# A saturated solution of Mg(OH)_2 in water has pH = 10.32. How do you calculate the Ksp of Mg(OH)_2?

Jun 15, 2016

You can do it like this:

#### Explanation:

Magnesium hydroxide is sparingly soluble and in solution the solid is in equilibrium with the aqueous ions:

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

We are told the $p H$ so we can find $\left[O {H}^{-}\right]$.

$p H + p O H = 14$

$\therefore p O H = 14 - p H = 14 - 10.3 = 3.7$

$\therefore - \log \left[O {H}^{-}\right] = 3.7$

From which $\left[O {H}^{-}\right] = 2 \times {10}^{- 4} \text{ ""mol/l}$

From the equation you can see that $\left[M {g}^{2 +}\right]$ must be half of this.

$\therefore \left[M {g}^{2 +}\right] = \frac{2 \times {10}^{- 4}}{2} = {10}^{- 4} \text{ ""mol/l}$

The expression for ${K}_{s p}$ is given by:

${K}_{s p} = \left[M {g}^{2 +}\right] {\left[O {H}^{-}\right]}^{2}$

Putting in the numbers:

${K}_{s p} = {10}^{- 4} \times {\left(2 \times {10}^{- 4}\right)}^{2} = 4 \times {10}^{- 12} {\text{ ""mol"^(3)."l}}^{- 3}$