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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
>>>Chop Here>>>>
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 > 2SO3
    SO3 + H2O > 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 > 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 > 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

Ŀ
  flask        mass of KpH     volume of NaOH    conc. of NaOH    
Ĵ
   1             .12             1.2 mL             0.00071      
Ĵ
   2             .14             1.5 mL             0.00103      




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:

ͻ
 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  
Ķ
 2.    25 mL             4.8 mL (0.0048L)     4.176 x 10-6  
ͼ 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
