Help with determining concentration of magnesium hydroxide for a specific pH level This seems like such an obvious and elementary-level question for someone who has taken high school chemistry, but yet I'm having a difficult time solving it. I need to find the right amount of $\ce{Mg(OH)2}$ to have a $\pu{250 mL}$ solution with a $\mathrm{pH}$ of $8$ and $9.$
I first started by trying to determine the concentration of $\ce{Mg(OH)2}$ needed to obtain a solution with a certain pH. Since $\ce{Mg(OH)2}$ is a weak base, we must use a RICE table to determine its concentration at equilibrium.
$$
\begin{array}{lcccc}
  &amp; \ce{Mg(OH)2 &amp;&lt;=&gt; &amp; Mg^2+ &amp;+ &amp; 2OH-} \\
\text{I} &amp; x &amp; &amp; 0 &amp;&amp; 0 &amp; \\
\text{C} &amp; x &amp; &amp; +x &amp;&amp; +2x \\
\text{E} &amp; x &amp; &amp; x &amp;&amp; 2x
\end{array}
$$
However, the equation involving $K_\mathrm{sp}$ only involves the $x$ values in the RICE table above, not the actual concentration. How would I calculate the $\mathrm{pH}$ level for $\ce{Mg(OH)2}$ with varying molarity? I thought it was a weak base, so I'd have to use a RICE table.
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 $\mathrm{pH}~9.$