Question

Aqueous ammonia is used as precipitating reagent for $A{l}^{3+}$ ions as $Al{\left(OH\right)}_3$ rather than aqueous NaOH because :

• A. $N{H}_4^{+}$ is a weak base
• B. NaOH is a strong base
• C. NaOH forms $[Al{\left(OH\right)}_4{]}^{-}$ ions
• D. NaOH forms $[Al{\left(OH\right)}_2{]}^{+}$ ions

Answer 1

The correct option is B NaOH is a strong base

$NaOH$ is a strong base. When it is used as precipitating reagent for $A{l}^{3+}$, initially it forms a colourless white gel precipitate of $Al(OH{)}_3$ which dissolves in excess of $NaOH$ by forming colourless complex of ${\left[Al\right(OH{)}_6]}^{3-}$.

But in case of $N{H}_3$ used as precipitating reagent, colourless white gel precipitate of $Al(OH{)}_3$ is formed which is not further affected by the excess of $N{H}_3$.