Chemistry - Help with determining concentration of magnesium hydroxide for a specific pH level

Solution 1:

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 $\mathrm{pH}~9.$

Solution 2:

There is an easy way to do what OP wants, assuming OP wants to prepare solutions at $\pu{25 ^\circ C}$. So, OP can prepare saturated solution of $\ce{Mg(OH)2}$ solution:

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

Since $K_\mathrm{sp}$ of $\ce{Mg(OH)2}$ is $\pu{5.61 \times 10^{-12} M3}$, you can find the solubility of $\ce{Mg(OH)2}$ at $\pu{25 ^\circ C}$ ($s$):

$$K_\mathrm{sp} = s \times (2s)^2 = 4s^3 \ \Rightarrow \ s = \left(\frac{K_\mathrm{sp}}{4}\right)^{\frac13} = \left(\frac{\pu{5.61 \times 10^{-12} M3}}{4}\right)^{\frac13} = \pu{\pu{1.12 \times 10^{-4} M}}$$

Thus, $[\ce{Mg^2+}] = \pu{1.12 \times 10^{-4} M}$ and $[\ce{OH-}] = 2 \times \pu{1.12 \times 10^{-4} M} = \pu{2.24 \times 10^{-4} M}$.

$$\therefore \ \mathrm{pOH} = -\log {[\ce{OH-}]} = -\log (\pu{2.24 \times 10^{-4} M}) = 3.65$$ Thus, $\mathrm{pH} = 14.00 - 3.65 = 10.35$. This means the $\mathrm{pH}$ of saturated $\ce{Mg(OH)2}$ solution is a little higher than what OP anticipated. The dilution of the saturated solution with deionized water do the trick as demonstrated in following example:

Suppose you want to make $\pu{250 mL}$ of $\ce{Mg(OH)2}$ solution with $\mathrm{pH} = 8.00$. Thus, $\mathrm{pOH} = 14.00 - 8.00 = 6.00$. Thus, $[\ce{OH-}] = \pu{1.00 \times 10^{-6} M}$. For the calculation for the dilution, you can use $c_1V_1 = c_2V_2$ equation.

In OP's case, $c_1 = \pu{2.24 \times 10^{-4} M}$, $c_2 = \pu{1.00 \times 10^{-6} M}$, and $V_2 = \pu{250 mL}$, the volume of anticipated solution with $\mathrm{pH} = 8.00$. The unknown $V_1$ is the volume of saturated $\ce{Mg(OH)2}$ solution ($\mathrm{pH} = 10.35$) needed to be diluted:

$$c_1V_1 = c_2V_2 \ \Rightarrow \ V_1 = \frac{c_2V_2}{c_1} = \frac{\pu{1.00 \times 10^{-6} M} \times \pu{250 mL}}{\pu{2.24 \times 10^{-4} M}} = \pu{1.12 mL}$$

Thus, you can measure $\pu{1.12 mL}$ of saturated $\ce{Mg(OH)2}$ solution into $\pu{250 mL}$ volumetric flask and diluted it with DI water to the $\pu{250 mL}$ line mark. After shaking well to get homogeneous solution, its $\mathrm{pH}$ should be anticipated $8$ (or closer to $8$ based on the accuracy of the measurements).

Note: It would be better if you can measure the $\mathrm{pH}$ of saturated solution before do the calculations. That's because, the factors such as temperature influence the realtime $\mathrm{pH}$.

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Solution 3:

$\ce{Mg(OH)2}$ is a strong base since it is ionic in nature; it usually dissociates completely and so it's degree of dissociation is one.

For weaker salts, the concentration values you assigned for the ions, $x$ and $2x$ respectively, do depend on the molarity of the $\ce{Mg(OH)2}.$ You have to define $x$ in the terms of its degree of dissociation $(\alpha),$ and the concentration $(c)$ as $c\alpha.$ The RICE table would come out as

$$ \begin{array}{lcccc} & \ce{Mg(OH)2 &<=> & Mg^2+ &+ &2 OH-} \\ \text{Initial} & c & & 0 && 0 \\ \text{Change} & -c\alpha & & +c\alpha && +2c\alpha \\ \text{Equilibrium} & c - c\alpha & & c\alpha && 2c\alpha \end{array} $$