# FE Review for Environmental Engineering

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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

Use periodic table to get molecular weight Convert mg/L to mol/L
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

Problem Strategy Solution
“in an Acid/Base reaction, n is the # of hydrogen ions that a molecule transfers”

Problem Strategy Solution

Problem Strategy Solution
“in a precipitation reaction, n is the valence of the element”

Example Solution Convert 200 mg/L HCl to ppm

Example Solution 200 mg/L = 200 ppm

Problem Strategy Solution Convert 300 ppm Mg 2+ to mg/L as CaCO3
Convert 30 mg/L Mg2+ as CaCO3 to mg/L Note: MW Mg2+ is g/mol

Problem Strategy Solution
Determine the molecular weight of the species Determine n Equate EW=MW/n Apply equation

Problem Strategy Solution Convert 300 ppm Mg 2+ to mg/L as CaCO3
300 ppm = 300 mg/L EW Mg2+ = 24.31/2 = g/eq (300)(50/12.16) = 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

Balance the following chemical equations: CaCl2 + Na2CO3CaCO3 + NaCl
Example Solution Balance the following chemical equations: CaCl2 + Na2CO3CaCO3 + NaCl C6H12O6 + O2 CO2 + H2O NO2+H2O HNO3 + NO

CaCl2 + Na2CO3CaCO3 + 2NaCl C6H12O6 + 6O2 6CO2 + 6H2O
Example Solution CaCl2 + Na2CO3CaCO3 + 2NaCl C6H12O6 + 6O2 6CO2 + 6H2O 3NO2+H2O 2HNO3 + NO

What is the pOH if [OH-] = 10-8? What is the pH if [OH-] = 10-8?
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 = = 6 What is the [H+] if [OH-] = 10-5? [H+]= = 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.

Problem Strategy Solution Evaluate the governing equation

Problem Strategy Solution

Problem Strategy Solution Find the hydrogen ion concentration and hydroxide ion concentration in tomato juice having a pH of 4.1

Problem Strategy Solution Review how to convert
[H+]=10-pH mol/L = 7.94 x 10-5 mol/L Review governing equation

Problem Strategy Solution end of example

Problem Strategy Solution 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.

????? Problem Strategy Solution The problem is asking:
However, this expression has two unknowns. Therefore, we need a second equation. ?????

Problem Strategy Solution The problem is asking: Second Equation

Problem Strategy Solution Recall, pH=7 means [H] = 10-7

Problem Strategy Solution Therefore:

Problem Strategy Solution

Consider the problem of removing nitrogen from municipal wastewater
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
Problem Strategy Solution 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)

Problem Strategy Solution 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.

Highly Low Soluble Solubility Want to consider [NH3]/[NH4+]
Problem Strategy Solution Highly Soluble Low Solubility Want to consider [NH3]/[NH4+] Should we decrease this or increase this?

Highly Low Soluble Solubility Increase it. How can we do this? Problem
Strategy Solution Highly Soluble Low Solubility Increase it. How can we do this?

Highly Low Soluble Solubility Reduce [H+] Increase pH. Problem
Strategy Solution Highly Soluble Low Solubility Reduce [H+] Increase pH.

We want to derive an equation with pH as an independent variable.
Problem Strategy Solution We want to derive an equation with pH as an independent variable.

Problem Strategy Solution Let’s start here:

Problem Strategy Solution

Problem Strategy Solution

Problem Strategy Solution ----- end of example.

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 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+

Now we need to convert to mg/L CaCO3
Problem Strategy Solution Now we need to convert to mg/L CaCO3

Problem Strategy Solution I will leave it up to you to
check calculations for H+ and OH- end of problem

The solubility product for the dissociation of
Problem Strategy Solution The solubility product for the dissociation of Mg(OH)2 is 9 x Determine the concentration of Mg2+ and OH- at equilibrium.

Problem 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

Problem Strategy Solution 1. Write the equation for the reaction.
2. The solubility product equation is:

Problem Strategy Solution
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. .....end of example

Problem Strategy Solution 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

Problem Strategy Solution
1. Identify the two governing equations (Ksp and Kw) 2. Recognize that [OH-] = pH 3. Substitute to derive an equation [Mg2+] = f(pH)

1. What are your two governing equations?
Problem Strategy Solution 1. What are your two governing equations? 2. Two unknowns, and two equations.

3. Given the pH, we know [H+].
Problem Strategy Solution 3. Given the pH, we know [H+]. 4. Solve for [OH-]2

5. Substitute into 1st governing equation, and solve for [Mg2+].
Problem Strategy Solution 5. Substitute into 1st governing equation, and solve for [Mg2+].

Problem Strategy Solution 6. Substitute value of pH given in the problem statement, then convert to mg/L. NOTE: units in [ ] are moles per liter!

Problem Strategy Solution 7. For a pH of 11, the solubility is mg/L. For a pH of 12 the solubility is mg/L. Work these solutions on your own. ..... 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.

Problem Strategy Solution Rearrange the Ideal Gas Law to solve for the concentration of 1,4-DCB in the air

Problem Strategy Solution 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:

Problem Strategy Solution
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. Use the ideal gas law to solve for V given n=12 moles

Problem Strategy Solution
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 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.

Note: STP is 273.13°K and 101.325 kPa (0°C and 1 atm).
Example Solution Show that one mole of any ideal gas will occupy L at standard temperature and pressure (STP). Note: STP is °K and kPa (0°C and 1 atm).

Use the ideal gas law to solve for volume. Note: J = N. m Pa = N/m2
Example Solution Use the ideal gas law to solve for volume. Note: J = N. m Pa = N/m2 ......end of example

Similarly, if we consider the volume at 25°C
Example Solution Similarly, if we consider the volume at 25°C ......end of example

Convert 80 mg/m3 of SO2 in 1 m3 of air, 25° C, 103.193 kPa to ppm
Example Solution Convert 80 mg/m3 of SO2 in 1 m3 of air, 25° C, kPa to ppm

Example Solution

Problem Strategy Solution A 1 m3 volume tank contains a gas mixture of moles of oxygen, 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 kPa?

Problem Strategy Solution Convert temperature
Use the ideal gas law to determine the pressure of each gas Apply Dalton’s Law

Problem Strategy Solution .....end of example

The Henry’s law constant for oxygen at 25° C is 1.29 x 10-3 mol/L-atm.
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.

Example Solution 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)

Problem Strategy Solution Zero-Order

First-Order Problem Strategy Solution
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 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 = hr-1

Problem 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

Problem Strategy Solution 1. 95% conversion means C= 0.05Co
2. Zero-order

Problem Strategy Solution 3. First order

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.

1. First order reaction is C=Coe-kt 2. If 99% is removed, C=0.01Co
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 accepts first-order degradation rate constants 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?

Problem Strategy Solution
Apply this equation to each individual compound

Problem Strategy Solution

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 radioactive disintegration per second).

Problem Strategy Solution 1. Determine k
2. Apply the equation C=Coe-kt

Problem Strategy Solution

Problem Strategy Solution
A biological wastewater treatment process is known to exhibit first-order kinetics with a temperature correction factor equal to 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.

Solve for t given C=Coe-kt For 75% conversion, C=0.25Co
Problem Strategy Solution Correct k Solve for t given C=Coe-kt For 75% conversion, C=0.25Co

Problem Strategy Solution

Problem Strategy Solution

Problem Strategy Solution