CEE 680

 

16 November 2000

SECOND EXAM

 

Closed book, two pages of notes allowed.

 

Answer all questions. Please state any additional assumptions you made, and show all work.

 

 

Problem #1: Carbonate System.

(50%) Two raw drinking waters are mixed as they enter the headworks of a water treatment plant. The two are characterized as follows:

 

Water

Flow (MGD)

Alkalinity

(mg/L as CaCO3)

pH

#1

20

70

6.98

#2

10

220

8.85

 

A. What will the pH of the blended water be immediately after mixing?

 

B. What will the pH of the blended water be after it has reached equilibrium with the bulk atmosphere?

 

 

 

 

Problem #2: Complexation

1. (50%) You are trying to control algal growths in a lake containing water that is well buffered at neutrality (pH 7.00). Accordingly, you have just added 10 M of Cu+2 for this purpose. However, you discover that the intended effect did not occur. Careful analysis of this lake water showed a substantial presence of EDTA. The total EDTA concentration (YT, the sum of all species) was measured as 20 M.

 

A. How much free copper actually existed? Assume the system is at equilibrium, and ignore any copper hydrolysis species.

 

B. How much manganese (+II) would have to be introduced into this water to reduce the degree of copper binding by EDTA to 50% of its total concentration (i.e., CuEDTA = 0.5*CuT)?

 

C. Describe in qualitiative terms how your approach to solving this problem would differ if the ligand was not EDTA but some monodentate ligand such as ammonia. Lets presume that you do not have access to MINEQL or any similar computer code.

 

 

Important Equilibria

 

Equilibrium

Log K

 

Y-4 + H+ = HY-3

10.26

 

Y-4 + 2H+ = H2Y-2

16.42

 

Y-4 + 3H+ = H3Y-

19.09

 

Y-4 + 4H+ = H4Y

21.09

 

Y-4 + 5H+ = H5Y+

22.59

 

Y-4 + 6H+ = H6Y+2

22.59

 

Y-4 + Cu+2 = CuY-2

18.80

 

Y-4 + Mn+2 = MnY-2

13.87

 

 

 


Selected Acidity Constants (Aqueous Solution, 25C, I = 0)

NAME

FORMULA

pKa

Perchloric acid

HClO4 = H+ + ClO4-

-7 STRONG

Hydrochloric acid

HCl = H+ + Cl-

-3

Sulfuric acid

H2SO4= H+ + HSO4-

-3 (&2) ACIDS

Nitric acid

HNO3 = H+ + NO3-

-0

Hydronium ion

H3O+ = H+ + H2O

0

Trichloroacetic acid

CCl3COOH = H+ + CCl3COO-

0.70

Iodic acid

HIO3 = H+ + IO3-

0.8

Bisulfate ion

HSO4- = H+ + SO4-2

2

Phosphoric acid

H3PO4 = H+ + H2PO4-

2.15 (&7.2,12.3)

o-Phthalic acid

C6H4(COOH)2 = H+ + C6H4(COOH)COO-

2.89 (&5.51)

Citric acid

C3H5O(COOH)3= H+ + C3H5O(COOH)2COO-

3.14 (&4.77,6.4)

Hydrofluoric acid

HF = H+ + F-

3.2

Aspartic acid

C2H6N(COOH)2= H+ + C2H6N(COOH)COO-

3.86 (&9.82)

m-Hydroxybenzoic acid

C6H4(OH)COOH = H+ + C6H4(OH)COO-

4.06 (&9.92)

p-Hydroxybenzoic acid

C6H4(OH)COOH = H+ + C6H4(OH)COO-

4.48 (&9.32)

Nitrous acid

HNO2 = H+ + NO2-

4.5

Acetic acid

CH3COOH = H+ + CH3COO-

4.75

Propionic acid

C2H5COOH = H+ + C2H5COO-

4.87

Carbonic acid

H2CO3 = H+ + HCO3-

6.35 (&10.33)

Hydrogen sulfide

H2S = H+ + HS-

7.02 (&13.9)

Dihydrogen phosphate

H2PO4- = H+ + HPO4-2

7.2

Hypochlorous acid

HOCl = H+ + OCl-

7.5

Boric acid

B(OH)3 + H2O = H+ + B(OH)4-

9.2 (&12.7,13.8)

Ammonium ion

NH4+ = H+ + NH3

9.24

Hydrocyanic acid

HCN = H+ + CN-

9.3

p-Hydroxybenzoic acid

C6H4(OH)COO- = H+ + C6H4(O)COO-2

9.32

Phenol

C6H5OH = H+ + C6H5O-

9.9

m-Hydroxybenzoic acid

C6H4(OH)COO- = H+ + C6H4(O)COO-2

9.92

Bicarbonate ion

HCO3- = H+ + CO3-2

10.33

Monohydrogen phosphate

HPO4-2 = H+ + PO4-3

12.3

Bisulfide ion

HS- = H+ + S-2

13.9

Water

H2O = H+ + OH-

14.00

Ammonia

NH3 = H+ + NH2-

23

Methane

CH4 = H+ + CH3-

34