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CEE 680 |
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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
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Input
Data |
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pK1 = |
6.3 |
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pKw = |
14 |
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K1 = |
5.01187E-07 |
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pK2 = |
10.3 |
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K2 = |
5.01187E-11 |
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pK3 = |
50 |
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K3 = |
1E-50 |
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Alk = |
120 |
mg/L = |
0.0024 |
equ/L |
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1 = |
1 |
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pH = |
7.35 |
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Flow = |
20 |
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pH |
H+ |
alpha-0 |
alpha-1 |
alpha-2 |
alpha-3 |
OH- |
CT |
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7.35 |
4.47E-08 |
0.081748 |
0.917223 |
0.001029 |
2.3E-46 |
2.24E-07 |
0.002611 |
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Treated
Water #1 (Closed System) |
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New
Target pH = |
9.1 |
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pH |
H+ |
alpha-0 |
alpha-1 |
alpha-2 |
alpha-3 |
OH- |
CT |
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9.1 |
7.94E-10 |
0.001489 |
0.939249 |
0.059263 |
7.46E-43 |
1.26E-05 |
0.002611 |
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Required
total Alkalinity = |
0.002774 |
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Alkalinity
to be added = |
0.000374 |
equ/L = |
18.7 |
mg/L as
CaCO3 |
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15.0 |
mg/L as
NaOH |
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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
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pKH= |
1.5 |
KH= |
0.031623 |
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ppCO2 = |
3.5 |
pCO2 = |
0.000316 |
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New
Target pH = |
9.1 |
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pH |
H+ |
alpha-0 |
alpha-1 |
alpha-2 |
alpha-3 |
OH- |
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9.1 |
7.94E-10 |
0.001489 |
0.939249 |
0.059263 |
7.46E-43 |
1.26E-05 |
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Required
total Alkalinity = |
0.007118 |
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Alkalinity
to be added = |
0.004718 |
equ/L = |
235.9 |
mg/L as
CaCO3 |
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188.7 |
mg/L as
NaOH |
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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.
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ZnL |
Log b1 = 5.7 |
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ZnL2 |
Log b2 = 11.1 |
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ZnL3 |
Log b3 = 16.1 |
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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:
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Constants |
Log(Step-wise "K") |
Log(Overall "Beta") |
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1st |
5.7 |
5.7 |
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2nd |
5.4 |
11.1 |
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3rd |
5.0 |
16.1 |
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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)
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NAME |
FORMULA |
pKa |
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Perchloric acid |
HClO4 = H+ + ClO4- |
-7 STRONG |
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Hydrochloric acid |
HCl = H+ + Cl- |
-3 |
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Sulfuric acid |
H2SO4= H+ + HSO4- |
-3 (&2) ACIDS |
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Nitric acid |
HNO3 = H+ + NO3- |
-0 |
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Hydronium ion |
H3O+ = H+ + H2O |
0 |
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Trichloroacetic acid |
CCl3COOH = H+ + CCl3COO- |
0.70 |
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Iodic acid |
HIO3 = H+ + IO3- |
0.8 |
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Bisulfate ion |
HSO4- = H+ +
SO4-2 |
2 |
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Phosphoric acid |
H3PO4 = H+ + H2PO4- |
2.15 (&7.2,12.3) |
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o-Phthalic acid |
C6H4(COOH)2 = H+ + C6H4(COOH)COO- |
2.89 (&5.51) |
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Citric acid |
C3H5O(COOH)3= H+ + C3H5O(COOH)2COO- |
3.14 (&4.77,6.4) |
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Hydrofluoric acid |
HF = H+
+ F- |
3.2 |
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Aspartic acid |
C2H6N(COOH)2= H+ + C2H6N(COOH)COO- |
3.86 (&9.82) |
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m-Hydroxybenzoic acid |
C6H4(OH)COOH = H+ + C6H4(OH)COO- |
4.06 (&9.92) |
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p-Hydroxybenzoic acid |
C6H4(OH)COOH = H+ + C6H4(OH)COO- |
4.48 (&9.32) |
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Nitrous acid |
HNO2 = H+ + NO2- |
4.5 |
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Acetic acid |
CH3COOH = H+ + CH3COO- |
4.75 |
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Propionic acid |
C2H5COOH = H+ + C2H5COO- |
4.87 |
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Carbonic acid |
H2CO3 = H+ + HCO3- |
6.35 (&10.33) |
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Hydrogen sulfide |
H2S = H+ + HS- |
7.02 (&13.9) |
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Dihydrogen phosphate |
H2PO4- = H+ + HPO4-2 |
7.2 |
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Hypochlorous acid |
HOCl = H+
+ OCl- |
7.5 |
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Boric acid |
B(OH)3 + H2O = H+ + B(OH)4- |
9.2 (&12.7,13.8) |
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Ammonium ion |
NH4+ = H+ + NH3 |
9.24 |
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Hydrocyanic acid |
HCN = H+
+ CN- |
9.3 |
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p-Hydroxybenzoic acid |
C6H4(OH)COO- = H+ + C6H4(O)COO-2 |
9.32 |
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Phenol |
C6H5OH = H+ + C6H5O- |
9.9 |
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m-Hydroxybenzoic acid |
C6H4(OH)COO- = H+ + C6H4(O)COO-2 |
9.92 |
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Bicarbonate ion |
HCO3- = H+ + CO3-2 |
10.33 |
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Monohydrogen phosphate |
HPO4-2 = H+ + PO4-3 |
12.3 |
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Bisulfide ion |
HS- = H+ + S-2 |
13.9 |
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Water |
H2O = H+ + OH- |
14.00 |
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Ammonia |
NH3 = H+ + NH2- |
23 |
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Methane |
CH4 = H+ + CH3- |
34 |
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