CEE 680

 

18 December 2006

 

FINAL EXAM

 

Closed book, three pages of notes allowed.

Answer Question A and either B or C.  Please state any additional assumptions you made, and show all work.  If you don’t 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.  (75%)

Gypsum (CaSO4.2H2O (s)) and Calcium Hydroxide (Ca(OH)2 (s)) are two important calcium-bearing minerals.

 

1.      Prepare a solubility diagram (log C vs pH) for a water in equilibrium with calcium hydroxide and Gypsum.  Assume the water has 10-2 M total sulfates.  Show all soluble species along with the CaT line and indicate where precipitation will occur and the type of precipitate.

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 sulfate (SO4T), the y-axis.  Assume a total soluble calcium concentration of 10 mM.  Note that this is a very simple system with only 2 soluble species.

3.      Describe in a qualitative way how the Calcium solubility would change if we added carbonate species.

 

 

 

Part II:. Answer either B or C

B. Redox (25%)

Monochloramine (NH2Cl) is a widely used disinfectant for drinking waters and wastewaters.  Its final reduced form is usually chloride and ammonia.  While monochloramine is both a disinfectant and an oxidant, it is not a very powerful one.  You have been asked to comment on the possibility that application of monochloramine to a water containing naturally-occurring bromide could oxidize that anionic species to hypobromous acid..

1.      Write a balanced equation for the oxidation of bromide to hypobromous acid by monochloramine

2.      Determine the stoichiometry of this reaction (e.g., mg-NH2Cl/mg-Br).

3.      Determine the log K for this reaction

4.      Calculate the equilibrium ratio of HOBr to bromide at pH 7, and in the presence of 10-4 M monochloramine, 10-4.5 M total free ammonia, and 10-3 M chloride.

 

 

C. Metal Matching (25%)

Write the letter of the correct answer in the box at left.  Each answer is used only once[1].

Correct Answer

Questions

 

Possible Answers

 

1. This metal has a pKso with carbonate of 8.34

 

A.  Iron

 

2. The pe0 for the reduction of the divalent form of this metal to its zero valent form is about -12.9

 

B.  Copper

 

3. The pe0 for the reduction of the divalent form of this metal to its zero valent form is about -4.0

 

C.  Arsenic

 

4. This divalent metal has an α0 of about 0.92 at pH 9

 

D.  Lead

 

5. This divalent metal has an α0 of about 0.05 at pH 9

 

E.  Zinc

 

6. This divalent metal has an α0 of about 0.76 at pH 9

 

F.  Aluminum

 

7. This metal has the highest ligand affinity of all in the Irving-Williams series

 

G.  Nickel

 

8. This metal does not form complexes with EDTA

 

H.  Cadmium

 

9. This metal forms oxo ions that are predomant under natural aquatic conditions

 

I.  Calcium

 

10. This metal forms strong complexes with fluoride, but does not normally undergo oxidation or reduction reactions in solution

 

J.  None of the above

 

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

½HOBr + ½H+ + e- = ½Br- + ½H2O

+1.33

8

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

+2.07

9

Al+3 + 3e-  =  Al(s)

-1.68

10

S(s) + 2H+ + 2e-  =  H2S (g)

+0.17

11

½NH2Cl + H+ +e- = ½Cl- + ½NH4+

+1.40

12

Zn+2  +  2e-   =   Zn(s)

-0.76

13

Ni+2  +  2e-   =   Ni(s)

-0.24

14

Pb+2  +  2e-   =   Pb(s)

-0.13

15

Cu+2  +  2e-   =   Cu(s)

+0.34

16

Hg2+2  +  2e-   =   2Hg(l)

+0.91

17

Fe+2  +  2e-   =   Fe(s)

-0.44

 

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

 

 

Selected Acidity Constants

(Aqueous Solution, 25°C, 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

 


[1] The pe0 is the negative log of the electron activity for the reaction under standard state conditions.  For this question, you should consult the thermodynamic data attached to this exam.