If an aqueous solution of #"Mg"("OH")_2# had a #"pH"# of #10.52# at #25^@ "C"#, what is the solubility of #"Mg"^(2+)#?

Given #"pH of Mg(OH)"_2 = 10.52#

#pOH = 14-10.52#

#"pOH of Mg(OH)"_2 = 14-10.52#

#"pOH of Mg(OH)"_2 = 3.48#

#("moles of OH")/("Liter") = 10^(-pOH)#

#("moles of OH")/("Liter") = 10^(-3.48)#

#(3.31xx10^-4 "moles of OH")/("Liter")#

From the chemical formula, we observe that there are 2 moles of OH for every mole of #"Mg(OH)"_2#

#(3.31xx10^-4" moles of OH")/"Liter"(1" mole of Mg(OH)"_2)/(2" moles of OH") = (1.66xx10^-4" moles of Mg(OH)"_2)/"Liter"#

Moles per Liter is the same as molarity.

I got #1.64 xx 10^(-4) "M"#.

Since #"Mg"("OH")_2# is placed into water, the #["H"^(+)]# we observe is within the context of the autoionization of water

#K_w = ["H"^(+)]["OH"^(-)] = 10^(-14)# at #25^@ "C"#

for

#"H"_2"O" rightleftharpoons "H"^(+)(aq) + "OH"^(-)(aq)#.

Recall the definition of #"pH"#:

#"pH" = -log["H"^(+)]#

And so, first one can find the molar concentration of #"H"^(+)# in solution at #25^@ "C"#:

#["H"^(+)] = 10^(-"pH") = 10^(-10.52) "M"#

Consequently, one can obtain the molar concentration of hydroxide:

#["OH"^(-)] = K_w/(["H"^(+)]) = (10^(-14) "M"^2)/(10^(-10.52) "M")#

#= 10^(-3.48) "M"#

The dissociation of magnesium hydroxide is:

#"Mg"("OH")_2(s) stackrel("H"_2"O"(l))(rightleftharpoons) "Mg"^(2+)(aq) + 2"OH"^(-)(aq)#

The #"Mg"("OH")_2# is NOT a strong base (it is less soluble in water than #"AgCl"#!), so we may need to already know that for #"Mg"("OH")_2#,

#K_(sp) = 1.8 xx 10^(-11)#.

And so, its equilibrium mass action expression is:

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

#= 1.8 xx 10^(-11) = ["Mg"^(2+)] (10^(-3.48) "M")^2#

Since #"Mg"^(2+)# is #1:1# with #"Mg"("OH")_2(s)#, we have:

#color(blue)(["Mg"("OH")_2(aq)] -= ["Mg"^(2+)]) = K_(sp)/(["OH"^(-)])^2#

#= (1.8 xx 10^(-11))/((10^(-3.48) "M")^2)#

#= color(blue)(1.64 xx 10^(-4) "M")#

METHOD CHECK

Now suppose we also tried by another method without knowing the #K_(sp)#; since we know #["OH"^(-)]# and that #"OH"^(-)# is #2:1# with #"Mg"("OH")_2(aq)#, we could also recognize that

#color(blue)(["Mg"("OH")_2(aq)]) = ["OH"^(-)] xx ("1 mol Mg"("OH")_2(aq))/("2 mol OH"^(-))#

#= 1/2 (10^(-3.48) "M")#

#= color(blue)(1.66 xx 10^(-4) "M")#

which is also quite close to what we got with the first method.