Alum (aluminum sulfate) is used as a flocculant to floc an anionic material in a water solution at pH = 4.0 and temperature = 130 degrees F. The solubility of the alum is noted as 87 grams per 100 cc's of liquid (assuming std temp & pressure).
Assume 1 liter of solution....based on this, how would I determine the actual solubility of the aluminum ion (how much dissolved vs. not) at say a pH of 3.0 vs. 4.0 vs. 5.0.
Typical aluminum chemistry suggests that lower pH increases solubility given a fixed temperature of 130 degrees F ---- I also heretell that the optimal pH for alum chemistry is in the range of 4.5 to 5.0.
But I do not understand this and want to know how to derive the soluble vs. insoluble Al at these various pH ranges.
I am ultimately trying to drive the following:
Al(SO4)3 + 6H2O = 2Al(OH)3 + 3H2SO4
My theory being that if I have more insoluble alum sulfate, I will produce more aluminum hydroxide (assuming sufficient OH)