Experiment 9:10 - Lab Report PDF

Preview

Summary

Lab Report...

Description

Experiment 9/10: Volumetric/Vinegar Analysis

Hypothesis If a solution of sodium hydroxide is standardized, the molar concentration of the base can be found. By finding the molar concentration of the base, the moles of a strong acid, such as vinegar, can be determined. This will allow for the determination of the % composition of the acid.

Materials and Methods Please refer to Experiments 9 and 10 on pages 135-148 of Laboratory Manual for Principles of General Chemistry by J. A. Beran (10th Ed.). Deviations from the published procedure are: 1. The same molar concentration from Part A, was used in Part B.

Data

Calculations A.3. Moles of KHC8H4O4 (mol) = [Tared mass of KHC8H4O4 (g)] / [Molar Mass of KHC8H4O4 (g/mol)] 0.2003 g / 204.23 g/mol = 0.000981 mol A.6. Volume of NaOH dispensed (mL) = [Buret reading of NaOH, final (mL)] – [Buret reading of NaOH, initial (mL)] 21.85 mL – 0.10 mL = 21.75 mL A.7. Moles of NaOH = Moles of KHC8H4O4 (mol) * (1 mol NaOH/1 mol KHC8H4O4) 0.000981 mol

A.8. Molar concentration of NaOH (mol/L) = [Moles of NaOH] / [Volume of NaOH dispensed (L)] 0.000981 mol / 0.02175 L = 0.0451 M A.9. Average molar concentration of NaOH (mol/L0 = [Molar concentration Trial 1 + Molar concentration Trial 2] / 2 (0.0451 M + 0.0376) / 2 = 0.0414 M

B.4. Mass of unknown vinegar (g) = [Mass of flask + unknown vinegar (g)] – [Mass of flask (g)] 91.8373 g - 86.9928 g = 4.8445 g B.7. Volume of NaOH dispensed (mL) = [Buret reading of NaOH, final (mL)] – [Buret reading of NaOH, initial (mL)] 72.18 mL - 2.15 mL = 70.03 mL B.8. Molar concentration of NaOH (mol/L), Part A = 0.0451 M B.9. Moles of NaOH dispensed (mol) = [Molar concentration of NaOH (mol/L), Part A] * [Volume of NaOH dispensed (L)] 0.0451 M * 0.07003 L = 0.003158 mol B.10. Moles of CH3COOH in vinegar (mol) = Moles of NaOH dispensed (mol) * (1 mol CH3COOH/ 1 mol NaOH) 0.003158 mol B.11. Molar mass of CH3COOH (g/mol) = From periodic table 60.052 g B.12. Mass of CH3COOH in vinegar (g) = [Moles of CH3COOH in vinegar (mol)] * [Molar mass of CH3COOH (g/mol)] 0.003158 mol * 60.052 g = 0.1896 g B.13. Percent by mass of CH3COOH in vinegar (%) = [Mass of CH3COOH in vinegar (g)] / [Mass of unknown vinegar (g)] * (100) (0.1896 / 4.8445) * 100 = 3.91%

Results In Part A, the molar concentration of the base, NaOH, was found to be 0.0451 M. This then allowed for the determination of the moles of CH3COOH in the strong acid, vinegar, to be

0.003158 mol. With the number of moles of CH3COOH, it was found that the mass of CH3COOH in the vinegar was 0.1896 g. From this it was determined that the percent by mass of CH3COOH in the vinegar sample was 3.91% in Trial 1.

Post-Lab Questions Exp 9: 4. Part A. The mass of KHC8H4O4 is measured to the nearest milligram; however, the volume of water in which it is dissolved is never of concern – water is even added to the wall of the Erlenmeyer flask during the titration. Explain why water added to the KHC8H4O4 has no effect on the data, whereas the water added to the NaOH solution may drastically affect the data. a. Water added to the flask has no effect on the data because the concentration of KHC8H4O4 isn’t used in the determination of moles of NaOH and ultimately in the mass of CH3COOH. However, water added to the NaOH would dilute the solution and would therefor drastically alter the results. This is because if water is added, the NaOH will be far less concentrated (will have a lower molarity), and more NaOH will be needed to reach the equivalence point and show a light pink color in the flask. 5. Part B.2. The wall of the Erlenmeyer flask is occasionally rinsed with water from the wash bottle during the analysis of the acid solution. Will this technique result in the molar concentration of the acid solution being reported as too high, too low, or unaffected? Explain. a. The molar concentration of the acid will be reported too low since the water will dilute the solution. Molarity is determined as mol/L. The moles of the solution will remain the same, as it’s determined by the moles of acid added; but the volume will increase, as it is determined by the volume of solvent, water. Therefore, by increasing the volume by adding water, the molarity of the solution will decrease. 6. Parts A.6. and B.2. For the standardization of the NaOH solution in Part A.6, the endpoint was consistently reproduced to a faint pink color. However, the endpoint for the titration of the acid solution in Part B.2 was consistently reproduced to a dark pink color. Will the reported molar concentration of the acid solution be too high, too low, or unaffected by the differences in the colors of the endpoints. Explain. a. If the endpoint was reproduced to a dark pink color, then the molar concentration of the acid was too low for the amount of NaOH added. In order to be a faint pink color, there needs to be equal number of moles of acid and base in this experiment. So, if the solution is dark pink that indicates that there is too much base and too little acid, so more acid must be added. In conclusion, the molar concentration of the acid will be too low if the titration was reproduced to a dark pink color.

Exp 10: 5. Part A.3. The buret is filled with the NaOH titrant and the initial volume reading is immediately recorded without waiting the recommended 10-15 seconds. However, in Part B.1, the 10-15 second time lapse does occur before the reading is made. Does this technique error result in an increase, a decrease, or have no effect on the reported percent acetic acid in the vinegar? Explain.

a. This technique error results in an increase in the reported percent acetic acid in the vinegar. Since the reaction takes time to occur, by swirling the acid/base in the flask you are ensuring there are no bubbles in the flask and all the products are mixed. Because the person didn’t wait the 10-15 seconds to record the volume of NaOH, they didn’t accurately find the difference between the two readings. This will result in a greater amount of NaOH believed to be dispensed than there actually was, causing the reported acetic acid to be too high. 6. Part B.1. The endpoint of the titration is overshot! Does this technique error result in an increase, a decrease, or have no effect on the reported percent acetic acid in the vinegar? Explain. a. By overshooting, there is more base in the solution than there should have been if the titration was performed properly. Because there is an increase the amount of base in the flask, there must be an increase in the amount of acid in order to reach equivalence point. By adding more acid, this would result in an increase in the reported percent acid in the vinegar to neutralize the base. 7. Part B.1. The wall of the flask is periodically rinsed with the previously boiled, deionized water from the wash bottle. Does this titrimetric technique result in an increase, a decrease, or have no effect on the reported percent acetic acid in the vinegar? Explain. a. By rinsing the walls of the flask with DI water, there will be no effect on the percent acetic acid in the vinegar. This is because the water has no effect on the number of H3O and OH molecules present in order to reach the equivalence point. 8. A drop of NaOH titrant, dispensed from the buret, adheres to the wall of the Erlenmeyer flask but is not washed into the vinegar with the was bottle. Does this error in technique result in the reported percent of acetic acid being too high, too low, or unaffected? Explain. a. The reported percent acetic acid will be too high. This is because there would be too little NaOH added than believed, so there will be a higher concentration of acid in the flask. This would ultimately result in a greater amount of acetic acid in the vinegar....