Vinegar contains acetic acid, CH3COOH. Titration of 5.00g of vinegar with 0.100M NaOH requires 33.0 cm^3 to reach the equivalence point. (a) What is the weight percentage of CH3COOH in vinegar? (b) If the vinegar has a density of 1.005 g/cm^3, what is the molarity of CH3COOH in vinegar?
The Correct Answer and Explanation is:
(a) Weight percentage of CH₃COOH in vinegar
Step 1: Moles of NaOH usedVolume of NaOH=33.0 cm3=0.0330 L\text{Volume of NaOH} = 33.0 \, \text{cm}^3 = 0.0330 \, \text{L}Volume of NaOH=33.0cm3=0.0330LMolarity of NaOH=0.100 mol/L\text{Molarity of NaOH} = 0.100 \, \text{mol/L}Molarity of NaOH=0.100mol/LMoles of NaOH=0.100×0.0330=0.00330 mol\text{Moles of NaOH} = 0.100 \times 0.0330 = 0.00330 \, \text{mol}Moles of NaOH=0.100×0.0330=0.00330mol
Step 2: Moles of CH₃COOH neutralized
Acetic acid reacts with NaOH in a 1:1 molar ratio:CH₃COOH+NaOH→CH₃COONa+H₂O\text{CH₃COOH} + \text{NaOH} \rightarrow \text{CH₃COONa} + \text{H₂O}CH₃COOH+NaOH→CH₃COONa+H₂OMoles of CH₃COOH=0.00330 mol\text{Moles of CH₃COOH} = 0.00330 \, \text{mol}Moles of CH₃COOH=0.00330mol
Step 3: Mass of CH₃COOH in sample
Molar mass of CH₃COOH:=12.01×2+1.008×4+16.00×2=60.05 g/mol= 12.01 \times 2 + 1.008 \times 4 + 16.00 \times 2 = 60.05 \, \text{g/mol}=12.01×2+1.008×4+16.00×2=60.05g/molMass=0.00330×60.05=0.19817 g\text{Mass} = 0.00330 \times 60.05 = 0.19817 \, \text{g}Mass=0.00330×60.05=0.19817g
Step 4: Weight percentageMass of vinegar=5.00 g\text{Mass of vinegar} = 5.00 \, \text{g}Mass of vinegar=5.00gWeight %=(0.198175.00)×100=3.96%\text{Weight \%} = \left( \frac{0.19817}{5.00} \right) \times 100 = 3.96\%Weight %=(5.000.19817)×100=3.96%
(b) Molarity of CH₃COOH in vinegar
Step 1: Volume of vinegar sample
Given mass = 5.00 g, density = 1.005 g/cm³:Volume=5.001.005=4.9751 cm3=0.0049751 L\text{Volume} = \frac{5.00}{1.005} = 4.9751 \, \text{cm}^3 = 0.0049751 \, \text{L}Volume=1.0055.00=4.9751cm3=0.0049751L
Step 2: MolarityMoles of CH₃COOH=0.00330 mol\text{Moles of CH₃COOH} = 0.00330 \, \text{mol}Moles of CH₃COOH=0.00330molMolarity=0.003300.0049751=0.6636 mol/L\text{Molarity} = \frac{0.00330}{0.0049751} = 0.6636 \, \text{mol/L}Molarity=0.00497510.00330=0.6636mol/L
Final Answers:
- (a) Weight percentage of CH₃COOH = 3.96%
- (b) Molarity of CH₃COOH = 0.664 mol/L
Explanation
Acetic acid, known by its chemical formula CH₃COOH, serves as the active acidic component in vinegar. To determine its concentration, titration is performed using sodium hydroxide (NaOH), a strong base. In the process, a known volume of NaOH is used to neutralize a known mass of vinegar, allowing calculation of the acid content.
The neutralization reaction between acetic acid and sodium hydroxide follows a 1:1 molar ratio. Therefore, the number of moles of NaOH used directly reflects the moles of CH₃COOH present. By multiplying the volume of NaOH in liters by its molarity, the number of moles involved in the reaction is established.
Once the moles of acetic acid are known, converting this to grams is achieved through multiplication by the molar mass of acetic acid. Comparing this mass to the initial mass of the vinegar sample enables calculation of the weight percentage of acetic acid. The result reflects how much of the vinegar’s mass is due to acetic acid content.
To determine molarity, the volume of vinegar is required. This is calculated using the density of the solution and its known mass. Converting grams to volume using the density relationship, and then expressing volume in liters, allows the use of the standard molarity formula: moles of solute per liter of solution.
These calculations collectively determine both the weight percentage and molarity of acetic acid in vinegar, offering insight into its acidic strength and concentration in the sample.
