CEE 680 16 November 2000

# SECOND EXAM

Closed book, two pages of notes allowed.

# 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 µM 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 µM.

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, 25°C, 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