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Offline slayer

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Re: dominant equilibrium of KI
« Reply #2 on: December 11, 2006, 05:21:55 PM »
the equilibrium reaction where the Ka or Kb is greater than the other.

If Ka process yields a larger Keq than a Kb process, the resulted dominating equilibrium would be that of a Ka process. Therefore the reaction must be that of a Ka process, where an H3O+ is produced.

Offline Borek

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Re: dominant equilibrium of KI
« Reply #3 on: December 11, 2006, 06:56:19 PM »
Not sure if I get what you are trying to say. For me situation is dominated by KI creation.

Take a look at more concentrated solution (1M) - while KI creation is still responsible for the largest changes in concetrations, solution gets slightly acidic - that's because KOH creation starts to play a role. Small, but measurable (with pKb = 0.5, see reference listed at this pKa and pKb page).
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Offline mdlhvn

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Re: dominant equilibrium of KI
« Reply #4 on: December 12, 2006, 11:13:42 AM »
Hi Borek!
The software you used to determine the concentrations and activities of ions in solution is very interesting.

Could you tell me where can I find this software?

Offline Borek

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Re: dominant equilibrium of KI
« Reply #5 on: December 12, 2006, 11:44:18 AM »
It is an early beta version of equilibrium calculator that I am working on. No idea when it will be ready.
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Offline mdlhvn

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Re: dominant equilibrium of KI
« Reply #6 on: December 13, 2006, 12:29:53 AM »
Anyway, thank you very much for your information. I will try searching on Internet.


Offline slayer

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Re: dominant equilibrium of KI
« Reply #7 on: December 13, 2006, 01:38:48 PM »
So, its this isnt it?

KI (aq) + 2H2O (l) <---> K+ (aq) + I- (aq) + H3O+ (aq) + OH- (aq) ?

EDIT:

where K and I are spectators.

therefore the final equilibrium is this:
2H2O (l) <---> H3O+ (aq) + OH- (aq)
where Keq = Kw?

Offline Borek

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Re: dominant equilibrium of KI
« Reply #8 on: December 13, 2006, 01:43:12 PM »
therefore the final equilibrium is this:
2H2O (l) <---> H3O+ (aq) + OH- (aq)
where Keq = Kw?

To be honest - I have no idea what you prof will accept as the correct answer and I don't think there IS a correct answer to this question.

IMHO most likely s/he thinks about water autodissociation, just because s/he is not aware of other equilibria present in the solution.
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Offline AWK

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Re: dominant equilibrium of KI
« Reply #9 on: December 14, 2006, 03:00:23 AM »
therefore the final equilibrium is this:
2H2O (l) <---> H3O+ (aq) + OH- (aq)
where Keq = Kw?

To be honest - I have no idea what you prof will accept as the correct answer and I don't think there IS a correct answer to this question.

IMHO most likely s/he thinks about water autodissociation, just because s/he is not aware of other equilibria present in the solution.
From Borek data you can read that for 0.1 M solution over 96 % of KI is ionized, and for 1 M solution percent of dissociation is diminished (~70 %), but only dissociation is a dominant process for both concentrations.
AWK

Offline mdlhvn

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Re: dominant equilibrium of KI
« Reply #10 on: December 14, 2006, 07:28:43 AM »
I've never heard the Keq of the dissociation of KI. It seems that it soluble nearly completely.

Also, the equations (from borek):

K+ + OH-------->KOH

I-   +  H+-------->HI

hardly occur because KOH is a strong base and HI is strong acid.

Offline Borek

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Re: dominant equilibrium of KI
« Reply #11 on: December 14, 2006, 08:06:04 AM »
I've never heard the Keq of the dissociation of KI. It seems that it soluble nearly completely.

This is complex formation. Kf = 1.6

Quote
K+ + OH-------->KOH

I-   +  H+-------->HI

pKb = 0.5
pKa = -9.5

For most practical purposes both can be considered dissociated 100%, but that doesn't mean they always are.
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Offline mdlhvn

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Re: dominant equilibrium of KI
« Reply #12 on: December 14, 2006, 12:31:01 PM »
Nice information  ;) .

Please let me know where can I find Keq of other substances. In some book I have there are only Keq of bad electrolytic substances.

Thanks!  ;)

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