# How do you calculate the solubility (moles/L) of Mg(OH)2 in water.The Ksp=5.0*10^-11?

Apr 23, 2017

We interrogate the equilibrium..........and get,

$S = 2.32 \times {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

We interrogate the equilibrium..........

$M g {\left(O H\right)}_{2} \left(s\right) r i g h t \le f t h a r p \infty n s M {g}^{2 +} + 2 H {O}^{-}$

And K_"sp"=[Mg^(2+)][""^(-)OH]^2

And now if we call the solubility of $M g {\left(O H\right)}_{2}$ $S$, then by definition, $S = \left[M {g}^{2 +}\right]$, and 2S=""^(-)OH.

And thus ${K}_{\text{sp}} = S \times {\left(2 S\right)}^{2} = 4 {S}^{3}$

And so (finally!), S=""^(3)sqrt(K_"sp"/4)

S=""^(3)sqrt((5.0xx10^-11)/4)=2.32xx10^-4*mol*L^-1

And of course we can convert this to a $\text{mass solubility}$, by multiplying $S$ by the molar mass..........i.e.

$58.32 \cdot g \cdot m o {l}^{-} 1 \times 2.32 \times {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1 = 1.35 \times {10}^{-} 2 \cdot g \cdot {L}^{-} 1.$

What is the $p H$ of a saturated solution? This is a common follow up question.