Conductometric Analysis Of Barium Hydroxide Of An Unknown Molarity PDF

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Conductometric Analysis Of Barium Hydroxide Of An Unknown Molarity
Introduction, Materials, Methods, Data, Results, Discussion, Conclusion....

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Experiment: Conductometric Analysis Of Barium Hydroxide Of An Unknown Molarity Analytical Chemistry Date Performed: March 10, 2020 Date Submitted: April 2, 2020

Title: Conductometric Titration of Barium hydroxide Introduction: A conductometric titration is a method that is used to find the concentration of an unknown solution by analyzing the amount of analyte within a given mixture. In this experiment the goal was to find the molarity of Barium hydroxide (BaOH)2. Barium hydroxide was the solution with an unknown molarity and Sulfuric acid was the titrant. Three trials were conducted in which Sulfuric acid was added to Barium Hydroxide until ana equivalence point was reached. The average molarity was found to be 0.37M. Materials: ● ● ● ● ● ● ●

Ring stand 50 mL Burette 250 mL beaker Distilled Water 10 mL of Ba(OH)2 Conductance probe 50 mL of sulfuric acid

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10mL graduated cylinder 50mL graduated cylinder Phenolphthalein indicator Magnetic stir bar and stirrer

Methods: Procedure **Safety precautions: Barium hydroxide is caustic be careful not to spill on skin or clothing. Sodium hydroxide is a strong acid, handle with care. 1. Using a graduated cylinder measure out 10 mL of Ba(OH)2 and 100 mL Distilled water and place into a 250mL beaker. Then add1 drop of phenolphthalein. 2. Place the beaker containing the Ba(OH)2 solution on the magnetic stirrer and insert the magnetic stirring bar and conductivity probe. 3. Place the 0.200 M H2SO4 solution into the burette. 4. Turn on the computer and open LoggerPro, open the corresponding file from the Chemistrywithvenier. Record the conductance value for zero.

5. Begin to drip the H2SO4 solution approximately 1-2mls and record the conductance value for each. Continue until the solution turns light pink. 6. Repeat steps 1-5, 3 times making sure to wash the conductance probe with distilled H2O between each trial. 7. Place all the solutions into the corresponding waste containers, wash the materials and place back into the corresponding place in the lab. Data: Table A1: A line graph of mols of Sulfuric acid and conductivity for trial 1.

Table A2: A line graph of mols of Sulfuric acid and conductivity for trial 2.

Table A3: A line graph depicting the mols of Sulfuric acid and conductivity for trial 3.

Results: According to table A1 the equivalence point was reached at 16.5mL of Sulfuric acid added giving a conductance of 31.02539 (approx. 31.03) For table A1 the equivalence point was reached at 15 mL of added Sulfuric acid giving a conductance of 41.3672 (approx. 41.37). Lastly

according to table A3 the equivalence point was reached at 13mL of Sulfuric acid added giving conductance of 37.2305 (approx. 37.23). Discussion: Based on the data the average mL required to reach the equivalence point was 14.83 approx. 15 mLs of Sulfuric acid and the average conductance value was 36.54. From the data the conductance values for trial 1 and 3 are close, 31.03 and 37.23 respectively, and trial 2 would be the outlier. When it comes to the amount of Sulfuric added to reach the equivalence point trial 1 and 2 had closer values, 16.5mL and 15mL respectively and trial 3 was the outlier being 13mL. Conclusion: The experiment concluded that the average molarity of Barium hydroxide used in the experiment was 0.37M. The experiment also concluded that the average conductance value was 36.54. References: HARRIS, D. A. N. I. E. L. C. (2019). Quantitative Chemical Analysis. S.l.: W H FREEMAN Tau.Ac.Il , 2020, https://www.tau.ac.il/~chemlaba/Files/conductometry-titrations.pdf. Accessed 22 Mar 2020.

Appendix: H2SO4 + Ba(OH)2 →BaSO4 + 2H2O 2H+ + SO42- + Ba2+  + 2OH-2 →BaSO4 + 2H2O Molarity = Mols of solute/Liters of solution 0.200 M = x mols/ 0.018 L = 0.0036 mols of sulfuric acid * 1 Mol of sulfuric acid / 1 Mol of Barium Hydroxide = 0.0036 mols of Barium Hydroxide / 0.01 L of Barium Hydroxide solution = 0.36 M of Barium Hydroxide...