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#.
Risposta:
#"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.