# If Mg(OH)2 is insoluble, why does the reaction of MgO and water produce a slightly alkali solution? [closed]

The equation of the reaction is: Mg(OH)2 + H2O -> Mg(OH)2. Why is it that this reaction produces a solution of around pH 9? There are no OH- ions produced.

## closed as off-topic by Mithoron, Tyberius, A.K., airhuff, Todd MinehardtNov 16 '18 at 20:05

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• That is not the definition of "insoluble." – Zhe Nov 14 '18 at 14:08
• There are lots of hydroxide ions produced. Most of them are reprecipitated as solid magnesium hydroxide. Magnesia is not only a base in water, but a moderately strong one. – Oscar Lanzi Nov 14 '18 at 16:12

Because Ksp of Mg(OH)2 is not zero ! First of all understand that solubility (diffusion of ions from solid to liquid phase) has some equilibrium constant. And that K (eq.constant ) of any reaction can't equal zero. So, for this small but finite K, some ions Mg and OH diffuse into the water and make the solution Alkaline !

As you noted

$$\ce{MgO + H2O -> Mg(OH)2}$$

but the Magnesium Hydroxide is not totally insoluble.

$$\ce{Mg(OH)2 <=> Mg^2+ + 2 OH-}\tag{1}$$ $$\mathrm{K}_{\mathrm{sp}} = 5.61\times10^{-12}= \ce{[Mg^2+][OH-]^2}\tag{2}$$

The dissolution of $$\ce{OH-}$$ from the $$\ce{Mg(OH)2}$$ is in equilibrium with the autodissociation of water:

$$\ce{H+ + OH- <=> H2O}\tag{3}$$

with

$$\mathrm{K}_\mathrm{w} = 1.00\cdot10^{-14} = \ce{[H+][OH-]}\tag{4}$$

We will ignore $$\ce{CO2}$$ from the atmosphere making $$\ce{CO3^{2-}}$$...

Let's assume that a "considerable" amount of $$\ce{Mg(OH)2}$$ dissolves, but not all, so that we can ignore the initial amount of $$\ce{H^+ \text{ and }OH-}$$ from the autoionization of the pure water. Then using $$\ce{[OH-] = 2[Mg^{2+}]}$$ we can substitute into equation (1) for $$\ce{[OH-]}$$ and get

$$\ce{[Mg^{2+}]} = \sqrt[3]{\dfrac{5.61\times10^{-12}}{4}} = 1.1194\cdot10^{-4}\tag{5}$$

which means that

$$\ce{[OH-]} = 2\times1.1194\cdot10^{-4} = 2.2388\cdot10^{-4}\tag{6}$$

This is basic enough that the autodissociation of water can be ignored, so our assumption was valid. We can calculate $$\ce{[H+]}$$ from equation (4):

$$\ce{[H+]} = \dfrac{1.00\cdot10^{-14}}{\ce{[OH-]}} = \dfrac{1.00\cdot10^{-14}}{2.2388\cdot10^{-4}} = 4.4667\cdot10^{-11}$$

and

$$\mathrm{pH} = -\log{(4.4667\cdot10^{-11})} = 10.35002 \ce{->[Rounding]} 10.350$$

An insoluble base means that the compound $$\ce{XOH}$$ remains as its own phase (usually solid) in the medium, while some of it dissolves in the water as $$\ce{X+}$$ and $$\ce{OH-}$$. In this case, the $$pH$$ of the medium is dictated by the solubility product ($$K_s = [\ce{X+}][\ce{OH-}]$$), instead of the acid/base constant as for something like $$\ce{NH3}$$ - where both itself and its conjugated acid $$\ce{NH4+}$$ exist in solution and in equilibrium and the $$pH$$ will be defined by the relation $$K_b = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}$$. Note that because the base itself is insoluble, it does not appear in the equilibrium constant because it has no chemical activity in the aqueous medium.

In a complete counter scenario, something like $$\ce{NaOH}$$ is considered a soluble base, but in this case the conversion into ionic species in solution is total, hence the base is also considered strong. For those, the pH is controlled by the concentration of the base itself ($$pH = 14 - pOH$$ with $$pOH = -\log([\ce{NaOH}]$$).

All the considerations above of "pH determination" are valid for systems where the base is the only solute in water.