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 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.  (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.  (35%)

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. (35%)

 

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

 

 

Development of hydroxide solubility diagram (already provided).

 

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:

 

 

 

 

Answer to A.1.

 

And for the carbonate, we must consider:

 

ZnCO30  =  Zn+2 + CO3-2

  -5.30

Zn(OH)2 (s)  =  Zn+2 + 2OH-

-15.55

ZnCO3 (s)  =  Zn+2 + CO3-2

-10.26

 

 

b. Using the equilibria for zinc carbonate from above:

 

b1. we get the following for the free aquo ion:

And for 10 mM total carbonates:

 

For Carbonate Calculations:

For pH<6.3, , or

For pH=6.3-10.3, , or

For pH>10.3, , or

 

            So for pH<6.3, we have:

,  @pH<6.3

 

            And for pH=6.3-10.3, we have:

,  @pH=6.3-10.3

 

            And for pH>10.3, we have:

,  @pH>10.3

 

b2. and for the monohydroxide:

            from the Zn(OH)2(s) based calculations, we know that:

            And now substituting in the ZnCO3(s)-based free zinc equation for pH<6.3, we have:

, @pH<6.3

 

            And for pH=6.3-10.3, we have:

, @pH=6.3-10.3

 

            And for pH>10.3, we have:

, @pH>10.3

 

 

b3. now for the dihydroxide:

            from the Zn(OH)2(s)-based calculations, we know that:

 

            And now substituting in the ZnCO3(s)-based ZnOH equation for pH<6.3, we have:

, @pH<6.3

 

            And for pH=6.3-10.3, we have:

, @pH=6.3-10.3

 

            And for pH>10.3, we have:

, @pH>10.3

 

 

B4. now for the trihydroxide:

            from the Zn(OH)2(s)-based calculations, we know that:

 

            And now substituting in the ZnCO3(s) based Zn(OH)2 equation for pH<6.3, we have:

, @pH<6.3

 

            And for pH=6.3-10.3, we have:

, @pH=6.3-10.3

 

            And for pH>10.3, we have:

, @pH>10.3

 

 

b5. now for the tetrahydroxide:

            from the Zn(OH)2(s)-based calculations, we know that:

 

            And now substituting in the ZnCO3(s) based Zn(OH)3 equation for pH<6.3, we have:

, @pH<6.3

 

            And for pH=6.3-10.3, we have:

, @pH=6.3-10.3

 

            And for pH>10.3, we have:

, @pH>10.3

 

 

 

 

 

 

 

 

 

Now combining, we find that the hydroxide controls at pHs of 10.15 and above.  Below this level, the carbonate is less soluble.

 

 

 

Answer to A.2.

 

Type A lines

First, lets consider the boundary between the aquo ion and the monohydroxide

When they’re present at equal concentrations:

or:

And this gives us A1:

pH = 8.96

 

following the same approach, we get the remaining A lines

Gives us A2:

pH =7.94

but since this is less than A1, it tells us that the monohydroxide never predominates, so we need to look at the boundary between the free zinc and the dihydroxide

 

At the boundary,

Gives us the revised A1:

pH =8.45

 

 

Gives us the revised A2:

pH =11.50

 

 

Gives us A3:

pH =12.80

 

 

 

 

 

 

Type B lines

First let’s examine the hydroxide equilibria. By looking at the Zn(OH)2 solubility graph, its appears that only free zinc (Zn+2) and the tetrahydroxide (Zn(OH)4-2) are the dominant soluble species in equilibrium with the hydroxide precipitate at a total soluble zinc concentration of 10-4 M.

 

B1:

 

So, first let’s look at the Zn+2/Zn(OH)2 precipitate boundary.  From the solubility product equation:

Of course, this is just a re-derivation of the equation that was already provided as part of the hydroxide solubility calculations:

 

now adopting a total Zn of 0.1 mM, we get:

or:

pH =8.225

 

this line is valid as the A lines show Zn+2 to be dominant at this pH, albeit just barely

 

B2:

 

Now let’s look at the Zn(OH)4-2/Zn(OH)2 precipitate boundary.  From the given solubility calculations, we have:

Once again, adopting a total Zn of 0.1 mM, we get:

or:

pH =12.375

 

This falls below the A3 line, and well into the Zn(OH)3-1 zone, so it’s not valid

 

 

So, now we need to consider the Zn(OH)3-1/Zn(OH)2 precipitate boundary.  From the given solubility calculations, we have:

Once again, adopting a total Zn of 0.1 mM, we get:

 

or:

pH =11.95

 

This falls between the A2 and A3 lines, so we’re in the Zn(OH)3-1 zone, and the line is valid

 

 

 

Inspection of the carbonate/hydroxide solubility diagram leads one to conclude that free zinc is the only soluble species that needs to be considered for the B lines defining carbonate solubility.

 

So, lets look at the Zn+2/ZnCO3 boundary:  From the solubility calculations we had developed three line segments.  We can conveniently take the equation for each, retaining the total carbonate term in its general form:

 

B3a:

            For pH<6.3, we had:

            And setting the free zinc equation to 0.1 mM, we get

,  @pH<6.3

 

B3b:

            And for pH=6.3-10.3, we have:

,  @pH=6.3-8.45

 

However, this is only valid up to pH 8.45, the limit of free zinc’s predominance (A1 line)

 

 

 

 

We could go on and graph the lines defining the border between the various zinc hydroxides, but it’s clear that they won’t be necessary.  This was determined from inspection of the carbonate/hydroxide solubility graph.  It’s also clear from the emerging predominance diagram (above).  The zinc carbonate lines already extend into the zinc hydroxide precipitate zone.  This border is defined by type C lines.

 

 

 

Type C lines

Finally, lets consider the boundary between the two solid phases.  This line will start on the low pH side, were the B lines intersect (pH=8.225).  This is well within the zone where bicarbonate is the dominant carbonate species.

 

First, from the hydroxide solid solubility as given, we have:

And from the carbonate solubility we had calculated:

 

C1b:

            For the bicarbonate zone; pH=6.3-10.3:

                        Substituting in the hydroxide-based equation, we get:

,  @pH=6.3-10.3

 

C1c:

            And for the carbonate zone;  pH>10.3, we determined:

                        Substituting in the hydroxide-based equation, we get:

,  @pH>10.3

 

The complete lines are as follows:

 

And removing extra line segments and zinc species that do not exist at 0.1 mM concentration (e.g., those “under” the precipitates, we get the final diagram:

 

 

 

 

 

Part II:. Answer either B or C

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 µg/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) (10%)

2.     Determine the stoichiometry of this reaction (e.g., mg-BrO3-/mg-chlorine as Cl2). (5%)

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. (10%)

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. (5%)

 

 

Answer to B.1.

Use the following redox constants:

Equ#

Half Cell Reaction

DEo (Volts)

7

½HOBr + ½H+ + e- = ½Br- + ½H2O

+1.33

10

½HOCl + ½H+ + e- = ½Cl- + ½H2O

+1.48

12

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

+1.45

 

Now there is one reduction reaction and two oxidation reactions.  The two oxidation reactions must be combined so that the intermediate (HOBr) cancels out.  This also requires that we determine the average potential per electron for the two oxidation steps.

 

 

Factor

Equ#

Half Cell Reaction

DEo (Volts)

3

10

1½HOCl + 1½H+ + 3e- = 1½Cl- + 1½H2O

+1.48

-1

7

½Br- + ½H2O = ½HOBr + ½H+ + e-

1/3 (-1.33)

-½

12

½HOBr + H2O = ½BrO3- +2½H+ +2e-

2/3 (-1.45)

 

 

1½HOCl + ½Br- = ½BrO3- + 1½Cl- + 1½ H+

+0.07

 

Or simplifying by making it per bromate ion:

3HOCl + Br- = BrO3- + 3Cl- + 3H+

 

Answer to B.2.

 

Stoichiometry:

Or:

 

 

Answer to B.3.

 

If we define the equation as:

1½HOCl + ½Br- = ½BrO3- + 1½Cl- + 1½ H+

This is a 3-electron transfer, so:

 

This means that following is true at equilibrium:

 

Since the concentration of chloride in sea water is about 0.5 M, bromide is about 1 mM, and the typical hydrogen ion concentration is 10-8, the equilibrium bromate concentration would be 47,200 M.  In other words, this is a reaction that essentially goes to completion, from the standpoint of thermodynamics.

 

And if the HOCl residual was 0.7 mg/L

 

However, if bromide dimishes to low concentrations due to other reactions, the bromate may be quite small.

 

Answer to B.4.

 

Returning to the prior equation for bromate formation

 

 

Now its reasonable to assume that the chlorine will be mostly used up in oxidizing bromide to bromate.  Then you can solve for the equilibrium residual chlorine concentration.

 

If this is the case, the [BrO3-] will be essentially equal to the HOCl dose, and the remaining bromide will still be in the form of bromide.

 

 

 

Now do the calculation for oxidation of bromide to hypobromous acid by hypochlorous acid

 

 

 

Factor

Equ#

Half Cell Reaction

DEo (Volts)

1

10

½HOCl + ½H+ + e- = ½Cl- + ½H2O

+1.48

-1

7

½Br- + ½H2O = ½HOBr + ½H+ + e-

-1.33

 

 

½HOCl + ½Br- = ½HOBr + ½Cl-

+0.15

 

Or simplifying by making it per hypobromous acid molecule:

HOCl + Br- = HOBr + Cl-

 

 

If we define the equation as:

HOCl + Br- = HOBr + Cl-

This is a 2-electron transfer, so:

 

This means that following is true at equilibrium:

 

Now rearrange, solving for the hypobromous acid concentration

 

 

Use the previously calculated equilibrium HOCl value, the raw sea wter bromide level, and the chloride level.

 

 

 

 

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

½HOCl + ½H+ + e- = ½Cl- + ½H2O

+1.48

11

½OCl- + H+ + e- = ½Cl- + ½H2O

+1.64

12

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

+1.45

13

S(s) + 2H+ + 2e-  =  H2S (g)

+0.17

14

½NH2Cl + H+ +e- = ½Cl- + ½NH4+

+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, 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] 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.