(a). #7.12xx10^(-8)"g"#
(b). #2.28xx10^(-28)"g"#
Part A:
Aluminium hydroxide dissociates:
#Al(OH)_(3(s))rightleftharpoons#
#Al_((aq))^(3+)+3OH_((aq))^(-)# #color(red)((1))#
So:
#K_(sp)=[Al_((aq))^(3+)][OH_((aq))^(-)]^3=3xx10^(-34)mol^3.l^(-3)# #color(red)((2))#
We can let#[Al_((aq))^(3+)]# #="s"#
We can see that #[OH_((aq))^(-)]="3s"#
So:
#sxx(3s)^(3)=3xx10^(-34)#
So:
#27s^(4)=3xx10^(-34)#
From which:
#s=1.826xx10^(-9)"mol/l"#
#M_r=78#
#s=1.826xx10^(-9)xx78=142.3xx10^(-9)"g/l"#
So to get the solubility in 500ml#rArr#
#=142.3/2xx10^(-9)=7.12xx10^(-8)"g"#
Part B
Now we are trying to dissolve the compound in a solution that already has a lot of #OH^-# ions.
From #color(red)((1))# we can see that Le Chatelier's Principle predicts that increasing #[OH_((aq))^-]# like this will shift the position of equilibrium to the left thus reducing the solubility of the compound.
This is known as "The Common Ion Effect" as the #OH^-# ions are common to both solutions.
We can now make an assumption that will make things a lot easier for ourselves. Because #K_(sp)# is so small we can assume that the concentration of #OH^(-)# from the #Al(OH)_3# is tiny compared with that from the #Ba(OH)_2#.
So we can set #[OH_((aq))^(-)]# as equal to #0.04xx2=0.08"mol/l"#
From #color(red)((2))rArr#
#3xx10^(-34)="s"xx(0.08)^(3)#
From which:
#s=5.85xx10^(-30)"mol/l"#
#s=5.85xx10^(-30)xx78 = 4.563xx10^(-28)"g/l"#
You can see here how it has been greatly reduced.
So the solubility in 500ml will be half that:
#=2.28xx10^(-28)"g"#
