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Formation of Aluminum Hydroxide from Aluminum Sulfate and Water

Assume we have a 1000L tank of water and very dilute weak acids. To this tank we add and dissolve 260 grams AL2(SO4)3 (solid granules) --- the resultant pH is 4. The intention is to form AL(OH)3 solids/particles in this reaction.

Question 1: How much Al(OH)3 will be formed in this reaction at pH=4.0

Question 2: If the pH were raised to 5.5 with NaOH, there is an appearance of more AL(OH)3 being formed ---- why? And, how much more aluminum hydroxide would be formed at this higher pH?

Question 3: It also appears that not all of the alum (aluminum sulfate) dissolves in water and therefore limits the max. creation of aluminum hydroxide --- why?

Solution Preview

Assume we have a 1000L tank of water and very dilute weak acids. To this tank we add and dissolve 260 grams AL2(SO4)3 (solid granules) --- the resultant pH is 4. The intention is to form AL(OH)3 solids/particles in this reaction.

The reaction equation involved can be given as:

Al2(SO4)3 + 6 H2O  2 Al(OH)3 + 3 H2SO4

Question 1: How much Al(OH)3 will be formed in this reaction at pH=4.0

Molar mass of Al2(SO4)3 = 2 x 27 + 3( 32 + 4(16) )
= 342 gmol-1
Therefore, moles of Al2(SO4)3 = mass/molar mass = 260 g/342 gmol-1
= 0.76 moles
Since 1 mole Al2(SO4)3 produces = 2 mole of Al(OH)3
So, ...

Solution Summary

Solution neatly answers 3 questions on aluminum formation with explanation, including topics like how much product formed, pH increase and limiting factors.

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