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.

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 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)

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 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

## 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.