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Solubility of Alum Amongst Paper By-Products

Problem Statement:

Chemical XYZ is added to a substrate at a defined rate of 1.17% of the dry weight of the substrate; paper makers alum (Al2(SO4)3-14H2O) is added to the substrate at a rate of 45% of the Chemical XYZ rate; the dry weight of the substrate is given as 1460 lbs/1000 ft2

The molecular weight of the alum is given as:

2 x Al = 54 g/mol
3 x S = 96 g/mol 594 g/mol total
28 x H = 28 g/mol
26 x O = 416 g/mol

Assuming 100% retention of the alum, what is the weight % Al+3 present in the substrate after alum addition and convert the weight % of Al+3 to ppm:


a) 1465 lbs/1000 ft2 = 1.465 lbs/ft2
b) Chemical XYZ add rate = 1.17% x 1.465 lbs/ft2 = .01714 lbs/ft2
c) Alum add rate = 45% of Chemical XYZ = 45% x .01714 lbs/ft2 = .007713 lbs/ft2
d) Al+3 = 54 g/mol /594 g/mol Al2(SO4)3-14H2O = 9.09% Al+3 by weight
e) ppm Al+3 = (.007713 lbs x 9.09%) / (1.465 lbs + .01714 lbs + .007713 lbs) = .000237 lb/lb = (.000237 lb/lb) x 106 = 237 ppm total mass basis


However, if in reality, the solubility of the alum is given as 87 g/100 cc's and we assumed a 10 liter solution of water, and the preceding masses of material were present or added to the 10 liter water solution (e.g. 1.465 lbs of "stuff"; .01714 lbs of XYZ; and .007713 lbs alum):

How would we then determine how much Al+3 actually dissolved? What is the dissolved and un-dissolved weight fraction of Al+3 present in the solution? And, what would the equivalent ppm value of the Al+3 (dissolved & un-dissolved) be as a function of the total solution?



However, in some ways I don't quite get this phenomena --- the alum is "partially soluble" not completely I'm grappling to intuitively understand why the .007713 g alum would completely dissolve. I get it on one level but don't get it on another ---- e.g. how can a partially soluble material completely dissolve?


Dr M:

I could use your help!!

I have attached a problem statement in which I believe I have correctly determined or answered the question. The question as you will see involves determing the weight fraction and ppm level of Al+3 as a function of alum addition to a given substrate. Please take a look at my answer and let me know if o.k or otherwise please correct.

Then the second part to the question provides a known solubility of alum as 87 g/ 100 cc's and asks us to determine again the weight fraction and ppm of Al+3 (both dissolved and undissolved) in a 10 liter solution of water.

Solubility equations always seem to stump do I proceed and answer on this second part to the question. Please try to be especially clear on the solubility equation!

see attached.

Many thanks



Solution Summary

Solubility of Alum amongst Paper By-products is discussed.