Aluminium hydroxide is not soluble in water at all, however I was doing some reading and read that it does in fact dissolve in a solution of sodium hydroxide. This just seemed a bit strange to me, so I was wondering if anyone could explain why this is the case.
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That's because $\ce{Al(OH)3}$ is amphoteric in nature, it acts as both acid and base. When you drop a chunk of $\ce{Al(OH)3}$ into a solution of $\ce{NaOH}$, you would get a soluble salt, which is sodium aluminate, and some extra water.
$$\ce{Al(OH)3 + NaOH -> Na[Al(OH)4]}$$
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1$\begingroup$ Out of curiosity, is there any way to tell if a substance is amphoteric? $\endgroup$ – Etched Jul 11 '17 at 3:41
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$\begingroup$ @Etched Yes, but you need to analyze the structure carefully. In the case of $\ce{Al(OH)3}$, it acts as a base since the $\ce{Al-O}$ bonds are broken, releasing $\ce{OH-}$. But at the same time, it can also accept $\ce{OH-}$ due to the presence of empty electron orbitals that can grasp onto electrons. This is just a little description, if you need more info, please ask a new question. $\endgroup$ – Pritt says Reinstate Monica Jul 11 '17 at 3:46
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4$\begingroup$ Another major point is that saying"Aluminium hydroxide is not soluble" isn't entirely correct. When a chemist says "not soluble" what he is really saying is that the solubility is very low. // I'd write the reaction as $$\ce{Al(OH)3 + NaOH -> Na^+[Al(OH)4]^- }$$ $\endgroup$ – MaxW Jul 11 '17 at 3:59