Determination of the Concentration of Acetic Acid in Vinegar Lab Exercise 4 CHEM 1106 9/19/12
Purpose: Standardize a sodium hydroxide solution using a primary standard acid. Determine the molarity and the percent by mass of acetic acid in vinegar by titration with the standardized sodium hydroxide solution. Introduction: Vinegar is a dilute solution containing acetic acid. Since vinegar has a low pH, it can be titrated with a base. Titration is a method used in order to ascertain the amount of a constituent in a solution by measuring the volume of a known concentration of a reagent required to complete a reaction with it, typically using a burette. The equivalence point of a titration occurs when chemically equivalent amounts of acid and base are present. In this experiment, the equivalence point occurs when the moles of acid in the solution equals the moles of base added in the titration. A rapid change in pH with the small addition of acid or base is the indicator of acid-base equivalency.
Accompanying material: Vinegar Battery Statement of the Problem
Using a pH meter and a graph of pH plotted versus the volume of base added, the equivalence point can be accurately found by finding the point that is in the middle of the vertical part of the curve. Once the equivalence point of the titration is known, the concentration of the sodium hydroxide can be determined. Sodium hydroxide is used to titrate vinegar so that the concentration of the vinegar is determined. The percentage of acetic acid in the solution can be determined from the concentration of the vinegar.
Part A – Standardization of a Sodium Hydroxide Solution:
Prepare 150 mL of approximately 0. M sodium hydroxide solution from solid NaOH. The solution can be prepared in a beaker, it is not necessary to use a graduated cylinder or a volumetric flask because the NaOH solution will be standardized.
Weigh a 150-mL beaker and record the mass to the nearest 0. 001 g. Add approximately 0.5 grams of potassium hydrogen phthalate (KHP) to the beaker. Record the mass of the beaker and KHP to the nearest 0. 001 g. Calculate the mass of KHP by difference and record it. Add approximately 50 mL of distilled water to the beaker. Stir the solution until the KHP has dissolved completely.
Record a titration curve using the MeasureNet pH probe and drop counter. (See Appendix F)
Repeat steps 2 and 3.
From the plots, determine the volume of NaOH required to neutralize the KHP solution in each titration. Record the volumes.
Calculate the molarity of sodium hydroxide. Part B – Determination of Acetic Acid Concentration in Vinegar
Transfer 2. 0 mL of vinegar to a clean, dry 150 mL beaker using a 10-mL volumetric pipet. Add sufficient water, 50 mL, to cover the pH electrode tip during the titration.
Record a titration curve using the MeasureNet pH probe and drop counter. See Appendix F)
From the plots, determine the volume of NaOH required to neutralize vinegar in each titration. Record the volumes.
Calculate the molarity of acetic acid in vinegar.
Calculate the percent by mass of acetic acid in vinegar. Data: Part A – Standardization of a Sodium Hydroxide Solution Mass of beaker97. 47 g Mass of beaker + KHP97. 99 g Mass of KHP0. 52 g Volume of NaOH to neutralize the KHP solution7. 755 mL Molarity of sodium hydroxide 0. 535 M NaOH Part B – Determination of the Concentration of Acetic Acid in Vinegar Volume of NaOH required to neutralize vinegar3. 18 mL Molarity of acetic acid in vinegar0. 8515 M CH3COOH Percent by mass of acetic acid in vinegar5. 115%.
Equations: (1) Molarity (M) = moles of solute/liter of solution (2) Percent solute= (grams of solute/grams of solution) x 100% (3) NaOH (aq) + CH3COOH (aq) NaCH3CO2 (aq) + H2O (l) (4) pH = -log[H3O+] (5) KHC8H4O4 (aq) + NaOH (aq) KNaC8H4O (aq) + H2O (l)
Calculations: Part A – Standardization of a Sodium Hydroxide Solution Calculate the mass of KHP: (Mass of Beaker + KHP) – Mass of Beaker = Mass of KHP 97. 99g – 97. 47g = 0. 52g
Calculate the molarity of sodium hydroxide: Moles of KHP = g/MW = 0. 52g/204. 22g = 0. 002546 mol KHP
Equation 5 = 1:1 ratio 0. 002546 mol KHP x 1 mol NaOH/1mol KHP = 0. 002546 mol NaOH
Equation 1 = 0. 002546 mol/0. 004755 L = 0. 535 M NaOH
Part B – Determination of the Concentration of Acetic Acid in Vinegar Calculate the molarity of acetic acid in vinegar: 3. 318 mL/1000 = 0. 00318 L NaOH 0. 00318 L NaOH x 0. 535 mol/1L NaOH = 0. 001703 mol NaOH
Equation 3 = 1:1 ratio 0. 001703 mol NaOH x 1 mol CH3COOH/1 mol NaOH = 0. 001703 mol CH3COOH 2. 0 mL CH3COOH/1000 = 0. 0020 L CH3COOH
Equation 1 = 0. 01703 mol CH3COOH/0. 0020 L soln. = 0. 8515 M CH3COOH
Calculate the percent by mass of acetic acid in vinegar: 0. 0020 L CH3COOH x 0. 8515/1 L soln. = 0. 001703 mol CH3COOH 0. 001703 mol CH3COOH x (60. 06g CH3COOH/1 mol CH3COOH) = 0. 1023g CH3COOH 2. 0 mL CH3COOH x (1g CH3COOH/1 mol CH3COOH) = 2. 0 mL CH3COOH soln.
Equation 2 of CH3COOH = (0. 1023 g CH3COOH/2. 0g CH3COOH) x 100% = 5. 115%
Final Answer: Molarity of vinegar:0. 8515 M CH3COOH
Percent mass: 5. 115% CH3COOH From the plots determine the volume of NaOH required to neutralize the KHP solution in each titration.
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