CEE 680

 

18 December 2008

 

FINAL EXAM

 

Closed book, three pages of notes allowed.

Answer both Question A and Question B. Please state any additional assumptions you made, and show all work. If you dont have time to complete a section, please describe how you would solve the problem (without using a computer program such as MINEQL).

 

 

Part I: Answer Question A

A. Solubility & Predominance. (70%)

Zinc carbonate (ZnCO3(s)) and Zinc Hydroxide (Zn(OH)2 (s)) are two important solid phases that may control zinc solubility in water[1]. In the attached pages is a detailed solution leading to a zinc hydroxide solubility diagram[2]. Please use this to help solve the following problems.

 

1.     Prepare a solubility diagram (log C vs pH) for a water that is potentially in equilibrium with zinc hydroxide and zinc carbonate. Assume the water has 10-2 M total carbonates (i.e., 10 mM CT). Show all soluble species along with the ZnT line and indicate where precipitation will occur and the type of precipitate. Please feel free to use any of the hydroxide calculations in developing the answer to part 1 or part 2. It could save you some time.

2.     Prepare a predominance diagram, showing the precipitates and major soluble species (in areas where there are no precipitates). As would be typical for a problem of this type, make pH the x-axis, and log total carbonate (CO3T), the y-axis. Assume a total soluble zinc concentration of 10-4 M (i.e., 0.1 mM ZnT). Again you may find that using the solubility diagram provided or the one you did for part 1 can help in directing your work for part 2.

 

Equilibrium[3]

Log K

ZnOH+ = Zn+2 + OH-

-5.04

Zn(OH)2o = ZnOH+ + OH-

-6.06

Zn(OH)3-1 = Zn(OH)2o + OH-

-2.50

Zn(OH)4-2 = Zn(OH)3-1 + OH-

-1.20

Zn(OH)2 (s) = Zn+2 + 2OH-

-15.55

ZnCO3 (s) = Zn+2 + CO3-2

-10.26

 

 

 

Part II:. Answer Question B

B. Redox (30%)

It has long been known that ozonation of a drinking water or wastewater containing bromide will lead to the formation of bromate. Because of its status as a suspected human carcinogen, bromate concentrations in finished drinking water have been regulated by the US EPA at 0.010 mg/L (10 g/L). For this reason, the bromate regulation has inadvertently resulted in dimished use of ozone with waters naturally high in bromide.

With more frequent use of marginal drinking water supplies, the treatment of waters with high chloride and bromide levels is becoming more common. Given the concern over bromate formation from ozonation, these more saline sources are treated almost exclusively with chlorine. Only systems with ozonation processes are required to monitor for bromate. However, recently the possibility that chlorine can oxidize bromide to bromate has been raised.

 

1.     Write a balanced equation for the oxidation of bromide to bromate by reaction with aqueous chlorine (use dominant species at pH 7)

2.     Determine the stoichiometry of this reaction (e.g., mg-BrO3-/mg-chlorine as Cl2).

3.     Determine the equilibrium constant (K) for this reaction and comment on what thermodynamics tells you about whether this is a favorable reaction in pure sea water (assume 0.5 M chloride, 1 mM bromide, and pH 8) when the chlorine residual is 0.7 mg/L. In the interest of simplifying this problem, assume that all active chlorine is in the form of HOCl.

4.     Now estimate the residual equilibrium HOBr concentration under these conditions. Assume the same conditions as in #3. Assume that all of the bromine atoms are either in bromate or bromide. Consider that there is a very large molar excel of bromide as compared to the added HOCl.

 

 


Some important equilibrium constants:

Equilibria

Log K

Mg(OH)2 (s) = Mg+2 + 2OH-

-11.6

Fe+3 + H2O = FeOH+2 + H+

-2.19

Mg+2 + H2O = MgOH+ + H+

-11.44

MgCO3 (s) = Mg+2 + CO3-2

-7.5

CaCO3(s) = Ca+2 + CO3-2

-8.34

Ca(OH2)(s) = Ca+2 + 2OH-

-5.19

CaSO4.2H2O(s) = Ca+2 + SO4-2 + 2H2O

-4.62

CaOH+ = Ca+2 + OH-

-1.15

AlOH+2 = Al+3 + OH-

-9.01

CdOH+ = Cd+2 + OH-

-3.92

CoOH+ = Co+2 + OH-

-4.80

CuOH+ = Cu+2 + OH-

-6.00

FeOH+ = Fe+2 + OH-

-4.50

HgOH+ = Hg+2 + OH-

-10.60

NiOH+ = Ni+2 + OH-

-4.14

PbOH+ = Pb+2 + OH-

-6.29

ZnOH+ = Zn+2 + OH-

-5.04

 

Some important half-cell reactions

Equ#

Half Cell Reaction

DEo (Volts)

1

O2(g) + 4H+ + 4e- = 2H2O

+1.23

2

Mn+3 + e- = Mn+2

+1.51

3

Mn+4 + e- = Mn+3

+1.65

4

MnO4- + 8H+ + 5e- = Mn+2 + 4H2O

+1.49

5

Fe+3 + e- = Fe+2

+0.77

6

Cu+2 + e- = Cu+

+0.16

7

HOBr + H+ + e- = Br- + H2O

+1.33

8

O3 (g) + 2H+ + 2 e- = O2 (g) + H2O

+2.07

9

Al+3 + 3e- = Al(s)

-1.68

10

HOCl + H+ + e- = Cl- + H2O

+1.48

11

OCl- + H+ + e- = Cl- + H2O

+1.64

12

BrO3- +5H+ +4e- = HOBr + 2H2O

+1.45

13

S(s) + 2H+ + 2e- = H2S (g)

+0.17

14

NH2Cl + H+ +e- = Cl- + NH4+

+1.40

15

Zn+2 + 2e- = Zn(s)

-0.76

16

Ni+2 + 2e- = Ni(s)

-0.24

17

Pb+2 + 2e- = Pb(s)

-0.13

 


Properties of Selected Elements

Element

Symbol

Atomic #

Atomic Wt.

Valence

Electronegativity

Aluminum

Al

13

26.98

3

1.47

Bromine

Br

35

79.904

1,3,5,7

2.74

Calcium

Ca

20

40.08

2

1.04

Carbon

C

6

12.01

2,4

2.50

Chlorine

Cl

17

35.453

1,3,5,7

2.83

Copper

Cu

29

63.54

1,2

1.75

Hydrogen

H

1

1.01

1

2.20

Magnesium

Mg

12

24.31

2

1.23

Manganese

Mn

25

54.94

2,3,4,6,7

1.60

Nitrogen

N

7

14.0047

3,5

3.07

Oxygen

O

8

16.00

2

3.50

Potassium

K

19

39.10

1

0.91

Sodium

Na

11

22.99

1

1.01

Strontium

Sr

38

87.62

2

0.99

Sulfur

S

16

32.06

2,4,6

2.44

Zinc

Zn

30

65.39

2

1.65

 

 

Selected Acidity Constants

(Aqueous Solution, 25C, I = 0)

NAME

FORMULA

pKa

Perchloric acid

HClO4 = H+ + ClO4-

-7

Hydrochloric acid

HCl = H+ + Cl-

-3

Sulfuric acid

H2SO4= H+ + HSO4-

-3

Nitric acid

HNO3 = H+ + NO3-

0

Bisulfate ion

HSO4-= H+ + SO4-2

2

Phosphoric acid

H3PO4 = H++ H2PO4-

2.15

o-Phthalic acid

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

2.89

p-Hydroxybenzoic acid

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

4.48

Nitrous acid

HNO2 = H+ + NO2-

4.5

Acetic acid

CH3COOH = H+ + CH3COO-

4.75

Aluminum ion

Al(H2O)6+3= H++ Al(OH)(H2O)5+2

4.8

Carbonic acid

H2CO3 = H+ + HCO3-

6.35

Hydrogen sulfide

H2S = H+ + HS-

7.02

Dihydrogen phosphate

H2PO4-= H+ + HPO4-2

7.2

Hypochlorous acid

HOCl = H+ + OCl-

7.5

Hypobromous acid

HOBr = H+ + OBr-

8.71

Ammonium ion

NH4+= H+ + NH3

9.24

Bicarbonate ion

HCO3-= H+ + CO3-2

10.33

Monohydrogen phosphate

HPO4-2 = H+ + PO4-3

12.3

 

Development of hydroxide solubility diagram.

 

a. Using the equilibria for zinc hydroxide from above:

we get the following for the free aquo ion:

 

From this we use the hydroxide equilibria to get hydroxide species concentrations:

 

and for the monohydroxide:

 

now for the dihydroxide:

 

now for the trihydroxide:

 

now for the tetrahydroxide:

 

 



[1] Hydrozincite is also an important solid, but it will not be considered here for the purpose of simplicity.

[2] Note that this considers only mononuclear species, as the known polynuclear species are never dominant

[3] For the purposes of this problem, we are ignoring all soluble zinc carbonate species.