Question

Calculate the `pH` at which `Mg(OH)_2` begins to precipitate from a solution containing `0.1` `M` `Mg^(2+)` ions? `K_(sp)` for `Mg(OH)_2` = `1.0` x `10^-11`.

Answer 1

`"pH" = 9`

Explanation:

The idea here is that you need to use magnesium hydroxide's solubility product constant to determine what concentration of hydroxide anions, `"OH"^(-)`, would cause the solid to precipitate out of solution.

As you know, the dissociation equilibrium for magnesium hydroxide looks like this

`"Mg"("OH")_text(2(s]) rightleftharpoons "Mg"_text((aq])^(2+) + color(red)(2)"OH"_text((aq])^(-)`

The solubility product constant, `K_(sp)`, will be equal to

`K_(sp) = ["Mg"^(2+)] * ["OH"^(-)]^color(red)(2)`

Rearrange to find the concentration of the hydroxide anions

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

Plug in your values to get

`["OH"^(-)] = sqrt((1.0 * 10^(-11))/0.1) = sqrt(1.0 * 10^(-10)) = 10^(-5)"M"`

As you know, you can use the concentration of hydroxide anions to find the solution's pOH

`color(blue)("pOH" = - log(["OH"]^(-)))`

`"pOH" = - log(10^(-5)) = 5`

Finally, use the relationship that exists between pOH and pH at room temperature

`color(blue)("pOH " + " pH" = 14)`

to find the pH of the solution

`"pH" = 14 - 5 = color(green)(9)`

So, for pH values that are below `9`, the solution will be unsaturated. Once the pH of the solution becomes equal to `9`, the solution becomes saturated and the magnesium hydroxide starts to precipitate.