Study Notes on Titration of Acetic Acid in Vinegar

Titration of Acetic Acid in Vinegar

Introduction to Titration

• Titration is a laboratory method used to determine the concentration of an unknown solution by reacting it with a solution of known concentration.

Problem Statement

• A 10.0 mL sample of vinegar is titrated.
• The titration requires 30.0 mL of 0.20 M NaOH (sodium hydroxide) solution.

Calculations Required

a) Molarity of Acetic Acid in Vinegar

• Formula for molarity (M) is given by: M = \frac{n}{V}
  • Where:
    • $M$ = molarity of the solution (mol/L)
    • $n$ = moles of solute
    • $V$ = volume of solution in liters (L)

Step 1: Calculate moles of NaOH used

• The moles of NaOH can be calculated using its molarity and volume:
  • Molarity of NaOH = 0.20 M
  • Volume of NaOH = 30.0 mL = 0.0300 L
• n{NaOH} = M{NaOH} \times V_{NaOH}
• Substituting the values:
  n_{NaOH} = 0.20 \, \text{mol/L} \times 0.0300 \, \text{L} = 0.0060 \, \text{mol}

Step 2: Stoichiometry of the Reaction

• The reaction between acetic acid (CH₃COOH) and sodium hydroxide (NaOH) is:
  \text{CH}3\text{COOH} + \text{NaOH} \rightarrow \text{CH}3\text{COONa} + \text{H}_2\text{O}
• The stoichiometry is 1:1, meaning 1 mole of acetic acid reacts with 1 mole of NaOH.
• Therefore, 0.0060 moles of NaOH corresponds to 0.0060 moles of acetic acid.

Step 3: Calculate Molarity of Acetic Acid

• The volume of the acetic acid in the vinegar sample is 10.0 mL = 0.0100 L.
• Using the moles of acetic acid:
  M{CH3COOH} = \frac{n{CH3COOH}}{V{CH3COOH}}
• Substituting the values:
  M{CH3COOH} = \frac{0.0060 \, \text{mol}}{0.0100 \, \text{L}} = 0.60 \, \text{M}

b) Mass/Mass Percent Concentration of Acetic Acid in Vinegar

• Mass/mass percent concentration is calculated using:
  \text{Percent} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100

Step 1: Calculate Mass of Acetic Acid

• To find the mass of acetic acid, use the molar mass of acetic acid:
• Molar mass of acetic acid (CH₃COOH):
  • C: 2 × 12.01 g/mol = 24.02 g/mol
  • H: 4 × 1.008 g/mol = 4.032 g/mol
  • O: 2 × 16.00 g/mol = 32.00 g/mol
• Total molar mass = 24.02 + 4.032 + 32.00 = 60.052 g/mol
• Mass of acetic acid:
  ext{mass}{CH3COOH} = n{CH3COOH} \times \text{molar mass}{CH3COOH}
• Substituting the values:
  ext{mass}{CH3COOH} = 0.0060 \, \text{mol} \times 60.052 \, \text{g/mol} = 0.3603 \, \text{g}

Step 2: Calculate Mass of the Solution

• The total volume of the solution (vinegar) takes into account the density of acetic acid:
• Density of acetic acid = 1.01 g/cm³ = 1.01 g/mL.
• The mass of the vinegar solution containing 10.0 mL:
  \text{mass}_{solution} = \text{density} \times \text{volume}
• For 10.0 mL of vinegar:
  \text{mass}_{solution} = 1.01 \, \text{g/mL} \times 10.0 \, \text{mL} = 10.10 \, \text{g}

Step 3: Calculate Percent Concentration

• Now substitute the mass of acetic acid and mass of solution into the percent formula:
  \text{Percent}{CH3COOH} = \frac{0.3603 \, \text{g}}{10.10 \, \text{g}} \times 100
• Calculating the mass/mass percent:
  \text{Percent}{CH3COOH} \approx 3.57 \%

Summary of Results

• a) Molarity of acetic acid in vinegar: 0.60 M
• b) Mass/mass percent concentration of acetic acid in vinegar: 3.57%