Lab Report #1

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University of Wisconsin, Milwaukee *

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Chemistry

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Apr 3, 2024

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Lynn Szmurlo Chem 221-803 2/2/2023 Acid-Base Titration: Determination of Acetic Acid in Commercial Vinegar OBJECTTIVES The purpose of this experiment is to determine the amount of acetic acid within a sample of Commercial Vinegar. This will be determined through a series of titrations. REAGENT INFORMATION Name: Chemical Formula: Molecular Weight: Physical State: Hazards: Acetic Acid CH3COOH 60.05 g/mol Liquid - Flammable - Causes skin burns and eye damage Sodium Hydroxide NaOH 39.97 g/mol Liquid - Causes severe skin burns and eye irritation Potassium Hydrogen Phthalate C8H5KO4 204.22 g/mol Solid - May form explosives when exposed to air - Eye irritant Phenolphthalein C20H14O4 318.3 g/mol Liquid - May cause eye, skin, or respiratory irritation -Flammable PROCEDURES Standardization of titrant : Roughly 4 grams of potassium hydrogen phthalate (KHP) was obtained and separated into 4, 0.7gram-0.9 gram samples. These samples were placed into separate 250 mL Erlenmeyer flasks and approximately 50 mL of deionized(ultrapure) water was added to each flask to dissolve the KHP. After the KHP was dissolved, 2 drops of

phenolphthalein indicator were added. The solutions were then titrated with NaOH until the indicator was a faint pink, this was the denotation of the endpoint. The concentration of the NaOH was then calculated. Determination of Acetic Acid: An unknown sample of vinegar was obtained. Then, 50 ml of the vinegar was pipetted into a clean 250 mL volumetric flask and deionized water was added to the mark. After mixing thoroughly, four- 50 mL samples of the solution were placed into Erlenmeyer flasks and 2 drops of the phenolphthalein indicator were added. Another 25 mL of deionized water was added to the samples. The four samples were then titrated with NaOH until the endpoint was observed. All data was recorded in a lab notebook. PRINCIPLES/METHODS Typical commercial vinegar contains about 5 grams of acetic acid for every 100 milliliters of vinegar. This is analyzed using a method called titration. Titration is a method to determine the concentration of an unknown substance by adding a solution of known concentration and determining when the unknown solution reaches its “endpoint”. Phenolphthalein was used as the indicator to see the endpoint of the titration. A strong base, sodium hydroxide, was used in this titration because the vinegar itself is a weak acid. Because a strong base was added to this weak acid, the acid was neutralized by the base and the conjugate acid was formed. The conjugate base formed was the acetate ion, and the conjugate acid was water. The overall equation for this reaction is: CH3COOH + OH- <----> CH3COO- +H2O Sodium hydroxide is a strong base that can absorb CO2 from the atmosphere which can change the concentration of the NaOH overtime: CO2 +OH- <---> HCO3-. Bicarbonate (HCO3-)

and Carbonate (CO3-2) ion will act as a buffer during the titration and cause a gradual change in the pH at the endpoint. A gradual change is not desired for a titration, instead an abrupt change is more ideal to be able to see clearly when the endpoint has been acquired. DATA Table 1: Standardization of Titrant Units Trial 1 Trial 2 Trial 3 Trial 4 Mean Volume of Titrant added mL 29.1 28.9 29.6 32.4 30 Weight of KHP g 0.7511 0.7501 0.7552 0.7759 0.7580 Tabe 2: Determination of Acetic Acid Units Trial 1 Trial 2 Trial 3 Trial 4 Mean Volume of titrant added mL 35.8 38.8 40.2 48.4 40.8 RESULTS AND CALCULATIONS Determine the concentration of NaOH: Trial 1 [0.7511g KHP/ (204.22 g/mol KHP)] (1 mol NaOH/ 1 mol KHP) (1/29.1mL NaOH) (1000mL/ L NaOH) = mol NaOH/ L= 0.126 M NaOH Trial 2 [0.7501g KHP/ (204.22 g/mol KHP)] (1 mol NaOH/ 1 mol KHP) (1/28.9mL NaOH) (1000mL/ L NaOH) = mol NaOH/ L= 0.127 M NaOH Trial 3 [0.7552g KHP/ (204.22 g/mol KHP)] (1 mol NaOH/ 1 mol KHP) (1/ 29.6mL NaOH) (1000mL/ L NaOH) = mol NaOH/ L= 0.124 M NaOH Trial 4 [0.7759g KHP/ (204.22 g/mol KHP)] (1 mol NaOH/ 1 mol KHP) (1/32.4mLNaOH) (1000mL/ L NaOH) = mol NaOH/ L= 0.117 M NaOH Mean (0.126 M+ 0.127 M+ 0.124 M+ 0.117 M)/4= 0.123 M Standard Deviation [[(0.126-0.123) ^2 + (0.127- 0.123) ^2 + (0.124-0.123) ^2 +(0.117-0.0123) ^2]/ 4-1] ^½= 0.0045 % RSD (0.0045/0.123) x100= 3.69%

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