# 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.

• Hydroxide reacts with HCl, you know... – Mithoron Jul 5 '17 at 20:10

$K_\mathrm{sp}$ 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

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• "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 $\ce{H2O}$." // WRONG. // $\ce{Mg(OH)2}$ is a solid and thus not in solution. // There is $\ce{OH-}$ in $\ce{HCl}$, albeit at a very low concentration. – MaxW Jul 6 '17 at 15:51
• @MaxW yeah like 10 femto molar in a 1M solution. That's totally negligeable and brings very little to the understanding of the problem IMHO. If you think otherwise please explain why. – user46680 Jul 6 '17 at 16:07