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(C) Potassium alum is $KAl(SO_4)_2 \cdot 12H_2O$. When $NaOH$ is added to it,initially $Al(OH)_3$ white precipitate is formed: $Al^{3+} + 3OH^- \right

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(C) Potassium alum is $KAl(SO_4)_2 \cdot 12H_2O$. When $NaOH$ is added to it,initially $Al(OH)_3$ white precipitate is formed: $Al^{3+} + 3OH^- \rightarrow Al(OH)_3 \downarrow$. On adding an excess of $NaOH$,the $Al(OH)_3$ precipitate dissolves to form a soluble complex,sodium tetrahydroxoaluminate: $Al(OH)_3 + OH^- \rightarrow [Al(OH)_4]^-$. Thus,the final result is a clear solution.

When an excess of $NaOH$ solution is added to a solution of potassium alum,what do we obtain?

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