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.