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ENV 3001 Introduction to Environmental Engineering Lecture II (c

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ENV 3001 Introduction to Environmental Engineering Lecture II (c Powered By Docstoc
					ENV 3001 Introduction to Environmental Engineering
Lecture II (c) Chemical Foundations Instructor: Dr. Ni-Bin Chang Spring, 2006

HW#3
It is due on next the coming Friday. Email to nibinchang@gmail.com

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.

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.

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.


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.

CO2 Reactions in Water
The dissolution of carbon dioxide in water: [ CO ] CO 2(gas) CO2(aq) K=
2

( aq)

[ CO2 ] ( gas)

= 10-1.47

<1% is hydrated to form carbonic acid: CO2 + H2O H2CO3 K =
[ H 2 CO3 ] [ CO2 ] ( aq)[ H 2 O] = 10- 2.80

CO2 Reactions in Water
Some of the carbonic acid dissociates into bicarbonate and hydrogen ions which lowers the pH: H2CO3 +H [ H CO ] As the pH rises, bicarbonate increases to 100% at a pH of 8.3. Above this, it declines by dissociating into carbonate: [ H ][ CO ] -2 + H+ HCO3 CO3 = 10 K =
2 3

HCO3-

+

K1=

[ H + ][ HCO - ] 3

= 10-6.35

+

3

-10.33

2

[ HCO - ] 3

Precipitation
The reaction between calcium and carbonate to form sparingly soluble calcite (calcium carbonate). CaCO3 (s) Ca2+ + CO32K=
2 [ Ca2+ ][ CO3 - ]

[ CaCO3 ] s
= [ Ca2+ ][ CO3 ] = 10-8.34 K SP

Soda pop chemistry

www.saddleback.cc.ca.us/faculty/thuntley/ms20/seawaterprops2/sld013.htm

Inorganic - C Equilibria
H2CO3 HCO3CO32-

pH
Note – 100% CO2 for pH< ~ 4.5; 100% bicarbonate for pH ~ 8 and 100% carbonate for pH > ~12

http://waterontheweb.org/curricula/ws/

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

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

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-] + 2[CO32-] +[OH-] - [H+]

Total Acidity (eq/L) = = 2[H2CO3] + [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)

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 >9, 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

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

Acidity of Rainfall in the United States in the Period 17–31 March 1973 (ANON 1974).

National Academy of Science, 2005

Acidity (as pH) in the Rainfall in the US in 1996

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.

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

Example Problem 2.9
Thus the concentration of H+ added is: [H+] = 40 moles/4,200,000 L=9.5 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.

Inorganic - C Equilibria
H2CO3 HCO3CO32-

pH
Note – 100% CO2 for pH< ~ 4.5; 100% bicarbonate for pH ~ 8 and 100% carbonate for pH > ~12

http://waterontheweb.org/curricula/ws/

Example Problem 2.9
Using equation (2.23) and substituting for [H+] = 10-6.5 M:
K1= ( 10-6.5 )[ HCO - ] 3 [ H 2 CO3 ] = 10 -6.35

[H2CO3] = 0.71[HCO3-] 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

Example Problem 2.9
0 0

4 x 10-4 eq/L = [HCO3-] + 2[CO32-] + [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] = 0.71[HCO3-] [H2CO3] = 0.71 x 4 x 10-4 =2.8 x 10-4 M

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 10-6=3.9 x 10-4M [H2CO3] = 2.8 x 10-4 +9.5 x 10-6 =2.9 x 10-4 M

Example Problem 2.9
K1= [ H + ](3.9x 10-4 ) (2.9x 10-4 ) = 10-6.35

[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.

Inorganic - C Equilibria
H2CO3 HCO3CO32-

pH
Note – 100% CO2 for pH< ~ 4.5; 100% bicarbonate for pH ~ 8 and 100% carbonate for pH > ~12

http://waterontheweb.org/curricula/ws/

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.

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.

Solubility Product
AaBb(s) <--------> aAb+ + bBaThe equilibrium constant is:
K= [ Ab+ ] [ B a - ] [ Aa Bb ] s
a b

(2.31)

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 = [Ab+]a[Ba-]b

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.

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

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.

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.”

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.

Supersaturated Solution
Crystallization from Supersaturated Solutions of Sodium Acetate http://genchem.chem.wisc.edu/demonst rations/Gen_Chem_Pages/11solutionsp age/crystallization_from_super.htm http://www.sdnhm.org/kids/minerals/gr ow-crystal2.html

http://www.sdnhm.org/fieldguide/minerals/index.html

Theoretical Solubility of Copper Hydroxides

mg/L

Precipitation region

Theoretical Solubility of Nickel Hydroxides

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 (Ba+2) in water that contains 10-3 M sulfate, SO42-? At 25oC, the KSP for BaSO4 is 1.0 x 10-10.

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) <-------> Ba2+ + SO42If 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

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 = [Ba2+][SO42-] = y (y + 10-3) and using the quadratic formula:
y= - 10-3 + 10-6 + 4(1.0x 10-10 ) 2

y= 1.0 x 10-7 M

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.

Example Problem 2.11
Solution: From the Ksp and the concentration of OH-, the concentration of Cd2+ can be calculated. Cd(OH)2 (s) <--------> Cd2+ + 2OHKsp= [Cd2+][OH-]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)2=2.0 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)2=2.0 x 10-6 M (pH = 10)

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.

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.

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.
.

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.

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 simple desorption method that is employed to lower the ammonia content of a wastewater stream.

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

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.

Gas Law
Ideal Gas Law: P, V, T Dalton’s Law: Partial Pressure in a mixed gas Roult’s Law: volatile liquid Henry’s Law: volatile liquid

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.

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

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
n V  P RT
3

m V



P( MW ) RT

g/ m =

ppm x MW x 1000 24.45

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.

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:

Example Problem
n V
m V  L  atm gmol  K



P RT

m V
m V



P( MW ) RT

1.0atm (28g / gmol) .08209 298K

 1.145 g / L or 1.145 mg / ml

Thus, the concentration of CO in air (in mg / m3 ) is:
CO  35 ml CO m air
3

x

1.145 mg CO ml CO

 40 mg / m

3

Dalton's Law of Partial Pressure
Dalton's Law can be expressed mathematically by the following equations. PT = P1 + P2 + P3 + ..... Pi = ΣPi or VT = Σ 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)

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: ni=number of moles of gas i nT =total number of moles of gas

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.

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.

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.

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.

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.

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= Henry’s law constant, mole/L- atm
Pa=HaXa Ha= Henry’s 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

Henry‘s Constant for Environmental Significant Gases (25oC)
Gas KH x 105, mole fraction/atm
2.28 6.10 1.42

KH x 104, mol/L-atm
12.8 338 7.90

O2 CO2 H2

CH4
N2 NO

2.43
1.16 3.48

13.4
6.48 2.00

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.

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.

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) Q=43.8 L/sec Mass Flowrate=(0.0064 g/L)(43.8 L/sec) =11.17 g/sec

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
C ppm = (197,200 g/ m3 )(24.45) (34g/mole) (1000)  142 ppm

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/latm)]= 0.00184 atm

Example Problem 2.12
From the ideal gas law: Cg, mole/L = n/V=Pg/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

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.


				
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