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CHEM 122 Expt #9 Analysis of % Acetic Acid in Vinegar

CHEM 122 Expt #9 Analysis of % Acetic Acid in Vinegar. CCBC-Catonsville. Outline. Part A: Standardization of NaOH soln: HCl + NaOH  H 2 O + NaCl To determine the molarity of the NaOH solution to be used in Part B.

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CHEM 122 Expt #9 Analysis of % Acetic Acid in Vinegar

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  1. CHEM 122 Expt #9Analysis of % Acetic Acid in Vinegar CCBC-Catonsville

  2. Outline Part A: Standardization of NaOH soln: HCl + NaOH  H2O + NaCl To determine the molarity of the NaOH solution to be used in Part B. (3 acceptable trial with very pale pink endpoint and vol HCl: vol NaOH within 0.04) Part B: Analysis of Vinegar HC2H3O2 + NaOH  H2O + NaC2H3O2 Using the average molarity of NaOH from Part A, determine the percent acetic acid in vinegar (prefer 3 acceptable trials).

  3. NaOH Initial Reading 1.58 mL Final Reading 19.89 mL 18.31 mL Part A: Standardization of NaOH soln: A 25.00-mL sample of standard 0.2000 M HCl is titrated with a NaOH(aq )solution. The total volume of NaOH solution used to reach the end point was 18.31 mL. What is the molarity of the NaOH solution? HCl (aq)+ NaOH (aq)H2O (l) +NaCl (aq) 25.00 mL18.31 mL 0.2000 M? M DO NOT USE M1V1 = M2V2 because this is NOT a dilution problem. HCl 25.00 mL 0.2000M

  4. NaOH Initial Reading 1.58 mL Final Reading 19.89 mL 18.31 mL Repeat to get 3 trials with acceptable endpoints. Find average molarity of NaOH to be used in Part B. HCl 25.00 mL 0.2000M

  5. NaOH Initial Reading Final Reading 16.87 mL Part B: Analysis of Vinegar: A 5.00-mL sample of vinegar is titrated using the NaOH solution described in question #2. It requires 16.87 mL of NaOH to react with all of the acetic acid (HC2H3O2) in the vinegar sample. What is the percent by weight of acetic acid in the vinegar? Assume the density of vinegar to be 1.02 g/mL) HC2H3O2(aq)+ NaOH(aq) H2O(l) +NaC2H3O2 5.00 mL16.87 mL ? % 0.2731 M (Av of 3 trials) vinegar 5.00 mL ? %

  6. NaOH Initial Reading Final Reading 16.87 mL 0.2731M NaOH HC2H3O2(aq)+ NaOH(aq) H2O(l) + NaC2H3O2 vinegar 5.00 mL ? %

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