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?

• There might very well be an issue with the value of $K_{\rm s}$ you're using - are you certain you've copied it correctly from the book? Also, see this relevant post on the vagaries of the solubility constant for $\ce{AlOH3}$. – Todd Minehardt Sep 15 '15 at 16:51
• @ToddMinehardt It's $1\cdot 10^{-33}$ for sure. I have double checked. – Arlic Sep 15 '15 at 17:02

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

• "Since the solution is acidic you have to consider [OH−] concentration to be 10−7" It shouldn't cointain then $[OH−] = \frac{K_w}{[H3O^+]}=10^{-9.5}$ or i think wrong ? Anyway, then why Atkins says in his book that answer is $3.0⋅10^{−5}? mol/l$ – Arlic Sep 15 '15 at 15:26
• Clearly you can't make an acidic solution by dissolving Al(OH)3. Either the question is wrong or you're holding back essential facts. And I assume latter to be the case – slhulk Sep 15 '15 at 17:15
• It may as well be the case where temperature is not 298K (25°C). So you ought to subtract the pH value from the specific pKw value given – slhulk Sep 15 '15 at 17:19