# Yellow precipitate in SnCl2 solution

I have a few small bottles of 0.5M $\ce{SnCl2}$ solution, however solution was prepared about 4 months ago, and there is a yellow precipitate on the bottom of each bottle. Working hypothesis is that this yellow precipitate is $\ce{Sn(OH)Cl}$, and reaction is:

$$\ce{SnCl2 (aq) + H2O (l) ⇌ Sn(OH)Cl (s) + HCl (aq)}$$

I added a few drops of $\ce{HCl}$ to push the equilibrium towards left-hand side, but so far only small part of yellow precipitate is dissolved. Should I add plenty of $\ce{HCl}$ then? If you have any idea on how much, please let me know.

Wikipedia says:

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).

do I read it right that I have to add the same volume of $\ce{HCl}$ (0.5M) to my 0.5M $\ce{SnCl2}$? (well, they didn't say that, but sort of hinted)

Thank you very much in advance!

UPDATE: I added decent amount of $\ce{HCl}$, and I still have some yellow precipitate (I think less, than before, but still a rather large amount), and my solution is not transparent, it is whitish-opaque. $\ce{SnCl2}$ solution has to be transparent.

• I'd rather suspect oxidation. – Ivan Neretin Aug 2 '16 at 15:15
• If you suspect oxidation by the air (oxygen), then I think it is unlikely, as bottles were full of solution and closed tightly - there was no air (oxygen) access to the solution. – Sleepy Hollow Aug 2 '16 at 15:18
• Maybe a silly question, but are you thoroughly agitating it? Increasing surface area always helps the speed of reactions. – ringo Aug 8 '16 at 16:43
• Thank you for asking, @ringo! I spent some tie agitating it and was unable to dissolve all precipitate I had. However, both experiments I needed this solution for, went very well. So I guess it was OK. Experiments I conducted: (1) youtube.com/watch?v=Btaf9AUnjDo (2) youtube.com/watch?v=AklooXzB1w8. Here is a picture of my tin hedgehog: facebook.com/… – Sleepy Hollow Aug 13 '16 at 22:14
• Other than forming the basic salt, it will particularly form $\ce{H[SnCl3]}$, hydrogen trichloridostannate(II) if the concentration of HCl is in excess or will form an aqua salt, $\ce{[Sn(H2O)Cl2]}$ if the conc. of SnCl2 is more than that of water. These are the undesirable side products and will resist the precipitate to dissolve and show whitish opaque color. – Nilay Ghosh Oct 31 '16 at 10:38