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    Acid Rain: Cause And Effects And Issues Essay

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    Introduction Acid rain has become an environmental concern of global importancewithin the last decade. With the increasing environmental awareness of the”unhealthy” 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 plant’s 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.

    This process reverses all the previous reactions and forms lead at theanode and lead peroxide at the cathode. Proposed Problemi) The concentration of sulfuric acid is 0. 0443 mol/L. The pH is: No. mol of hydrogen ions = 0. 0443 mol/L x 2 = 0.

    0886 mol/L hydrogen ionspH = – log [H]= – log (0. 0886)= – (-1. 0525)= 1. 05Therefore, pH is 1. 05. ii) The amount of base needed to neutralize the lake water is: volume of lake = 2000m x 800m x 50m= 800,000,000 m3 or 8×108 m3 since 1m3=1000L, therefore 8×1011 L 0.

    0443 mol/L x 8×1011 = 3. 54 x 1010 mol of H2SO4 in water # mol NaOH = 3. 54 x 1010 mol H2SO4 x 2 mol NaOH 1 mol H2SO4= 7. 08 x 1010 mol of NaOH needed Mass of NaOH = 7. 08 x 1010 mol NaOH x 40 g NaOH1 mol NaOH = 2.

    83 x 1012 g NaOHor 2. 83 x 109 kg NaOH Therefore a total of 2. 83 x 1012 g of NaOH is needed to neutralize the lake water. iii) The use of sodium hydroxide versus limestone toneutralize the lake water: Sodium hydroxide: Sodium hydroxide produces water when reacting withan acid, it also dissolves in water quite readily.

    When using sodiumhydroxide to neutralize a lake, there may be several problems. One problemis that when sodium hydroxide dissolves in water, it gives off heat andthis may harm aquatic living organisms. Besides this, vast amounts ofsodium hydroxide is required to neutralize a lake therefore large amountsof this substance which is corrosive will have to be transported. This isa great risk to the environment if a spill was to occur.

    The following equation shows that water is produced when using sodiumhydroxide. 2NaOH + H2SO4 ??> Na2 SO4 + 2H2O Limestone: Another way to neutralize a lake is by liming. Liming oflakes must be done with considerable caution and with an awareness that theaquatic ecosystem will not be restored to its original pre-acidic stateeven though the pH of water may have returned to more normal levels. Whenlimestone dissolves in water it produces carbon dioxide. This could be aproblem since a higher content of carbon dioxide would mean a loweredoxygen content especially when much algae growth is present.

    As a result,fish and other organisms may suffer. Limestone also does not dissolve asreadily as sodium hydroxide thus taking a longer period of time to reactwith sulfuric acid to neutralize the lake. The equation for theneutralization using limestone is as follows: Ca CO3 + H2SO4 ??> CaSO4 + H2O. iv) The effect of the Acid or excessive Base on the plant and animal life: You will probably find that there aren’t many aquatic living organismsin waters that are excessively basic or acidic. A high acidic or basiccontent in lakes kill fishes and other aquatic species. Prolonged exposureto acidic or excessively basic conditions can lead to reproductive failureand morphological aberration of fish.

    A lowered pH tends to neutralizetoxic metals. The accumulation of such metals in fish contaminates foodchains of which we are a part as these metals can make fish unfit for humanconsumption. Acidification of a lake causes a reduction of the productionof phytoplankton (which is a primary producer) as well as in theproductivity of the growth of many other aquatic plants. In acidicconditions, zooplankton species will probably becompletely eliminated. Inaddition, bacterial decomposition of dead matter is seriously retarded inacidified lake waters.

    Other effects of acidic conditions areanoverfertilization of algae and other microscopic plant lifecausing algaeblooms. Overgrowth of these consumes quickly most of the oxygen in waterthus causing other life forms to die from oxygen starvation. When there are excessive base or acid in waters, not only do aquaticorganisms get affected but animals who depend on aquatic plants to survivewill starve too, since few aquatic plants survive in such conditions. Therefore each organism in the aquatic ecosystem is effected by excessivebasic or acidic conditions because anything affecting one organism willaffect the food chain, sending repercussions throughout the entireecosystem.

    v) The factors that govern this plant’s location, if this plant employs40% of the towns people: The major factors that would govern this plant’s location would bewhether there is ready access to raw materials; whether the location isclose to major transportation routes; whether energy resources are readilyavailable and if there is an adequate water supply in the area. Since thisplant would employ 40% of the towns people, the plant should be close tothe town while still far enough so that in case of any leakage of theplant, the town will be within a safe distance of being severely affected. The factor of whether the general living conditions in the area aresuitable for the workers should also be considered as well. Additional Commentsa) The situation of pollution in the Great Lakes and process being used to start cleaning it up–comments: Everyday, roughly 3630 kilograms of toxic chemicals enter the lakes, nearby land and air.

    Pollution of the Great Lakes has become an increasingly serious problem. Just in Lake Ontario, hundreds of thousands of tons of contaminants have been deposited over the years. These include DDT, PCBs, mercury, dioxins and mirex, a pesticide. About 4. 6 million people depend on Lake Ontario alone for drinking water.

    The environmental problem of greatest concern to Lake Ontario neighbours is water-discharged toxic chemicals and industrial air pollutants. Not only is this occurring in Lake Ontario but the other Great Lakes as well. The lakes probably have all these poisonous chemicals in them: salts drained from urban streets, coliform bacteria from the sewage civilization plus a selection of substances such as phosphorus, polychlorinated biphenyls and heavy metals. It is reported that the toxic chemicals in the Great Lakes basin are a health risk linked to brain damage, birth defects and cancer.

    All the predator species at the top of the food chain have shown health problems as a result of toxic chemicals building up in their bodies. Chemicals that exist in low levels in the air and water accumulate as they move up through the food chain. At present 35 million humans who live around the horridly polluted five Great Lakes face increasing health risks from environmental contaminants. Millions of people in the Great Lakes are exposed to hazardous chemicals. They drink them in the contaminated water, eat them concentrated in the flesh of the fish and breathe them in the air. Mulroney said that the risks are too high and that we cannot afford anymore risks.

    He said pollution problems could be fought under a three-stageplan over the next decade:1) A “toxic freeze” banning new polluters from puttingup pipes or smokestacks in the region2) An attack on “non-point sources” of pollution, suchas run-off from streets and farms where groundwateris loaded with pesticides. 3) A crackdown on existing polluters when their smokeand sewer-discharge permits come up for renewal,requiring them to scale down their pollution. Consumers can also help by demanding pesticide-free food. International agreements have been made to clean up the Great Lakes. Canada’s federal Conservative government has announced in 1989 to spend$125 million over five years on Great Lakes cleanup. By one estimate, itmay cost as much as $100 billion to retrieve the purity of the Great Lakesonce had.

    b) The treatment of water for drinking and water purifiers one can purchase–comments: As the people’s uncertainty to the quality of our drinking water increases, many more people are buying water treatment devices and purifiers. Even though most treated tap water is fit to drink, people are losing faith in the government to keep it that way. therefore purifier leave become increasingly popular among consumers. However each of the most popular cleansing methods has some disadvantages. Many filters use some form of “activate” carbon.

    However, few carbon filters alone do a very good job of reducing heavy metals such as lead even though the smallest sink-tap charcoal strainer will make cloudy water look and taste a bit better. Distillation units turn water to steam and recondense it to a cleaner state. This process has its disadvantages, too for they can also pass along harmful chemicals with low boiling points into the water. Another water treatment device is the reverse-osmosis device which uses sophisticate membranes to separate pure water from impure. Even though this is effective, three gallons of water for every good one produced is generally wasted. Some machines zap germs with lethal doses of ultraviolet light.

    A specific example of a water filter is the NSA 3000HM high density filter. This filtration unit is designed to remove lead, iron, sulfur and manganese from your drinking water supply. Still another example is a water treatment system called the NSA Bateriostatic water treatment system. This system removes chlorine, bad taste and odours, reduces undissolved particles (sediment, discolouration, etc.

    ) and inhibits bacteria growth. 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 NaOH????????????????????????????????????????????????????????????????????????1? . 12 ?1.

    2 mL ?0. 00071????????????????????????????????????????????????????????????????????????2? . 14 ?1. 5 mL ?0. 00103????????????????????????????????????????????????????????????????????????flask 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 of??surfuric acid sol’n(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

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