# 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 $$\mathrm{pH}$$.

## 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

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