CEE 680 14 November 2006

# SECOND EXAM

Closed book, two pages of notes allowed.

# 1. Carbonate System.

(40% for both parts) You’re treating a drinking water that contains 25 mg/L of alkalinity.  The initial pH is 6.2, and you wish to raise this to 8.5 for purposes of corrosion control.

A.   (20%) 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.   (20%) 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

Original Water

 Input Data pK1  = 6.3 pKw = 14 K1 = 5.01187E-07 pK2 = 10.3 K2 = 5.01187E-11 pK3 = 50 K3 = 1E-50 Alk = 25 mg/L  = 0.0005 equ/L 1 = 1 pH = 6.2 Flow = 60 pH H+ alpha-0 alpha-1 alpha-2 alpha-3 OH- CT 6.2 6.31E-07 0.557292 0.442673 3.52E-05 5.57E-49 1.58E-08 0.001131

 New Target pH = 8.5 pH H+ alpha-0 alpha-1 alpha-2 alpha-3 OH- CT 8.5 3.16E-09 0.006173 0.978322 0.015505 4.9E-44 3.16E-06 0.001131 Required total Alkalinity = 0.001144 Alkalinity to be added = 0.000644 equ/L = 32.2 mg/L as CaCO3 25.8 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 = 8.5 pH H+ alpha-0 alpha-1 alpha-2 alpha-3 OH- 8.5 3.16E-09 0.006173 0.978322 0.015505 4.9E-44 3.16E-06 Required total Alkalinity = 0.001638 Alkalinity to be added = 0.001138 equ/L = 56.9 mg/L as CaCO3 45.5 mg/L as NaOH

# 2. Complexation

(40% for both parts) Aqueous cyanide forms strong complexes with many metals.  The following two part problem concerns complexes with cadmium and zinc.

A. (20%) On the attached blank graph template, sketch out a set of alpha curves (vs log[ CN- ]) for the Cadmium-cyanide system.  Point out they key graphical features that you used to make this sketch (e.g. intersection points).  Assume that the pH is high enough so that there is no significant formation of HCN.  Use the following stability constants from Benjamin’s text.  Note that these are all overall formation constants.

 CdL Log b1 = 5.32 CdL2 Log b2 = 10.37 CdL3 Log b3 = 14.83 CdL4 Log b4 = 18.29

B. (20%) Attached is an accurate graph of alpha values (vs log[ CN- ]) for the Zinc-cyanide system.    Using this graph determine the species composition when the total zinc concentration is 0.1 mM and the total cyanide concentration is 0.5 mM.  Assume once again that the pH is high enough so that there is no significant formation of HCN.

# Solution to #2a

First calculate the k values from the difference of the various betas:

 log value log value k1 = 5.32 B1 = 5.32 k2 = 5.05 B2 = 10.37 k3 = 4.46 B3 = 14.83 k4 = 3.46 B4 = 18.29

Now plot these recognizing that the intersection points occur where the free ligand concentration is equal to the k value.

# Solution to #2b

To clarify, alpha graph was made based on the following 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 applied the complexation equations that are based on the beta's and the free ligand concentration.

and

Now to solve this, we recognize that the equilibrium based n-bar is:

And the mass balance based n-bar is:

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

Below are the “exact” values.  What you read from the graph will be close, but with only single to double digit precision.

# 3. Multiple Choice.

(20%) Answer all 10 of the following questions.  Indicate which of the options is the best choice.

1. The sum of total acidity and total alkalinity on any given sample is equal to

a. the UV absorbance

b. twice the total carbonate

c. the value one

d. half of the hardness

e. zero

2. When a solution spontaneously absorbs CO2 from the atmosphere it:

a. resuts in higher total carbonate

b. drops in pH

c. approaches equilibrium

d. all of the above

e. none of the above

3. Phenolphthalein

b. is rarely used because noone can spell it

c. complexes calcium forming an insoluble salt

d. is the drug of choice for malaria

e. changes from colorless to red as pH increases

4. H2CO3*:

a. is composed mostly of aqueous CO2

b. is conservative in closed systems

c. is am ampholyte

d. all of the above

e. none of the above

5. Ion pairs:

a. are always charged

b. are larger than Bartlett pears

c. are almost impossible to separate

d. are outer-sphere complexes

6. The coordination number:

a. is usually 6 or less

b. is related to the charge on the central atom

c. is a function of the size of the ligand

d. all of the above

e. none fo the obove

7. The buffer intensity of the open carbonate system:

a. is independent of the alkalinity

b. is independent of the CT

c. is always higher than the pCO2

d. is at a minimum where the pH = pK1

8. Detergent “builders” are used to:

a. help solubilize grease

b. complex trace metals

c. take hardness cations from the surfactants

d. elevate the acidity

e. reduce the caloric content

9. EDTA

a. stands for ethylene dinitro tetraacetic acid

b. is most commonly used as a pH buffer

c. is a higly potent carcinogen

d. all of the above

e. none of the above

10. The Irving Williams Series

a. is a means of estimating alkalinity

b. describes the inverse proportionality of acidity to alkalinity

c. includes a number of books, such as The Chapman Report, and The Prize

d. provides a comprehensive description of ligand structure

e. follows the increase in ligand affinity from Mn(II) to Cu(II)

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) Citric acid C3H5O(COOH)3= H+  + C3H5O(COOH)2COO- 3.14 (&4.77,6.4) Hydrofluoric acid HF = H+  + F- 3.2 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 Methane CH4 = H+ + CH3- 34

Problem 2A

Problem 2B