What decides who gets the ion?

[ning]

Sorry for the dumb title, but how better to say it?

This is something that's been bugging me for a long time. Maybe some bee could help me out?

Let's say we have two bases, like Ca(OH)2 and Na2CO3. (this example from soapmaking!) When brought together in solution, they exchange ions to give CaCO3 and NaOH. Fine.

I was taught that this process is driven by the redox potential of the positive ions involved in the reaction. However, as I have come to realize, this is only half the story.

Why do Ca(OH)2 and Na2CO3 exchange ions? Does the metal with the higher redox potential want to get the stronger acid ion, to become the weaker base? Does this imply that LiOH is a stronger base than NaOH?

Once I reacted sodium oxalate with CaCl2, and it gave NaCl and calcium oxalate--this seems to violate the idea above! Wouldn't CaCl2 bee the weaker base?
Then I added NaOH to a solution of CaCl2, and Ca(OH)2 precipitated! What the hell??

The only thing I could think of is that perhaps some equilibrium is formed, and the poor solubility of Ca(OH)2 drove the reaction forward even if it would not be electrochemically favorable. But that seems so...ugh...

Can any bee shed some light on this matter for poor confused ning? What factors decide whether two salts exchange ions in solution? I much appreciate it!

[Osmium]

> I was taught that this process is driven by the redox potential
> of the positive ions involved in the reaction.

It has nothing to do with redox potential.

> What factors decide whether two salts exchange ions in solution?

In the examples you gave it's solubility. The least soluble compound will precipitate.

[moo]

Read about solubility product.

[ning]

So then, all the ions just disassociate and drift about freely in solution, and when they form less soluble combinations, that's the product?

So what about Ca(OEt)2 + 2 NaOH --> 2 NaOEt + Ca(OH)2 ? Same deal?

So then, in that case, couldn't a person, say, produce aluminum ethoxide by refluxing aluminum with ethanol, then react it with NaOH to yield NaOEt? Since it would seem that aluminum hydroxide is rather insoluble in anything much.

[Rhodium]

So then, all the ions just disassociate and drift about freely in solution, and when they form less soluble combinations, that's the product?

Yes.

So what about Ca(OEt)2 + 2 NaOH --> 2 NaOEt + Ca(OH)2 ? Same deal?

Yes.

So then, in that case, couldn't a person, say, produce aluminum ethoxide by refluxing aluminum with ethanol, then react it with NaOH to yield NaOEt? Since it would seem that aluminum hydroxide is rather insoluble in anything much.

Theorethically, yes - but Aluminum is a somewhat special case, as it may form sodium aluminates with varying ethoxide and hydroxide groups to each molecule, and if you have ever tried to filter aluminum hydroxide from anything, then you know this isn't a practical idea.

[starlight]

Methlyamine acetate is a useful Knoevenagel condensation catalyst. It is relatively easy to prepare a methanolic solution of Methylamine acetate of the required concentration without much water in by mixing methanolic solutions of Methylamine.HCL, and KOH, filtering and adding equimolar GAA.

However when trying to prepare solutions in IPA, it is more difficult due to the lower solubility of the KOH and MeAm.HCl in IPA. The IPA need to be warmed in order  to get enough of the salts into solution. I have always wondered if this leads to more Methylamine being lost when the solutions are mixed. So what could be done is to dissolve the KOH in warm IPA, and add equimolar GAA to make KOAc (similar solubility to KOH in alcohol). This could then be mixed with the MeAm.HCl solution, causing KCl to fall out and forming the non-volatile methlyamine acetate directly.

Edit: It works a treat! 2.63g (40mmol) 85% KOH was dissolved in 50ml Hot IPA. To this was added 2.3ml GAA. 2.72g (40mmol) of MeAm.HCL (40mmol) was dissolved in 60ml Hot IPA. The two hot solutions were mixed in a cold flask. A voluminous precipitate of KCL formed that was vacuum filtered off.