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Why is the product of $\ce{Al(OH)3 + NaOH -> Na[Al(OH)4]}$ and not $\ce{NaAlO3H2 + H2O}$? Is this an acid-base reaction?

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    $\begingroup$ Think about it, $\ce{Al(OH)3}$ has a vacant pair of electrons, desperately looking for anything that would generously give it a pair of electrons. With $\ce{NaOH}$ around, possessing free $\ce{OH-}$, why not? $\endgroup$ – Pritt says Reinstate Monica Jun 17 '17 at 10:02
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    $\begingroup$ Note that aluminum hydroxide is not fully acidic. It is actually amphoteric in nature i.e. It dissolves in acid, forming $\ce{[Al(H2O)6]^3+}$ (hexaaquaaluminium) or its hydrolysis products and dissolves in strong alkaline solutions, forming $\ce{[Al(OH)4]-}$ (tetrahydroxidoaluminate). $\endgroup$ – Nilay Ghosh Jun 17 '17 at 10:23
  • $\begingroup$ Sorry bout my last comment. It's vacant orbital not vacant pair of electrons. $\endgroup$ – Pritt says Reinstate Monica Jun 17 '17 at 15:12
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All reaction sources: Chemiday

As I said in the comment section that aluminium hydroxide is amphoteric i.e. It dissolves in acid, forming $\ce{[Al(H2O)6]^3+}$ (hexaaquaaluminium) or its hydrolysis products and dissolves in strong alkaline solutions, forming $\ce{[Al(OH)4]−}$ (tetrahydroxidoaluminate). So, when it is dissolved in base, it forms a sodium aluminohydroxo complex salt. Why? Because as @Pritt mentioned, aluminium has vacant orbital for which free electron can reside there forming a complex salt.

$$\ce{Al(OH)3 + NaOH(conc.) → Na[Al(OH)4]}$$

$$\ce{Al(OH)3 + 3NaOH → Na3[Al(OH)6]}$$

No worries. There is also an acid-base reaction where the two combines to form a salt and water but this occurs at very high temperature and you won't find this type of reaction occurring in laboratory. This reaction can only be find in industries.

$$\ce{Al(OH)3 + NaOH ->[1000 C] NaAlO2 + 2H2O}$$

If only aluminium was used, there would be evolution of hydrogen gas from a very similar exothermic reaction. Note for simplicity, the species is written $\ce{NaAlO2}$ but it actually contains $\ce{[Al(OH)4]−}$ ion or perhaps the $\ce{[Al(H2O)2(OH)4]−}$ all due to presence of vacant orbital(see sodium aluminate).

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  • $\begingroup$ OK to make it clearer, I would like to point out that this is indeed an acid-base reaction between a Lewis acid and a Lewis base. $\endgroup$ – Tan Yong Boon Jun 19 '17 at 10:11

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