CEE 680

 

20 November 2008

SECOND EXAM

 

Closed book, two pages of notes allowed.

 

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

 

 

1. Carbonate System.

 

 (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 ?

 

 

Solution to Problem #1

 

Part A.

This is a closed system problem.  Therefore the total carbonate concentrations (CT's) must be determined and treated as conservative.  Likewise the alkalinities are conservative, and then the final pH can be determined from the blended CT 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

alpha-1

alpha-2

6.50

0.5855

0.0000866

8.85

0.9650

0.03195

 

so, for water #1:

CT = {(25/50,000) - (10-7.50) + (10-6.50)}/(0.5855 + 2*0.0000866) = 0.0008543 M

and for water #2:

CT = {(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

CT = (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 – CT) is 0.00240–0.002369 = -0.0001472

and (Alk – 2CT) is 0.00240–2(0.002369) = –0.0026277

 

 

 

or

pH = 7.459

2.     HCO3->> CO3-2, pH >> pK1 and Alk is large compared to [OH] and [H]

 

which becomes:

 

 

But if pH>>pK1, 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

 

 

Part B.

 

This is now an open system problem.

 

There are many ways to solve this, depending on the extent of simplifying assumptions you’re 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 CT 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 = CT

 

Under conditions where pK1 << pH<< pK2, the following is approximately true:

 

CT = [HCO3-]

 

And from the equilibrium equation we can conclude:

 

[HCO3-] = K1 [H2CO3*]/[H+]

 

And since this is an open system, we can say:

 

[H2CO3*] = KH pCO2

 

which becomes, for the bulk atmosphere at 25oC

 

[H2CO3*] = 10-5

 

we can combine and get the equation for alkalinity (or bicarbonate) vs H+ in an open system

 

Alk = [HCO3-] = K1 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

 

 

Solution to part C

This is a closed system.  Recall that:

 

We also know that:

Alki = 0.0023333 eq/L

 

Since caustic soda does not contain carbonates, CT 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

 

 

 

 

2. Complexation

 

(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 Beryllium-Fluoride 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.

 

Solution to Problem #2

 

Part A

 

 

 

 

 

 

 

The mass balance n-bar curve is shown in the upper figure as a solid red line.  The two n-bar 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

α-value

Species

Conc (M)

α0

0.035

Be+2

3.5x10-6

α1

0.45

BeF+

4.5x10-5

α2

0.45

BeF2

4.5x10-5

α3

0.068

BeF3-

6.8x10-6

α4

0

BeF4-2

~0

 

 

 

Part B.

 

Recognize that at pH 2.2, we are a full log unit below the pKa for HF.  This means that the ratio of hydrogen-bound 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 re-write the mass-balance-based n-bar equation:

 

 

 

This revised n-bar 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

α-value

Species

Conc (M)

α0

0.42

Be+2

4.2x10-5

α1

0.53

BeF+

5.3x10-5

α2

0.051

BeF2

5.1x10-6

α3

0

BeF3-

~0

α4

0

BeF4-2

~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+ + 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

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

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-

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