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This is the equation for the formation of Al(OH)3:

6NaOH (aq) + Al2(SO4)3 (aq) �¨ 2Al(OH)3 (s) + 3Na2SO4 (aq)

Would any side products be formed if NaOH was added quickly into Al2(SO4)3?

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I don't think so. If too much NaOH is added, the Al(OH)3 dissolves with the formation of Al(OH)4^-

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