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ARRoGANT CoURiERS WiTH ESSaYS
Grade Level: Type of Work Subject/Topic is on:
[ ]6-8 [ ]Class Notes [A report on the causes/ ]
[ ]9-10 [ ]Cliff Notes [effects/and other issues]
[x]11-12 [x]Essay/Report [involvong acid rain. ]
[ ]College [ ]Misc [ ]
Dizzed: 07/94 # of Words:2,500 School: co-ed/public State: NY
ФФФФФФФФФ>ФФФФФФФФФ>ФФФФФФФФФ>Chop Here>ФФФФФФФФФ>ФФФФФФФФФ>ФФФФФФФФФ>ФФФФФФФФФ
Introduction
Acid rain has become an environmental concern of global importance
within the last decade. With the increasing environmental awareness of the
"unhealthy" condition of our planet earth the concern about acid rain has
not lessened.
In brief, acid rain is rain with pH values of less than 5.6. When
dealing with acid rain one must study and understand the process of making
Sulfuric acid. In this project we will take an in depth look into the
production of sulfuric acid, some of its uses and the effects of it as a
pollutant in our environment.
Sulfuric Acid Industry in Ontario
Among the many plants in Ontario where sulfuric acid is produced, there
are three major plant locations that should be noted on account of their
greater 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 each
manufacturing plant. Some of these factors that have to be considered when
deciding 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 of
a plant. The following will explain in greater detail why these factors
should 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 a
careful examination and a compromise among all of the factors that have
been discussed above.
Producing Sulfuric Acid
Sulfuric acid is produced by two principal processes-- the chamber
process and the contact process.
The contact process is the current process being used to produce
sulfuric acid. In the contact process, a purified dry gas mixture
containing 7-10% sulfur dioxide and 11-14% oxygen is passed through a
preheater to a steel reactor containing a platinum or vanadium peroxide
catalyst. The catalyst promotes the oxidation of sulfur dioxide to
trioxide. This then reacts with water to produce sulfuric acid. In
practice, sulfur trioxide reacts not with pure water but with recycled
sulfuric acid.The reactions are:
2SO2 + O2 ФФ> 2SO3
SO3 + H2O ФФ> H2SO4
The product of the contact plants is 98-100% acid. This can either be
diluted to lower concentrations or made stronger with sulfur trioxide to
yield oleums. For the process, the sources of sulfur dioxide may be
produced from pure sulfur, from pyrite, recovered from smelter operations
or by oxidation of hydrogen sulfide recovered from the purification of
water gas, refinery gas, natural gas and other fuels.
Battery Acid Industry
Many industries depend on sulfuric acid. Among these industries is the
battery acid industry.
The electric battery or cell produces power by means of a chemical
reaction. A battery can be primary or secondary. All batteries, primary or
secondary, work as a result of a chemical reaction. This reaction produces
an 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 charged
electrode, called the cathode, a negatively charged electrode, called the
anode, and a chemical substance, called an electrolyte, in which the
electrodes are immersed. In either a wet or dry cell, sufficient liquid
must be present to allow the chemical reactions to take place.
Electricity is generated in cells because when any of these chemical
substances is dissolved in water , its molecules break up and become
electrically charged ions. Sulfuric acid is a good example. Sulfuric acid,
H2SO4, has molecules of which consist of two atoms of hydrogen, one of
sulfur and four oxygen. When dissolved in water, the molecules split into
three parts, the two atoms of hydrogen separate and in the process each
loses an electron, becoming a positively charged ion (H+). The sulfur atom
and the four atoms of oxygen remain together as a sulfate group (SO4), and
acquire the two electrons lost by the hydrogen atoms, thus becoming
negatively charged (SO4--). These groups can combine with others of
opposite charge to form other compounds.
The lead-acid cell uses sulfuric acid as the electrolyte. The
lead-acid storage battery is the most common secondary battery used today,
and is typical of those used in automobiles. The following will describe
both the charging and discharging phase of the lead-storage battery and how
sulfuric acid, as the electrolyte, is used in the process. The lead
storage battery consists of two electrodes or plates, which are made of
lead and lead peroxide and are immersed in an electrolytic solution of
sulfuric 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 by
the following process. At the sulfate ion that is present in the solution
from the sulfuric acid. At the cathode, meanwhile, the lead peroxide
accepts two electrons and releases the oxygen; lead oxide is formed first,
and then lead joins the sulfate ion to form lead sulfate. At the same
time, four hydrogen ions released from the acid join the oxygen released
from the lead peroxide to form water. When all the sulfuric acid is used
up, the battery is "discharged" produces no current. The battery can be
recharged by passing the current through it in the opposite direction.
This process reverses all the previous reactions and forms lead at the
anode and lead peroxide at the cathode.
Proposed Problem
i) 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 ions
pH = - log [H]
= - log (0.0886)
= - (-1.0525)
= 1.05
Therefore, 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 8x108 m3
since 1m3=1000L, therefore 8x1011 L
0.0443 mol/L x 8x1011 = 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 NaOH
1 mol NaOH
= 2.83 x 1012 g NaOH
or 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 to
neutralize the lake water:
Sodium hydroxide: Sodium hydroxide produces water when reacting with
an acid, it also dissolves in water quite readily. When using sodium
hydroxide to neutralize a lake, there may be several problems. One problem
is that when sodium hydroxide dissolves in water, it gives off heat and
this may harm aquatic living organisms. Besides this, vast amounts of
sodium hydroxide is required to neutralize a lake therefore large amounts
of this substance which is corrosive will have to be transported. This is
a great risk to the environment if a spill was to occur.
The following equation shows that water is produced when using sodium
hydroxide.
2NaOH + H2SO4 ФФ> Na2 SO4 + 2H2O
Limestone: Another way to neutralize a lake is by liming. Liming of
lakes must be done with considerable caution and with an awareness that the
aquatic ecosystem will not be restored to its original pre-acidic state
even though the pH of water may have returned to more normal levels. When
limestone dissolves in water it produces carbon dioxide. This could be a
problem since a higher content of carbon dioxide would mean a lowered
oxygen content especially when much algae growth is present. As a result,
fish and other organisms may suffer. Limestone also does not dissolve as
readily as sodium hydroxide thus taking a longer period of time to react
with sulfuric acid to neutralize the lake. The equation for the
neutralization using limestone is as follows:
Ca CO3 + H2SO4 ФФ> 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 organisms
in waters that are excessively basic or acidic. A high acidic or basic
content in lakes kill fishes and other aquatic species. Prolonged exposure
to acidic or excessively basic conditions can lead to reproductive failure
and morphological aberration of fish. A lowered pH tends to neutralize
toxic metals. The accumulation of such metals in fish contaminates food
chains of which we are a part as these metals can make fish unfit for human
consumption. Acidification of a lake causes a reduction of the production
of phytoplankton (which is a primary producer) as well as in the
productivity of the growth of many other aquatic plants. In acidic
conditions, zooplankton species will probably becompletely eliminated. In
addition, bacterial decomposition of dead matter is seriously retarded in
acidified lake waters. Other effects of acidic conditions arean
overfertilization of algae and other microscopic plant lifecausing algae
blooms. Overgrowth of these consumes quickly most of the oxygen in water
thus causing other life forms to die from oxygen starvation.
When there are excessive base or acid in waters, not only do aquatic
organisms get affected but animals who depend on aquatic plants to survive
will starve too, since few aquatic plants survive in such conditions.
Therefore each organism in the aquatic ecosystem is effected by excessive
basic or acidic conditions because anything affecting one organism will
affect the food chain, sending repercussions throughout the entire
ecosystem.
v) The factors that govern this plant's location, if this plant employs
40% of the towns people:
The major factors that would govern this plant's location would be
whether there is ready access to raw materials; whether the location is
close to major transportation routes; whether energy resources are readily
available and if there is an adequate water supply in the area. Since this
plant would employ 40% of the towns people, the plant should be close to
the town while still far enough so that in case of any leakage of the
plant, the town will be within a safe distance of being severely affected.
The factor of whether the general living conditions in the area are
suitable for the workers should also be considered as well.
Additional Comments
a) 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 any
more risks. He said pollution problems could be fought under a three-stage
plan over the next decade:
1) A "toxic freeze" banning new polluters from putting
up pipes or smokestacks in the region
2) An attack on "non-point sources" of pollution, such
as run-off from streets and farms where groundwater
is loaded with pesticides.
3) A crackdown on existing polluters when their smoke
and 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, it
may cost as much as $100 billion to retrieve the purity of the Great Lakes
once 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 and
many machines as suggested by the above examples combine several
approaches.
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 Joint
Commission, announced to study to determine the polluting
effects on the Great Lakes urban development and agricultural
land use, find remedies and estimate cleanup costs; Canada
and the United States signed a Great Lakes Quality Agreement.
1974: Canadians say the cleanup financed by Washington
is already running far behind the schedule
envisaged when the agreement was signed.
1978: Canada and the United States agreed to the goal
of zero discharge of pollution. 1987: the
goal made in 1978 is made again, this means both
countries agreed to work toward completely
eliminating persistent toxic pollutants, not just
the amount being discharged by industry; Mulroney
also proposed that the U.S. slash industrial
sulfide and nitrogen oxide emissions by half
before 1994.
The Canada-U.S. International Joint Commission meets every two years to
discuss 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.
Bibliography
Encyclopedias
Collier Encyclopedia, volume 3, U.S.A.: MacMillan
Educational 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.
Articles
Cleaning 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 and
Mail, Oct. 16, 1989.
"N.Y. Accused of Overlooking Pollution in Lake", Toronto
Star, 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 and
Mail, Oct. 14, 1989.
Others
Countdown 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 the
Lines, 1980.
TITRATION LABORATORY
Purpose: 1) to prepare 0.1 mol/L NaOH solution.
2) to standardize the NaOH solution in part 1, using
potassium hydrogen phthalate.
3) to determine the unknown molarity of a H2SO4 solution using
standardized solution.
Part 1 - Prepare 0.1 mol/L NaOH solution
Observations:
Data:
mass of NaOH + paper tray = 4.58 g
mass of paper tray = 3.46 g
mass of NaOH pellets = 1.12 g
Calculation:
Number of mole of NaOH = mass of NaOH pellets = 1.12g = 0.028mol g.
mol mass of NaOH 40g
Conclusion:
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, using
potassium hydrogen phthlate
Observations:
Data:
mass of vial + KpH = 22.19g
mass of vial + KpH after
transfer to 1st flask = 22.19g
mass of vial + KpH after
transfer to 2nd flask = 21.93g
mass of vial + KpH after
кФФФФФФФФФФТФФФФФФФФФФФФФФФФФФФТФФФФФФФФФФФФФФФФФФТФФФФФФФФФФФФФФФФФФФП
Г flask Г mass of KpH Г volume of NaOH Г conc. of NaOH Г
УФФФФФФФФФФХФФФФФФФФФФФФФФФФФФФХФФФФФФФФФФФФФФФФФФХФФФФФФФФФФФФФФФФФФФД
Г 1 Г .12 Г 1.2 mL Г 0.00071 Г
УФФФФФФФФФФХФФФФФФФФФФФФФФФФФФФХФФФФФФФФФФФФФФФФФФХФФФФФФФФФФФФФФФФФФФД
Г 2 Г .14 Г 1.5 mL Г 0.00103 Г
РФФФФФФФФФФСФФФФФФФФФФФФФФФФФФФСФФФФФФФФФФФФФФФФФФСФФФФФФФФФФФФФФФФФФФй
flask 1
To calculate the concentration of NaOH (mol/L) the number of moles of
KpH have to be calculated. No. of mol of KpH = 0.12
204g/mol
= 5.9 x 10-4 mol
The ratio of KpH to NaOH is 1:1
Therefore, the no. of NaOH = 5.9 x 10-4mol.
The equation being used is: KpH + NaOH --> KHC8H3NaO4+H2O
The following equation is used to calculate the concentration of NaOH.
c = n n=number of mol = 5.9 x 10-4mol
v v=volume = 1.2 x 10-3
c = 5.9 x 10-4mol c=concentration = ?
1.2 x 10-3L
c = 0.492 mol/L
Therefore, the NaOH solution in Flask 1 is 0.492 mol/L.
flask 2
No. of mol of KpH = 0.14
204g/mol
= 6.9 x 10-4 mol
The ratio of KpH to NaOH is 1:1
Therefore, the no. of NaOH = 6.9 x 10-4mol.
c = n n=number of mol = 6.9 x 10-4mol
v v=volume = 1.5 x 10-3
c = 6.9 x 10-4mol c=concentration = ?
1.5 x 10-3L
c = 0.46 mol/L
Therefore, 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/L
Conclusions:
Questions:
1. The equation for the neutralization of potassium hydrogen phthalate
solution with NaOH solution is:
NaOH + KHC8H4O4 --> KHC8H3O4Na + H2O
2. 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 using
standardized NaOH solution.
Observations:
ЩЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭбЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭбЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЛ
К Volume of known Г Volume of known Г Molarity of К
Кsurfuric acid sol'n(mL)Г conc. of NaOH(mL) Г sulfuric acid К
ЬЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭиЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭиЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЭЙ
К 1. 25 mL Г 4.5 mL (0.0045L) Г 3.915 x 10-6 К
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К 2. 25 mL Г 4.8 mL (0.0048L) Г 4.176 x 10-6 К
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To find the molarity of unknown sulfuric acid solution:
Equation of reaction:
2NaOH + H2SO4 --> Na2SO4 + 2H2O
General equation solve:
Ca Va = Cb Vb a = acid
na nb b = 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 L
n = ?
c = ?
#mol of H2SO4 = 2.1 x 10-3 x 1 H2SO4 2
NaOH
= 1.05 x 10-3
Solving the equation:
Ca Va = Cb Vb
na nb
Ca x 0.025 L = 0.476 mol/L x 0.0045 L
1.05 x 10-3 2.1 x 10-3 mol
Ca = 0.476 x 0.0045 x 1.05 x 10-3
2.1 x 10-3 x 0.025
Ca = 2.25 x 10-6
5.25 x 10-5
Ca = 0.0429 mol/L
Flask 2
Ca x 0.025 L = 0.476 mol/L x 0.0048 L
1.14 x 10-3 2.28 x 10-3
Ca = 0.476 x 0.0048 x 1.14 x 10-3
2.28 x 10-3 x 0.025
Ca = 0.0457 mol/L
The average molarity of H2SO4 solution = flask 1 + flask 2
2
= 0.0429 + 0.0457
2
= 0.0443 mol/L