Lab Report #1 - Joanna Le

Lab Report #1 Acid- Base Titration of Acetic Acid in Vinegar with Sodium Hydroxide

Le 1 Joanna Le Professor Dr. Gray Chemistry 1A - Section 3 16 November 2023 INTRODUCTION : Titration is used to determine the molar concentration of an unknown solution by slowly adding a solution of known concentration to a known volume of a solution of unknown concentration. When the two solutions mix and undergo a chemical reaction, an indicator present in the solution will reveal a sudden physical change called the End Point. In this experiment, the acetic acid (CH 3 COOH) in commercial vinegar will be titrated with a solution of known concentration of standardized sodium hydroxide (NaOH). The balanced chemical reaction (CH 3 COOH (aq) + NaOH (aq) → CH 3 COONa (aq) + H 2 O (l) ) demonstrates that exactly one mole of NaOH must be added to one mole of acetic acid to complete the reaction; this point when equal amounts of acid and base are present is called the Equivalence Point. The indicator added to the acid, Phenolphthalein, will remain colorless until the equivalence point has been crossed; this is when the solution turns basic and the phenolphthalein becomes light pink. To remain as close to the equivalence point as possible, the base will be added drop by drop via a microburette, and the titration will be stopped at the appearance of the first permanent light pink color. PURPOSE : To calculate the concentration of acetic acid in vinegar through an acid- base titration. MATERIALS : Please refer to page 5 of the lab handout. PROCEDURE : Please refer to pages 6-8 of the lab handout.

Le 3 CALCULATIONS : PART 1: Standardization of Aqueous Sodium Hydroxide Solution Moles of KHP (C 8 H 5 O 4 K) Molarity of KHP (C 8 H 5 O 4 K) = ?𝑎?? ?? ?𝐻? * ???𝑎? ?𝑎?? ?? ?𝐻? = 0.5102g * 204.22 g/mol = 0.002498 mol = ????? 𝑉????? = 0.002498 ??? 0.02500 ? = 0.09992 M Molarity of NaOH Average Molarity of NaOH = ???𝑎?𝑖?𝑦 ?𝐻? * 𝑉????? ?𝐻? 𝑉????? ?𝑎?𝐻 Trial 1 = (0.09992 ?) * (0.00101 ?) 0.00093 ? = 0.11 M Trial 2 = (0.09992 ?) * (0.00101 ?) 0.00095 ? = 0.11 M Trial 3 = (0.09992 ?) * (0.00102 ?) 0.00083 ? = 0.12 M = (0.11?) * (0.11?) * (0.12?) 3 = 0.11 M PART 2: Titration of Acetic Acid in Vinegar with the Standardized Sodium Hydroxide Solution CH 3 COOH (aq) + NaOH (aq) → CH 3 COONa (aq) + H 2 O (l) Equivalence Point: moles of NaOH = moles of CH 3 COOH Average Molarity of NaOH (from Part 1) = 0.11 M Trial 1 Trial 2 Trial 3 Volume of NaOH (L) 0.00098 L 0.00090 L 0.0010 L Moles of NaOH = 𝐴?? ??? ?𝑎?𝐻 * 𝑉????? ?𝑎?𝐻 (0.11 M * 0.00098 L) = 0.00011 mol (0.11 M * 0.00090 L) = 0.000099 mol (0.11 M * 0.0010 L) = 0.00011 mol

Le 4 Volume of CH 3 COOH (L) 0.00010 L 0.00012 L 0.00011 L Moles of CH 3 COOH 0.00011 mol 0.000099 mol 0.00011 mol Molar Mass of CH 3 COOH (g/mol) 60.06 g/mol 60.06 g/mol 60.06 g/mol Mass of CH 3 COOH (g) = ????? 𝐴𝐴 * ???𝑎? ?𝑎?? 𝐴𝐴 (0.00011 mol * 60.06 g/mol) = 0.0066 g (0.000099 mol * 60.06 g/mol) = 0.0059 g (0.00011 mol * 60.06 g/mol) = 0.0066 g Mass of Vinegar Sample (g) 0.0987 g 0.1139 g 0.1124 g Mass Percent of Acetic Acid of Vinegar (%) = ( ?𝑎?? ?? 𝐴𝐴 ?𝑎?? ?? 𝑉𝑖???𝑎? ) * 100% ( (0.0066 g / 0.0987 g) * 100%) = 6.7% ( (0.0059 g / 0.1139 g) * 100%) = 5.2% ( (0.0066 g / 0.1124 g) * 100%) = 5.9% Average Mass Percent of Acetic Acid in Vinegar = (6.7%) * (5.2%) * (5.9%) 3 = 5.9% RESULTS/DISCUSSION : The results showed that the molar concentration of acetic acid in commercial vinegar varied between 5% to 7% with an average of 5.9%. My trials 2 and 3 were closer in precision than my trial 1. This could be due to how my trials were carried out. For my first trial, my burette had a few issues titrating and I used the wrong glassware; instead of a erlenmeyer flask, I used a volumetric flask. The type of glassware could have possibly affected the mixing of my solutions, hence why the mass percent of acetic acid of vinegar was much higher than the other two trials. For my second and third trials, I was not only more careful, but I grew more comfortable with making sure I titrated one drop at a time. However, because my mass percentages were not accurate and overall, not precise, some experimental errors must have occurred. A possibility could have been that I added a drop or a few too many since my trial 1 and trial 3 solutions were not a light pink, but a brighter pink than my trial 2. Instead, I should have added partial drops when it got closer to reaching the endpoint. Other human factors could be misreading the meniscus when measuring the solutions and how the solution inside the burettes sometimes created air bubbles which altered my titration measurements.