What is the conjugate base of [Al(H_2O)_3(OH) | vedclass

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(B) conjugate base is formed by the removal of a proton $(H^+)$ from an acid.For the complex $[Al(H_2O)_3(OH)_3]$,the removal of one $H^+$ ion from on

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$[Al(H_2O)_2(OH)_4]^-$

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(B) conjugate base is formed by the removal of a proton $(H^+)$ from an acid.For the complex $[Al(H_2O)_3(OH)_3]$,the removal of one $H^+$ ion from one of the water ligands results in the formation of an $OH^-$ group.Reaction: $[Al(H_2O)_3(OH)_3] \rightarrow [Al(H_2O)_2(OH)_4]^- + H^+$.Therefore,the conjugate base is $[Al(H_2O)_2(OH)_4]^-$,which corresponds to option $B$.

What is the conjugate base of $[Al(H_2O)_3(OH)_3]$?

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