Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: slayer on December 11, 2006, 04:45:36 PM
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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?
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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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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.
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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 (http://www.chembuddy.com/?left=BATE&right=dissociation_constants) page).
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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?
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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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Anyway, thank you very much for your information. I will try searching on Internet.
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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?
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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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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.
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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.
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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
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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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! ;)