Report Sheet
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Table 2. Color Observations for Control Solutions.
0.01 M
NaOH
0.001 M
NaOH
0.0001 M
NaOH
0.00001 M
NaOH
Light green
Time: 0:00
Violet
Violet
Reddish-violet
Time: 0:30
Violet
Violet
Reddish-violet
Time: 1:00
Violet
Violet
Reddish-violet
Time: 1:30
Violet
Violet
Reddish-violet
Time: 2:00
Violet
Violet
Blue
Time: 2:30
Violet
Violet
Blue
Time: 3:00
Violet
Reddish-violet
Light green
Time: 3:30
Violet
Reddish-violet
Green
Time: 4:00
Violet
Reddish-violet
Time: 4:30
Violet
Reddish-violet
Yellow
Yellow-orange
Time: 5:00
Violet
Reddish-violet
Yellow
Yellow-orange
Time to
Green
(if applicable)
Not applicable
Not applicable
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Yellow
3:20
Light green
Light green
Light green
Light green
Green
Green
Green-yellow
Yellow
2:30
Table 3. Color Observations for Test Solutions.
0.01 M
NaOH
Time: 0:00
Violet
0.001 M
NaOH
0.0001 M
NaOH
0.00001 M
NaOH
Violet
Reddish-violet
Light green
Violet
Yellow
Orange
Time: 0:30
Violet
Time: 1:00
Violet
Reddish-Violet
Yellow
Orange-red
Time: 1:30
Violet
Reddish-Violet
Yellow
Orange-red
Yellow
Orange-red
Time: 2:00
Reddish-violet
Blue
Time: 2:30
Reddish-violet
Light green
Yellow-orange
Orange-red
Time: 3:00
Reddish-violet
Green
Yellow-orange
Orange-red
Time: 3:30
Reddish-violet
Green
Yellow-orange
Orange-red
Time: 4:00
Reddish-violet
Green
Yellow-orange
Orange-red
Time: 4:30
Reddish-violet
Green
Yellow-orange
Orange-red
Time: 5:00
Reddish-violet
Yellow-orange
Orange-red
Time to
Green
Not applicable
Green-yellow
0:05
2:15
0:01
1. The lab experiment instructs you to react 0.21g of NaHCO3 with excess CH3COOH. How
much CO2(g) in mL would this reaction generate if all the sodium bicarbonate reacts fully?
(Assume the room temperature is 25 °C.)
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2. Now, look at the color changes that occurred in your wells.
a. What trends do you notice? Think about the changes you see in one well over time,
and then think about the differences you see between wells.
All of the wells in the test group underwent a color change. The wells in the control group also experienced color changes at a slower rate because
parafilm is semi-permeable to gases. Both groups experienced similar trends of degree and rate of color change. The degree in which a color
change occurred depends on the initial concentration of NaOH. The wells with less concentration of NaOH underwent more color changes and
thus experienced a greater change in pH. The rates at which the color change initially occurred is inversely proportional to the concentration of the
NaOH. The wells with the lower molarity of NaOH underwent color changes more rapidly before remaining one color for a longer period of time.
The higher the molarity of NaOH, the less of a color change and the longer it took for a color change to occur.
b. What is happening at the molecular level to explain your observations? In other
words, what’s a scientific model that’s representative of your data? (It’s okay if this is
very general.)
On a molecular level, as time passes the concentration of CO2 in the wells is increasing to reach equilibrium as per Henry’s law. The CO2 that
dissolves into the water then reacts with the water to produce carbonic acid. Carbonic acid then dissociates in the water to produce H+ and
HCO3-. This can then further dissociate to produce another H+ and CO32-. However, this second dissociation occurs to a smaller degree because,
in accordance with the common ion effect, the H+ concentration in the water from the previous dissociation pushes the equilibrium position of
the second dissociation further towards remaining HCO3-. This dissociation of carbonic acid increases the concentration of H+ in the solution
and thus makes the solution more acidic.
The produced H+ then reacts with the NaOH in the water. The NaOH has dissociated into Na+ and OH-, the OH- making the solution basic. The
OH- reacts with the produced H+ to form water. This neutralizes the acidic H+ and basic OH-. Therefore as more H+ is produced by the CO2
entering solution and reacting with water, more of the basic NaOH is neutralized and the pH becomes more neutral. As carbonic acid continues
to form, the solution becomes neutral and then starts to become acidic. Therefore, as time passed, the concentration of H+ increased and the
concentration of OH- decreased, causing the solution to become more acidic.
The presence of NaOH does not affect the amount of CO2 that enters solution nor the equilibrium position of the reaction with water or
dissociation of carbonic acid. Hence, an equal concentration of H+ is produced in all four wells. Thus, the starting molarity of the NaOH affects
how many moles of OH- that the H+ from the CO2 must react with and neutralize before the solution undergoes a change is pH, to become
neutral, and how acidic the solution will become after five minutes.
3. Explain why the different concentrations of NaOH in each of the wells takes a different
amount of time to react with the CO2.
Different concentrations of NaOH do not affect the equilibrium position of the CO2 (aq), but as the molarity of NaOH increases the
reaction rate decreases. This is because as the concentration of NaOH increases, the availability of water to react with the CO2
decreases. In the dissociation of NaOH, water molecules surround the Na+ and OH- ions. These ion-dipole attractions make it
harder for the CO2 to interact and react with the water, and there are less lone water molecules available to react with CO2. The
NaOH does not react with the CO2, it reacts with the carbonic acid produced when the CO2 reacts with water. Therefore, in order
for the NaOH to 'react with' the CO2, the CO2 must first enter aqueous solution and react with water, and the presence of NaOH
dissociated in the water decreases reaction rate of CO2 with H2O.
4. Draw the Lewis structures of CO2, H2CO3, HCO3− and CO32−. Rank these in order of
increasing attraction to water molecules. Explain your choice. What evidence do you have
that supports your predictions?
Because water is polar the partial charges of water are attracted to the partial charges of other compounds, therefore a compound's
attraction to the water molecules depends on the polarity of the compound. Because water molecules have hydrogen bonding, they
are most attracted to the H2CO3, which has 2 hydrogen bonds that can attract water. HCO3- is less attractive to water because it
only has one hydrogen bond. The CO2 and CO3- are both much less attracted to water because they have neither hydrogen
bonding nor are they polar. However, CO3- is more attracted to water because it has strong London dispersion force attraction than
CO2 because CO3- has more electrons and so it is more likely to form a temporary dipole, which is attractive to water, than CO2.
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How can real-time analysis be used to reduce pollution?
Two-thirds of all SOx produced come from electric power plants. Many power plants use coal as
a fuel source. Coal is not a pure carbon – when it is burned the sulfur in it combines with oxygen
to produce SO2. As you’ve learned this week, this emitted SO2 can eventually produce acid rain.
However, SO2 can be removed or ‘scrubbed’ from the exhaust of coal power plants by spraying a
wet slurry of limestone (calcium carbonate, CaCO3) into a large chamber that contains the SO2
exhaust:
2CaCO3(s) + 2SO2(g) + O2(g) → 2CaSO4(s) + 2CO2(g)
One of the byproducts of this reaction, calcium sulfate (CaSO4), can be used to make wallboard
and cement and has a role in agricultural and construction applications.3
5. The pH meter in the scrubber for a coal-burning electric power plant records a pH drop for the
calcium carbonate slurry from 10 to 8. The pH meter in the scrubber for a methane-burning
power plant records a drop from 10 to 9.8. Which plant (coal or methane) produced more SO2
exhaust? Explain your reasoning.
Because both plants began with a pH of 10, it can be concluded that the pH of the calcium carbonate slurry by itself has a
pH of 10. The reaction of the slurry calcium carbonate with SO2 forms CO2, which can then react with the water of the wet
slurry to form carbonic acid and lower the pH. Thus, the pH of the slurry in the coal plant experienced a greater change
because it reacted with more SO2 exhaust and the methane plant experienced a smaller change in pH because there was less
SO2. Therefore, the coal plant produced more SO2 exhaust than the methane plant.
6. When used, limestone scrubbers prevent the release of approximately 95% of the SO2 produced
from power plants. However, this reaction produces other byproducts. What is one potential
disadvantage of this particular type of scrubber? Explain your choice.
One potential disadvantage of this particular type of scrubber is that it produces CO2. Carbon dioxide is
another greenhouse gas that can cause acid rain. Therefore, while the problem of SO2 acid rain is
greatly diminished, the threat of carbonic acid rain is increased. While some of the CO2 reacts with the
water in the slurry, the remaining gas enters the atmosphere where it can react with water in the
atmosphere, acidify the rain, and rain down to acidify the ocean and damage organisms.
7. Scrubbers are a way to remediate (remove) pollution after it has already been formed. Is that
consistent with green chemistry? Why or why not?
Scrubbers are not consistent with green chemistry. The first principle of green chemistry states that
prevention is better than trying to clean up the waste after it has been created. Five percent of the
original emissions is much better, but as energy needs continue to rise five percent can still result in
millions of tons of harmful gases entering the atmosphere. Additionally, scrubbing is not energy
efficient as it is an extra process that must be carefully monitored and requires power to complete,
and it also does not promote renewable feedstocks because it encourgages the continued use of fossil
fuels.
3
Duke Energy, https://www.duke-energy.com/our-company/environment/air-quality/sulfur-dioxide-scrubbers
(accessed July 19, 2018)
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