Discussion Forum



1)

 The dissolution of Al(OH)3 by a solution of NaOH results in the formation of 


A) $[Al(H_{2}O)_{4}(OH)_{2}]^{+}$

B) $[Al(H_{2}O)_{3}(OH)_{3}]^{}$

C) $[Al(H_{2}O)_{2}(OH)_{4}]^{-}$

D) $[Al(H_{2}O)_{6}(OH)_{3}]^{}$

Answer:

Option C

Explanation:

Al(OH)3  dissolves in NaOH solution to  give  $Al(OH)_4^-$  ion which is supposed to have the octahedral complex species 

$[Al(OH)_4(H_{2}O)_{2}]^{-}$ in aqueous solution 

$Al(OH)_3+NaOH(aq)\rightarrow Al(OH)_4 (H_{2}O)_{2}- (aq)+Na+(aq)$