(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 500mlrArr
=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"