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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:

Answer:

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

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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?

Answer:

????

However, in some ways I don't quite get this phenomena --- the alum is "partially soluble" not completely soluble....so 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?

JJ

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 me.....how 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

JJ

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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:

Answer:

a) 1460 lbs/1000 ft2 = 1.460 lbs/ft2
b) Chemical XYZ add rate = 1.17% x 1.460 lbs/ft2 = .01708 ...

Solution Summary

Solubility of Alum amongst Paper By-products is discussed.

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