# 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.

"pH" = 9

#### Spiegazione:

The idea here is that you need to use magnesium hydroxide's costante di solubilità del prodotto 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), sarà uguale a

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

Rearrange to find the concentration of the hydroxide anions

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

Inserisci i tuoi valori per ottenere

["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 a temperatura ambiente

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

to find the pH of the solution

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

So, for pH values that are sotto 9, the solution will be insaturi. Once the pH of the solution becomes equal to 9, la soluzione diventa saturato and the magnesium hydroxide starts to precipitate.