Report

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 0.005491 0.00549N b) 150 m g / L H 2 SO4 MW 21 32.06 416 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 0.001532 .00306N c) 100 m g / L Ca HCO3 2 “in a precipitation reaction, n is the valence of the element” MW 40 2(1) 212 616 162g mol EW 81 g eq 100mg L m olarity 0.000617M mg g (1000 g )162 mol norm ality 0.0006172 .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 + Na2CO3CaCO3 + NaCl C6H12O6 + O2 CO2 + H2O NO2+H2O HNO3 + NO Example Solution CaCl2 + Na2CO3CaCO3 + 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 7.94 105 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 The problem is asking: NH 3 100 NH 3 NH 4 However, this expression has two unknowns. Therefore, we need a second equation. ????? The problem is asking: 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 200NH 4 3 NH 3 100 NH 3 NH 4 NH 3 100% NH 3 200NH 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 5 106 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 17.262 pH Mg 10 10282 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 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 20C (68F) 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 20C ? 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 5.3 104 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 mole8.3143 J K mole27316 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 ole8.3143J K mole 273 25 K 1000L m3 101.325kPa1000Pa 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 80g 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 moles8.314 J K mole298 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 1.2910 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 Lhr First-Order C Co e kt C 50 ln ln Co 120 k t 4hr ln0.417 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 0.95150m 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 hr3600 s hr hr 3. First order C Co e kt 0.05Co Co e kt ln0.05 0.219hr 1 t t 13.68 hr 3 m V Qt 0.5 24,622 m3 s 13.68hr3600s 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 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? 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 day1 t1 / 2 69 days 0.693 0.693 0.003 day1 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 radioactive disintegration per second). 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