CEE 680 20 November 2008

# SECOND EXAM

Closed book, two pages of notes allowed.

# 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:

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- 34