CEE 680 8 November 2001

# SECOND EXAM

Closed book, two pages of notes allowed.

# Problem #1: Carbonate System.

(50% for both parts) You’re treating a drinking water that contains 120 mg/L of alkalinity.  The initial pH is 7.35, and you wish to raise this to 9.10 for purposes of corrosion control.

A.   (25%) How much of a caustic soda (NaOH) dose do you need to add to accomplish this if the water goes right into the water main?

B.   (25%) Answer the above question, but this time assume you have a large finished water reservoir after caustic addition that allows the water to reach equilibrium with the atmosphere before entering the distribution system.

# Solution to #1

A. Recognize that alkalinity is conservative, and in a closed system, so is the total carbonates.  Calculate total carbonates (CT) for the original water, using the equation for alkalinity.  This first requires that you calculate the alpha values for the initial pH.

Next, determine the new alkalinity using this same equation, but inserting the new desired pH, and recalculating the alphas

 Input Data pK1  = 6.3 pKw = 14 K1 = 5.01187E-07 pK2 = 10.3 K2 = 5.01187E-11 pK3 = 50 K3 = 1E-50 Alk = 120 mg/L  = 0.0024 equ/L 1 = 1 pH = 7.35 Flow = 20 pH H+ alpha-0 alpha-1 alpha-2 alpha-3 OH- CT 7.35 4.47E-08 0.081748 0.917223 0.001029 2.3E-46 2.24E-07 0.002611

 Treated Water #1 (Closed System) New Target pH = 9.1 pH H+ alpha-0 alpha-1 alpha-2 alpha-3 OH- CT 9.1 7.94E-10 0.001489 0.939249 0.059263 7.46E-43 1.26E-05 0.002611 Required total Alkalinity = 0.002774 Alkalinity to be added = 0.000374 equ/L = 18.7 mg/L as CaCO3 15.0 mg/L as NaOH

B. Here we use the open system term for CT in place of the constant.  In other words, CT is no longer a conservative parameter

 pKH= 1.5 KH= 0.031623 ppCO2 = 3.5 pCO2 = 0.000316 New Target pH = 9.1 pH H+ alpha-0 alpha-1 alpha-2 alpha-3 OH- 9.1 7.94E-10 0.001489 0.939249 0.059263 7.46E-43 1.26E-05 Required total Alkalinity = 0.007118 Alkalinity to be added = 0.004718 equ/L = 235.9 mg/L as CaCO3 188.7 mg/L as NaOH

# Problem #2: Complexation

(50% for both parts) Zinc forms a series of complexes with aqueous cyanide.

A. (40%) Sketch out a set of alpha curves (vs log[ CN- ]) for the Zinc-cyanide system.  Assume that the pH is high enough so that there is no significant formation of HCN.  Use the following stability constants from Stumm and Morgan.  Note that these are all overall formation constants.

 ZnL Log b1 = 5.7 ZnL2 Log b2 = 11.1 ZnL3 Log b3 = 16.1 ZnL4 Log b4 = 19.6

Using this determine the species composition when the total zinc concentration is 0.1 mM and the total cyanide concentration is 0.5 mM.

B. (10%) How would the answer to this problem change if the pH was 7.3?  Explain qualitatively in words why this makes a difference, and which way it would shift the equilibria.  Be quantitative if you can.

# Solution to #2

To clarify the various constants:

 Constants Log(Step-wise "K") Log(Overall "Beta") 1st 5.7 5.7 2nd 5.4 11.1 3rd 5.0 16.1 4th 3.5 19.6

Next we can apply the complexation equations that are based on the beta's and the free ligand concentration.

and

and

where "M" is Zinc (Zn), and "L" is cyanide (CN-).

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