CEE 680

 

8 November 2001

SECOND EXAM

 

Closed book, two pages of notes allowed.

 

Answer all questions.  Please state any additional assumptions you made, and show all work.

 

 

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