Acid Rain Cause And Effects And Issues Essay Example For Students

Acid Rain: Cause And Effects And Issues Essay Introduction Acid rain has become an environmental concern of global importancewithin the last decade. With the increasing environmental awareness of theunhealthy condition of our planet earth the concern about acid rain hasnot lessened. In brief, acid rain is rain with pH values of less than 5.6. Whendealing with acid rain one must study and understand the process of makingSulfuric acid. In this project we will take an in depth look into theproduction of sulfuric acid, some of its uses and the effects of it as apollutant in our environment. Sulfuric Acid Industry in Ontario Among the many plants in Ontario where sulfuric acid is produced, thereare three major plant locations that should be noted on account of theirgreater size. These are: (1) Inco. Sudbury, (2) Noranda Mines Ltd. Welland, and (3) Sulfide Ontario There are a number of factors which govern the location of eachmanufacturing plant. Some of these factors that have to be considered whendeciding the location of a Sulfuric Acid plant are:a. Whether there is ready access to raw materials;b. Whether the location is close to major transportation routes;c.Whether there is a suitable work force in the area for plant construction and operation;d.Whether there is sufficient energy resources readily available;e. Whether or not the chemical plant can carry out its operation without any unacceptable damage to the environment. Listed above are the basic deciding factors that govern the location ofa plant. The following will explain in greater detail why these factorsshould be considered. 1) Raw Materials The plant needs to be close to the raw materials that are involved in the production of sulfuric acid such as sulfur, lead, copper, zinc sulfides, etc.. 2) Transportation A manufacturer must consider proximity to transpor- tation routes and the location of both the source of raw materials and the market for the product. The raw materials have to be transported to the plant, and the final product must be transported to the customer or distributor. Economic pros and cons must also be thought about. For example, must sulfuric plants are located near the market because it costs more to transport sulfuric acid than the main raw materials, sulfur. Elaborate commission proof container are required for the transportation of sulfuric acid while sulfur can be much more easily transported by truck or railway car. 3) Human Resources For a sulfuric acid plant to operate, a large work force will obviously be required. The plant must employ chemists, technicians, administrators, computer operators, and people in sales and marketing. A large number of workers will also be required for the daily operation of the plant. A work force of this diversity is therefore likely to be found only near major centres of population. 4) Energy Demands Large amounts of energy will also be required for the production of many industrial chemicals. Thus, proximity to a plentiful supply of energy is often a determining factor in deciding the plants location. 5) Environmental Concerns Most importantly, however, concerns about the environment must be carefully taken into consideration. The chemical reaction of changing sulfur and other substances to sulfuric acid results in the formation of other substances like sulfur dioxide. This causes acid rain. Therefore, there is a big problem about sulfuric plants causing damage to our environment as the plant is a source of sulfur emission leading to that of acid rain. 6) Water Supplies Still another factor is the closeness of the location of the plants to water supplies as many manufacturing plants use water for cooling purposes. In addition to these factors, these questions must also be answered: Is land available near the proposed site at a reasonable cost? Is the climate of the area suitable? Are the general living conditions in the area suitable for the people involved who will be relocating in the area? Is there any suggestions offered by governments to locate in a particular region? The final decision on where the sulfuric acid plant really involves acareful examination and a compromise among all of the factors that havebeen discussed above. Producing Sulfuric Acid Sulfuric acid is produced by two principal processes the chamberprocess and the contact process. The contact process is the current process being used to producesulfuric acid. In the contact process, a purified dry gas mixturecontaining 7-10% sulfur dioxide and 11-14% oxygen is passed through apreheater to a steel reactor containing a platinum or vanadium peroxidecatalyst. The catalyst promotes the oxidation of sulfur dioxide totrioxide. This then reacts with water to produce sulfuric acid. Inpractice, sulfur trioxide reacts not with pure water but with recycledsulfuric acid.The reactions are: 2SO2 + O2 2SO3 SO3 + H2O H2SO4 The product of the contact plants is 98-100% acid. This can either bediluted to lower concentrations or made stronger with sulfur trioxide toyield oleums. For the process, the sources of sulfur dioxide may beproduced from pure sulfur, from pyrite, recovered from smelter operationsor by oxidation of hydrogen sulfide recovered from the purification ofwater gas, refinery gas, natural gas and other fuels. Battery Acid Industry Many industries depend on sulfuric acid. Among these industries is thebattery acid industry. The electric battery or cell produces power by means of a chemicalreaction. A battery can be primary or secondary. All batteries, primary orsecondary, work as a result of a chemical reaction. This reaction producesan electric current because the atoms of which chemical elements are made,are held together by electrical forces when they react to form compounds. A battery cell consists of three basic parts; a positively chargedelectrode, called the cathode, a negatively charged electrode, called theanode, and a chemical substance, called an electrolyte, in which theelectrodes are immersed. In either a wet or dry cell, sufficient liquidmust be present to allow the chemical reactions to take place. Electricity is generated in cells because when any of these chemicalsubstances is dissolved in water , its molecules break up and becomeelectrically charged ions. Sulfuric acid is a good example. Sulfuric acid,H2SO4, has molecules of which consist of two atoms of hydrogen, one ofsulfur and four oxygen. When dissolved in water, the molecules split intothree parts, the two atoms of hydrogen separate and in the process eachloses an electron, becoming a positively charged ion (H+). The sulfur atomand the four atoms of oxygen remain together as a sulfate group (SO4), andacquire the two electrons lost by the hydrogen atoms, thus becomingnegatively charged (SO4). These groups can combine with others ofopposite charge to form other compounds. The lead-acid cell uses sulfuric acid as the electrolyte. Thelead-acid storage battery is the most common secondary battery used today,and is typical of those used in automobiles. The following will describeboth the charging and discharging phase of the lead-storage battery and howsulfuric acid, as the electrolyte, is used in the process. The leadstorage battery consists of two electrodes or plates, which are made oflead and lead peroxide and are immersed in an electrolytic solution ofsulfuric acid. The lead is the anode and the lead peroxide is the cathode. When the battery is used, both electrodes are converted to lead sulfate bythe following process. At the sulfate ion that is present in the solutionfrom the sulfuric acid. At the cathode, meanwhile, the lead peroxideaccepts two electrons and releases the oxygen; lead oxide is formed first,and then lead joins the sulfate ion to form lead sulfate. At the sametime, four hydrogen ions released from the acid join the oxygen releasedfrom the lead peroxide to form water. When all the sulfuric acid is usedup, the battery is discharged produces no current. The battery can berecharged by passing the current through it in the opposite direction. Anyalization of Pearl from the 'Scarlet Letter' Essay Each of these processes can reduce impurities in your water supply andmany machines as suggested by the above examples combine severalapproaches. c) BRIEF OUTLINE OF THE KEY EVENTS IN THE U.S.-CANADA RELATIONS WITH RESPECT TO CLEANING UP THE GREAT LAKES:1972: the U.S. chairman of the International JointCommission, announced to study to determine the pollutingeffects on the Great Lakes urban development and agriculturalland use, find remedies and estimate cleanup costs;Canadaand the United States signed a Great Lakes Quality Agreement. 1974: Canadians say the cleanup financed by Washingtonis already running far behind the scheduleenvisaged when the agreement was signed. 1978: Canada and the United States agreed to the goalof zero discharge of pollution.1987: thegoal made in 1978 is made again, this means bothcountries agreed to work toward completelyeliminating persistent toxic pollutants, not justthe amount being discharged by industry; Mulroneyalso proposed that the U.S. slash industrialsulfide and nitrogen oxide emissions by halfbefore 1994. The Canada-U.S. International Joint Commission meets every two years todiscuss pollution and other issues concerning the Great Lakes, At present,they are making a ten-year headline for the Great Lakes to be cleaned up. BibliographyEncyclopediasCollier Encyclopedia, volume 3, U.S.A.: MacMillanEducational Company, New York, 1984. Encyclopedia of Industrial Chemical Analysis, volume 18,U.S.A.: John Wiley Sons, Inc, 1973. Science Technology Illustrated: The World Around U.S.,Volume 3, U.S.A: Encyclopedia Britannica Inc, 1984. ArticlesCleaning Up By Cleaning Up Newsweek: Feb. 27, 1989. Deadline Urged for Cleanup of Great Lakes, Toronto Star,Oct. 14, 1989. Great Afflictions of the Great Lakes, The Globe and Mail,Oct. 14, 1989. Great Lakes Pollution as a Political Issue, The Globe andMail, Oct. 16, 1989. N.Y. Accused of Overlooking Pollution in Lake, TorontoStar, Feb. 26, 1990. Pact On Great Lakes Cleanup Not Working, Greenpeace Says,Globe and mail, July 19, 1989. The Clean Water Industry Grows on Fear, Uncertainty,Toronto Star, Jan. 28, 1990. Information Scarce On Great Lakes Chemicals, The Globe andMail, Oct. 14, 1989. OthersCountdown Acid Rain, Facts: Ministry of the Environment,1989. Sanderson, Kimberly, Acid Forming Emissions, Canada:Environment Council of Edmonton, Alberta, 1984. The New How It Works, volume 2, Westport Connecticut; H.S. Stuttman Inc., 1987. Weller, Phil., Acid Rain: Silent Crisis, Canada: Between theLines, 1980. TITRATION LABORATORYPurpose: 1) to prepare 0.1 mol/L NaOH solution. 2) to standardize the NaOH solution in part 1, usingpotassium hydrogen phthalate. 3)to determine the unknown molarity of a H2SO4 solution usingstandardized solution. Part 1 Prepare 0.1 mol/L NaOH solutionObservations:Data:mass of NaOH + paper tray = 4.58 gmass of paper tray = 3.46 gmass of NaOH pellets= 1.12 gCalculation:Number of mole of NaOH = mass of NaOH pellets = 1.12g = 0.028mol g. mol mass of NaOH40gConclusion:Questions:1. When the NaOH pellets are left in the atmosphere, it reacts with the gases and absorbs water (moisture) in the air making it unable to neutralize too well. 2. The gram mole mass of a substance is the mass in gram of 1 mol of that substance. 3. The solution of NaOH must be standardized in order to accurately calculate the concentration of the acid. Part 2 Standardize the NaOH solution prepared in Part 1, usingpotassium hydrogen phthlateObservations:Data:mass of vial + KpH = 22.19gmass of vial + KpH aftertransfer to 1st flask= 22.19gmass of vial + KpH aftertransfer to 2nd flask= 21.93gmass of vial + KpH after flask?mass of KpH? volume of NaOH ? conc. of NaOH1? .12 ?1.2 mL ?0.000712? .14 ?1.5 mL ?0.00103flask 1To calculate the concentration of NaOH (mol/L) the number of moles ofKpH have to be calculated. No. of mol of KpH = 0.12 204g/mol= 5.9 x 10-4 molThe ratio of KpH to NaOH is 1:1Therefore, the no. of NaOH = 5.9 x 10-4mol. The equation being used is: KpH + NaOH KHC8H3NaO4+H2OThefollowing equation is used to calculate the concentration of NaOH. c = nn=number of mol = 5.9 x 10-4molvv=volume = 1.2 x 10-3 c = 5.9 x 10-4molc=concentration = ?1.2 x 10-3L c = 0.492 mol/LTherefore, the NaOH solution in Flask 1 is 0.492 mol/L. flask 2No. of mol of KpH = 0.14204g/mol= 6.9 x 10-4 molThe ratio of KpH to NaOH is 1:1Therefore, the no. of NaOH = 6.9 x 10-4mol. c = nn=number of mol = 6.9 x 10-4molvv=volume = 1.5 x 10-3 c = 6.9 x 10-4molc=concentration = ?1.5 x 10-3L c = 0.46 mol/LTherefore, the NaOH solution in Flask 2 is 0.46 mol/L. The average molarity of NaOH solution = flask 1 + flask 2 2= 0.492 + 0.46 mol/L 2= 0.476 mol/LConclusions:Questions:1. The equation for the neutralization of potassium hydrogen phthalate solution with NaOH solution is: NaOH + KHC8H4O4 KHC8H3O4Na + H2O2. The primary error in this titration process is that it is very easy to go over the endpoint. We can improve this by being very careful when letting the NaOH solution into the acidic solution. Especially when we see that the pink colour is starting to stay we should allow only part drops of the NaOH solution into the acidic solution to make certain that we do not go over the endpoint. 3. The endpoint of a titration is the point at which the number of moles of hydroxide ion added is the same as the number of moles of hydrogen ion originally present in the flask. The difference between the stoichiometric point and endpoint of a reaction is that the stoichiometric point is exactly the point at which the number of moles of hydroxide ion is equal to the number of moles of hydrogen ion while the endpoint is usually a little over this point when the solution has turned pink. 4. Phenolphthalein was chosen as the indicator of this titration because phenolphthalein is a dye that is colourless in acidic solutions but shows-up bright red or pink in basic solutions. 5. An indicator is a compound that detects the presence of acids and bases by changing to different colours. Part 3 To determine the unknown molarity of a H2SO4 solution usingstandardized NaOH solution. Observations: Volume of known ? Volume of known?Molarity ofsurfuric acid soln(mL)? conc. of NaOH(mL) ?sulfuric acid 1.25 mL?4.5 mL (0.0045L) ?3.915 x 10-6 2.25 mL?4.8 mL (0.0048L) ?4.176 x 10-6 To find the molarity of unknown sulfuric acid solution: Equation of reaction:2NaOH + H2SO4 Na2SO4 + 2H2O General equation solve:Ca Va = Cb Vba = acidnanbb = base Flask 1 For NaOH (base):v = 4.5 x 10-3 L c = 0.476 mol/L n = ?n = v c= 4.5 x 10-3 x 0.476= 2.1 x 10-3 mol NaOH For H2SO4 (acid):v = 0.025 Ln = ?c = ?#mol of H2SO4 = 2.1 x 10-3 x 1 H2SO4 2NaOH= 1.05 x 10-3 Solving the equation:Ca Va = Cb VbnanbCa x 0.025 L = 0.476 mol/L x 0.0045 L1.05 x 10-32.1 x 10-3 molCa = 0.476 x 0.0045 x 1.05 x 10-3 2.1 x 10-3 x 0.025Ca = 2.25 x 10-65.25 x 10-5Ca = 0.0429 mol/L Flask 2Ca x 0.025 L = 0.476 mol/L x 0.0048 L1.14 x 10-32.28 x 10-3Ca = 0.476 x 0.0048 x 1.14 x 10-32.28 x 10-3 x 0.025Ca = 0.0457 mol/L The average molarity of H2SO4 solution = flask 1 + flask 2 2= 0.0429 + 0.0457 2= 0.0443 mol/L