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Q. Statement 1: $Al(OH)_{3}$ is insoluble in $NH_{4}OH$ but soluble inNaOH
Statement 2: NaOH is a stronger base.

The p-Block Elements

Solution:

Both are correct statements but statement 2 is not the correct explanation of statement 1.
$Al(OH)_{3}$ forms soluble complex with NaOH and not with
$NH_{4}OH.$
$Al(OH)_{3} + OH^{-} \to [Al(OH)_{4}]^{-}$