Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Perci on December 25, 2024, 08:39:18 AM
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Hi! I'm doing a level chemistry and I'm currently revising the transition metals topic. I understand ligand substitution of the equilibrium equations but I just can't understand why ammonia substitutes water in the way it does...
[Cu(H2O)6]2+ + 4NH3 :requil: [Cu(NH3)4(H2O)2]2+ + 4H2O
So, if the above is what occurs when you add excess ammonia
the position of equilibrium of:
[Cu(H2O)6]2+ + 2NH3 ::equil:: [Cu(H2O)4(OH)2] + 2NH4+
shifts to the left (correct me if I'm wrong)
So, I get what's happening ;but why?
Why does the ammonia substitute 4 of the water ligands?
In my head, it's not due to the chelate effect as it isn't entropically favourable, as the number of mols is equal on both sides of the equation.
Additionally, why partial substitution and not complete?
Lastly, why does it substitute to become [Cu(NH3)4(H2O)2] 2+ and not Cu(NH3)4(OH)2 ?
Sorry, I know I've asked a lot of questions there and that the answers are complicated and I definitely don't need to know the answers for my exam, but I'm just wondering as my teachers won't go into detail about it.
Thank you ! (also Merry Christmas!) :)
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So, if the above is what occurs when you add excess ammonia
the position of equilibrium of:
[Cu(H2O)6]2+ + 2NH3 ::equil:: [Cu(H2O)4(OH)2] + 2NH4+
shifts to the left (correct me if I'm wrong)
No, it shifts to the right.
Predicting stability of complexes is not for the faint of heart ;) Google for the ligand field theory, it should be a good starting point.
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Why does it shift to the right? I’m not trying to be facetious but on chemguide it says the equilibriums interact with each other (but I do understand in general that if you increase the concentration of the reactants it would normally shift the equilibrium to the right). (I’ve attached a screenshot below)
Also thank you ! I’ll go down the rabbit hole of ligand field theory (instead of doing actual revision hahaha) :)
(also sorry if I haven’t formatted this correctly ie. replying to your comment, it’s my first time posting on here!)
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[Cu(H2O)6]2+ + 2NH3 ::equil:: [Cu(H2O)4(OH)2] + 2NH4+
For this isolated reaction it shift to the right as already mentioned.
Building of precipitate.
But already shown the precipitate is not stable in access of ammonia because water will be exchange by ammonia.
This is a different reaction and has nothing to do for the equillibrium of the first reaction.
Dissolving of the precipitate
[Cu(H2O)4(OH)2] + 4 NH3 => [Cu(NH3)4(H2O)2]2+ + 2 OH-
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[Cu(H2O)6]2+ + 2NH3 ::equil:: [Cu(H2O)4(OH)2] + 2NH4+
For this isolated reaction it shift to the right as already mentioned.
But already shown the precipitate is not stable in access of ammonia because water will be exchange by ammonia.
This is a different reaction and has nothing to do for the equillibrium of the first reaction.
Dissolving of the precipitate
[Cu(H2O)4(OH)2] + 4 NH3 => [Cu(NH3)4(H2O)2]2+ + 2 OH-
Yes, but the equilibriums interact with each other so in excess it would shift left as (what you’ve said) the other reaction takes place when the precipitate dissolves (?)
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No you have to see every reaction separate.
For [Cu(H2O)6]2+ + 2NH3 ::equil:: [Cu(H2O)4(OH)2] + 2NH4+
It shift to right if reactants ammonia or copper aquo ions increased.
You dont get [Cu(H2O)6]2+ anymore there is no shift to left only get more precipitate. Means shift to right.
The behaviour of the precipitate in higher alcaline and ammonia environment is different It has its own reaction as described already.
In practice, if you drop ammonia solution to a copper sulfate solution you will get a milky blue precipitate. If you continue it dissolves again under building dark blue colour.
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No you have to see every reaction separate.
For [Cu(H2O)6]2+ + 2NH3 ::equil:: [Cu(H2O)4(OH)2] + 2NH4+
It shift to right if reactants ammonia or copper aquo ions increased.
You dont get [Cu(H2O)6]2+ anymore there is no shift to left only get more precipitate. Means shift to right.
The behaviour of the precipitate in higher alcaline and ammonia environment is different It has its own reaction as described already.
In practice, if you drop ammonia solution to a copper sulfate solution you will get a milky blue precipitate. If you continue it dissolves again under building dark blue colour.
Sorry, I understand this but please could you look at the screenshot from chemguide ? (thank you !)
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Yes you have two reaction in competition.
One guides to the precipitate and one goes direct to the Ammin complex depending on the amount of ammonia.
But this doesn't mean the equillibrium is shifted., because the precipitate will not react backwards to the aquo complex, it also build up the ammin complex.
[Cu(H2O)6]2+ + 4NH3 :requil: [Cu(NH3)4(H2O)2]2+ + 4H2O excess ammonia
[Cu(H2O)6]2+ + 2NH3 ::equil:: [Cu(H2O)4(OH)2] + 2NH4+ less ammonia gives precipitate
[Cu(H2O)4(OH)2] + 4 NH3 => [Cu(NH3)4(H2O)2]2+ + 2 OH- adding more NH3 to get also ammin complex
There is no reason that the back reaction, you called shift to left takes place.
[Cu(H2O)4(OH)2] + 2NH4+ ::equil:: [Cu(H2O)6]2+ + 2NH3 will not take place.
[Cu(H2O)6]2+ cannot built up in ammonia environment.