Why is magnesium sulfate acidic? Why are my calculations showing that it is basic?

Let's say I was dissolving $\pu{1 M}~\ce{MgSO4}$ into water and I wanted to find its $\mathrm{pH}$. I would go right to the reaction of the dissociation of $\ce{Mg(OH)2}$:


$$\ce{Mg(OH)2 <=> Mg^2+ + 2OH-}$$


The $K\_\mathrm{sp}$ expression for this reaction is:


$$K\_\mathrm{sp}=[\ce{Mg^2+}][\ce{OH-}]^2$$


According to wikipedia (see sources at bottom), the $K\_\mathrm{sp}$ for $\ce{Mg(OH)2}$ is $\pu{5.61\*10^{-12}}$


$$\pu{5.61\*10^{-12}}=[\ce{Mg^2+}][\ce{OH-}]^2$$


My molarity of $\ce{MgSO4}$ will be the same as that of $\ce{Mg^2+}$, which is $\pu{1 M}$. Substitute that in.


$$\pu{5.61\*10^{-12}}=[\ce{OH-}]^2$$


And then solving for $[\ce{OH-}]$ we get:


$$[\ce{OH-}]=\pu{2.37\*10^{-6} M}$$


Therefore:


$$\mathrm{pOH}=5.63$$
and
$$\mathrm{pH}=8.37$$


So my final solution of $\ce{MgSO4}$ will be basic. However, on stack exchange it says that the $\mathrm{pH}$ of $\ce{MgSO4}$ is between 5.5 and 6.5. The post does not explain why, but that the general consensus is that that is the range of the $\mathrm{pH}$.


What did I do wrong?


Sources:
<https://en.wikipedia.org/wiki/Magnesium_hydroxide>


[Is magnesium sulfate basic, neutral or acidic?](https://chemistry.stackexchange.com/questions/49825/is-magnesium-sulfate-basic-neutral-or-acidic)




Another aspect to be taken into account is the difference between measured solubility and solubility as calculated from the solubility product. Salts like $\ce{MgSO\_4}$ are always much more soluble than what can be calculated from the solubility product. I don't have the numerical values for $\ce{MgSO\_4}$, but I know the corresponding values for $\ce{CaSO\_4}$, which is its neighbor in the periodic table. Calculated from $\ce{K\_{sp}}$, the solubility is $4.7$ mM. The measured value, obtained by titration, is $18$ mM, as stated in J. Chem. Educ. $77, 12$, Dec. $2000$, p.$1558$


This discrepancy is due to hydrolysis. An important part of the $\ce{Ca^{2+}}$ ions react with water to produce ions like $\ce{[Ca(OH)]^{+}}$ according to the equation $$\ce{Ca^{2+} + H\_2O -> [Ca(OH)]^+ + H^+}$$ $\ce{Mg^{2+}}$ should react the same way.$$\ce{Mg^{2+} + H\_2O -> [Mg(OH)]^+ + H^+}$$ Simultaneously, an important part of the sulfate ions are reacting according to the following equation $$\ce{SO\_4^{2-} + H\_2O -> HSO\_4^- + OH^-}$$ And there is also the possibility that some $\ce{CaSO\_4}$ ou $\ce{MgSO\_4}$ gets dissolved without being dissociated. The relative extent of these two last reactions is the reason why $\ce{MgSO\_4}$ gives non neutral solutions in water.


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Your answer lies not in the solubility product but in the acidity constants of sulfuric acid and the basicity constants of $\ce{Mg(OH)2}$, though the solubility product may well come into play. You would need to solve a system of 6 equations, one for each of the:


* 2 $\mathrm{p}K\_\mathrm{a}$ values for the acid
* Two $\mathrm{p}K\_\mathrm{b}$ values for the base
* Proton condition or charge neutrality
* And as you will be groping in $\mathrm{pH}$, one to be sure you don't exceed the solubility product.


As sulfuric acid is 'stronger' as an acid than $\ce{Mg(OH)2}$ is as a base, you would expect the answer to be slightly on the "acid side" of neutral $\mathrm{pH}$.


 2