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solomon levi
05-20-2009, 12:09 AM
I had made some notes at the library a while back on what
dissolves in what...
Here's an interesting item.
It said stannic (IV) oxide reacts with NaOH to form
stannate ions; no hydroxide.

So what does that mean? Is it monoatomic without precipitating?

It seems a pretty simple way to make a good tin medicine.


I added some tin oxide to an NaOH solution. It spread out as milk
but it appears to have all fallen to the bottom upon resting. I don't
think it dissolved at all.
????

Aleilius
05-20-2009, 12:42 AM
Yeah, stannates are formed when you dissolve tin oxide in a highly alkaline solution of NaOH or KOH. This is a property called "amphoterism." Most metal oxides will dissolve in highly alkaline solutions. Copper is a good example, lead is another example, antimony will function the same way.

Start with a tin chloride salt and then you can precipitate with an alkali via the normal precipitation method for m-states.



I added some tin oxide to an NaOH solution. It spread out as milk
but it appears to have all fallen to the bottom upon resting. I don't
think it dissolved at all.

I think this is because your sodium hydroxide solution isn't concentrated enough. The pH needs to be pretty high.

solomon levi
05-20-2009, 07:16 PM
Okay, I'll raise the pH.

But what I was thinking is there's no need to do the chloride m-state
method if the tin oxide becomes stannate ions... you can't get
smaller than an ion! That's a single atom, right? True M-state without
the tedious drop-by-drop and pH measuring!