# 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.

$$\begin{array}{lcccc} & \ce{Mg(OH)2 &<=> & Mg^2+ &+ & 2OH-} \\ \text{I} & x & & 0 && 0 & \\ \text{C} & x & & +x && +2x \\ \text{E} & x & & x && 2x \end{array}$$

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.

• Mg(OH)2 is a strong electrolyte/base. It is an ionic compound. Do you mean to say that it is sparingly soluble as the reason for using the RICE table – Safdar Faisal Jul 12 '20 at 9:59

## 4 Answers

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 $$\mathrm{pH}$$.

• Dear Mathew, It would be better if you don't attach units to the equilibrium constants. – M. Farooq Jul 12 '20 at 22:23
• @M. Farooq : Thanks for your kind comment. I'll keep it in mind for future perspectives. I just want to show how we get concentration term at the end for $s$. Thanks again. ;-) – Mathew Mahindaratne Jul 13 '20 at 0:06

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 $$\mathrm{pH}~9.$$

$$\ce{Mg(OH)2}$$ 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

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

• However Mg(OH)2 is sparingly soluble. dont you need to consider that? – Safdar Faisal Jul 12 '20 at 12:16
• Rolled back edits; is it okay now? – Harry Holmes Jul 12 '20 at 12:28
• @HarryHolmes No, why would you rollback edits? Your image is not readable as it is rotated 90° and also lacks readability in comparison with the markup. – andselisk Jul 12 '20 at 12:32
• @andselisk I don't deserve that edit though.. – Safdar Faisal Jul 12 '20 at 12:32
• @andselisk: I meant the completely stupid answer after my second edit; Safdar's markup formatting's great. – Harry Holmes Jul 12 '20 at 12:33

Now much base of any kind you need to get $$\mathrm{pH} = 7$$? None, the water itself is enough.

For $$\mathrm{pH}$$ levels near 7 (8 is pretty much eligible) you want to account for the $$\ce{OH-}$$ ions that come from water.

$$[\ce{OH-}] = 1 \times 10^{-6} (\mathrm{pH} = 8)$$

$$[\ce{OH-}][\ce{H+}] = 1 \times 10^{-14}$$ (water at room temp)

$$[\ce{OH-}] = [\ce{H+}] + 2[\ce{Mg^2+}]$$ (electric neutrality)

(everything in molar concentrations)

... solve it for $$[\ce{Mg^2+}]$$.

For $$\mathrm{pH} = 9$$ you can skip the water $$[\ce{OH-}]$$ and the error will be less than 1%. This error may or may not be acceptable in your context.

You may next want to check if $$\ce{Mg(OH)2}$$ is at all soluble that much.