Calculating solubility in water with other species present

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:


1. Why does Atkins assume that $\ce{Al(OH)3}$ doesn't affect $\mathrm{pH}$?
2. Which one is correct?




Since the solution is acidic you have to consider $\ce{[OH- ]}$ concentration to be $10^{-7}$


Then solubility product: $$k\_\text{sp} = \ce{[Al^3+ ] [OH- ]^3}$$


And if $x$ amount dissociated: $$k\_\text{sp} = x \cdot (3x)^3$$


Then we have,


$$10^{-33} = 27 \cdot x^4$$


So $$x = 2.466 \cdot 10^{-9}~\mathrm{mol/L}$$


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