Here's our balanced equation:
#Mg(OH)_2(s) rightleftharpoons Mg^(2+)(aq) + 2OH^(-)(aq)#
From this equation, we can see that #1# mole of #Mg(OH)_2# corresponds to #1# mole of #Mg^(2+)# and #2# moles of #OH^(-)#.
Now, to find #K_(sp)#, we'd have to use this expression:
#K_(sp) = [Mg^(2+)][OH^(-)]^2#
(If you're a bit unclear on how we this expression was derived, check out this awesome webpage on #K_(sp)#!)
Now, to find #K_(sp)#, we'd need to find the concentrations of #Mg^(2+)# and #OH^(-)#. That's where the solubility given to us in the question comes in handy:
The question tells us that the solubility for #Mg(OH)_2# is #"0.00064 g/100 mL"#.
We'd need to convert this value into #"mol/L"#, which is the unit used for concentrations in #K_(sp)# calculations.
First, let's find the number of moles in #"0.00064g"# of #Mg(OH)2#:
#"0.00064 g"/"molar mass" = ("0.00064 g"/(24.31 + 2xx16.00 + 2xx1.008 " g/mol"))#
#= 1.097 xx 10^(-5)# #"mol"#
#1.097 xx 10^(-5)# #"mol/100 mL"# would equate to #1.097 xx 10^(-4)# #"mol/1000 mL"# (which is #"mol/L"#).
With this, we can find the #"mol/L"# concentrations of the #K_(sp)# expression components!
We've already established, with our balanced equation, that #1# mole of #Mg(OH)_2# corresponds to #1# mole of #Mg^(2+)# and #2# moles of #OH^(-)#.
So, #1.097 xx 10^(-4)# moles of #Mg(OH)_2# will correspond to #1.097 xx 10^(-4)# moles of #Mg^(2+)# and #2xx(1.097 xx 10^(-4)) = 2.20 xx 10^(-4)# moles of #OH^(-)#.
Now, we can just plug these values for #Mg^(2+)# and #OH^(-)# into our #K_(sp)# expression and solve!:
#K_(sp) = [Mg^(2+)][OH^(-)]^2#
#K_(sp) = (1.097 xx 10^(-4))(2.20 xx 10^(-4))^2#
#K_(sp) = 5.3 xx 10^(-12)#
