(a) How much aluminium hydroxide will dissolve in 500ml of water at 25∘C given that Ksp=3×10−34? (b) How much will dissolve in 500ml of a solution of 0.04M Ba(OH)2 ?

1 Answer
Michael
Apr 21, 2015

(a). 7.12×10−8g

(b). 2.28×10−28g

Part A:

Aluminium hydroxide dissociates:

Al(OH)3(s)⇌

Al3+(aq)+3OH−(aq) (1)

So:

Ksp=[Al3+(aq)][OH−(aq)]3=3×10−34mol3.l−3 (2)

We can let[Al3+(aq)] =s

We can see that [OH−(aq)]=3s

So:

s×(3s)3=3×10−34

So:

27s4=3×10−34

From which:

s=1.826×10−9mol/l

Mr=78

s=1.826×10−9×78=142.3×10−9g/l

So to get the solubility in 500ml⇒

=142.32×10−9=7.12×10−8g

Part B

Now we are trying to dissolve the compound in a solution that already has a lot of OH− ions.

From (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 Ksp 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 0.04×2=0.08mol/l

From (2)⇒

3×10−34=s×(0.08)3

From which:

s=5.85×10−30mol/l

s=5.85×10−30×78=4.563×10−28g/l

You can see here how it has been greatly reduced.

So the solubility in 500ml will be half that:

=2.28×10−28g

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