Titrating NaOH and acetic acid in vinegar

Question

If the vinegar used in the lab had 5.89 g of acetic acid per 100. mL of vinegar, what volume (in mL) of 0.0998 M NaOH would you use to reach the endpoint of the titration?

Answer 1

Hello Amy

 

The main chemical in vinegar that is reacting with NaOH in the titration is acetic acid (HC₂H₃O₂).  The balanced chemical reaction is:

 

HC₂H₃O₂ (aq) + NaOH (aq) → NaC₂H₃O₂ (aq) + H₂O (l)

 

Starting with 5.89 g of HC₂H₃O₂, calculate the # of moles of NaOH that will be required to completely neutralize it using stoichiometry.  Molar mass of HC₂H₃O₂ = 60.05 g/mol.  Mole ratio of HC₂H₃O₂ to NaOH is 1:1.

 

5.89 g HC₂H₃O₂ x (1 mol HC₂H₃O₂ / 60.05 g HC₂H₃O₂) x (1 mol NaOH / 1 mol HC₂H₃O₂) = 0.0981 mol NaOH

 

Volume of NaOH needed to reach the end point is....

 

Molarity = mol / L

 

L = mol / Molarity

 

   = 0.0981 / 0.0998 = 0.983 L

 

Volume in mL = 0.983 L x (1000 mL / 1 L) = 983 mL