Why wont Al(OH)3 form in this reaction?

Question

I am in grade 11. We were learning qualitative salt analysis and my teacher wrote a reaction on board - $$\ce{Al^3+ +H2O\rightarrow[Al(H2O)6]^3+}.$$

Now as far as structure of the complex is concerned i get it that $\ce{H2O}$ is a ligand and $\ce{H2O}$ donates lone pair to form that complex and $\ce{H2O}$ acts as Lewis base. My question is why wont the following reaction be possible - $$\ce{Al^3+ + H^+ OH^- \rightarrow Al(OH)3 + H+_{(aq)}}$$

I have a limited knowledge of coordination chemistry. So am I missing something?

Answer 1

The equations written by Aleph are all wrong, mainly because they are not stoichiometrically correct. They should have been written $$\ce{Al^{3+} + 6 H2O -> [Al(H2O)6]^{3+}}$$ $$\ce{Al^{3+} + 3 H2O -> Al(OH)3 + 3 H^+}$$ Anyway, the first reaction produces a complex hydrated cation. And the second equation ignores the complexation of the aluminium ion. In reality, as Poutnik mentioned, the ion $\ce{Al(H2O)6^{3+}}$ produced in the first equation is successively deprotonated according to the series of three equations $$\ce{[Al(H2O)6]^{3+} + H2O -> [Al(H2O)5(OH)|^{2+} + H3O+}$$ $$\ce{[Al(H2O)5(OH)|^{2+} + H2O -> [Al(H2O)4(OH)2|^{+} + H3O+}$$ $$\ce{[Al(H2O)4(OH)2|^{2+} + H2O -> [Al(H2O)3(OH)3| + H3O+}$$ The last product is sometimes simplified as $\ce{Al(OH)3}$

Answer 2

It is rather $$\ce{[Al(H2O)_n(OH)_{6-n}]^{(n-3)+}(aq) + H2O(l) <=> \\ <=> [Al(H2O)_{n-1}(OH)_{7-n}]^{(n-4)+}(aq) + H3O+(aq)}.$$ Coordinated water molecules become more acidic than free ones.