How to separate Sodium ion from Stannous Hydroxide ion?

 

By boiling it in H2O?

 

Why stannouses valency in this molecule is (VI) ?

Edited November 6, 201411 yr by Romix

Sn's valency is 4, not 6 - the Na+ ions take care of the rest.

As for separation, I would suggest mixing with an acid to change the pH - hydrous Sn(OH)2 or Sn(OH)4 (can't remember which) should precipitate in a certain pH range.

Sn's valency is 4, not 6 - the Na+ ions take care of the rest.

As for separation, I would suggest mixing with an acid to change the pH - hydrous Sn(OH)2 or Sn(OH)4 (can't remember which) should precipitate in a certain pH range.

 

It will precipitate out from pH between 2 and 11

 

Ok, if I boil boards in NaOH.

All the solder should dissolve in it.

Forming Na₂[sn(OH)₆] and Na₂[Pb(OH)₄]

 

Na₂[sn(OH)₆] will decompose to Na2SnO3 + 3H2O

And I'm shore it will! Last time I heated aluminum chloride on my hotplate, it decomposed it, and temperatures for AlCl3 are much higher.

Na₂[Pb(OH)₄] decomposes to Na2PbO3 at the temperature of 300°C, about the same as for AlCl3.

 

I left with solution of Na2SnO3, Na2PbO3 and NaOH?

Edited December 19, 201411 yr by Romix

What about boiling it on low flame.

And when all dissolve, dilute it with distill water, would Sn and Pb oxides precipitate out?

i cant understand Why stannouses valency in this molecule is (VI) ??

Although it is in the 4th main group and thus you would expect it to lose four elektrons to gain octet configuration, it has empty orbitals to which filled orbitals of other atoms can bond with. So normally you would expect Sn(OH)₄ as a neutral compound, in basic environments it can take up to 2 additional bonds, effectively gaining an elektron from each of them to obtain Sn(OH)₆^2-. Note that the valency of Sn is still (IV).

Edited June 17, 201510 yr by Fuzzwood