Lab 14: Titration Lab

 Summary: In this lab we performed a titration in which we found the molarity of an acid by the amount of base that it took to neutralize the acid. It was a very frustrating lab since a little too much base would cause the experiment of completely go awry. It was an interesting lab and I liked the stirring thing.

Why is the percent ionization such a low number?
The percent ionization is such a low number because acetic acid is a weak acid and therefore does not fully ionize. Therefore most of the H+ protons stay connected and do not bond with water to form hydronium. The percent ionization was 0.46%.

Data

	Trial 1	Trial 2
Volume of NaOh used	24.4 mL	23.7 mL
Volume of of Acid used	7 mL	7 mL
pH of Acid	2.4	2.4
Molarity of Acid	0.871 M	0.846 M
Molarity of Base	0.250 M	0.250 M

Lab 13: Solubility

Introduction: Solubility is the property of one substance dissolving in another. In this lab we were given a solute and were asked to figure out what the solute was. This was based on the solubility of the salt in 10 mL of water at 50 degrees Celsius. By dissolving a certain amount of solute in the water and looking if there is any solute left undissolved we can figure out what solute it based on the graph given. Our solute was Potassium Nitrate.

Procedure:

1. Measure out 13 grams of solute between three weigh boats in 5 grams, 5 grams, and 3 grams.
2. Measure out 10 mL of distilled water into a graduated cylinder and then pour into a beaker
3. Plug in hot plate and turn it on
4. Fill up a large beaker with water and place on hot plate, check the temperature with a thermometer until the water reaches of 50 degrees Celsius.
5. Place small beaker in large beaker filled with water and heat indirectly until small beaker has a temperature of 50 degrees Celsius (take care that no water from the large beaker enters the small beaker)
6. Pour 5 grams of solute in and stir with glass stirrer keeping the water at 50 degrees Celsius (this may require taking off the heat, just check the temperature with the thermometer every so often)
7. If the solute dissolves repeat step 6 pouring in the other 5 grams of solute
8. If solute dissolves repeat step 6 again pouring in the final 3 grams

Observations (Qualitative Data):  I observed that when the first amount of solute was poured in it started as cloudy but then the solution became clear and no particles of solute were left at the bottom. When the second amount of solute was added it remained cloudy when stirred, and when the solution settled there was particles of solute at the bottom of the beaker. That meant that the solution was saturated.

Quantitative Data:

Grams of Water	10 grams
Grams of Solute (Trial 1)	5 grams
Grams of Solute (Trial 2)	10 grams
Grams of Solute (Trial 3)	13 grams
Temperature of Water	50 °C
Trial in which Water was Saturated	Trial 2

Conclusion: Since the water was saturated with 10 grams of solute of at 50 °C that means according to the graph we were given the solute was Potassium Nitrate. We knew that the solution was saturated with 10 g of solute at 50 °C because there were particle of the solute at the bottom of the beaker, meaning it had not dissolved all the way making it saturated. That differed from when 5 g of solute were dissolved in the solution at 50 °C because then the solution had now particles floating around and was completely clear, so the solute was dissolved. So, the solute could not be Sodium Chloride because sodium Chloride would have saturated the solution at 5 g and it could not be Sodium Nitrate because the solute would have dissolved all the way if there was 10 g of solute and the solute was Sodium Nitrate. This means that the solute could only be Potassium Nitrate. Certain amounts of solute dissolve in solutions at certain temperatures because the transfer of energy from breaks the intermolecular forces, and the strength of the forces depend on the molecule. For Ions the more heat the more solute can be dissolved for gases the opposite is true.

Lab 12: Alka Seltzer and the Ideal Gas Law

Summary: Today in class we learned about gasses and how the move and act in certain environments. In this experiment we found the data needed to find the ideal amount of gas in the balloon and the actual gas. This showed the ideal gas law is ideal in that all gas occupy's a certain area completely.

Analysis Questions:
1. Some mistakes that could have occurred are not having an accurate volume of how much space the carbon dioxide took up, as well as not all carbon dioxide going into the balloon.

2. The amount of carbon dioxide in the water could affect the amount of atoms of carbon dioxide and therefore the amount of moles or n making n too small.

3. The volume of the balloon is 1072.67 mL.

4. The difference between the volume obtained by filling the balloon minus the answer from #3 is 6.33 mL. Which is fairly close, however I think that the volume obtained by filling the balloon was more accurate because the balloon is not a perfect sphere, so measuring it by by filling it with water you get a more accurate volume because the water takes up the actual space inside the balloon.

5. The difference between ideal gas and real gas is ideal gas occupies one space alone, and none of it would dissolve in water or somehow be released, but real gas would.

6. No, because not all of the gas was present in the balloon when we found out the grams in the balloon.

Advanced Questions:
1. The amount of gas that should be collected in total from 3 tablets of Alka Seltzer is 2.06 grams of carbon dioxide.

2. The percent yield of the sample of carbon dioxide collected was 93.7%.

3. Some of the carbon dioxide would have dissolved in the water making the actual amount of moles slightly less than the theoretical amount of moles.

Data

Mass of Alka Seltzer Powder	9.14 grams
Circumference of Balloon	39.9 cm
Volume of Water that fits in Balloon	1079 mL
Room Temperature	20 °C
Barometric Pressure	759.2 mmHg
CO2   Pressure	741.2 mmHg
Moles of CO2 in Balloon	0.0437 moles of CO2
Mass of CO2	1.93 grams

Lab 11B: Calories in Food Lab

Summary: The purpose of this the lab was to use what we used about energy in class today to measure the calories in food through the energy they transferred in the form of heat into water.

Lab 10: Evaporation and Intermolecular Attractions

Data

Substance	Maximum or Initial Temp (°C)	Minimum or Final Temp. (°C)	∆ Temp. (°C) (- and + indicates whether temps. increased or decreased)
Methanol	18.9°C	3.2°C	-15.7°C
Ethanol	19.6°C	10.6°C	-9.0°C
n-Butanol	20.3°C	18.2°C	-2.1°C
Glycerin	20.3°C	23.3°C	+3.0°C
Water	19.9°C	14.8°C	-5.1

Pre-Lab Questions:

2. The substances that we tested had a vast range of the difference in temperature, in order from the substances that had the biggest decrease in temperature to the ones which had the smallest decrease in temperature it goes Methanol with a decrease of 15.7°C,  Ethanol with a decrease of 9.0°C, water with a decrease of 5.1°C, n-Butanol with a decrease of  2.1°C, and Glycerin which actually gained 3.0°C. The temperature increase or decrease were due to intermolecular forces. Each substance posses hydrogen bonds, which are the attraction between the partially positive positive hydrogen and partially negative atoms, which in this case is the oxygen in all these substances. More hydrogen bonds the more energy it takes to evaporate, and a less acute drop in temperature. So, the substance with the most hydrogen bonds was Glycerin with 3 potential places for hydrogen bonds per molecule, and Glycerin was the only substance to increase in temperature during the lab.

3. Hydrogen bonds only explain part of the difference in temperature seeing as three substances, Methanol, Ethanol, and n-Butanol, had the same amount of potential places for hydrogen bonds and yet had differences in the decreases in temperature. This is due to London Dispersion Forces. London Dispersion Forces create a temporary poles on molecules called a momentary dipole-dipole. This is due to the fact that electrons move around the nucleus and when more electrons move to one side that side becomes more negative and the other becomes more positive, this also forces the molecules around the molecule with the momentary dipole-dipole also to become momentary dipole-dipole. It takes more energy (and therefore a smaller decrease in temperature during evaporation) in molecules that have more molar mass. This is due to the fact more molar mass means more electrons, which mean a stronger momentary dipole-dipole. That explain why n-Butanol, which has the biggest mass had the smallest decrease in temperature.

4. The number of -OH’s correlated to evaporation because of that is where hydrogen bonds will form. This is expected because hydrogen bonds usually form between atoms with higher electronegativity such as oxygen, which is often partially negative in many molecules, and hydrogen. 1 -OH means one potential hydrogen bond could be formed, because there is one partially negative atom and one hydrogen which will for the hydrogen bond. This is why water has two places for bonding because it has two hydrogens which will form bonds, and more bonds means more energy needed for evaporation, and a less severe temperature decrease.