Does Ksp depend on the solvent?

E.g. would the solubility and Ksp of $\ce{Mg(OH)2}$ be the same in water as it would be in HCl? Doesn't solubility depend on the polarity of the solute and solvent, so wouldn't different amounts of a solute dissolve in different solvents. Wouldn't this give different Ksp values? However, I only ever see a single Ksp for anything.


Also, if $\ce{Mg(OH)2}$ is said to completely **dissolve** in $\ce{HCl}$, is it implied that is completely dissociates into $\ce{Mg^{2+}}$ and $\ce{OH^-}$ ions, or does it only exist as small solid pieces of $\ce{Mg(OH)2}$ floating around in the $\ce{HCl}$? I am assuming the latter, as the Ksp of $\ce{Mg(OH)2}$ is small.


I have a question in my book where $\ce{Mg(OH)2}$ is dissolved in $\ce{HCl}$ and then back-titrated with $\ce{NaOH}$. I can't seem to understand why, if the $\ce{Mg(OH)2}$ dissolves completely, we would need to do a back titration. I thought a back titration was useful when the solute doesn't dissociate completely.




$K\_\mathrm{sp}$ for a compound is solvent specific and can change depending on the $\mathrm{pH}$ of solution and on the intrinsic properties of the materials involved. 


As $\ce{Mg(OH)2}$ is a base, it will not be in the form of $\ce{OH-}$ in $\ce{HCl}$; it will instead be either $\ce{H3O+}$ or $\ce{H2O}$.


If a salt is said to completely dissolve, it usually means that ions are fully solvated, thus completely surrounded by solvent.


But, remember that if the $\ce{OH-}$ are acidified, the counter ion for the magnesium will become chloride ions. 


Lastly, if the solvent changes the nature of the compound, the $K\_\mathrm{sp}$ will be that of the different compound it has changed into.


Solubility of:  

Magnesium Hydroxide (.0090g / Liter)  

Magnesium Chloride (542g / Liter) 


Source: [Solubility Table](http://www.saltlakemetals.com/SolubilityProducts.htm)


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