CE4501 Environmental Engineering Chemical Processes
Problem Set 2 - SOLUTIONS
Fall 2008
Due: Friday, 10/3 by 5 p.m.
Solutions will be posted on the Web. Problem sets will be graded for completeness, and one
problem (selected at random) will be graded in detail. Each problem set contributes 2.5%
towards your final grade.
1.
Problem 6.8, p. 118 in the text.
NaHCO3 is added to water. Generate a species list diagram for this system.
SOLUTION:
We must first define the system. Here it will be defined as closed, and the amount of NaHCO3
added is small enough that it will all dissolve.
Starting components:
NaHCO3(s)
H2O
HCO3-
Na+
CO32-
H+
OH-
H2CO3
2. Problem 6.9, p. 118 in the text.
Write a complete mathematical model for the system above. The system is closed, and all
NaHCO3(s) dissolves.
SOLUTION:
1. Define system: closed; all bicarbonate dissolves
2. List species (see above)
a. H2O, H+, OH-, Na+, HCO3-, CO32-, H2CO3 (7)
3. Identify 7 equations for the system
a. Kw = {H+}{OH-}/{H2O(l)}
b. K1 = {HCO3-}{H+}/{H2CO3}
c. K2 = {CO32-}{H+}/{HCO3-}
d. CT = [Na+]
e. CT = [H2CO3] + [HCO3-] + [CO32-]
f. [H+] + [Na+] = [OH-] + [HCO3-] + [CO32-]
g. {H2O(l)} = 1
3. If you wanted to predict the pH of rain water in which you had found Ca2+, Na+, Cl-,
HCO3-, SO42-, and NO3-, list the chemical species you would consider and the equations
that you would use to solve for the pH.
SOLUTION:
1. Define the system: closed (rain sample has been collected and bottled)
2. List species:
a. Starting substances:
H2O, Ca2+, Na+, Cl-, HCO3-, SO42-, NO3b. Initial hydrolysis:
H+, OH-, H2CO3, CO32-, HSO4-, HNO3, HCl
c. Further hydrolysis
H2SO4, NaHCO3, NaCO3-, NaSO4-, CaOH+,
CaHCO3+, CaCO3, CaCl+, CaSO4,
The total above is 23. However, HCl, H2SO4 and HNO3 are strong acids that dissociate
nearly completely; the concentrations of these species will be vanishingly small. This
leaves us with 20 unknowns for which we need 20 equations.
1. Kw = {H+}{OH-}/{H2O}
2. K1 = {H+}{HCO3-}/{H2CO3}
3. K2 = {H+}{CO32-}/{HCO3-}
4. Ka2 = {H+}{SO42-}/{HSO4-}
5. Kf1 = {CaHCO3+}/{Ca2+}{HCO3-}
6. KNaHCO3= {NaHCO3}/{Na+}{HCO3-}
7. KNaCO3 = {NaCO3-}/{Na+}{CO32-}
8. KNaSO4 = {NaSO4-}/{Na+}{SO42-}
9. KCaOH = {CaOH+}/{Ca2+}{OH-}
10. KCaCO3 = {CaCO3}/{Ca2+}{CO32-}
11. KCaCl = {CaCl+}/{Ca2+}{Cl-}
12. KCaSO4 = {CaSO4}/{Ca2+}{SO42-}
13. CT-Ca = [Ca2+] + [CaHCO3+] + [CaOH+] + [CaCO3] + [CaCl+] + [CaSO4]
14. CT-Na = [Na+] + [NaHCO3] + [NaCO3-] + [NaSO4-]
15. CT-Cl = [Cl-] + [CaCl+]
16. CT-S = [HSO4-] + [SO42-] + [CaSO4] + [NaSO4-]
17. CT-N = [NO3-]
18. CT-C = [H2CO3] + [ HCO3-] + [CO32-] + [CaHCO3+] + [CaCO3] + [NaHCO3] +
[NaCO3-]
19. [H+] + 2[Ca2+] + [Na+] + [CaHCO3+] + [CaOH+] + [CaCl+] = [OH-] + [Cl-] + [NO3-] +
2[SO42-] + [HSO4-] + [HCO3-] + 2[CO32-] + [NaCO3-] + [NaSO4-]
20. {H2O} = 1
-------------------------------------------------------------------------------------------------4. Problem 7.2, p. 138 in the text.
At 30oC, Kw = 10-13.83. Determine the equilibrium pH of pure water at this temperature.
SOLUTION:
Kw = {H+}{OH-}/{H2O} = {H+}{OH-}
In pure water, {H+} = {OH-}
Kw = {H+}2 = 10-13.83
Î {H+} = 10-6.92
Î pH = 6.92
---------------------------------------------------------------------------------5. Problem 7.12, p. 139 in the text.
Find the equilibrium pH of orange juice when only the first dissociation of citric acid is
considered. (Ignore activity corrections)
SOLUTION:
1. System is closed with no solids present.
2. Identify components and species:
H3C, K+, H+, OH-, H2C- (5 specices) Assume K3C, K2HC, KH2C, KC2- etc. are
unimportant if present.
3. Identify constraints (5 equations)
a. Kw = {H+}{OH-} = 10-14
b. Ka1= {H+}{H2C-}/{H3C} = 10-3.14
c. CT-citrate = [H2C-] + [H3C] = (8 g/L)/(192 g/mole) = 0.042 M = 10-1.38 M
d. CT-K = [K+] = (1.6 g/L)/(39 g/mole) = 0.041 M = 10-1.39 M
e. [H+] + [K+] = [OH-] + [H2C-]
4. Solve
a. Assume solution is acidic so OH- is negligible in charge balance
Î [H2C-] = [H+] + [K+]
[H3C] = CT-citrate – [H2C-] = CT-citrate – [H+] – CT,K = 0.001 - [H+]
Ka1 = {H+}({H+} + 0.041)/(0.001-{H+})
0.001Ka1 – Ka1{H+} = {H+}2 + 0.041{H+}
{H+}2 + {H+}(0.041+Ka1) – 0.001Ka1 = 0 = {H+}2 + 0.04175{H+} -10-6.14
{H+} = (- 0.04175 + (0.041752 - 4·(-10-6.14))0.5 )/2
{H+} = 10-4.76 Î pH = 4.76
----------------------------------------------------------6. Problem 8.1, p. 168 in the text.
Find the equilibrium pH of a 10-3 M NaCN solution using graphical means.
SOLUTION:
Ka = {H+}{CN-}/{HCN} = 10-9.2 or pKa = 9.2
{CN-} = KaCT/({H+} + Ka)
pH
0
2
4
6
8
10
12
14
0
2
4
pC
6
8
10
OH
12
H
CN
14
HCN
16
Given the proton condition: [HCN] + [H+] = [OH-]
The solution must be where the HCN line crosses the OH- line at pH ~ 10.2.
The charge balance equation ([H+] + [Na+] = [OH-] + [CN-]) has no single solution point.
------------------------------------------------------------------------------------------------------------7. Problem 8.8, p. 169 in the text.
Write proton conditions.
A. NH4Cl
Solution: The proton reference level is NH4+, Cl- and H2O
Species with more protons: HCl, H+
Species with fewer protons: NH3, OHProton Condition: [HCl] + [H+] = [NH3] + [OH-]
B. Na2S
Solution: The proton reference level is Na+, S2-, H2O
Species with two more protons: H2S
Species with one more proton: HS-, H+
Species with fewer protons: OHProton Condition: 2[H2S] + [HS-] + [H+] = [OH-]
C. KH2PO4
Solution: The proton reference level is K+, H2PO4-, H2O
Species with one more proton: H3PO4, H+
Species with one less proton: HPO42-, OHSpecies with two fewer protons: PO42Proton Condition: [H3PO4] + [H+] = [HPO42-] + [OH-] + 2[PO42-]
D. (NH4)2CO3
Solution: The proton reference level is NH4+, CO32-, H2O
Species with two more protons: H2CO3
Species with one more proton: HCO3-, H+
Species with one less proton: NH3, OHProton Condition: 2[H2CO3] + [HCO3-] + [H+] = [NH3] + [OH-]
-------------------------------------------------------------------------------------------------------------8. Problem 18.5, p. 414 in the text.
The MCL for chloroform in drinking water is 100 μg/L. The occupational exposure level for
chloroform in air is 10 ppm by weight (2.54x10-6 atm). If treated water in an enclosed drinking
water reservoir is at the MCL for chloroform, is the OEL for chloroform in air exceeded?
Assume that the air volume of the enclosed reservoir is very large.
SOLUTION:
We are asked to find the concentration of chloroform in air that will be in equilibrium with 100
μg/L in water.
KH = 10-0.6 mole/L-atm
CHCl3(g) ÍÎ CHCl3(aq)
M.W. = 12 + 1 + 3·35.5 = 199.5
(100 μg/L)/(199.5 μg/μmole) = 0.837 μM
[CHCL3(g)] = [CHCl3(aq)]/KH = (0.837x10-6 mole/L)/(10-0.6 mole/L-atm)
[CHCL3(g)] = 3.33x10-6 atm
This value is higher than the OEL of 2.54x10-6 atm
9. Problem 18.7, p. 414. Use a pC-pH diagram and neglect the species S2-.
A ground water containing 0.5 mg/L as S of total dissolved sulfide at pH 7.5 is aerated to
volatilize H2S. If the H2S concentration in the atmosphere in contact with the aerated ground
water is 0.75 ppm by volume and the total system pressure is 1 atm, what will be the total sulfide
concentration in the water after equilibrium has been reached? Assume the pH of the water
remains constant.
SOLUTION:
First, convert the mixing ratio to a concentration in atm.
vi pi
=
V P
p
0.75
= i
6
10
1atm
pi = 7.5 x10−7
Next, use Henry’s Law constant to calculate the equilibrium concentration of H2Saq.
[H S ]
mol
= 2 ( aq )
K H = 10−1.01
L − atm
pH 2 S
[ H 2 S ]( aq ) = K H ⋅ pH 2 S = 10−1.01
mole
⋅ 7.5 x10−7 atm
L − atm
mol
L
This, however, is not the total concentration of reduced sulfur. We also need to calculate the
amount present as HS-.
{H + }{HS − }
= 10−7
Ka =
{H 2 S( aq ) }
[ H 2 S ]( aq ) = 7.3x10−8
{HS − } = K a
{H 2 S( aq ) }
{H + }
= 10−7 ⋅
7.3 x10−8
10−7.5
{HS − } = 2.4 x10−7
STotal = [ H 2 S ] + [ HS − ]
STotal = (0.73 + 2.4) x10−7 = 3.13 x10−7
mole
L
pH
0
2
4
6
8
10
12
14
0
2
4
OH
pC
6
H
H2S
8
HSpH
10
12
14
The pC-pH diagram can also be used to determine the results. Because the gas phase H2S
concentration is fixed, the concentration of H2Saq will also remain fixed and independent of pH.
Thus the H2S line is horizontal. The line for HS- will intersect the H2S line at the pKa (both
components have equal concentration at that pH), and it will have a slope of minus 1. We are
told to find the concentrations at pH = 7.5, the vertical line on the graph.
10. a. If the alkalinity of the wastewater treatment sample that you have worked with in the
lab is 2.5 meq/L, what must the pCO2 in the sample be to cause the measured pH of 6.65?
b. If the sample is equilibrated with room air (pCO2 = 400 μatm), what will be the
resultant pH? Include a pC-pH diagram with your solution of this problem; you may
either use the diagram to solve the problem, or simply to depict the situation.
SOLUTION:
a. Species present: H2O, H+, OH-, CO2(g), H2CO3*, HCO3-, CO32- (6)
b. Equations: Kw = {H+}{OH-}
KH = 10-1.46 = {CO2(aq)}/{CO2(g)}
Ka1 = {H+}{HCO3-}/{H2CO3*} = 10-6.25
Ka2 = {H+}{CO32-}/{HCO3-} = 10-10.3
pH = 6.65
[HCO3-] = 2.5x10-3 mol/L
c. Rearranging to solve:
{OH-} = Kw/{H+}
pCO2(g) = {CO2(aq)}/KH = [H2CO3*]/KH
{H2CO3*}= {H+}{HCO3-}/Ka1 = (10-6.65)(2.5x10-3)γ1/10-6.25
⎛
0.005 ⎞
2
⎜⎜ −0.5⋅1 ⋅
⎟
1+ 0.005 ⎟⎠
⎝
If I ~ 5 mM, then γ 1 = 10
= 0.927
*
*
{H2CO3 }= [H2CO3 ] = [CO2(aq)] = (10-6.65)(2.5x10-3)0.927/10-6.25 = 9.2x10-4 M
pCO2(g) = (9.2x10-4 M)/10-1.46 = 0.0266 atm = 26.6 matm
For the second question, let’s use a pC-pH diagram. This will be an open system with pCO2 =
400 μatm. From the KH equation, we can solve for H2CO3* = 10-1.46(400x10-6atm) = 1.39x10-5 M
Or pH2CO3 = 4.86
{HCO3-} = Ka1{H2CO3*}/{H+} or pHCO3 = pKa1 + pH2CO3* - pH = 6.25+4.86 – pH =11.11 - pH
{CO32-} = Ka2{HCO3-}/{H+} or pCO3 = pKa2 + pHCO3- - pH
pH
0
2
4
6
8
10
12
14
0
2
4
OH
pC
6
H+
H2CO3*
8
pHCO3pCO3
10
12
14
The proton reference level is HCO3-, H2O so the proton condition is:
[H+] + [H2CO3*] = [OH-] + [CO32-]
This equation is satisfied at the position of the circle, approximately pH 8.5.