ENV 3001 Introduction to Environmental Engineering

1
ENV 3001 Introduction to Environmental Engineering
Lecture II (c) Chemical Foundations Instructor
Dr. Ni-Bin Chang Spring, 2006
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HW3

• It is due on next the coming Friday.
• Email to nibinchang_at_gmail.com

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Buffers

• Many solutions exist in nature which are capable
  of withstanding the addition of strong acids and
  bases with little change in pH.
• These solutions are called buffers.

4
Buffers

• They are generally combinations of weak acids and
  their salts.
• For example, a combination of sodium bicarbonate
  (NaHCO3) and sodium carbonate ((Na)2CO3) will
  form a buffered solution with a pH near the pKA
  of bicarbonate, 6.35.
• The ability of a buffered solution is not
  infinite, however, and works best within 1 pH
  unit of the system pKA.

5
The Carbonate System

• One of the most important buffering systems in
  nature is the carbonate system, composed of
• carbon dioxide (CO2),
• carbonic acid (H2CO3),
• bicarbonate (HCO3-) and
• carbonate (CO32-) ions.

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The Carbonate System

• The total water system surrounding the planet
  Earth is called the hydrosphere.
• It includes freshwater systems, oceans,
  atmosphere vapor, and biological waters.
• The Arctic, Atlantic, Indian, and Pacific oceans
  cover 71 of the Earth surface, and contain 97
  of all water.

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CO2 Reactions in Water

• The dissolution of carbon dioxide in water
• CO 2(gas) CO2(aq)
• lt1 is hydrated to form carbonic acid
• CO2 H2O H2CO3

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CO2 Reactions in Water

• Some of the carbonic acid dissociates into
  bicarbonate and hydrogen ions which lowers the
  pH
• H2CO3 HCO3- H
• As the pH rises, bicarbonate increases to 100
  at a pH of 8.3. Above this, it declines by
  dissociating into carbonate HCO3-
  CO3-2 H

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Precipitation

• The reaction between calcium and carbonate to
  form sparingly soluble calcite (calcium
  carbonate).
• CaCO3 (s) Ca2 CO32-

10
Soda pop chemistry
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Inorganic - C Equilibria
Note 100 CO2 for pHlt 4.5 100 bicarbonate
for pH 8 and 100 carbonate for pH gt 12
http//waterontheweb.org/curricula/ws/
12
Buffer

• The ability of a natural water to withstand pH
  changes is measured by its acidity or alkalinity.

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Inorganic - C Major Sources and Sinks

• Sources
• Atmospheric CO2 (invasion)
• Respiration and other aerobic and anaerobic
  decomposition pathways in the water and sediments
• Groundwater from soil decomposition products
• Groundwater from volcanic seeps
• Sinks
• pH dependent conversions to bicarbonate and
  carbonate
• Precipitation of CaCO3 and MgCO3 at high pH
• Photosynthesis

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CO2 Chemistry Alkalinity

• Alkalinity the ability of water to neutralize
  acid a measure of buffering capacity or acid
  neutralizing capacity (ANC)
• Total Alkalinity (eq/L) HCO3- 2CO32-
  OH- - H
• Total Acidity (eq/L) 2H2CO3 HCO3-
  H - OH-
• Total Alkalinity typically measured by titration
  with a strong acid.
• The units are in mg CaCO3/L for reasons relevant
  to drinking water treatment (details in Chapter
  6)

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Alkalinity and Water Treatment

• Advanced wastewater treatment (domestic sewage)
• Phosphorus nutrient removal by adding lime
  (Ca(OH)2) or calcium carbonate (CaCO3)
• As pH increases gt9, it precipitates adsorbed
  PO4-3
• Settle and filter the effluent to obtain 90-95
  removal
• Used for particle (TSS) removal also
• Drinking water treatment
• For TSS removal prior to disinfection

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Acid Rain Impacts

• Acid-rain mitigation to whole lakes
• Lime or limestone added as powdered slurry to
  increase impacted lake pH value
• Also broadcast aerially to alkalize entire
  watersheds

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Acidity of Rainfall in the United States in the
Period 1731 March 1973 (ANON 1974).
National Academy of Science, 2005
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Acidity (as pH) in the Rainfall in the US in 1996
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Example Problem 2.9

• A 1-m deep lake with a surface area of 4,000 m2,
  has an initial pH of 6.5 buffered by the
  carbonate system (alkalinity 20 mg/L as CaCO3).
• Calculate the pH of the lake after receiving 5
  cm of acid rain with a nitric acid (a strong
  acid) concentration of 2x10-4 M.

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Example Problem 2.9

• Solution
• First calculate the increase in the concentration
  of H in the lake due to the addition of the acid
  rain
• 0.05 m rain x 4000 m2 surface area 200 m3
  200,000 L of acid rain
• 200,000 L x 0.0002 moles/L HNO3
• 40 moles H added
• The final lake volume 4,000 m2 x 1 m 200,000
  L 4,200 m3 4,200,000 L

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Example Problem 2.9

• Thus the concentration of H added is
• H 40 moles/4,200,000 L9.5 x 10-6 M
• Because the pH is at 6.5, the concentration of
  carbonate is negligible and the buffering is
  provided by bicarbonate and carbonic acid (see
  next slide).
• Now, calculate the concentration of bicarbonate
  and carbonic acid before the addition of the acid
  rain.

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Inorganic - C Equilibria
Note 100 CO2 for pHlt 4.5 100 bicarbonate
for pH 8 and 100 carbonate for pH gt 12
http//waterontheweb.org/curricula/ws/
23
Example Problem 2.9

• Using equation (2.23) and substituting for H
  10-6.5 M
• H2CO3 0.71HCO3-
• Alkalinity 20 mg/L as CaCO3 (Given condition)
• Converting to eq/L (using equation 2.30)
• 0.020 g/L/50 g/eq 4 x 10-4 eq/L

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Example Problem 2.9
0
0

• 4 x 10-4 eq/L HCO3- 2CO32- OH- -
  H
• Neglecting both carbonate and hydroxide
  concentrations at the low pH of 6.5,
• HCO3- - H 4 x 10-4 M
• But H 10-6.5 M 3.16x 10-7 M
• HCO3- 4 x 10-4 M
• We already have had H2CO3 0.71HCO3-
• H2CO3 0.71 x 4 x 10-4 2.8 x 10-4 M

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Example Problem 2.9

• With the addition of 9.5 x 10-6 M H in the acid
  rain, 9.5 x 10-6 M of bicarbonate is converted to
  carbonic acid with the following resulting
  concentrations
• HCO3- 4 x 10-4 - 9.5 x 10-63.9 x 10-4M
• H2CO3 2.8 x 10-4 9.5 x 10-6 2.9 x 10-4 M

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Example Problem 2.9

• H 3.32 x 10-7M
• pH 6.48
• Note that because of the effective buffer system
  present, the pH depression was limited to 0.02
  units.

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Inorganic - C Equilibria
Note 100 CO2 for pHlt 4.5 100 bicarbonate
for pH 8 and 100 carbonate for pH gt 12
http//waterontheweb.org/curricula/ws/
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Solubility Product

• Another example of the application of the
  equilibrium concept is the solubility of solids.
• The solubility product constant is an equilibrium
  constant which describes the dissolution of a
  solid into ions in aqueous solutions.

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Solubility Product

• It is widely used in designing methods to remove
  toxic metal ions dissolved in water.
• Arsenic removal systems are needed for removal of
  arsenic to less than 10 ppb when groundwater
  system is part of the drinking water system.
• The treatment process may also remove other heavy
  metals including lead, iron and manganese.

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Solubility Product

• AaBb(s) lt--------gt aAb bBa- (2.31)
• The equilibrium constant is
• Since the concentration of AaBb(s) is 1 M, by
  definition, the solubility product constant can
  be written as follows (for a system at
  equilibrium)
• KSP AbaBa-b

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Solubility Product

• Write the expression for the solubility constant
  of mercury (I) sulfate, Hg2SO4
• Hg2SO4 (s) ? Hg2 2(aq) SO42-(aq)
• Ksp (Hg2 2) (SO42-)
• So the product of the concentration of the ions,
  raised to the power of its coefficient in the
  solubility equation, is a constant.

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Solubility Product

• KSP is a function in terms of temperature and
  pressure.
• The solubility constant, Ksp, has a fixed value
  at a given temperature (normally at 25 Degrees
  Celsius)
• A solution can be described as being
• saturated,
• unsaturated, or
• Supersaturated

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Solubility Product

• An unsaturated solution is not at equilibrium and
  can dissolve more solid.
• A saturated solution is at equilibrium.
• A saturated solution implies that the system
  cannot dissolve more solid unless the temperature
  or pressure is changed.

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Solubility Product

• Supersaturated solutions are those solutions that
  contains more solute than it would if the
  dissolved solute were in equilibrium with the
  undissolved solute.
• Supersaturated solutions are those solutions that
  are above the solubility limits.
• Supersaturated solutions are meta stable.

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Solubility Product

• A supersaturated solution can be created by
  dissolving a solid at an elevated temperature and
  then allowing it to cool.
• Once cooled, precipitation may not occur,
  although the solution is not at equilibrium.
• Precipitation will occur if the reaction vessel
  is shaken or otherwise disturbed.

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Supersaturated Solution

• Crystallization from Supersaturated Solutions of
  Sodium Acetate
• http//genchem.chem.wisc.edu/demonstrations/Gen_Ch
  em_Pages/11solutionspage/crystallization_from_supe
  r.htm
• http//www.sdnhm.org/kids/minerals/grow-crystal2.h
  tml

http//www.sdnhm.org/fieldguide/minerals/index.htm
l
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Theoretical Solubility of Copper Hydroxides
mg/L
Precipitation region
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Theoretical Solubility of Nickel Hydroxides
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(No Transcript)
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Example Problem 2.10

• Calculate the solubility of barium sulfate,
  BaSO4, in pure water at 25o C in mg/l.
• What is the equilibrium concentration of barium
  (Ba2) in water that contains 10-3 M sulfate,
  SO42-?
• At 25oC, the KSP for BaSO4 is 1.0 x 10-10.

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Example Problem 2.10

• Solution
• Let x equal the number of moles/l of BaSO4 which
  will dissolve in pure water as follows
• BaSO4(s) lt-------gt Ba2 SO42-
• If x moles of BaSO4 dissolve, then x mole of Ba2
  and x moles of SO42- will enter the solution.
• KSP 1.0 x 10-10 Ba2 SO42- x2
• x 1.0 x 10-5 moles/L

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Example Problem 2.10

• If the water initially contains 10-3 M SO42-,
  after y moles/l of BaSO4 dissolve, the solution
  will contain (y 10-3) M SO42- and y moles/l of
  Ba2. Then
• KSP 1.0 x 10-10 Ba2SO42-
• y (y 10-3)
• and using the quadratic formula

y 1.0 x 10-7 M
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Example Problem 2.11

• (a) Calculate the concentration of cadmium as the
  pH of a solution decreases from 11 to 10 to 9.
• Assume that the solubility of cadmium is
  controlled only by hydroxide. The Ksp for
  cadmium hydroxide (Cd(OH)2) at 250C is 2.0 x
  10-14.
• (b) The national groundwater drinking water
  standard for cadmium is 0.005 mg/l. Calculate
  the minimum hydroxide concentration required to
  meet the groundwater standard.

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Example Problem 2.11

• Solution
• From the Ksp and the concentration of OH-, the
  concentration of Cd2 can be calculated.
• Cd(OH)2 (s) lt--------gt Cd2 2OH-
• Ksp Cd2OH-2
• At pH 11, the concentration of OH- is 10-3 M,
  so the concentration of Cd2 is
• Cd2 2.0 x 10-14/ (10-3)22.0 x 10-8 M (pH
  11)
• At pH 10, the concentration of OH- is 10-4 M,
  so the concentration of Cd2 is
• Cd2 2.0 x 10-14/ (10-4)22.0 x 10-6 M (pH
  10)

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Example Problem 2.11

• At pH 9, the concentration of OH- is 10-5 M, so
  the concentration of Cd 2 is
• Cd 2 2.0 x 10-14/ (10-5)2 2.0 x 10-4 M (pH
  9)
• As you can see, as the pH decreases (as the
  hydroxide concentration gets lower) the cadmium
  concentration (a toxic heavy metal) gets much
  higher.

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Example Problem 2.11

• In order to calculate the concentration of
  hydroxide when cadmium is 0.005 mg/L, first the
  concentration must be expressed in units of
  moles/L.
• Cd 2 5x 10-6 g/L / 112.4 g/mole 4.45 x 10-8
  moles/L
• From the Ksp, the minimum hydroxide concentration
  can be calculated
• OH- (2.0 x 10-14/4.45 x 10-8)0.5 2.1 x 10-2 M
• Note This corresponds to a pH of 10.8. So, to
  control Cd in water, we would need to add a base,
  like NaOH, to raise the pH to at least 10.8.

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Gas Law and Environmental Applications

• Stripping of nuisance gases, such as ammonia,
  hydrogen sulfide, etc involves the transfer of
  these gases from water or wastewater to the
  atmosphere.
• The transfer of oxygen into water to support
  aquatic life and the biological degradation of
  organic compounds.
• .

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Hydrogen Sulfide

• Individuals living near a wastewater treatment
  plant, a gas and oil drilling operation, a farm
  with manure storage or livestock confinement
  facilities, or a landfill may be exposed to
  higher levels of hydrogen sulfide.
• Just a few breaths of air containing high levels
  of hydrogen sulfide gas can cause death.
• Lower, longer-term exposure can cause eye
  irritation, headache, and fatigue.

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Ammonia Impacts

• The US EPA (1984) reports LC50 (Lethal
  Concentration ) of ammonia in the aquatic
  environment starting at 0.53 mg/L NH3.
• Ammonia stripping is a simpledesorption method
  that is employed to lower the ammonia content of
  a wastewater stream.

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Ammonia Stripping

• lime or caustic soda is added to the wastewater
  until the pH reaches 10.8 to 11.5.
• NH4 (aq) OH-(aq) H2O NH3 (gas)
• The reaction is highly dependent on the pH. At
  high pH (above pH 9), the ammonia (NH3) is
  liberated from the wastewater into the gas phase

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Ammonia Stripping

• Counterflow stripping columns (air stripper) are
  the most common design.
• By a combination of pH adjustment and
  temperature, we calculate the tower size for any
  required efficiency.

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Gas Law

• Ideal Gas Law P, V, T
• Daltons Law Partial Pressure in a mixed gas
• Roults Law volatile liquid
• Henrys Law volatile liquid

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Ideal Gas Law

• The Ideal Gas Law can be expressed more
  universally as shown below
• PV nRT (2.34)
• where
• n number of moles of gas present, and
• R the universal gas constant.
• The units of the universal gas constant, R, are a
  function of the units of the pressure, volume,
  and temperature terms.

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Ideal Gas Law

• The Ideal Gas Law can be used to determine the
  volume of one mole of gas at Standard Temperature
  and Pressure (STP 25o C, 1 atm) as follows.
• V/n RT/P
• V/n (0.08205 L-atm/mol-K)(25 273 oK)/
• (1 atm) 24.45 L/gmole

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Ideal Gas Law

• The Ideal Gas Law can also be used to convert
  between two popularly used expressions of gas
  concentrations, ppm (by volume) and µg/m3.
• An example of this conversion using the above
  calculation is shown in the following equation

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Dalton's Law of Partial Pressure

• In a mixture of gases, each gas exerts pressure
  independently of other gases present.
• The partial pressure exerted by each gas is
  proportional to the amount (by volume) of gas
  present.

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Example Problem

• The 1-hour National Ambient Air Quality Standard
  for CO is 35 ppm. Calculate the corresponding
  concentration in mg/m3, at 250 C and atmospheric
  pressure.
• Solution
• For gases in air, ppm refers to volume fraction.
• So 35 ppm is equivalent to 35 ml of CO per
  million ml of polluted air, or 35 ml of CO per m3
  of air.
• The mass density of pure CO in mg/ml can be
  derived from the ideal gas law

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Example Problem
Thus, the concentration of CO in air (in mg / m3
) is
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Dalton's Law of Partial Pressure

• Dalton's Law can be expressed mathematically by
  the following equations.
• PT P1 P2 P3 ..... Pi SPi or VT S
  Vi
• Where
• Pi the partial pressure exerted by gas i
  (if gas i were the only gas present in the total
  volume VT)
• Vi the partial volume occupied by gas i
  (at the total pressure PT)

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Dalton's Law of Partial Pressure

• From the Ideal Gas Law
• Pi niRT/vT or Vi ni RT/PT
• PT nTRT/vT or VT nT RT/PT
• Pi/PT ni/nT or Vi/VT ni/nT
• Where
• ninumber of moles of gas i
• nT total number of moles of gas

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Raoult's Law

• A volatile liquid or mixture of liquids may be in
  equilibrium with the gas phase above it.
• The partial pressure of the component in the gas
  phase will be directly proportional to the mole
  fraction of the component in the liquid mixture.
• It also is proportional to the volatility of the
  component as measured by its vapor pressure.

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Raoult's Law

• Raoult's Law may be expressed by the following
  equation
• Pi XiPV
• Where
• Pi partial pressure of component i in the
  gas phase,
• Xi mole fraction of i in the liquid mixture,
  
• PV vapor pressure of component i.

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Applications of Raoult's Law

• Raoult's Law can be used to predict the vapor
  phase concentration of components of gasoline
  spilled into the subsurface when the gasoline is
  present as a floating pool.
• It is the case of nonaqueous phase liquid on the
  water table in site remediation.

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Henry's Law

• Often, environmental pollutants are present in
  very dilute solutions and Raoult's Law is not
  applicable.
• For dilute solutions, a variation on Raoult's Law
  called Henry's Law can be applied.

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Henry's Law

• Henry's Law states that, under equilibrium
  conditions, the concentration of a gas dissolved
  in a liquid is proportional to its concentration
  in the gas that is in contact with the liquid.

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Henry's Law

• The proportionality constant is called Henry's
  Constant and takes on many different units,
  depending on the units of the gas and liquid
  concentration terms.
• Caq KHPg
• KH Henrys law constant, mole/L- atm
• PaHaXa
• Ha Henrys law constant, atm/mole fraction
• Pa partial pressure of the solute a in the
  gas phase
• Xa mole fraction of solute a in the liquid
  phase

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Henrys Constant for Environmental Significant
Gases (25oC)
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Example Problem 2.12

• A drinking water must be treated to control
  taste and odor due to the presence of 6.4 mg/L of
  H2S.
• It is proposed to remove the H2S from the water
  by transferring it to an air stream in a
  stripping tower.
• Within the stripping tower, water flows downward
  at 40 million gallons/day (MGD) and air flows
  upward at 120,000 standard cubic feet per min
  (scf/min).
• The temperature is 25oC and the pressure is 1
  atm.

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Example Problem 2.12

• First, calculate the air concentration of H2S in
  µg/m3 and ppm assuming that the H2S is completely
  removed from the water.
• Next calculate the equilibrium concentration of
  H2S in the air assuming the water concentration
  remains at 6.4 mg/L.

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Example Problem 2.12

• Solution
• First, calculate the H2S mass flow rate in g/sec.
• Mass Flowrate C (g/L) x Q (L/sec)
• Q(40 MGD)(106 gal/MG)(3.785 L/gal)/(86,400
  sec/d)
• Q43.8 L/sec
• Mass Flowrate(0.0064 g/L)(43.8 L/sec)
• 11.17 g/sec

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Example Problem 2.12

• Now calculate the concentration of the H2S in the
  air, assuming that all H2S is transferred to gas.
• Cgas, µg/m3(Mass Flowrate, µg/sec)/(QAir m3/sec)
• QAir, m3/sec(120,000 scf/min)/(60
  sec/min)(35.31 ft3/m3) 56.6 m3/sec
• C (11.17 x 10 6 µg/sec)/ 56.6 m3/sec
• 197,200 µg/m3

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Example Problem 2.12

• Calculate the concentration of H2S in the air if
  it were in equilibrium with the incoming water
  using Henry's Law.
• Henry's Constant for H2S is 0.1022 mol/L-atm.
• Pg KH Caq
• Pg (0.0064 g/L)/(34 g/mole)(0.1022
  mole/l-atm) 0.00184 atm

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Example Problem 2.12

• From the ideal gas law
• Cg, mole/L n/VPg/RT(0.00184atm)/0.08206)(298)
  
• 7.52 x 10-5 moles/L
• Cg (7.52 x 10-5)(0.08206)(298)106/1
• Cg 1840 ppm

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Example Problem 2.12

• The equilibrium gas concentration of H2S (1840
  ppm) is much higher than the actual exit gas
  concentration (142 ppm) achieved by completely
  stripping the H2S from the water.
• Since the required exit gas concentration is much
  lower than the concentration achievable under
  equilibrium conditions, the air flow is probably
  sufficient to remove the gas.