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.

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    $\begingroup$ 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 $\endgroup$ Jul 12 '20 at 9:59

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 $\mathrm{pH}$.

  • 1
    $\begingroup$ Dear Mathew, It would be better if you don't attach units to the equilibrium constants. $\endgroup$
    – M. Farooq
    Jul 12 '20 at 22:23
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    $\begingroup$ @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. ;-) $\endgroup$ 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

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

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    $\begingroup$ However Mg(OH)2 is sparingly soluble. dont you need to consider that? $\endgroup$ Jul 12 '20 at 12:16
  • $\begingroup$ Rolled back edits; is it okay now? $\endgroup$
    – harry
    Jul 12 '20 at 12:28
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    $\begingroup$ @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. $\endgroup$
    – andselisk
    Jul 12 '20 at 12:32
  • $\begingroup$ @andselisk I don't deserve that edit though.. $\endgroup$ Jul 12 '20 at 12:32
  • $\begingroup$ @andselisk: I meant the completely stupid answer after my second edit; Safdar's markup formatting's great. $\endgroup$
    – harry
    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.


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