# (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 ?

Apr 21, 2015

(a). $7.12 \times {10}^{- 8} \text{g}$

(b). $2.28 \times {10}^{- 28} \text{g}$

Part A:

Aluminium hydroxide dissociates:

$A l {\left(O H\right)}_{3 \left(s\right)} r i g h t \le f t h a r p \infty n s$

$A {l}_{\left(a q\right)}^{3 +} + 3 O {H}_{\left(a q\right)}^{-}$ $\textcolor{red}{\left(1\right)}$

So:

${K}_{s p} = \left[A {l}_{\left(a q\right)}^{3 +}\right] {\left[O {H}_{\left(a q\right)}^{-}\right]}^{3} = 3 \times {10}^{- 34} m o {l}^{3.} {l}^{- 3}$ $\textcolor{red}{\left(2\right)}$

We can let$\left[A {l}_{\left(a q\right)}^{3 +}\right]$ $= \text{s}$

We can see that $\left[O {H}_{\left(a q\right)}^{-}\right] = \text{3s}$

So:

$s \times {\left(3 s\right)}^{3} = 3 \times {10}^{- 34}$

So:

$27 {s}^{4} = 3 \times {10}^{- 34}$

From which:

$s = 1.826 \times {10}^{- 9} \text{mol/l}$

${M}_{r} = 78$

$s = 1.826 \times {10}^{- 9} \times 78 = 142.3 \times {10}^{- 9} \text{g/l}$

So to get the solubility in 500ml$\Rightarrow$

$= \frac{142.3}{2} \times {10}^{- 9} = 7.12 \times {10}^{- 8} \text{g}$

Part B

Now we are trying to dissolve the compound in a solution that already has a lot of $O {H}^{-}$ ions.

From $\textcolor{red}{\left(1\right)}$ we can see that Le Chatelier's Principle predicts that increasing $\left[O {H}_{\left(a q\right)}^{-}\right]$ 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 $O {H}^{-}$ ions are common to both solutions.

We can now make an assumption that will make things a lot easier for ourselves. Because ${K}_{s p}$ is so small we can assume that the concentration of $O {H}^{-}$ from the $A l {\left(O H\right)}_{3}$ is tiny compared with that from the $B a {\left(O H\right)}_{2}$.

So we can set $\left[O {H}_{\left(a q\right)}^{-}\right]$ as equal to $0.04 \times 2 = 0.08 \text{mol/l}$

From $\textcolor{red}{\left(2\right)} \Rightarrow$

$3 \times {10}^{- 34} = \text{s} \times {\left(0.08\right)}^{3}$

From which:

$s = 5.85 \times {10}^{- 30} \text{mol/l}$

$s = 5.85 \times {10}^{- 30} \times 78 = 4.563 \times {10}^{- 28} \text{g/l}$

You can see here how it has been greatly reduced.

So the solubility in 500ml will be half that:

$= 2.28 \times {10}^{- 28} \text{g}$