Question

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

Answer 1

You can do it like this:

Explanation:

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

`Mg(OH)_(2)rightleftharpoonsMg^(2+)+2OH^(-)`

We are told the `pH` so we can find `[OH^-]`.

`pH+pOH=14`

`:.pOH=14-pH=14-10.3=3.7`

`:.-log[OH^-]=3.7`

From which `[OH^-]=2xx10^(-4)" ""mol/l"`

From the equation you can see that `[Mg^(2+)]` must be half of this.

`:.[Mg^(2+)]=(2xx10^(-4))/2=10^(-4)" ""mol/l"`

The expression for `K_(sp)` is given by:

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

Putting in the numbers:

`K_(sp)=10^(-4)xx(2xx10^(-4))^2=4xx10^(-12)" ""mol"^(3)."l"^(-3)`