Sciencemadness Discussion Board - Tin (II) Chloride

Tin (II) Chloride

MttLsp - 19-7-2010 at 13:33

Alright so I made some Stannous Chloride (SnCl2) by adding muriatic acid to lead free Solder Wire, which contained Tin and Antimony:
2HCl + Sn --> SnCl2 + H2
The reaction was ridiculously slow but if I cut up the Solder Wire into smaller pieces it would accelerate. Adding heat is out because a) HCl gas will be produced leaving me an excess of tin b) the stannous chloride may undergo hydrolysis and become Sn(OH)Cl or c) the antimony will decompose.

When I had a clear solution of tin chloride and hcl, I evaporated it by adding heat with an alcohol burner. I find this to be problematic since some of the SnCl2 might take on a hydroxide ion, as stated above. Is there a way to grow crystals from a stannous chloride solution that is slightly acidic using a seed crystal?

As a side note, I made my first piece of conductive glass out of a broken vase I had lying around. I heated the Stannous chloride to temperatures over 400 degrees C and let the gas, which contains Tin Oxide, make contact with a small piece of glass. The tin oxide formed a thin rainbow colored transparent film on the glass, and when I tested the resistance (2000k), I got readings as low as 0.08! Now I can make dye solar cells without spending 900 bucks on 50 slides of ITO glass. All I need is Solder Wire, Muriatic Acid, and a few microscope slides.

[Edited on 19-7-2010 by MttLsp]

[Edited on 20-7-2010 by MttLsp]

Panache - 20-7-2010 at 01:49

i think you may find that the hydroxide's of tin(II) have difficulty forming without any OH present, which is the case if you have excess acid, is this rainbow coloured glass the same as on older style high wattage spotting globes? i have one which is exactly as described.

Eclectic - 20-7-2010 at 03:46

You can use a ceramic crock pot To digest the solder and keep topping it up with Muriatic acid to get saturated SnCl2 solution. SnCl2 crystalizes just fine when this is cooled without any precipitate of SnO, as long as there is an excess of HCl.

MttLsp - 20-7-2010 at 18:19

Quote: Originally posted by Panache

i think you may find that the hydroxide's of tin(II) have difficulty forming without any OH present, which is the case if you have excess acid, is this rainbow coloured glass the same as on older style high wattage spotting globes? i have one which is exactly as described.

Yes I think it is probably the same material as your globe. And thanks for your input...I kind of figured that the Sn would have a hard time forming Sn(OH)Cl in the presence of acid. However I assumed that since the Hydrochloric Acid has a lower boiling point than that of water, hydrolysis of the Tin would occur after the HCl boiled off. But I did not take into account that the less concentrated the acid is, the higher the boiling point, and since the HCl reacted with the Tin it was a much lower concentration than when I first started, making it possible to evaporate it with heat. Once I get my temp. controlled hot plate I'll feel more comfortable doing this.

But hydrolysis occurs when a substance reacts with the Hydroxide ions in water right?

cnidocyte - 13-2-2011 at 09:09

I added some 20% hydrochloric acid (which I distilled from drain cleaner) to some lead free solder and let it stand overnight at room temperature but no reaction occured. Would adding some H2O2 help the HCl oxidise the tin?

not_important - 13-2-2011 at 11:07

20% hydrochloric acid is nearly the azeotropic (AKA constant boiling) proportions, doesn't fume when exposed to air,and boils around 109 C with the same ratio of HCl-H2O distilling off. It is concentrated enough that SnCl2 will not hydrolyse during concentration/evaporation.

Some of the Sb will go into solution with con hydrochloric, even in the cold. In an alloy a minor constituent (like the 5% Sb) is effectively very finely divided and reactions unimportant in bulk become significant.

blogfast25 - 13-2-2011 at 11:48

I've dissolved pure Sn in an excess of boiling 22 w% HCl, under reflux. It takes a while to do so, it's quite sluggish even at BP.

Obtaining the hydrate as a solid is a bit trickier. Carefully evaporate the excess HCl, maling sure you don't go too far or the SnCl2 WILL hydrolyse (happened to me!) SnCl2 is very water soluble sot it's hard to tell when to stop evaporating: stop when there's little liquid left, allows to cool and see if you get crystals (they may take a while to form). If not, evaporate a little further, and try again...

Alternatively use a combination of strong HCl and strongish HNO3 (the right combinations have been described in a few tin related threads on this forum).

Sn dissolves very, very quickly in this kind of ‘aqua regia’ but the nitric acid oxidises the tin to Sn [+IV], not Sn [+II]. Reduce this back with zinc (powder or granules) to Sn2+, then alkalise with strong ammonia (NH3). Sn will precipitate as Sn(OH)2.n H2O (stannous hydroxide) but Zn2+ remains in solution as Zn(NH3)2 (2+). Filter and wash thoroughly and the Sn(OH)2 is now the basis for most Sn [+II] salts, including the chloride, by dissolving it in the relevant acid…

[Edited on 13-2-2011 by blogfast25]

cnidocyte - 16-2-2011 at 15:30

Quote: Originally posted by blogfast25

I've dissolved pure Sn in an excess of boiling 22 w% HCl, under reflux. It takes a while to do so, it's quite sluggish even at BP.

I tried refluxing the HCl solution and the solder turned black but didn't dissolve like I was expecting. The solder 99.3% tin and 0.7% copper.

entropy51 - 16-2-2011 at 16:18

Quote: Originally posted by cnidocyte

I tried refluxing the HCl solution and the solder turned black but didn't dissolve like I was expecting. The solder 99.3% tin and 0.7% copper.

What's your point? Tin dissolves in HCl. This is the standard production of stannous chloride. All the books have it written. Many of us have done it. No mystery here. Either you don't have Sn, you don't have HCl, or your HCl is too dilute. Fix your problem and move on. You have not discovered a new phenomenon in chemistry.

Sedit - 16-2-2011 at 19:06

I may be off but just an idea that I would look into if I wanted to try this(and I do soon because SnCl2 is used to give a luster oil slick appearence to glass and ceramic glazes) I would try to heat the Tin in H2SO4 in an attempt to form the sulfate then precipitate it as the oxide using either (aq)NH3 or NaOH. Reacting the precipitated SnO with HCl should produce SnCl2 and water by SnO + 2HCl => SnCl2 + H2O.

I agree that it takes entirely to long to dissolve even with 35% H2O2 added. I did not try HNO3 yet however but I do need some tin salts to add to reduction firings in the kiln at 1200F because the look it creates is awsome.

cyanureeves - 16-2-2011 at 19:43

@cnidocyte it has got to be your hcl acid because i used 95% tin 5%copper powder dutchboy solder and it dissolved into a clear solution leaving a little bit of grey mud on the bottom. all i used as heat source was a scent candle warmer.the acid was industrial 33%. but does anybody know if the copper content will give me a false reading if i do a stannous chloride test on a gold solution?i heard somewhere that it would, but also isnt copper non reactive with hcl acid?

Eclectic - 17-2-2011 at 01:34

The grey or black mud left after dissolving lead free solder in boiling muriatic acid is most likely the alloying metals in the solder, which don't dissolve unless you have oxidizing conditions, such as lots of air or H2O2.

I've made SbCl3 solution dissolving the residue from Sn/Sb solder digestion in HCl/H2O2.

blogfast25 - 17-2-2011 at 09:33

Quote: Originally posted by cyanureeves

all i used as heat source was a scent candle warmer.the acid was industrial 33%. but does anybody know if the copper content will give me a false reading if i do a stannous chloride test on a gold solution?i heard somewhere that it would, but also isnt copper non reactive with hcl acid?

If any copper did enter the solution (as Cu++) your solution would be blueish/greenish tinged. In the absence of an oxidising agent HCl is a very poor solvent for Cu.

Quote: Originally posted by Eclectic

I've made SbCl3 solution dissolving the residue from Sn/Sb solder digestion in HCl/H2O2.

Are you sure it was Sb (+III) and not Sb (+V)? In oxidising conditions Sb goes easily to +V. See also my thread on KSbCl6. Sb (+III) is easily oxidised by permanganate, so if your solution can’t discolour dilute KMnO4 then it’s likely to be H3O+ + SbCl6(-), rather than SbCl3…

cnidocyte - 20-2-2011 at 06:11

Quote: Originally posted by cyanureeves

@cnidocyte it has got to be your hcl acid because i used 95% tin 5%copper powder dutchboy solder and it dissolved into a clear solution leaving a little bit of grey mud on the bottom. all i used as heat source was a scent candle warmer.the acid was industrial 33%.

I distilled the HCl myself and since HCl forms an azeotrope with water at around 20% I assumed thats the concentration but I don't know the concentration of HCl in the drain cleaner I distilled it from so I'm guessing its under 20% so I need to redistill to up the concentration.

Eclectic - 20-2-2011 at 06:18

@Blogfast:

Could be, but I had slight excess of the Sb mud which should keep the valence at III. All I was really aiming for was Sb dopant for the SnCl2 solution, which is also reducing.

blogfast25 - 20-2-2011 at 11:04

Yes, with an excess Sb you're likely to stay at +III.

Arthur Dent - 12-3-2011 at 12:03

To say that the process of producing SnCl2 with 32% HCl and metallic tin is slow is a formidable understatement!

About 10 days ago, I decided to get rid of a spool of tin/silver solder I've had for a long time (96% tin/4% silver)

Knowing that tin chloride is a reducing agent that helps precipitate the silver in metallic form, I figured that I would snip off the solder in small 2mm bits (20g), put that in a flask and pour 100 ml of HCl and i would have a lovely solution of tin chloride and some metallic silver at the bottom and i'd be done with that...

Not so! It's been 10 days and after some effervesence during the first few days, the reaction has been slower and slower until it's just dark grey bits that have barely dissolved, and a very modest pinhead-sized hydrogen bubble once in a while. At that rate, it will take a billion years to completely dissolve. I've even put that on the hotplate at low for a short while and it did effesversce vigorously, but not much dissolution of the metal bits in the flask...

I tested the pH of the solution and it's still strongly acidic, so I haven't run out of acid, it looks like the metal has mostly passivated

could this be caused by the silver?

Oh, and my acid is indeed 32% HCl and the solder according to the Kester data sheet specifies the correct amount of tin vs silver. Could it be that the rosin core of the tin solder affects the reaction? I have not the ghost of an idea what is in that rosin, only that it says it's not recommended for electronics.

Robert

blogfast25 - 12-3-2011 at 12:06

Robert:

Forget doing this without heating: at reflux with 32 % HCl progress is reasonable, at RT it's the slowest boat to China.

Remember they used to coat the inside of tin cans with tin for a reason!

[Edited on 12-3-2011 by blogfast25]

Arthur Dent - 12-3-2011 at 12:18

Okay then, I guess i'll pull out the ol' vigreux, which happens to be 24/40 like my flask, and put it to good use. i'll set the hotplate at low, and leave it runing for a couple of hours. Thanks blogfast25!

Edit: Here's my setup:

I added a bubbler at the top of the vigreux, with a flask full of NaOH to avoid any rogue gases from filling-up my lab.

Robert

[Edited on 12-3-2011 by Arthur Dent]

blogfast25 - 12-3-2011 at 14:07

In a few hours the SnCl2 should be yours... Mwhahaha...

Arthur Dent - 26-3-2011 at 11:02

The tin chloride synthesis worked fine, after decanting the solution, I got lovely pure white crystals. After refluxing and boiling the solution, I was left with a very fine dark gray dust at the bottom of the flask.

Remember: My original compound was Kester tin solder 96% tin - 4% silver.

So that fine, totally insoluble dust (even in hot 32% HCl) must be silver, right?

So I decided to thoroughly wash the dust with distilled water (4 times to make sure no traces of Cl compounds are left). and then I added some nitric acid to the wet slurry at the bottom of the flask. After some angry orange fumes, I was left with this:

What gives? Silver nitrate is supposed to be a colorless solution, and even if there were tin traces left, stannous nitrate is supposed to be colorless too, right?

So I had this turbid light purple solution that settled overnight to a slightly cloudy clear supernatant liquid with a very generous light purple precipitate.

Could this be caused by another metallic contaminant, or traces of chloride? The HNO3 is 42be lab grade stuff.

Robert

blogfast25 - 26-3-2011 at 11:12

Try again. I feel you didn't use enough nitric (what's the w% of 42 Beaume?) and what you have is partly reduced AgCl (it's very light sensitive - OBVIOUSLY). Filter off or decant off the supernatant, then add good dollop of nitric, until ALL dissolves. Heat a bit if needed. OUTSIDE!

Show us the crystals of SnCl2?

[Edited on 26-3-2011 by blogfast25]

Arthur Dent - 26-3-2011 at 12:02

42 Beaume is approx. 67% which is fairly concentrated. I added a good 10 ml to the silver residue which looked like just a small sprinkling of dark dust. Didn't measure it but to the eye, it was something like 1 g or less. Okay I'll try to add a bit more nitric acid to see if it dilutes the remaining gunk.

Robert

blogfast25 - 26-3-2011 at 13:56

Ok, 67 %, that's the 'real deal' (wish I had me some of that!)

Remember, AgCl is seriously insoluble (it's used as a gravitational lever in the wet analytical determination of silver - old style). So your acid has to carry out a displacement: AgCl(s) + HNO3(aq) --- > AgNO3(aq) + HCl(g).

Any NOx evolved is due to some AgCl having been photo reduced to Ag + 1/2 Cl2, the metal then reacts with the nitric, giving off the usual witches' brew.

Your purple insoluble really looks like AgCl that's seen just a bit too much light. T'is really the Dracula of chemical compounds!

Arthur Dent - 27-3-2011 at 06:27

Quote: Originally posted by blogfast25

Ok, 67 %, that's the 'real deal' (wish I had me some of that!)

Remember, AgCl is seriously insoluble (it's used as a gravitational lever in the wet analytical determination of silver - old style). So your acid has to carry out a displacement: AgCl(s) + HNO3(aq) --- > AgNO3(aq) + HCl(g).

Any NOx evolved is due to some AgCl having been photo reduced to Ag + 1/2 Cl2, the metal then reacts with the nitric, giving off the usual witches' brew.

Your purple insoluble really looks like AgCl that's seen just a bit too much light. T'is really the Dracula of chemical compounds!

Yeah, I agree that there was probably much more Silver Chloride than elemental Silver at the bottom of my flask before I added the Nitric acid. I'll filter the solution and save the filtrate and i'll try to see if I can react some of the dried precipitate with fresh HNO3 to further dissolve that stuff.

Robert

PS: Oh and the acid, I bought a 2.5l glass bottle over 25 years ago from Anachemia, still more than 3/4 full. Has lost nothing of its potency and still perfectly clear and colorless!

PPS: Here are the tin chloride crystals... I dropped a few drops of the saturated solution on a watchglass and let dry for a couple of hours. Lovely needlelike crystal lattice!

[Edited on 27-3-2011 by Arthur Dent]

blogfast25 - 28-3-2011 at 04:28

Yep, they're the real mccoy alright. Nice one!

Lambda-Eyde - 28-3-2011 at 04:44

AgCl dissolves in aqeous ammonia, giving the diaminesilver ion Ag(NH3

2+. Reduction with e.g. glucose will make a shiny silver mirror, which can be dissolved with HNO3.

[Edited on 28-3-2011 by Lambda-Eyde]

Arthur Dent - 28-3-2011 at 08:27

Question... Will it work if I use a vacuum buschner funnel with a #40 wheaton paper filter to dry off the precipitate?

Or will the precipitate clog-up the paper and the filter's frit? The idea of using ammonia to dissolve the AgCl is appealing. Perhaps I could use ammonia to wash-off the funnel after the filtering?

Robert

blogfast25 - 28-3-2011 at 08:50

Robert, gentle vacuum assisted filtering should work well. You will always lose some product on the filter, of course. Just make sure the precipitate is still wettish. If your filter is glass frit (but it doesn't appear to be) , you can carry out the acid digestion on it, it also resists heat well, so no probs there…

Getting rid of most of the water should greatly help dissolving it into 68 % HNO3. Remember: you’re displacing a very strong acid! Heat will also help: it will drive off the HCl, thereby pushing the equilibrium to the right (mass balance effect). Heat may also decompose the AgCl a bit, giving the nitric something to get its teeth into (the silver). Multiply pronged attack!

Arthur Dent - 28-3-2011 at 09:08

Quote: Originally posted by blogfast25

Robert, gentle vacuum assisted filtering should work well. You will always lose some product on the filter, of course. Just make sure the precipitate is still wettish. If your filter is glass frit (but it doesn't appear to be) , you can carry out the acid digestion on it, it also resists heat well, so no probs there…

Getting rid of most of the water should greatly help dissolving it into 68 % HNO3. Remember: you’re displacing a very strong acid! Heat will also help: it will drive off the HCl, thereby pushing the equilibrium to the right (mass balance effect). Heat may also decompose the AgCl a bit, giving the nitric something to get its teeth into (the silver). Multiply pronged attack!

I'll use a Kimax 30ml buschner funnel with a glass frit and drop a small 1" circle of wheaton filter paper on top. I'll use Nitric to clean the funnel.

Okay, i'll let the precipitate dry off a bit, as for the reaction, i'll do it outside because the mean orange clouds scare me

. When the reaction stabilizes, i'll bring it back inside and put it on my hotplate for a bit. I have a sheet of pure Silver (JT Baker) so I might drop a tiny piece to see if it dissolves equally fast in the Nitric

Robert

[Edited on 28-3-2011 by Arthur Dent]

blogfast25 - 28-3-2011 at 11:39

The ‘mean orange clouds’ should scare you:

Ag === > Ag+ + e

NO3(-) + 4 H+ + 3e ==== > NO + 2 H2O

Or: Ag + 1/3 HNO3(-) + H+ === > Ag+ + 1/3 NO + 2/3 H2O

(In reality the reaction products tend to depend on concentration, temperature and type of metal but NO is always part of the mix)

NO reacts immediately with air oxygen to NO2 (NO + ½ O2 === > NO2), which is the amber/reddish/brown gas (nitrogen dioxide) you’re seeing. It smells somewhat like chlorine and is probably even more toxic. Stay AWAY!

cyanureeves - 29-3-2011 at 04:54

i always get that stubborn bluish grey stuff whenever i dissolve sterling and it turns out to be silver. it never re- dissolves for me.i've never gotten a clear solution like nurd rage because sterling is an alloy. i bet you can turn it to metal silver with hcl acid and zinc.

hkparker - 30-3-2011 at 20:28

I have a tin(II) chloride question...

I've been trying to make some recently, and I was a little impatient with the reaction between my tin and HCl (I was using tin about the size of coins and hadn't heated my HCl. I'm probably going to just reflux is next time. Easier then what I just tried to do...). But before I do that i'm curious as to what happened with my last attempt...

Being impatient with my HCl I decided to add some HNO3 to get things going, and I figured I would convert any tin nitrate to chloride later. I also knew this would make tin(IV) but planned on reducing that. I was left with a clear solution of what I believe was SnCl6(2-). Jor told me that HNO3 oxidizes Sn to SnO2, but the presence of Cl- would likely form that complex. I neutralized with Na2CO3 and sure enough got a bright white precipitate of (what I believe was) SnO2. So I dissolved that in HCl to yield SnCl4. Heres where things got tricky. As I wanted to make SnCl2, I though adding tin metal would reduce it back. SnCl4 + 2Sn --> 2SnCl2. However immediately after adding the tin metal the solution went from clear, to yellow, to green, to a brown. Any Idea on what this brown solution could be?

blogfast25 - 31-3-2011 at 12:36

No but I'm guessing higher acid strength will remediate the problem...

Waffles SS - 27-6-2011 at 09:59

I have a another idea to produce Tin(II)chloride:
First Tin(II) sulfate by adding tin metal to copper sulfate solution:

Sn (s) + CuSO4 (aq) → Cu (s) + SnSO4 (aq)

Second adding calsium chloride to tin sulfate solution:

SnSO4 (aq) + CaCl2(aq) → SnCl2 (aq) + CaSO4(S)

Then filter calsium sulfate and evoparation solution under vaccum.

[Edited on 27-6-2011 by Waffles SS]

blogfast25 - 27-6-2011 at 13:06

Calcium sulphate is a little soluble in water. Your tin sulphate is likely to be contaminated with CaSO4...

LanthanumK - 27-6-2011 at 13:26

For some reason, leaded solder dissolves in cold HCl in about 1 week, leaving behind a generous serving of lead powder, which burns nice and toxic. Unleaded solder(?) hardly dissolves at all in HCl, so I just gave up attacking it that way.

Neil - 29-6-2011 at 07:42

Quote: Originally posted by Arthur Dent

Oh, and my acid is indeed 32% HCl and the solder according to the Kester data sheet specifies the correct amount of tin vs silver. Could it be that the rosin core of the tin solder affects the reaction? I have not the ghost of an idea what is in that rosin, only that it says it's not recommended for electronics.

Robert

Rosin is in "Rosin" - Rosin comes from tree sap.

If it is not electrical solder but rather metal working solder then instead of Rosin it has an acid flux which can be anything from urea to metal chlorides. If you want to find a specific one you often have to find the manufacture and search for an MSDS for "Acid Core" not "acid cored solder" they tend to list the core as "acid core" on the solder msds sheets.

In dissolving thick tin plumbing solder I found that winding long tight coils and tossing those in a old HCl bottle with some 32% let the reaction self heat and proceed easily, dissolving the whole roll in three days/three additions.

Waffles SS - 3-7-2011 at 07:46

Quote: Originally posted by blogfast25

Calcium sulphate is a little soluble in water. Your tin sulphate is likely to be contaminated with CaSO4...

Yes,
Barium sulfate is insoluble in water and we can use barium chloride instead of calcium chloride:

BaCl2(aq) + SnSO4(aq) = SnCl2(aq) + BaSO4(s)
Solubility of barium sulfate in water is 0.0002448 g/100 mL (20 °C)

Waffles SS - 5-7-2011 at 11:23

Difficult part is evoporation step tin(II)chloride decompose in hot water(hydrolyse occur even in more water)

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)

What we can do for evoporation step?evoporation under vaccum seems not work very well.How we can evoporate water?

Lambda-Eyde - 5-7-2011 at 11:25

Use an excess of HCl?

blogfast25 - 5-7-2011 at 12:43

Quote: Originally posted by Waffles SS

How we can evoporate water?

Carefully and by having an acid reserve at all times. It's a bit hit and miss for small quantities: boil in too far and you get hydrolysis, not far enough and the product doesn't crystallise. But it worked for me.

The product of hydrolysis is still soluble in HCl though, so you can try over and over again untill you get it right.

Worth exploring would be what works with some chloride hydrates like AlCl3.6H2O or ZrOCl2.8H2O: saturate a cold solution of the salt with HCl (gassing with HCl gas), the hydrate then precipitates.

[Edited on 5-7-2011 by blogfast25]

Waffles SS - 5-7-2011 at 21:41

"Have an acid reserve at all time" is good idea but how we can get rid of HCl in final crystal?(by this method there is some HCl in final product)I think we should wash final product with suitable solvent ,Tin(II) chloride is soluble in ethanol and acetone(i think we should find another solvent for this purpose)also it think result of tin chloride hydrolysis will react with HCl again and make to Tin(II) Chloride again

Sn(OH)Cl + HCl =SnCl2 +H2O

LanthanumK - 6-7-2011 at 03:02

When evaporated, much of the HCl fumes away, decreasing the amount of HCl in the final crystals.

Arthur Dent - 6-7-2011 at 03:19

It's pretty hard to crystallize. And it is indeed necessary to have an excess of HCl in the solution. Half of my yield was simply "evaporated" at room temperature on a glass plate, giving the familiar needle-like structures, that can be scraped-off the glass and be put in a vial (still a bit moist and acidic). That is a long process but it did yield fairly pure, snow white powder that dissolves readily in very little water.

The other half, I tried to evaporate on a hotplate at low temp with the solution in a large petri dish. Most of it hydrolized quite rapidly, turning into a beige, opaque, insoluble paste, but around the rim, a cluster of perfectly square crystals formed (about 1 to 1.5mm) which I salvaged one by one with tweezers and washed with acetone. They look like perfectly clear pickling salt crystals. This second method is hardly recommended because a lot of the Tin Chloride was lost when it hydrolized into tin hydroxide and oxide (about 75%).

Robert

Waffles SS - 6-7-2011 at 04:17

@Arthur dent(robert),
Tin chloride solution even is sensitive to air :

Quote:

Solutions of SnCl2 are also unstable towards oxidation by the air:

6 SnCl2 (aq) + O2 (g) + 2 H2O (l) → 2 SnCl4 (aq) + 4 Sn(OH)Cl (s)

"Evaporating at room temperature on a glass plate" means long time contact with air oxygen!.(are you sue you didnt get Tn(IV)chloride?)
also you "tried to evaporate on a hotplate at low temp with the solution in a large petri dish" and you saw insoluble paste?i think this is impossible to see insoluble paste if your solution contain HCl

Sn(OH)Cl (s) + HCl(aq) = SnCl2(aq) +H2O(l)

I am interested to try evaporation under vaccum but i am not sure the result change

[Edited on 6-7-2011 by Waffles SS]

LanthanumK - 6-7-2011 at 04:18

Last time I tried evaporating a SnCl2 solution, I experienced the sentence of doom for heatless evaporation: deliquescence.

Cloner - 6-7-2011 at 05:10

Quote: Originally posted by Waffles SS

@Arthur dent(robert),
Tin chloride solution even is sensitive to air :

"Evaporating at room temperature on a glass plate" means long time contact with air oxygen!.(are you sue you didnt get Tn(IV)chloride?)
[Edited on 6-7-2011 by Waffles SS]

The boiling point of tin(IV)chloride is much lower than the BP of tin(II)chloride so this is likely. For the purpose of making conducting glass this is fine though.

Waffles SS - 6-7-2011 at 05:36

Quote: Originally posted by Cloner

Quote: Originally posted by Waffles SS

The boiling point of tin(IV)chloride is much lower than the BP of tin(II)chloride so this is likely. For the purpose of making conducting glass this is fine though.

@cloner thanks but:
Boiling point of tin(IV) chloride is 114c
We are talking about evaporation" at room temperature" not at boiling point of water!

Arthur Dent - 6-7-2011 at 09:07

Yes indeed it is sensitive to air, but much more in the absence of HCl, so as I mentioned, my solution is evaporated at room temperature until the formation of needles (still moist a bit) and harvested at that moment. The vial containing the Tin Chloride has the characteristic smell of HCl.

As for the insoluble paste, when the HCl is completely driven off by heat, the Tin Chloride undergoes hydrolysis and becomes a cruddy, insoluble goop that leaves a precipitate in water. I've tried it twice and got the exact same results, under heat, a small quantity of the solution reaches the ideal level of water/HCl and turns into lovely, big crystals, the rest turns into tin hydroxide and oxide.

Robert

[Edited on 6-7-2011 by Arthur Dent]

blogfast25 - 6-7-2011 at 10:11

I have a pot of that stuff (SnCl2.2H2O - bought): it isn't very sensitive or deliquescent. Evaporation gets rid of any HCl.

In my experience the freshly hydrolysed stuff dissolves well in strong HCl: just don't keep heating it any further...

Waffles SS - 6-7-2011 at 11:35

Anyone know solubility of Tin(II) chloride in cold acetone or ethanol.

LanthanumK - 6-7-2011 at 12:17

Wiki states that it is soluble in acetone and ethanol; Handbook of Preparative Inorganic Chemistry states that it is "quite soluble" in acetone and absolute ethanol.

not_important - 6-7-2011 at 12:33

If you go to the Internet Archives or Google Books, and search for "solubilities" and the last names of Arthur Messinger Comey & Dorothy Anna Hahn, or Atherton Seidell, you will find books of solubilities for the downloading. Quite helpful for the home experimenter, especially when having to prepare ones own reagents.

Waffles SS - 6-7-2011 at 22:57

Thanks my dear friend Not_important(i am glad to see your post),
I think gasing Hydogen chloride in mixture of small pieces Tin metal and dry Acetone will solve our problem.Hydrogen Chloride react with Tin metal to produce Tin(II)Chloride and will dissolve quickly in acetone.In this case i dont think Tin(IV)Chloride produce and also there is no water and oxygen for hydrolyse and oxidize
also after evaporation acetone under vacuum we will have a nice crystal of Tin(II)chloride

[Edited on 7-7-2011 by Waffles SS]

blogfast25 - 7-7-2011 at 12:52

Quote: Originally posted by Waffles SS

Thanks my dear friend Not_important(i am glad to see your post),
I think gasing Hydogen chloride in mixture of small pieces Tin metal and dry Acetone will solve our problem.[Edited on 7-7-2011 by Waffles SS]

Not very likely at all, actually. The oxidation of Sn to Sn (II) is carried out by H3O+. Acetone is an aprotic solvent: no H3O+ is present even with much HCl.

Reaction of Sn with HCl gas would proceed only at high temperature, if at all... Anhydrous SnCl2 can be obtained by direct union of Sn metal and Cl2 gas.

LanthanumK - 8-7-2011 at 03:00

Sn + 2 Cl2 → SnCl4

What about this reaction?

blogfast25 - 8-7-2011 at 06:22

Quote: Originally posted by LanthanumK

Sn + 2 Cl2 → SnCl4

What about this reaction?

Only at higher temp., LaK. Also by leading Cl2 over/through molten SnCl2 (anh.) at elevated temps, IIRW...

Waffles SS - 8-7-2011 at 07:59

Thanks blogfast(it seems you are full of chemistry.you remember @sauron for me)
I think reaction of Tin metal and chlorine occur better if we use suitable solvent and i think this reaction will be done at room temperature

[Edited on 8-7-2011 by Waffles SS]

blogfast25 - 8-7-2011 at 12:07

Quote: Originally posted by Waffles SS

Thanks blogfast(it seems you are full of chemistry.you remember @sauron for me)
I think reaction of Tin metal and chlorine occur better if we use suitable solvent and i think this reaction will be done at room temperature

[Edited on 8-7-2011 by Waffles SS]

Yes but it has to be a protonable solvent and that really is water... water and... water! (there are others but very unlikely to be suitable)

Making some anh. SnCl2 from tin and chlorine gas is something one fine day I'll do.

Waffles SS - 8-7-2011 at 21:47

It seems wiki wrote wrong information:

"Anhydrous SnCl2 is prepared by the action of dry hydrogen chloride gas on tin metal"
http://en.wikipedia.org/wiki/Tin%28II%29_chloride

blogfast25 - 9-7-2011 at 05:05

Quote: Originally posted by Waffles SS

It seems wiki wrote wrong information:

"Anhydrous SnCl2 is prepared by the action of dry hydrogen chloride gas on tin metal"
http://en.wikipedia.org/wiki/Tin%28II%29_chloride

I'm not sure they're wrong but I feel that chlorine gas would work better...

dann2 - 9-7-2011 at 08:45

Tin and Chlorine gas make SnCl4 (Stannic Chloride).There are threads about this.

Tin and Aqua Reiga make SnCl4:5H20 (I believe).
HCl acid + Hydrogen Peroxide will not make SnCl4 but some sort of hydrated Oxide.

You can make SnCl2 using Tin metal and HCl (35% HCl acid).

Some info here may be useful

http://www.oxidizing.110mb.com/chlorate/makesncl4.html

[Edited on 9-7-2011 by dann2]

blogfast25 - 9-7-2011 at 12:47

dann2, Interesting point in the *.pdf about using KClO3 as an oxidiser (Sn (II) === > Sn (IV)), instead of HNO3, I hadn't thought of that. It's a simple route to K2SnCl6, for those who want it...

smaerd - 7-1-2013 at 12:33

Figured I'd post my experimental results here even though this is straightforward.

Experimental:
1.10g of Sn metal(99.96% pure) "shot" was loaded into a 50mL RBF. 20mL of conc. HCl was added. A vigreux condenser was placed ontop with a gas in-let adapter. The joints were greased to ensure an air-tight fit. Tubing was lead from the gas-inlet to an inverted funnel that is suspended in ice bath chilled beaker with tap-water inside. The heating mantle was turned on and reflux began.

5 minutes into reflux gas evolution is apparent and the inverted funnel bubbles lightly.

55 minutes later the reaction is still going and the inverted funnel water was tested with a pH strip to be pH 1. No leaks are apparent in the set-up tested with moist pH strips. Reflux is maintained to approximately half of the vigreux height with condensation slightly above this line.

40 minutes later, reaction still going, no leaks. Tin metal still visible.

3 hours later no tin metal remains. Flakes of suspected tin oxide are still apparent. Appears to be a nearly quantitative transformation. Heat is turned off and the solution will be used as is for later transformations.

Making Tin (II) Chloride

rtrombetta - 11-7-2013 at 02:58

Hi!
I know this is an old post, but I have some doubts about making Tin (II) Chloride and certainly one of you can help me with it.

I followed all the instructions found on the book Inorganic Chemical Preparations with a different proportion (100g of tin/ 150ml HCl).

I tried in two different ways. All of them used a hotplate to heat the baloon.

In the first, I mixed all into a volumetric baloon with a vigreaux column attached at the top and a hose at the top of the column directing the gas into a flask with NaOH solution. I don´t know why, but this method didn´t work. The Tin does not get dissolved by the HCl. The solution was heated for more than 48hours and nothing happened. I´m using a 99% Tin and 36% analytical grade HCl.

I then removed the vigreaux and in a few hours i got all the Tin dissolved following the instructions on the book. I then evaporated into a s.g. 1.985 and the solution was rested for a few hours but nothing happened. I put it into the refrigerator and got nice crystals. I then drain it using a funnel with a glass and put into a desiccator using calcium chloride as desiccant.

Today is 3rd day that the crystals are in the desiccator and apparently it gets more humidity than before. My question is... can I dry these crystals that holds a few portion of HCl in a desiccator or there is another way to do that ? I never worked with a desiccator, so I don´t know how much time it will let to get dryed.

I´m sorry about any mistake writing in english... my primary language is other.

Thank You!

[Edited on 11-7-2013 by rtrombetta]

watson.fawkes - 11-7-2013 at 05:25

Quote: Originally posted by rtrombetta

I followed all the instructions [...] with a different proportion (200g of tin/ 150ml HCl).
[...]
I then drain it using a funnel with a glass and put into a desiccator using calcium chloride as desiccant.

Pardon me if I doubt that you've followed all the instructions, given that you immediately state that you're not using the ratio of reagents recommended. The g/ml ratio in the reference is 2:3 and you're using 4:3. No wonder you had trouble dissolving all the tin, and why subsequent results differ. In addition, the desiccant recommended is sulfuric acid, not calcium chloride. I can't tell from your description to other ways that your actual procedure deviated from the instructions. I'd have to guess you didn't bother with the step to "feather" the tin (quotes in original), which increases its surface area and helps the reaction proceed faster. ("Feathering" is melting it and dropping from a moderate height, ~ 1 m, into a bucket filled with water, which you can see by chasing the note in the reference, where it point to this procedure on cadmium(!)).

blogfast25 - 11-7-2013 at 06:55

There should be no problem dissolving relatively pure tin in an excess of 36 % HCl, with reflux and on a steam bath, but it's not a fast process. It took me a few hours to dissolve a few grams of lump metal.

Obtaining the crystalline dihydrate (IIRW) is trickier though because this product is very soluble in water and has a tendency to hydrolyse. It's really a question of evaporating most (95 % or so) of the solution by gentle boiling, then removing the last bits of solvent by atmospheric evaporation or by means of vacuum.

Tin dissolves very quickly in HNO3 + HCl but that yields Sn (IV). Magnesium ribbon will reduce that back to Sn (II), which could then be separated from the Mg (II) with excess strong alkali. This will form soluble stannite, while the Mg(OH)2 precipitates. Filtering separates the two and careful acidification of the stannite filtrate precipitates Sn(OH)2 which can be redissolved in HCl to give SnCl2.

[Edited on 11-7-2013 by blogfast25]

rtrombetta - 11-7-2013 at 08:35

Sorry... I used 100g of tin, and not 200g like I mentioned before. I didn´t pay attention when I post it, so I will edit and correct the post...

And you are right, I´m not using H2SO4 but other desiccant and I really don´t know if it will work. I saw many desiccators using anhydrous calcium choride as a desiccant, so i tried it... and I "feathed" the tin before put them into the baloon.

And thanks for help... I will try to gently evaporate the portion of HCl that remains in the SnCl2 crystals.

Thanks again!

[Edited on 11-7-2013 by rtrombetta]