(a) How much aluminium hydroxide will dissolve in 500ml of water at 25^@C given that K_(sp)=3xx10^(-34)? (b) How much will dissolve in 500ml of a solution of 0.04M Ba(OH)_2 ?

1 Answer
Apr 21, 2015

(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"