I'm very confused about calculating solubility.
Everything is in temperature of $298\ \mathrm K$
From Atkins's book, exercise 16.83:
Based on solubility constant, calculate solubility of $\ce{Al(OH)3}$ in acid solution of $\mathrm{pH}=4.5$, $K_{\rm s}=1\cdot10^{-33}$.
Using stoichiometry and equilibrium constants I proceed as follows:
$\mathrm{pH}=4.5$ corresponds to $\mathrm{pOH}=9.5$ and
$$\ce{Al(OH)3 (s) \rightarrow Al^3+ (aq) + 3OH- (aq)}$$
$$K_{\rm s}=\ce{[Al^3+][OH- ]^3}$$
Let's call $\ce{[Al^3+]} = x$.
From stoichiometry and the assumption that $\ce{[OH- ]} >> 10^{-7}$ we know that $\ce{[OH- ]} \approx 3x+10^{-9.5}$.
We get a equation like
$$x(3x+10^{-9.5})^3=K_{\rm s}$$
which I solve as $x=2.39\cdot10^{-9}$, but Atkins's answer is $3.0\cdot10^{-5}$.
Atkins's solution looks like:
$$K_{\rm s}=\ce{[Al^3+][OH- ]^3}$$
$${K_{\rm s}\over[\ce{OH-}]^3}=[\ce{Al^3+}] = 3.0\cdot10^{-5}$$
It looks like Atkins assumes that $[\ce{OH-}]=10^{-9.5}$.
I have two questions:
Why does Atkins assume that $\ce{Al(OH)3}$ doesn't affect $\mathrm{pH}$?
Which one is correct?