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Topic: Reaction of KalSO42.12H2O (s) + KOH (aq) question  (Read 3047 times)

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Offline quarkz

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Reaction of KalSO42.12H2O (s) + KOH (aq) question
« on: September 25, 2011, 07:00:28 PM »
I figured it would be
KAlSO42 aq + 4 KOH aq ----> 2K2SO4 aq + KAlOH4 aq

Which would make the net ionic equation to nothing, therefore no reaction.

But in my lab, I initally added the KOH to the Alum solution and nothing happened. But as i added more drops of KOH, percipatation formed; this boggled my mind.

Why does adding more KOH produce a reaction even though the molecular forumla and therefore, the net ionic equation state that it doesnt have a rxn. Even the physical states of the moledcular forumla states there isnt a reaction. Please explain. Thanks

Offline Fluoroantimonicacid

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Re: Reaction of KalSO42.12H2O (s) + KOH (aq) question
« Reply #1 on: September 26, 2011, 01:07:03 AM »
The precipitate is Al(OH)3,I think.

Offline bidiboom

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Re: Reaction of KalSO42.12H2O (s) + KOH (aq) question
« Reply #2 on: September 26, 2011, 10:43:48 AM »

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