### Chemical Foundations - Civil and Environmental Engineering | SIU

```FE Review for
Environmental
Engineering
Problems, problems, problems
Presented by L.R. Chevalier, Ph.D., P.E.
Department of Civil and Environmental Engineering
Southern Illinois University Carbondale
FE Review for Environmental Engineering
CHEMICAL FOUNDATIONS
Problem
Strategy
Solution
Calculate the molecular weight, equivalent weight,
molarity and normality of the following:
a. 200 mg/L HCl
b. 150 mg/L H2SO4
c. 100 mg/L Ca(HCO3)2
Problem
Strategy
Solution
• Use periodic table to get molecular weight
• Convert mg/L to mol/L
• Determine n for each compound
• Apply equations
• EW = MW/n
• N = Mn
a) 200 m g / L HCl
“in an Acid/Base reaction, n
is the # of hydrogen ions that
a molecule transfers”
MW  (1)  35.45  36.45 g mol
EW  36.45 g equivalent
200mg L
m olarity
 0.00549M
mg
g
(1000 g )36.45 mol 
norm ality 0.005491  0.00549N
b) 150 m g / L H 2 SO4
MW  21  32.06  416  98.06 g mol
EW  98.06 / 2  49.03 g eq
150mg L
m olarity
 0.00153M
mg
g
(1000 g )98.06 mol 
norm ality 0.001532  .00306N
c) 100 m g / L Ca HCO3 2
“in a precipitation reaction, n is
the valence of the element”
MW  40  2(1)  212  616  162g mol
EW  81 g eq
100mg L
m olarity
 0.000617M
mg
g
(1000 g )162 mol 
norm ality 0.0006172  .00123N
Example
Solution
Convert 200 mg/L HCl to ppm
Example
Solution
200 mg/L = 200 ppm
Problem
Strategy
Solution
a)
Convert 300 ppm Mg 2+ to mg/L as CaCO3
b)
Convert 30 mg/L Mg2+ as CaCO3 to mg/L
Note: MW Mg2+ is 24.31 g/mol
•
•
•
•
Determine the molecular weight of the species
Determine n
Equate EW=MW/n
Apply equation
 50 
mg
mg

as CaCO3 
as species
 EW

L
L
species 

Problem
a)
Strategy
Solution
Convert 300 ppm Mg 2+ to mg/L as CaCO3
300 ppm = 300 mg/L
EW Mg2+ = 24.31/2 = 12.16 g/eq
(300)(50/12.16) = 1233.55 mg/L as CaCO3
Problem
Strategy
Solution
b) Convert 30 mg/L Mg2+ as CaCO3 to mg/L
(30 mg/L as CaCO3)(12.16/50) = 7.3 mg/L
Example
Solution
Balance the following chemical equations:
CaCl2 + Na2CO3CaCO3 + NaCl
C6H12O6 + O2 CO2 + H2O
NO2+H2O HNO3 + NO
Example
Solution
CaCl2 + Na2CO3CaCO3 + 2NaCl
C6H12O6 + 6O2 6CO2 + 6H2O
3NO2+H2O 2HNO3 + NO
Example
Solution
• What is the pH if [H+] = 10-3?
• pH = 3
• What is the pOH if [OH-] = 10-8?
• pOH = 8
• What is the pH if [OH-] = 10-8?
• pH = 14 - 8 = 6
• What is the [H+] if [OH-] = 10-5?
• [H+]=105-14 = 10-9 mol/L
Problem
Strategy
Solution
Derive a proof that in a neutral solution,
the pH and the pOH are both equal to 7.
• Evaluate the governing equation
 

H  OH   K w  10 14 @ 25 C
OH    H 
 H OH    H  H    H 


 10


14

7
H

10
 
pH  7


 2
Problem
Strategy
Solution
Find the hydrogen ion concentration and
hydroxide ion concentration in tomato
juice having a pH of 4.1
• Review how to convert
• [H+]=10-pH
• 10-4.1 mol/L = 7.94 x 10-5 mol/L
• Review governing equation
 H  OH   K


w
 10
14

@ 25 C
 H   10

 pH
 10
 4.1
5
 7.94  10 mol / L


14
H
OH

10
  
14
14
10
10

OH   H   7.94 105
 
 1.25  10 10 mol / L
...... end of example
What percentage of total ammonia (i.e.
NH3 + NH4+) is present as NH3 at a pH of
7? The pKa for NH4+ is 9.3.
NH 4  H   NH 3
Ka





NH
H
 9. 3
3
 10 
NH 

4
NH 3   100

NH 3   NH 4 
However, this expression has two unknowns.
Therefore, we need a second equation.
?????
NH 3   100

NH 3   NH 4 
Second Equation
K a  10
9.3

NH 3 H

NH 

4


K a  10
K a  10
9.3
9.3

NH 3 H



NH 

NH 10 

NH 

4
7
3

4
Recall, pH=7
means
[H] = 10-7
K a  10
Therefore:
9.3

NH 3 10

7
NH 


4
NH   200NH 

4
3
NH 3   100

NH 3   NH 4 

NH 3 

 100%
NH 3   200NH 3 
 0.5%
Problem
Strategy
Solution
Consider the problem of removing nitrogen from
municipal wastewater
• Remove nitrogen to prevent the stimulation of
algae growth
• Prevent excessive nitrate [NO3-] level in drinking
water from causing a potentially lethal condition
in babies known as methemoglobinemia
One way to remove is a process known as
ammonia stripping
• When organic matter decomposes, nitrogen is
first released in the form of ammonia
• NH3 - low solubility in water (ammonia)
• NH4+ - highly soluble in water (ammonium ion)

4

NH  H  NH 3
• By driving the equilibrium toward the right, less soluble
gas is formed and encouraged to leave the solution and
enter air stream in a gas stripping tower.
• This technique has been adapted for use in removing
VOC’s (volatile organic chemicals) from groundwater.
• How can the reaction be driven to the formation of
ammonia (NH3)?
• Need to decrease [H+] or increase the pH.
NH 4  H   NH 3
Ka





NH
H
9.3
3
 10 
NH 

4
Highly
Soluble
Low
Solubility

4

NH  H  NH 3
Ka





NH
H
9.3
3
 10 
NH 

4
Want to consider
[NH3]/[NH4+]
Should we decrease this
or increase this?
Highly
Soluble
Low
Solubility

4

NH  H  NH 3
Ka





NH
H
9.3
3
 10 
NH 

4
Increase it.
How can we do this?
Highly
Soluble
Low
Solubility

4

NH  H  NH 3
9.3
K a  10

NH 3 H

NH 

4


Reduce [H+]
Increase pH.
Problem
Strategy
Solution
We want to derive an equation with pH as an
independent variable.
Let’s start here:

NH H 
K  10 
NH 

NH 10 
NH   10

 NH 10

9.3
3

4
a

4
 pH
3
9.3
9.3 pH
3
Ka





NH
H
9.3
3
 10 
NH
NH 

NH 10 


4

4
 pH
3
10 9.3
 NH 3 10 9.3 pH

NH 3


NH 3 
fraction 
NH 3   NH 4 
NH 3

NH 3 
fraction 

NH 3   NH 4 

NH 3 

9.3 pH


NH 3   NH 3 10
1

9.3 pH
1  10
Fraction NH
3
1 .00
0.80
0.60
0.4 0
0.2 0
0.00
0
2
4
6
8
pH
----- end of example.
10
12
14
Summary Of Example Problem
• Nitrogen, in the form of ammonia (NH3) is removed chemically
from the water by raising the pH
• This converts ammonium ion (NH4+) into ammonia
• NH3 is then stripped from the water by passing large
quantities of air through the water
Problem
Strategy
Solution
A sample of water at pH 10 has 32.0 mg/L
of carbonate and 56.0 mg/L of bicarbonate
ion. Find the alkalinity as CaCO3.
Problem
1.
2.
3.
4.
Strategy
Solution
Determine the MW of HCO3- and CO3-2
Determine the EW of HCO3- and CO3-2
Convert the concentrations of HCO3- , CO3-2, H+ and OH- to mg/L as
CaCO3
Add the concentrations in mg/L as CaCO3 of HCO3- , CO3-2, and OH-, and
subtract H+
2
3
CO : MW  60, n  2, EW  30

3
HCO : MW  61, n  1, EW  61
Now we need to convert to mg/L CaCO3
CO32  (32.0 mg L )50 30   53.3 mg L as CaCO3
HCO  (56.0 mg L )50 61  45.9 mg L as CaCO3

3
H   5  106 mg L as CaCO3

OH  5.0 mg L as CaCO3
Alkalinity  53.3  45.9  5  104.2
I will leave it up to you to
check calculations for H+ and OH-
mg
L
as CaCO3
...... end of problem
Problem
Strategy
Solution
The solubility product for the dissociation of
Mg(OH)2 is 9 x 10-12. Determine the concentration
of Mg2+ and OH- at equilibrium.
Problem
1.
2.
3.
Strategy
Solution
Write the equation for the reaction
Write the solubility product equation
Recognize from Eqn. 1 the relationship between the number of
moles of Mg2+ and the number of moles of OH- resulting from the
dissociation of Mg(OH)2, and how this relates to Eqn 2
1. Write the equation for the reaction.
Mg OH  2  Mg  2OH
2
2. The solubility product equation is:
 Mg OH 
2
 2
 9  10
12

3. If x is the amount of Mg2+ resulting from the dissociation is given as x, then
the amount of OH- is equal to 2x.
 x 2 x  9  10 12
2
4 x  9  10
3
12
4
x  1.3  10 moles / L  Mg
4
2 x  2.6  10 moles / L  OH
.....end of example
Magnesium is removed from an industrial
waste stream by hydroxide precipitation at
a pH = 10. Determine the solubility of Mg2+
in pure water at 25° C and pKsp of 10.74.
Mg OH 2  s   Mg
2
 2OH

Problem
Strategy
Solution
1. Identify the two governing equations (Ksp and Kw)
2. Recognize that [OH-] = 10-14+pH
3. Substitute to derive an equation [Mg2+] = f(pH)
1. What are your two governing equations?
 Mg OH   10
OH  H   10
 2
2


10.74
14
2. Two unknowns, and two equations.
3. Given the pH, we know [H+].
 H   10

 pH
 10
10
4. Solve for [OH-]2
14
14
10
10

14 pH
OH   H   10 pH  10
 
OH 
 2
 10
28 2 pH
5. Substitute into 1st governing equation, and solve
for [Mg2+].
10.74
10
17.262 pH
Mg


10

 10282 pH
2
6. Substitute value of pH given in the problem
statement, then convert to mg/L. NOTE: units in [
] are moles per liter!
 Mg   10
10
24.3
2
 2.74 mol
 2.74 mol
L
g
L
3 mg
mg

10

44
.
2
mol 
g
L
Problem
Strategy
Solution
7. For a pH of 11, the solubility is 0.442 mg/L. For a
pH of 12 the solubility is 0.004 mg/L. Work these
..... end of example.
Problem
Strategy
Solution
The chemical 1,4-dichlorobenzene (1,4-DCB) is
used in an enclosed area. At 20C (68F) the
saturated vapor pressure of 1,4-DCB is 5.3 x 10-4
atm. What would be the concentration in the air
of the enclosed area (units of g/m3) at 20C ? The
molecular weight of 1,4-DCB is 147 g/mol.
Problem
Strategy
Solution
Rearrange the ideal gas law to solve for n/V [mol/L] and apply the
appropriate conversions.
Rearrange the Ideal Gas Law to solve for the
concentration of 1,4-DCB in the air
n
P
5.3 104 atm


L  atm 
V RT 
o
0
.
0821
293
K


m ol K 

g 

5 m ol  1000L 
  2.2 10

147

3
L  m 
m ol

g
 3.2 3
m


Anaerobic microorganisms metabolize organic
matter to carbon dioxide and methane gases.
Estimate the volume of gas produced (at
atmospheric pressure and 25° C) from the
anaerobic decomposition of 2 moles of glucose.
The reaction is:
C6 H12O6  3CH 4  3CO2
Recognize that each mole of glucose produces 3 moles of methane and 3
moles of carbon dioxide gases, for a total of 6 moles. Therefore, 2 moles of
glucose produces a total of 12 moles.
C6 H12O6  3CH 4  3CO2
Use the ideal gas law to solve for V given n=12 moles
Each mole of glucose produces 3 moles of methane and 3 moles of carbon dioxide
gases, for a total of 6 moles. Therefore, 2 moles of glucose produces a total of 12
moles. The entire volume is then
L  atm 


12 m ol 0.0821
 298 K
nRT
K  m ol

V

1 atm
P

 294L
Note: The volume of 1 mole of any gas is the same. Thus, 1 mole of
carbon dioxide gas is the same volume of 1 mole of methane gas.

Example
Solution
Show that one mole of any ideal gas will
occupy 22.414 L at standard temperature and
pressure (STP).
Note: STP is 273.13°K and 101.325 kPa (0°C
and 1 atm).
Use the ideal gas law to solve for volume.
Note: J = N. m Pa = N/m2
 N m 
mole
  K
K  mole
V 
 m3
N
m2
. K
1 mole8.3143 J K  mole27316

1000 L m3 

. kPa 1000 Pa kPa 
101325
 22.414 L
......end of example
Similarly, if we consider the volume at 25°C
 N m 
m ole
 K 
 K  m ole
V
 m3
N
m2

1m ole8.3143J K  mole  273 25 K


1000L m3 
101.325kPa1000Pa kPa


 24.45L
......end of example
Example
Solution
Convert 80 mg/m3 of SO2 in 1 m3 of air,
25° C, 103.193 kPa to ppm
ppm 
Vp
Va
 M p 
L  T2  101.325kPa 

 22.414




MW 
mole  273K 
P2



L
Va 1000
m3
 80g

g
V p  64.06 mol
ppm 

Va
 0.03 ppm SO2

L  298K   101.325kPa 
 22.414



 
m ole 273K  103.193kPa 

L
Va 1000 3
m
Problem
Strategy
Solution
A 1 m3 volume tank contains a gas mixture
of 18.32 moles of oxygen, 16.40 moles of
nitrogen and 6.15 moles of carbon dioxide.
What is the partial pressure of each
component in the gas mixture at 25°C and
101.3 kPa?
Problem
Strategy
Solution
• Convert temperature
• Use the ideal gas law to determine the pressure of each gas
• Apply Dalton’s Law
Convert temperature
T  25.0  273  298 K
PO2 
18.32 moles8.314 J K mole298 K 
10
. m3
 45389 Pa  45.39 kPa
PN2  40.63 kPa
PCO2  15.24 kPa
 P  45.39  40.63  15.24  1013. kPa
.....end of example
Example
Solution
Calculate the concentration of dissolved oxygen
(units of mol/L and mg/L) in a water equilibrated
with the atmosphere at 25° C.
The Henry’s law constant for oxygen at 25° C is 1.29 x
10-3 mol/L-atm.
Note: The partial pressure of oxygen in the
atmosphere is 0.21 atm.
m ol 

3
K H  PO2  1.2910
0.21atm
L  atm 

 4 m ol
 2.7 10
L
which you can convert to 8.7 mg/L
Problem
Strategy
Solution
A constant volume, batch chemical reactor achieves a reduction of
compound A from 120 mg/L to 50 mg/L in 4 hours. Determine the
reaction rate for both zero- and first-order kinetics. Clearly indicate the
units of k.
Problem
Strategy
Solution
Using the two boundary conditions
A = 120 mg/L at t=0
A = 50 mg/L at t=4 hrs
Determine k using:
C = Co - kt (zero order)
C = Coe-kt (first order)
Zero-Order
C   Co   kt
Co  C
k
t

120 50 mg L

4hr
 17.5 mg Lhr
First-Order
C   Co e
 kt
C 
50


ln 
ln

Co 
120


k 

t
4hr
ln0.417
1

 0.219 hr
4hr
Note the difference in units. The units associated with rate constants are specific for
the reaction order.
Problem
Strategy
Solution
Consider how the choice of a rate constant effects the design of a treatment
facility.
For Q = 0.5m3/s and an initial concentration of 150 mg/L, what size of reactor is
required to achieve 95% conversion assuming
a)
b)
Zero-order reaction
First order reaction
Use the values of k from the previous problem
Zero – order k = 17.5 mg/L·hr
First order k = 0.219 hr-1
Problem
1.
2.
3.
4.
Strategy
Solution
Note that 95% conversion means C = 0.05Co
Solve for t in each case
Recognize that Q = [L3/T] = Volume/time
Solve for volume V= Qt for each case
1. 95% conversion means C= 0.05Co
2. Zero-order
C  Co  kt
0.05Co  Co  kt
0.95150m g 
L
0.95Co

t

k
17.5 m g
L  hr
 8.14 hr

3
m
V  Qt  0.5
 14,657m 3


8.14 hr3600 s 
hr
hr
3. First order
C  Co e  kt
0.05Co  Co e  kt


ln0.05   0.219hr 1 t
t  13.68 hr

3
m
V  Qt  0.5
 24,622 m3
s 13.68hr3600s hr
Problem
Strategy
Solution
How long will it take the carbon monoxide
(CO) concentration in a room to decrease
by 99% after the source of carbon
monoxide is removed, and the windows
opened? Assume the first-order rate
constant for CO removal (due to dilution by
the in coming clean air) is 1.2 h-1.
Problem
Strategy
Solution
1. First order reaction is C=Coe-kt
2. If 99% is removed, C=0.01Co
Problem
Strategy
Solution
This is a first-order reaction, so use
[CO]=[COo]e-kt
When 99% of the CO is removed, [CO] = 0.01[COo]
0.01[COo] = [COo]e-kt
where k = 1.2 h-1
Solve for t = 3.8 h
Problem
Strategy
Solution
An engineer is modeling the transport of a
chemical contaminant in groundwater. The
individual has a mathematical model that only
and a handbook of with a table for “subsurface
chemical transformation half-lifes”. Subsurface
half-lives for benzene, TCE and toluene are
listed as 69, 231, and 12 days respectively.
What are the first-order rate constants for all
three chemicals?
0.693
t1/ 2 
k
0.693
k
t1/ 2
Apply this equation to each individual
compound
k benzene
kTCE
0.693
0.693


 0.01 day1
t1 / 2
69 days
0.693
0.693


 0.003 day1
t1 / 2
231 days
k toluene
0.693 0.693
1


 0.058 day
t1 / 2
12 days
Problem
Strategy
Solution
After the Chernobyl nuclear accident, the concentration of 137Cs in milk was
proportional to the concentration of 137Cs in the grass that cows consumed.
The concentration in the grass was, in turn, proportional to the
concentration in the soil. Assume that the only reaction by which 137Cs was
lost from the soil was through radioactive decay, and the half-life for this
isotope is 30 years. Calculate the concentration of 137Cs in cow’s milk after 5
years if the concentration in milk shortly after the accident was 12,000 Bq/L.
(Note: A Bequerel is a measure of radioactivity. One Bequerel equals one
Problem
Strategy
1. Determine k
2. Apply the equation C=Coe-kt
Solution
0.693
1
k
 0.023 yr
t1 / 2

137
Cs

t 5


137
Cs

t 0
e
 kt
 12,000Bq  0.023 / yr 5 yr 

e
L


 10,700 Bq L
Problem
Strategy
Solution
A biological wastewater treatment process is known to exhibit first-order kinetics
with a temperature correction factor equal to 1.023. For 20ºC, k=6.0 day-1.
Determine the required reaction time required to meet 75% conversion in the
summer and winter. Assume an average summer and winter temperature of 30ºC
and 0ºC respectively.
Problem
Strategy
Solution
Correct k
Solve for t given C=Coe-kt
For 75% conversion, C=0.25Co
kT  k 20 
(T  20 )

1.023
 6 day 1.023
1
k30  6 day
k0
1
( 30  20 )
( 0  20 )
1
 7.53day
1
 3.807day
C   Co e
 kt
C
 (1  0.75)Co 

ln 
ln
Co 
Co



t

k
k
 1.386

k

For 30 C
 1.386
t
 0.184 day  4.4 hr
1
 7.53 day

For 0 C
 1.386
t
 0.364 day  8.74 hr
1
 3.807day
```