Help with determining concentration of magnesium hydroxide for a specific pH level
If $c$ is the concentration of $\ce{Mg^2+}$ in the $\ce{Mg(OH)2}$ solution, the concentration $[\ce{OH-}] = 2c.$
At $\mathrm{pH}~9,$ $[\ce{OH-}] = \pu{1E-5 M},$ then $c = \pu{5E-6 M}.$ So you have to dissolve $\pu{1.25 μmol}$ $\ce{Mg(OH)2}$ in $\pu{250 mL}$ water. This is $\pu{71.3 μg}$ of $\ce{Mg(OH)2}.$ This is difficult to do in practice, as such a small amount is difficult to weigh. It may be done in two steps: first prepare a moderately concentrated solution, then dilute it to $\mathrm{pH}~9.$