CEE 680 

20 November 2008 
Closed book, two pages of notes allowed.
Answer all questions. Please state any additional assumptions you made, and show all work.
(55% for both parts) 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 
25 
6.50 
#2 
10 
300 
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?
C. How many mg/L of caustic soda (NaOH) must be added to the unequilibrated blended water in part A to raise the pH to 9.80 ?
This is a closed system problem. Therefore the total carbonate concentrations (C_{T}'s) must be determined and treated as conservative. Likewise the alkalinities are conservative, and then the final pH can be determined from the blended C_{T} and alkalinity.
for either water:
_{ }
first, I would determine the alpha's at the two pH's. Recall the general equations for a diprotic acid are:
_{ }
This results in the following values (assuming pKs of 6.3 and 10.3:
pH 
alpha1 
alpha2 
6.50 
0.5855 
0.0000866 
8.85 
0.9650 
0.03195 
so, for water #1:
C_{T} = {(25/50,000)  (10^{7.50}) + (10^{6.50})}/(0.5855 + 2*0.0000866) = 0.0008543 M
and for water #2:
C_{T} = {(300/50,000)  (10^{5.15}) + (10^{8.85})}/(0.9650 + 2*0.03195) = 0.0058246 M
now we need to calculate the blended alkalinities and total carbonates:
Alk = (2*25 + 300)/3/50,000 = 0.002333 equ/L
C_{T} = (2*0.0008453 + 0.0058246)/3 = 0.002511 M
Now calculate the pH from the earlier equation for total carbonates, and making any one of the following three sets of simplifying assumptions:
1. Alk is large compared to [OH] and [H]
_{ }
which becomes:
_{ }
_{ }
Now use the quadratic equation:
_{ }
which is simplified to:
_{ }
and (Alk C_{T}) is 0.002400.002369 = 0.0001472
and (Alk 2C_{T}) is 0.002402(0.002369) = 0.0026277
_{ }
or
pH = 7.459
2. HCO_{3}^{}>> CO_{3}^{2},
pH >> pK_{1} and Alk is large compared to [OH] and [H]
_{ }
which becomes:
_{ }
_{ }
But if pH>>pK_{1}, then we can ignore the first terms in the denominator of the alpha quotient. So this equation simplifies to:
_{ }
Now we can solve directly for H+:
_{ }
or
pH = 7.468
or,
3. Make no assumptions and solve for the
exact solution. This gives:
pH = 7.45885
This is now an open system problem.
There are many ways to solve this, depending on the extent of simplifying assumptions youre willing to try. Here are a few examples:
1. Assume that H^{+} and OH^{}
are insignificant
_{ }
Since this is an open system, we know what C_{T} is:
_{ }
And now:
_{ }
_{ }
_{ }
or
pH = 8.698
2. Assume that bicarbonate is the only
carbonate species of any importance.
Charge balance considerations dictate that:
Alk = C_{T}
Under conditions where pK_{1} << pH<< pK_{2}, the following is approximately true:
C_{T} = [HCO_{3}^{}]
And from the equilibrium equation we can conclude:
[HCO_{3}^{}] = K_{1} [H_{2}CO_{3}^{*}]/[H^{+}]
And since this is an open system, we can say:
[H_{2}CO_{3}^{*}] = K_{H} p_{CO2}
which becomes, for the bulk atmosphere at 25^{o}C
[H_{2}CO_{3}^{*}] = 10^{5}
we can combine and get the equation for alkalinity (or bicarbonate) vs H^{+} in an open system
Alk = [HCO_{3}^{}] = K_{1} 10^{5} /[H^{+}]
Then substituting and isolating H^{+}, we get
[H^{+}] = 10^{11.35}/Alk
and substituting in for the blended water alkalinity
[H^{+}] = 10^{11.35}/0.002333
[H^{+}] = 1.91 x 10^{9}
or
pH = 8.718
This is a closed system. Recall that:
We also know that:
Alk_{i
}= 0.0023333 eq/L
Since
caustic soda does not contain carbonates, C_{T} is a constant for this
case.
@ pH 9.80,
@ pH 8.5,

Alk 
= (0.77192+2*0.2278)*2.511x10^{3} 

Alk 
= 3.145x10^{3} equivalents/L 

delta
Alkalinity 
= 3.145x10^{3} equivalents/L  2.333x10^{3}
equivalents/L 


= 8.12x10^{4} equivalents/L 
and this
change in alkalinity must come from the caustic soda added:
Caustic Soda
Dose = 8.12x10^{4} equ. *
40g/equ.
Caustic Soda Dose = 32.5 mg/L
(45%
for both parts) Aqueous fluoride
forms strong complexes with many metals.
The following two part problem concerns complexes with Beryllium.
A.
(20%)
Attached is an accurate graph of alpha values (vs log[ Fr]) for the BerylliumFluoride
system (equilibria data shown below). Using
this graph determine the complete species composition when the total Beryllium
concentration is 0.10 mM and the total fluoride concentration is 0.40 mM. Ignore the possible formation of any other
complexes other than those from Be and F; also ignore any possible
precipitation reactions. Assume the
water is a neutral pH.
B.
(20%)
Describe in qualitative terms the impacts of pH on this system.
Estimate in quantitative terms what the concentrations of each of the
species would be if the pH were 2.2, instead of 7.0, and explain how you got
these.
The mass balance nbar curve is shown in the upper figure as a solid red line. The two nbar curves intersect at a Log[L] of about 3.6 (solid green arrow). At this point the various alpha values can be read off the graph. They are
Values from the graph as read at Log[L] = 3.6
Curve 
αvalue 
Species 
Conc (M) 
α_{0} 
0.035 
Be^{+2} 
3.5x10^{6} 
α_{1} 
0.45 
BeF^{+} 
4.5x10^{5} 
α_{2} 
0.45 
BeF_{2} 
4.5x10^{5} 
α_{3} 
0.068 
BeF_{3}^{} 
6.8x10^{6} 
α_{4} 
0 
BeF_{4}^{2} 
~0 
Recognize that at pH 2.2, we are a full log unit below the pKa for HF. This means that the ratio of hydrogenbound F to free F will be 9:1. In other words the amount of F that is not bound to Be will always be 10 times the free F (i.e., the free F plus HF). With this we can rewrite the massbalancebased nbar equation:
This revised nbar curve is shown in the upper figure as a dashed red line. The dashed green arrow indicates the intersection point. Values derived from the graph are shown in the table below.
Values from the graph as read at Log[L] = 4.6
Curve 
αvalue 
Species 
Conc (M) 
α_{0} 
0.42 
Be^{+2} 
4.2x10^{5} 
α_{1} 
0.53 
BeF^{+} 
5.3x10^{5} 
α_{2} 
0.051 
BeF_{2} 
5.1x10^{6} 
α_{3} 
0 
BeF_{3}^{} 
~0 
α_{4} 
0 
BeF_{4}^{2} 
~0 
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+ + SO42 
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 

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+
+ HPO42 
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 

Phenol 
C6H5OH = H+ + C6H5O 
9.9 

mHydroxybenzoic acid 
C6H4(OH)COO = H+ + C6H4(O)COO2 
9.92 

Bicarbonate ion 
HCO3 = H+
+ CO32 
10.33 

Monohydrogen phosphate 
HPO42 = H+ + PO43 
12.3 

Bisulfide ion 
HS = H+ + S2 
13.9 

Water 
H2O = H+ + 
14.00 

Methane 
CH4 = H+ + CH3 
34 
