Chemistry 113 Lab Reports
Chemistry 113 Lab Reports CHM 113
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This 9 page Bundle was uploaded by Pooja Dave on Thursday September 29, 2016. The Bundle belongs to CHM 113 at University of Miami taught by Dr. Tegan Eve in Fall 2015. Since its upload, it has received 4 views. For similar materials see General Chemistry 1 lab in Chemistry at University of Miami.
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Date Created: 09/29/16
Pooja Dave CHM 113, Section FY Conductimetric Titration and Gravimetric Determination Lab Report Introduction: Gravimetric analysis is used in determining the concentration of an unknown substance in a compound. It is one of the most reliable procedures in calculating the amount of an element in a sample of unknown concentration. There are five requirements for conducting this analysis: the compound must be pure and have a known stoichiometry; the precipitation reaction must be complete; the precipitation reagent should be specific and uncontaminated by other substances; the solid formed should be big enough so that it may be easily filtered; the molecular weight of solid should be high enough so that the precipitate is heavy enough to be weighed. In addition to gravimetric determination, conductimetric titration can be used in determining the concentration of an unknown substance. In this, the conductivity of a solution containing ions is monitored via a Conductivity Probe that sends out an electrical current. This then measures the drop in ion level as acid is added to the base, neutralizing the solution and hitting the equivalence point. From there the equivalence point can be used to find the concentration of the unknown solution that was titrated. Procedure: 1. Obtain 10mL of Ba(OH)₂ in a 100mL beaker. Add 30mL of distilled water in 10mL increments. 2. Calibrate the Drop Counter using the LabPro on the Vernier computer program. 3. Measure out approximately 60mL of 0.1 M H₂SO₄ into a 250mL beaker and pour into the 60mL reagent reservoir, draining a small amount so that the tip fills with H₂SO₄. 4. Place the beaker containing Ba(OH)₂ underneath the Drop Counter, lining up the center of the beaker with the Drop Counter slot and the tip of the reagent reservoir. Place magnetic stir bar in the beaker and the Conductivity probe through the slot so that it just touches the bottom of the beaker. 5. Begin titration and data collection. Continue until the graph shows an uptick in conductivity. 6. Examine graph and find the equivalence point, which is the lowest point on the graph. 7. Place beaker containing titrated solution on a hot plate and bring to a boil. 8. Weigh the cooled filter crucible and place on a vacuum filtration apparatus, attaching it with a hose to the vacuum. 9. Remove heated beaker from the hot plate once it reaches a boil and allow it to cool close to room temperature. 10.Spray about a millimeter of methanol into the crucible before pouring the solution into the crucible, being careful not to let the crucible overflow. Allow the filtration to sit until the solution is completely filtered. 11.Allow the crucible and filter paper to dry. At the end of the drying time, weigh the crucible with the barium sulfate and calculate the weight of the barium sulfate. Equations: Ba(OH)₂ (aq) + H₂SO₄ (aq) → BaSO₄ (s) + 2H₂O (l) moles of Ba(OH)₂ = moles of H₂SO₄ = moles of BaSO₄ moles of H₂SO₄ = (equiv point volume (L)) * (molarity of H₂SO₄ (M)) concentration of Ba(OH)₂ = (moles of Ba(OH)₂) / (volume of Ba(OH)₂ used) mass of BaSO₄ = mass of BaSO₄ and crucible – mass of crucible moles of BaSO₄ = mass of BaSO₄ / molecular weight of BaSO₄ Observations: During our experiment, the greatest error was in using the LabPro computer interface. There were cracks in the graph and moments when the conductivity probe was not receiving data. There was also difficulty in calibrating the drop counter, as there would be some drops that were not being counted. Data: Equivalence Point from Titration – 21.8mL Molarity of H₂SO₄ solution – 0.100 M Mass of Ba(OH)₂ from Titration - 0.00218 mol Volume of original Ba(OH)₂ - 40.0mL Molarity of Ba(OH)₂ - 0.055 M Mass of crucible and precipitate – 17.69 g Mass of crucible – 17.29 g Mass of precipitate – 0.397 g Concentration of Ba(OH)₂ from gravimetric – 0.058M Calculations: xmolesof H ₂SO₄ 0.1M of H₂SO₄ = .02182L Moles of H₂SO₄ = .002182 mol × 1molBa(OH)₂ .002182 mol H₂SO₄ 1mol H₂SO₄ = .002182 mol Ba(OH)₂ .002182molBa(OH)₂ Molarity of Ba(OH)₂ = = 0.055M from Titration .040LBa(OH)₂ Mass of precipitate = 17.689 g – 17.292 g = 0.397 g 1mol BaSO₄ × 0.397 g BaSO₄ 233.43gBaSO₄ = 0.0017 mol BaSO₄ × 1molBa(OH)₂ 0.0017 mol BaSO₄ 1molBaSO₄ = 0.0017 mol Ba(OH)₂ .0017mol Ba(OH)₂ Molarity of Ba(OH)₂ = .040L Ba(OH)₂ = 0.058 M from gravimetric Discussion and Conclusion: The equivalence point obtained from titration shows when the barium hydroxide solution was neutralized by the sulfuric acid. Collecting this data allows for the calculation of the moles of barium hydroxide, as the moles of acid equals the moles of the base at this point. With the known molarity of the sulfuric acid and the known stoichiometry, the concentration of barium hydroxide can be calculated. Then using gravimetric determination, the precipitate can be filtered and weighed, again being used to calculate the concentration of barium hydroxide as the moles of the precipitate equals the moles of barium hydroxide. Our data supports the idea that gravimetric determination is more reliable than conductivity titration. Majority of the errors in the experiment were done during the titration, making the data obtained from the titration more unreliable than the data obtained from the gravimetric determination. The molarity of barium hydroxide from the titration was 0.055 M while the molarity from gravimetric was 0.058M. The molarity is higher and more reliable from the gravimetric determination, making the actual molarity of barium hydroxide closer to 0.058M than 0.055M. Errors: The calibration of the drop counter may have been inaccurate. This would then affect the equivalence point and thus the molarity of the barium hydroxide, making it lower than it actually is. The collection of conductivity from the probe was not smooth, so some data may have been lost. This would have altered the equivalence point, making it and the concentration of barium hydroxide lower. All of the precipitate may not have been filtered into the crucible, making the mass of the precipitate and the concentration of barium hydroxide lower than it actually is. Some of the precipitate may have been too small to be caught by the filter paper, losing some mass in the precipitate. This would then have made the concentration of barium hydroxide lower. Pooja Dave CHM 113, Section FY Determination of Ka: Titration of a Weak Acid Introduction: The Ka value, the acid dissociation constant, represents the strength of an acid. The lower the Ka, the weaker the acid. There are two methods of finding the Ka value, both involving the usage of pH to find pKa. The first method is to measure the pH of a solution containing a known concentration of a weak acid. From there, the pKa may be calculated using the Henderson-Hasselbach equation. The second method is to measure the pH at half-neutralization point in titration of a weak acid with a strong base. Because the pH at half-neutralization point is equal to the pKa, the Ka can then be easily calculated. Titration of the acid by the strong base allows the unknown concentration of the weak acid and the pH to be measured for both methods. The equivalence point is found by looking at the point on the graph where the slope changes from increasing to decreasing. This means that it is also the inflection point so that first and second derivatives can be used in its calculation. In this experiment, we use acetic acid (HC₂H₃O₂) to show how the Ka value and pH are related. Procedure: 1. Obtain 20mL of weak acid solution in a 100mL beaker 2. Begin calibrating the drop counter using distilled water and LoggerPro 3. Fill the reagent reservoir with about 60mL NaOH, letting a little bit out so tip is filled with NaOH 4. Place the 20mL of weak acid under the drop counter 5. Insert the pH sensor and magnetic stirring bar inside the beaker and begin data collection 6. Continue data collection until the graph shows a large increase in pH 7. Changing the y axis to show second derivative, find the highest point on the graph to find the pH at the equivalence point 8. Divide the volume added at equivalence point by half to find the pH at half- neutralization point Equations: HC₂H₃O₂ + NaOH -> HOH + NaC₂H₃O₂ (Molarity of base) * (volume at equivalence point) = (molarity of weak acid) * (volume of acid originally) [ A⁻] pH = pKa + log ( [HA] ) pH½ = pKa pKa = -log(Ka) Observations: During our experiment, the greatest error was in calibrating the drop counter. Some drops were not counted properly by the LabPro computer interface, making the calibration of the counter inaccurate for recording the data. Data: Volume at Equivalence Point from Titration – 38.44mL Molarity of NaOH – 0.100M Molarity of Weak Acid originally - 0.1922M Volume at half-neutralization point – 19.25mL pH at Equivalence Point – 9.90 pH at half-neutralization point – 5.15 Ka from pH at Equivalence Point – 6.55 ×10−11 Ka from pH at half-neutralization point – 7.08 ×10 −6 Calculations: .1M )×(38.44mL )=[HA ]×(20mL ) [HA] original = 0.1922M 0.1M 0.1922M (¿) 9.90=pKa+log¿ pKa = 10.182 10.182 = -log(Ka) ×10 −11 Ka from equivalence point = 6.55 5.15 = pKa 5.15 = -log(Ka) Ka from half-neutralization point = 7.08 ×10 −6 Discussion and Conclusions: The equivalence point obtained from titration shows when the weak acid is neutralized by sodium hydroxide base, or the inflection point of the curve. On the graph created by collecting data, the equivalence point is contained in the nearly vertical portion. The point is half-way up the linear portion of the titration curve and can be calculated using the derivatives. The first derivative peaks at the equivalence point, while the second derivative passes through zero at this point. The volume and pH at this point can be used in determining the original concentration of the weak acid along with the Ka value by the Henderson- Hasselbach equation. Our lab shows that the equivalence point is at 38.44mL with a pH of 9.90. Another method is to use the half-neutralization point in calculating the Ka value. The half-neutralization point is halfway to the equivalence point. In our lab, this point was at approximately 19.25mL with a pH of 5.15. The Ka value at the equivalence point is 6.55e-11 while at the half-neutralization point it is 7.08e-6. The difference between the two values illustrates the reliability of the half-neutralization point method, as the actual Ka value for acetic acid is 1.77e-5. Errors: The calibration of the drop counter may have been inaccurate. This would affect the equivalence point and the concentration of weak acid and Ka, making them lower. The calculation of the half-neutralization point may have been wrong. Dividing the equivalence point by half may have given us a larger pH and a higher Ka value than in reality. The collection by the pH sensor may not have been accurate due to some interference with the sensor and the distilled water solution it was placed in. This would alter the data collected, making the pH level higher than it actually was. More NaOH may have entered the beaker than was counted, making the pH higher than it actually was. This then would have made the Ka higher. Graph: Pooja Dave CHM 113, Section FY Freezing Point Depression with Lauric Acid Introduction: The freezing point depression is used in determining the temperature at which the system freezes. The molecules within a substance get closer together as it solidifies, the exception being water as its molecules are farther away when solid. The addition of another substance creates an impurity and can alter these molecular interactions. Freezing point lowers as solutes are added. This freezing point is found within the flat portion of the graph Time vs. Temperature in Celcius. The flat portion represents the heat of fusion during the cooling process. The temperature does not change and the molecules are being packed tightly into a solid. The freezing point of Lauric acid changes with the addition of benzoic acid, as the molecules cannot be packed as tightly into the solid state. However, it can be estimated by finding the average freezing point constant using molality and the change in temperature. Molality is used because it does not change with volume which is temperature dependent, unlike molarity. The freezing point constant for both solution is averaged together to get an approximate value for pure Lauric acid. The differences in freezing point constant for the two solutions also show how different amounts of solute can lead to different freezing points. Procedure: 1. Set up LoggerPro, labeling the x and y axis Time in second and Temperature in Celcius and connecting the temperature probe to the data collector 2. Fill a 400mL beaker with water and heat up using a hot plate until approximately 60 to 80⁰C 3. Obtain a test tube containing a solution of 0.75 g of Benzoic Acid with 8.00 g of Lauric Acid 4. Place the test tube in water bath and melt the mixture 5. Once fully melted take it out of hot water bath and add the temperature probe, hitting collect 6. Place test tube in the room temperature water bath and stir as data has started collecting 7. Determine freezing point from the graph, looking for the point when the graph straightens out 8. Repeat steps 3 through 8 but with a test tube containing the solution with 8.00 g Lauric Acid with 1.50 g Benzoic Acid Equations: ΔT = K f molessolute m = masssolvent(kg) Observations: During our experiment, the greatest error was in the data collection. The beginning of the graph was not smooth, making freezing point harder to find. There was also a gap in time from when the test tube left the water bath to the time data started collecting. Data: Molecular Weight of Benzoic Acid = 122.12 g/mol Melting point of pure Lauric Acid = 43.2⁰C Moles of 0.75 g BA = 0.00614 mol Molality of 0.75g BA/8.00g LA = 0.768 mol/kg Change in Temperature = 3.2 ⁰C Kfof 0.75g BA/8.00g LA = 4.17 Moles of 1.5g BA = 0.0122 mol Molality of 1.5g BA/8.00g LA = 1.54 mol/kg Change in Temperature = 6.1 ⁰C Kfof 1.5g BA/8.00g LA = 3.84 Calculations: 0.75g BA/8.00g LA: 1mol × Moles of BA = 0.75g 122.12g = 0.00614 mol 0.00614mol Molality (m) = 0.008kg = 0.768 mol/kg ΔT = 40 – 43.2 = 3.2 ⁰C 3.2 ⁰C = (0.768) K f K = 4.17 f 1.50g BA/8.00g LA: × 1mol Moles of BA = 1.50g 122.12g = 0.0122 mol 0.0122mol Molality (m) = = 1.54 m 0.008kg ΔT = 37.1 – 43.2 = 6.1 ⁰C 6.1 ⁰C = (1.54) K f K f 3.84 Average K = f.003 Discussion and Conclusion: The average K value obtained from this experiment shows the value when f Lauric acid freezes. It represents the average freezing point constant for Lauric acid. This freezing point constant is the moment when the molecules of Lauric acid form close bonds, making it a solid. With the known melting point of Lauric acid and the molecular weight of Benzoic acid, the moles and change in temperature can be calculated. These two pieces of data can be applied to the equation to solve for freezing point constant. The differences between the solution with 0.75g Benzoic acid and the one with 1.50g Benzoic acid show how the K value chafges with more solute. The solution with 1.50g of BA had a lower K value fhan the one with 0.75g of BA. Because the K valfe is lower, the freezing point is lower meaning that it takes more energy for the solution to freeze. Our data then supports the idea that the one with the most solute interferes the most with the molecular arrangement during solid formation. Errors: The temperature probe was first placed in the heat bath to test its temperature. This may have interfered with the collecting of data when testing the temperature as Lauric acid freezes. The initial temperature then would have been higher than in reality, and the warm temperature probe may have delayed the freezing of Lauric acid. The time gap between taking the test tube out of the heat bath and collecting data may have cause data to be missed. This would then have altered the freezing point, making the initial temperature lower than it actually was. Some of the Lauric acid may not have completely melted when we took the test tube out of the water bath. This would cause the freezing point to be higher than what it actually was. Stirring the warm test tube in room temperature water bath may have disrupted the cooling of Lauric acid, as there were inconsistencies in the stirring. This would have delayed the freezing, making the freezing point lower than it actually is. Graph:
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