CEE 680 25 November 1996

SECOND EXAM

Closed book, two pages of notes allowed.

Problem #1: (40%) Carbonate System.

A groundwater that is at pH 8.4 and has an alkalinity of 150 mg/L as CaCO3 is pumped to the surface. It is immediately mixed in equal proportions with a pH 7.0 surface water which has an alkalinity of 300 mg/L as CaCO3. What will the pH of this water be at the moment of mixing?

Solution to Problem #1

This is a closed system problem. Therefore the total carbonate concentrations (CT's) must be determined and treated as conservative. Likewise the alkalinities are conservative, and then the final pH can be determined from the blended CT and alkalinity.

for either water:

first, I would determine the alpha's at the two pH's. Recall the general equations for a diprotic acid are:

This results in the following values (assuming pKs of 6.3 and 10.3:

 pH alpha-1 alpha-2 7 0.83331 0.000418 8.4 0.97988 0.012336

so, for the groundwater:

CT = {(150/50,000) - (10-5.6) + (10-8.4)}/(0.97988 + 2*0.012336) = 0.002984 M

and for the surface water:

CT = {(300/50,000) - (10-7) + (10-7)}/(0.83331 + 2*0.000418) = 0.007193 M

now we need to calculate the blended alkalinities and total carbonates:

Alk = (150 + 300)/2/50,000 = 0.0045 equ/L

CT = (0.002984 + 0.007193)/2 = 0.005088 M

and now calculate the pH after making some simplifying assumptions (based on the final pH being close to neutrality):

so:

[H+] = K1(CT-Alk)/Alk = 10-6.3(0.005088-0.0045)/0.0045 = 6.5 x 10-8

or

pH = 7.18

(N.B. the solution without simplifying assumptions is: pH = 7.1803)

Problem #2: (60%) Complexation

Folic acid (also known at Vitamin B10) is an important vitamin found in Liver, green leafy vegetables, brewer's yeast, oranges, and whole grains. It is involved in the production of red blood cells, cell growth and reproduction, intestinal activity, nucleic acid formation, and protein metabolism.

This compound has 7 nitrogen atoms, two carboxyl groups and one phenolic-OH. As a result, it has as many as 10 sites where t could bind to metal atoms. Little complexation data exists for this compound, but at least one study was conducted with aluminum:

Complexation of Folic Acid by Aluminum (Al+3) Footnote1

 Constants Step-wise Constants (Log K's) Overall Constants (Log Beta's) 1st 5.80 5.80 2nd 4.70 10.50 3rd 4.65 15.15

Part A.

Imagine you have a wastewater containing Folic Acid that is well buffered at pH 5.0. Furthermore, you wish to treat this wastewater by alum coagulation. In order for this process to be effective, you have to add enough alum to get the formation of an aluminum hydroxide precipitate. Aquatic chemistry calculations indicate that at the pH of this wastewater (5.0), precipitation occurs once the free Al+3 concentration reaches 10-6.5 M. To simplify this problem consider only the first complex from the above table (i.e., Al(FA)1).

Part B.

The principals of your firm, Bob Stumm and Clarence "Bud" Morgan have taken issue with your analysis. They believe that the second complex must also be considered. Again assuming a 0.01 M total Folic Acid concentration, re-do your analysis as follows:

Solution to Problem #2

For both sections of #2, we must assume that the [Al] (i.e.., the free Al+3) concentration is 10-6.5M.

#2.A.1.

Here we only concern ourselves with the formation of Al(FA)1. Therefore, our mass balance is:

FAT = [FA] + [Al(FA)1]

and the equilibrium is:

combining and substituting for [Al] we get:

10-5.80 = (FAT - [FA])/(10-6.5[FA])

and solving for [FA] we get:

[FA]/FAT = 1/(1 + (10+5.80*10-6.5)) = 0.834

This means that 83% of the folic acid is free, and the remaining 17% is bound to aluminum (as Al(FA)1) under these conditions.

#2.A.2.

To answer this question we must calculate the amount of aluminum bound to the folic acid, which requires a total folic acid concentration. Since:

Albound = [Al(FA)1] = FAT - [FA]

we need only calculate the free folic acid (i.e., [FA]) by rearranging the above equation:

[FA] = FAT/(1 + (10+5.80*10-6.5)) = 0.834*0.01 = 0.00834

and:

Albound = 0.01 - 0.00834 = 0.00166 M

Therefore the additional amount of aluminum coagulant required will be 1.66 mM as Al.

#2.B.1.

Now we need to concern ourselves with the formation of both Al(FA)1 and Al(FA)2. Our new mass balance is:

FAT = [FA] + [Al(FA)1] + [Al(FA)2]

and the equilibria are:

combining and substituting for [Al] we get:

[FA] = 0.001384 M

so the bound fraction is:

[FA]/FAT = 0.001384 M/0.01= 0.138

This means that 14% of the folic acid is free, and the remaining 86% is bound to aluminum (as Al(FA)1 and Al(FA)2) under these conditions.

#2.B.2.

To answer this question we must calculate the amount of each Al-FA species and add them. These can be determined from the equilibria.

[Al(FA)1] = 10+5.80 [Al][FA] = 10+5.80(10-6.5)0.001384 = 0.0002762 M

[Al(FA)2] = 10+10.50 [Al][FA]2 = 10+10.50(10-6.5)(0.001384)2 = 0.01917 M

The total bound Al is then the sum of these two which is 0.0194 M. So that the additional amount of aluminum coagulant required will be 19.2 mM as Al.

Footnote1

From: Nayan, R., & Dey, A.K. (1970). Z. Naturforsch. Teil B, 25: 1453., as cited in: Nordstrom, D.K., & May, H.M. (1989). Aqueous Equilibrium Data for Mononuclear Aluminum Species. In G. Sposito (Editor), The Environmental Chemistry of Aluminum (pp. 29-53). Boca Raton, FL: CRC Press, Inc.