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?