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+ Inorganic chemistry
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Mutegi Njeru

Synthesis of Potassium Trioxalatoaluminate (III) Trihydrate

David Grisell  Follow
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Potassium is not bound to anything. Product is a dissolved salt, fully dissociated.

Do you know what is a net ionic reaction?
Yes, I have done those so Potassium is a spectator that doesn't participate in the intermediate reactions?

Just to add to my last post:
I just learned that the cause is something called the Chelate Effect. Two O-'s from each of the three oxalates binds to Aluminum. The Al core has 3+ and the O-'s are 6- (since there are six of them). That is why we have K3 because K+ times three = 3+ to make the final charge 0. But what about the 4 OH-'s from the beginning? I was thinking that the H+'s the were removed from oxalic acid to make oxalate would join with the OH-'s to make water but not sure.
I have K3Al(C2O4)3 ∙ 3H2O + 4OH- + 6H+ as final products but something seems wrong.

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David Bernheim  Follow
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so Potassium is a spectator that doesn't participate in the intermediate reactions?

Yes.

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4OH- + 6H+

Can they exist in a solution at the same time?

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Andy Heilveil  Follow
Potassium is not bound to anything. Product is a dissolved salt, fully dissociated.

Do you know what is a net ionic reaction?

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Azn Faruqi  Follow
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Good starting point would be to write correct formulas for every compound involved.

For example, no such thing as K(Al(OH)3).
How about this? 2Al + 2KOH + 6H2O = 2K(Al(OH)4) + 3H2
Now I want 2K(Al(OH)4) + C2H2O4 ∙ 2H2O to make K3Al(C2O4)3 ∙ 3H2O but I'm unsure on how to proceed. I thought of doing a double replacement but that's clearly wrong. Judging by the product it appears that three oxalates bind to Aluminum and leave the Potassium untouched but I don't know why. How are three oxalates bound to Aluminum when each one has two O-?

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Dan Vasii  Follow
Quote from:
Quote from:
so Potassium is a spectator that doesn't participate in the intermediate reactions?

Yes.

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4OH- + 6H+

Can they exist in a solution at the same time?
I see so it is 2H+ and 4H2O. I understand it now. Thank you.

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Alex Pupp  Follow
Good starting point would be to write correct formulas for every compound involved.

For example, no such thing as K(Al(OH)3).

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