Sciencemadness Discussion Board

what is this white stuff

cyanureeves - 18-6-2013 at 13:08

i added ammonium hydroxide to tin chloride solution and right away got a white substance.what would be the name of that white compound? thank you.

bfesser - 18-6-2013 at 13:12

Hydrated <a href="http://en.wikipedia.org/wiki/Tin(II)_oxide" target="_blank">tin(II) oxide</a> <img src="../scipics/_wiki.png" />.
<a href="http://en.wikipedia.org/wiki/Tin(II)_hydroxide" target="_blank">
Quote:
There had been confusion as it had been believed that Sn(OH)<sub>2</sub> was precipitated when a tin(II) salt is reacted with an alkali hydroxide such as NaOH, but this product was determined analytically to be hydrated tin(II) oxide, being either 5 SnO &middot; 2 H<sub>2</sub>O or 3 SnO &middot; H<sub>2</sub>O, The structure of pure Sn(OH)<sub>2</sub> is not known. <img src="../scipics/_wiki.png" />
</a>

[Edited on 7/9/13 by bfesser]

papaya - 18-6-2013 at 13:12

Hydrated tin oxide/oxychlorides ?

bfesser - 18-6-2013 at 13:19

<strong>Papaya</strong>, I don't believe any oxychlorides are formed. What causes you to assert that they are?

[Edited on 7/8/13 by bfesser]

papaya - 18-6-2013 at 13:24

I don't state it, but if the SnCL2 is in excess to the base, what you expect then?

bfesser - 18-6-2013 at 13:29

So you have no evidence or reasoning, but you're telling him that tin(IV?) oxychloride is a product of the reaction? <em>Why are you even posting‽</em>

cyanureeves - 18-6-2013 at 13:35

thank you gentlemen. i was googling for tinaminechloride.stannouschlorideammonia. so hydrated tin oxide it is and i must say it sounds better than white stuff similar to "white stuff in south american cocaine manufacture video".i have been adding ammonia to copper salts all day almost just to make blue stuff and man it is exciting when immediate stuff happens.

Hexavalent - 18-6-2013 at 13:36

Quote: Originally posted by bfesser  
you're telling him that tin(IV?) oxychloride is a product of the reaction?


He used a question mark, which means that he isn't sure, but it is a possibility. There is a difference between suggesting something for discussion, which is perfectly valid, and stating something directly.

papaya - 18-6-2013 at 14:05

Quote: Originally posted by bfesser  
So you have no evidence or reasoning, but you're telling him that tin(IV?) oxychloride is a product of the reaction? <em>Why are you even posting‽</em>


I never stated tin (IV) oxychloride as it must not change the oxidation state, rather I put a wrong name for Sn(OH)Cl, must be read stannous hydroxy chloride I guess, which as stated on the wiki page you provided can form by SnCL2 hydrolysis. Note equilibrium, how it will shift when you react HCL forming on right side with ammonia ?. This is why I "speculated" this way, I'm very sorry for that, of course I cannot find more direct proof for that (that you'll never get a pure hydrate oxide), however even now this sounds very reasonable to me..

bfesser - 18-6-2013 at 14:15

<strong>Hexavalent</strong>, <strong>papaya</strong>; fair enough. My apologies for getting overexcited.

papaya - 18-6-2013 at 14:19

OK, but what do you think about my last post ? I mean Sn(OH)Cl formation, I'm in doubts now..

<!-- bfesser_edit_tag -->[<a href="u2u.php?action=send&username=bfesser">bfesser</a>: removed unnecessary quoting]

[Edited on 7/8/13 by bfesser]

bfesser - 18-6-2013 at 14:27

http://en.wikipedia.org/wiki/Tin(II)_chloride#Chemical_properties

[Edited on 6/18/13 by bfesser]

papaya - 18-6-2013 at 14:35

Yes, but I see a mismatch there, because
first it says what I was thinking:

Tin(II) chloride can dissolve in less than its own mass of water without apparent decomposition, but as the solution is diluted hydrolysis occurs to form an insoluble basic salt:

SnCl2 (aq) + H2O (l) is in equilibrium with Sn(OH)Cl (s) + HCl (aq)

Therefore if clear solutions of tin(II) chloride are to be used, it must be dissolved in hydrochloric acid (typically of the same or greater molarity as the stannous chloride) to maintain the equilibrium towards the left-hand side (using Le Chatelier's principle).

but then it says:

If alkali is added to a solution of SnCl2, a white precipitate of hydrated tin(II) oxide forms initially; this then dissolves in excess base to form a stannite salt such as sodium stannite

So at the end it doesn't say that with alkali addition you may end up partially with Sn(OH)Cl, why?

bfesser - 18-6-2013 at 15:55

I suggest we consult a source more reputable than Wikipedia. I'll <a href="viewthread.php?tid=19098&goto=search&pid=288540">request</a> the paper cited in the article, as well as consulting my personal library of books. This seems like something <strong>woelen</strong> may like to chime in on.

[Edited on 6/19/13 by bfesser]

woelen - 18-6-2013 at 22:57

I also think hydrated tin(II) oxide is a good option, but I'm quite sure that it will not be pure. Certainly, chloride ions are part of the precipitate as well, but I do not think that you will get a precisely stoichiometrically determined compound like Sn(OH)Cl. I would say, Sn(OH)xCly, with x + y = 2 and most likely x > 1, but at the latter I'm not 100% sure.

The problem with many kinds of precipitate reactions, especially with halides, is that non-stoichiometric compounds are formed. With colored ions one can see the difference. A nice example is precipitating copper(II) with hydroxide from a chloride solution and from a sulfate solution. From the former you get a greenish precipitate, from the latter you get a pure blue precipitate. The green precipitate contains a lot of chloride besides the hydroxide and this has a strong influence on the color. With tin you can't see the difference, because all of these precipitates are white.

Another interesting thing: Hydrous tin(II) oxide is white, but the anhydrous oxide is dark grey. I have some of the anhydrous oxide and this easily dissolves in hydrochloric acid and gives a colorless solution of tin(II) chloride.

bfesser - 22-6-2013 at 08:21

Thank you, <strong>woelen</strong>. Insightful as always!

By the way, the <a href="http://www.sciencemadness.org/talk/viewthread.php?tid=19098&page=19#pid288547">reference</a> has been posted for anyone interested.