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Tetrafluoroborate salts, such as lithium tetrafluoroborate, are reasonably stable in aqueous solutions. Does the same hold for sodium tetrachloroaluminate as well or does it end up as $$\ce{NaAlCl4 + 4H2O -> NaAl(OH)4 + 4HCl}?$$

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NaAlCl4 will hydrolyze according to the following reaction $$\ce{AlCl4^- + \ 2H2O -> Al(OH)2^+ + 2 H+ +4 Cl-}$$and if there is enough water it could goo further away like $$\ce{AlCl4^- + \ 3H2O -> Al(OH)3 + 3 H+ +4 Cl-}$$ But the reaction will never produce $\ce{NaAl(OH)4}$ as you think. The substance $\ce{NaAl(OH)4}$ can only be produced in very basic solution, and here the solution is acidic.

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  • $\begingroup$ Does it happen at STP or does the mixture have to be heated or something? $\endgroup$ – Just A Young Artist Jan 1 at 16:41
  • $\begingroup$ Then, can NaAl(OH)4 be generated if the HCl is boiled away? $\endgroup$ – Just A Young Artist Jan 1 at 16:54
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    $\begingroup$ This hydrolysis occurs at room temperature by simple dilution. And even if HCl is boiled away, it will not generate NaAl(OH)4. Al(OH)3 is the end point of this hydrolysis. To reach NaAl(OH)4, a large excess of OH- is necessary. $\endgroup$ – Maurice Jan 1 at 20:14
  • $\begingroup$ The sodium simply ends as NaCl. $\endgroup$ – Oscar Lanzi Jan 1 at 22:14

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