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








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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.




As you noted


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


but the [Magnesium Hydroxide](https://en.wikipedia.org/wiki/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 


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


 2

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 !


 1

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.


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