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#.

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.