Lab 14: Titration Lab

Last lab!

In this lab we determined the percent ionization of acetic acid in vinegar.  We filled a burette with 50 mL of NaOH (the base) that had a molarity of 0.25 M and slowly added drops to a flask with 7.6 mL vinegar (the acid) and water while stirring.  We added 5 drops of phenolphthalein to the solution as an indicator that would turn from clear to pink when the solution changed from an acidic form to a basic form.  The drops of NaOH created a momentary pink color in the solution, and once the solution reached a very light pink, we knew that it was the equivalence point between H+ ions and OH- ions.  We then recorded the volume of NaOH used by looking at the volume on the burette.

Titration setup

Analyte at equivalence point

Trial 1:

Molarity of acid: 0.86 M
Volume of acid: 7.60 mL

Molarity of base: 0.25 M
Volume of base: 26.0 mL

Trial 2:

Molarity of acid: 0.87 M
Volume of acid: 7.40 mL

Molarity of base: 0.25 mL
Volume of base: 25.7 mL

Average concentration of acetic acid: 0.87 M
[H3O+] = 10^(-pH) = 10^(-2.4) = 0.0040
Percent ionization: [H3O+] / molarity of acetic acid x 100 = 0.0040 / 0.87 = 0.46%

The percent ionization of acetic acid in vinegar is 0.46%.  This is such a low number because acetic acid is weak, which means that not all of the ions break apart and ionize, so there are not many H+ and OH- ions in the solution.

Lab 13: Solubility (A Very Salty Lab)

Introduction:

The goal of this inquiry lab was to create and implement our own procedure to identify an unknown solid using experimental data and known solubility data.  We used a graph of the solubility curves of three solids (NaCl, NaNO3, and KNO3) to help identify.  We were only allowed to use the given solid, 10 mL of water, and other equipment such as hot plates, balances, thermometers, flasks, and stirring rods.

Important terms:

-solute: the substance that is dissolved
-solvent: the substance that the solute is dissolving in
-solubility: the measure of the amount of solute that will dissolve in a given amount of solvent

Procedure:

1.  We chose 45°C as our testing temperature because the solubility varied greatly at that temperature, so it was easier to differentiate between the solids based on the amount of sample dissolved.

2.  We measured 10 mL of water and poured that into a small beaker.  We knew that that was 10 g because 1 gram of water is equal to 1 mL of water.

3.  Next, we filled a larger beaker about halfway with water to act as the hot bath for the small beaker, and placed that on a hot plate.

4.  While that was heating up, we used the solubility graph to find the solubility in grams for each salt at 45°C.  We divided that mass by ten because the mass given on the graph is per 100g of water, and we only had 10g water.

5.  We measured out 4.8 grams of NaCl, which was one gram more than the mass on the solubility line, so we would know for sure if the substance was above the solubility line and unsaturated or saturated.

6.  When the thermometer in bath on the hot plate reached 45°C, we added that mass of substance to the small beaker and placed that into the bath (bigger beaker).  After stirring with a stirring rod for a couple of minutes, we checked to see if the substance had dissolved, which it had, meaning it was an unsaturated solution.

7.  We then moved on to the next salt, KNO3, and added 3.5 more g to the beaker to obtain 8.3 total g of substance, which was one more gram than the mass on the solubility line for KNO3.

8.  Because the substance dissolved for both NaCl and KNO3, we knew that the mystery substance was NaNO3.  11.7 grams of NaNO3 did not dissolve in the 10 g of water, making it a saturated solution.

Data:

Trial 1 (NaCl test):

Temperature - 45°C
Mass of solute - 4.8 grams
Mass of solvent (water) - 10 grams
Dissolved

Trial 2 (KNO3):

Temperature - 45°C
Mass of solute - 8.3 grams
Mass of solvent (water) - 10 grams
Dissolved

Trial 3 (NaNO3):

Temperature - 45°C
Mass of solute - 11.7 grams
Mass of solvent (water) - 10 grams
Not dissolved

Conclusion:

The mystery substance we had was NaNO3.  We found this out because in the first trial, we measured 4.8 grams of solute into 10 grams of solvent at 45°C and the solute dissolved.  In the second trial, we measured 8.3 grams of solvent into the same amount of solvent at the same temperature and the solute again dissolved.  This left NaNO3 as the identity, because the solution would be saturated if it was NaCl or KNO3.  The relationship between the solubility of a substance and the temperature is that as solubility increases, temperature also increases.  For NaNO3, the saturation point at 45°C is 38 grams, and the saturation point at 100°C is 242 grams.  A challenge I encountered during the trials was making sure the temperature was right at 45°C when we started adding and stirring the substance, or else the results would be off.

Lab 12: Alka Seltzer and the Ideal Gas Law

In today's lab, we dissolved three Alka Seltzer tablets and collected the gas (carbon dioxide) given off from the citric acid and sodium bicarbonate. We placed a balloon filled with crushed Alka Seltzer on top of a test tube containing water and turned the balloon over to start the reaction.  After the fizzing ceased, we measured the circumference of the balloon, now filled with gas, and then used the ideal gas law learned in class today to determine the mass of the gas produced.

Analysis Questions:

1.  Discuss an area in this lab where experimental error may have occurred.

Experimental error may have occurred during the transfer of Alka Seltzer to the balloon, since a little bit of powder remained in the weigh boat, and some powder was spilled.  Also, when we filled the balloon with water to measure the circumference, we may have been off and the size of the balloon might not have been exactly the same as it was filled with CO2.  Error could have also taken place when we mixed the water in the test tube with the Alka Seltzer if we didn't wait long enough for the fizzing to stop when we measured the balloon.

2.  Choose one error from above and discuss if it would make "n" the number of moles of CO2 too big or too small.

Since some Alka Seltzer spilled, it would make the number of moles of CO2 too small, because not enough of the reaction would have occurred between the Alka Seltzer and the water, and therefore not enough carbon dioxide would have been collected in the balloon.

3.  Calculate the volume of the balloon mathematically.

C = 2πr

38 = 2πr

r = 19/π

V = (4/3)π * (19/π)^3 = 926.6 cm^3= 926.6 mL

4.  Are the two volumes close?  Which do you feel is more accurate and why?

Yes, the two volumes are close.  The volume found during the experiment is more accurate than the calculated volume, because the shape of the balloon isn't exactly a sphere, so using the formula for the volume of a sphere isn't precise.  However, the volume we found during the experiment isn't 100% accurate either because of the human error possibilities listed in the previous questions.

5.  Give two differences between a real gas and an ideal gas.

A real gas molecule interacts with other particles through repulsions and attractions.  An ideal gas molecule (imaginary) reacts independently of others and there are no interactive forces.  Another difference is that when real gas molecules collide between particles or the wall of the container, there is a net loss of energy, where as in ideal gases, there is no loss of energy during collisions.

6.  Would the CO2 you collected in this lab be considered ideal?  Why or why not?

No, the CO2 would not be considered ideal, because it lost energy from particle collisions and had interactive forces between its molecules.  It also occupied space and mass.

Advanced Questions:

1.  Calculate the mass of CO2 that should be collected per tablet.

Each tablet had 1000 mg C6H8O7 and 1916 mg NaHCO3.  We used three tablets and found that the citric acid produced 2.062 g CO2, and the sodium bicarbonate produced 3.011 g CO2.  Due to citric acid being the limiting reactant, the theoretical yield is 2.062 g CO2.

2.  What percent is the percent yield for the CO2 collected in your sample?

We divided the experimental mass of CO2, 1.86 g, by the calculated mass, 2.062 g, and obtained 90.2% as our percent yield.

3.  What effect does the fact that CO2 is water soluble have on your calculated "n" value?

The fact that CO2 is water soluble may have decreased our n value, since some of the CO2 may have dissolved in the water and so was not measured in the balloon.  This is why our actual results were smaller than the theoretical yield.

Lab 11B: Calories in Food

In this lab, we indirectly measured the amount of Calories per gram of three sample food items: a pecan, cashew, and cheese puff.  To do this, we used a calorimeter made of an Erlenmeyer flask filled with water on top of a metal can (which acted as a stove).  We ignited the food item, which was underneath the metal can, and measured the change in temperature of the water in the flask above.  By doing this, we were then able to calculate the amount of energy in the food tested because the heat gained by the water equals the heat lost by the food item.

A pecan burning in the calorimeter

Data Table and Calculations:

Questions:

1.  We measured a temperature change in the water.  The water absorbed the heat given off of the burning food, so we knew that the heat gained by the water would be equal to the heat released by the food sample.  Therefore, the temperature increase in the water is also the temperature change in the sample.

2.  We measured the energy released by the food sample, which is the same as the energy gained by the water, since the energy released heated the water.

3.  The small amount of energy that was not absorbed by the water could have escaped through the holes in the metal can as smoke and been released into the surrounding air.

4.  I was expecting the nuts to have more Calories per gram than the cheese puff because of previous knowledge, but it turned out to be the other way around.  This is most likely because our cheese puff didn't burn thoroughly and the mass of food burned was too small (0.04 g compared to 1.25 g and 1.38 g of nut burned).

Lab 10: Evaporation and Intermolecular Attractions

Pre-Lab Table:

Data Table:

Questions #2-4:

2.  Explain the differences in the difference in temperature of these substances as they evaporated.  Explain your results in terms of intermolecular forces.

The differences in temperature change of the five substances varied due to the difference in the strengths of intermolecular forces.  Methanol had the highest change and highest vapor pressure because it had the weakest intermolecular force, so it evaporated and cooled faster.  On the other hand, glycerin had the smallest temperature change (the initial temperature actually increased) and the slowest evaporation rate because its intermolecular force was the strongest --- it had three -OH groups (three hydrogen bonds).

3.  Explain the difference in evaporation of any two compounds that have similar molar masses.  Explain in terms of intermolecular forces.

Two compounds that had similar molar masses were methanol (32.042 g) and ethanol (46.068 g).  The change in temperature for methanol (15.9°C) was higher than that of ethanol (10.1°C).  This is because even though they both have one hydrogen bond, the molar mass of methanol is smaller, so the bond is weaker.  A weaker bond causes faster evaporation and also lower vapor pressure.  Ethanol's molar mass is larger, so its bond is stronger, causing slower evaporation and thus lower temperature change.

4.  Explain how the number of -OH groups in the substances tested affects the ability of the tested compounds to evaporate.  Explain in terms of intermolecular forces.

Methanol and ethanol both have one -OH group (hydrogen bond), but because methanol's molar mass is smaller, it's temperature change is greater and its evaporation rate is quicker.  Ethanol had the second highest temperature change.  Water has two hydrogen bonds, but it has the smallest molar mass, so it had the third quickest evaporation rate and third highest temperature change.  n-Butanol has one hydrogen bond, but because its molar mass is so high, the bonds are stronger and thus the temperature change is smaller and the evaporation rate is slower.  Lastly, glycerin has three -OH groups, or the strongest intermolecular force, causing an increase in temperature (smallest difference in temperature) and the slowest evaporation rate.  This is because the rate glycerin was absorbing heat from the condensation in the room was faster than the evaporation rate.