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:


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


([from Wikipedia](https://en.wikipedia.org/wiki/Tin(II)_chloride#Chemical_properties))) 


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.


