Question

What happens when you add excess KOH to potassium alum? There are two equations:

The first one: KAl(SO₄)₂ * 12H₂O(aq) + 3KOH(aq) = Al(OH)₂ (s) + 2K₂SO₄(aq) + 12H₂O(l)

What is the second equation? What is the equation for the solution turning cleae again?

When I added KOH to the alum of the alum, the solution became white and cloudy. The more KOH I added, the more cloudy and dense looking it became. However, after a certain amount of drops, the solution turned clear and translucent again, the way it was before adding the KOH.  

Answer 1

Answer :

Reaction between alum & aq.KOH,

KAl(SO4)2 * 12H2O(aq) + 3KOH(aq) = Al(OH)3 (s) + 2K2SO4(aq) + 12H2O(l)

Initially there cloudy appearance due to formation of Al(OH)3 (s) ppt.

When excess of KOH is added, there is formation of [Al(OH)₄]^- Aluminium tetrahydroxide comples which is

completely soluble in aqueous phase and hence cloudy appearances goes off & there is clear & translucent solution again.

The corresponding second reaction is,

Al(OH)₃ (s) + KOH K[Al(OH)₄]aq K+ (aq) + [Al(OH)₄]^- (aq).

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