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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4$\begingroup$ Think about it, $\ce{Al(OH)3}$ has a vacant orbital, 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 MonicaCommented Jun 17, 2017 at 10:02
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4$\begingroup$ 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 GhoshCommented Jun 17, 2017 at 10:23
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