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.
1 Answer
Explanation:
The idea here is that you need to use magnesium hydroxide's solubility product constant to determine what concentration of hydroxide anions,
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) = ["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