Question

The distinctive odor of vinegar is due to acetic acid, HC2H3O2. Acetic acid reacts with sodium hydroxide in the following fashion: HC2H3O2(aq) + NaOH(aq) H2O(l) + NaC2H3O2(aq). If 1.74 mL of vinegar requires 34.9 mL of 0.1041 M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.00 qt sample of this vinegar? (1 L = 1.0567 qt)

Answer 1

118.6 grams of acetic acid.

Step-by-step explanation:

C1V1/C2V2 = N1/N2

N1 is the number of moles of acetic acid in the equation of reaction

N2 is the number of moles of acetic acid in the equation of reaction

At equivalence point, N1 = N2

Therefore, C1V1 = C2V2

C1 is the concentration of acetic acid

V1 is the volume of acetic acid = 1.74 mL

C2 is the concentration of sodium hydroxide = 0.1041 M

V2 is the volume of sodium hydroxide = 34.9 mL

C1 = 0.1041×34.9/1.74 = 2.088 M

MW of acetic acid (HC2H3O2) = 1 + (12×2) + (1×3) + (16×2) = 1+24+3+32 = 60 g/mol

Mass of acetic acid = concentration × volume × MW

volume = 1 qt = 1 qt × 1 L/1.0567 qt = 0.9463 L

Mass of acid acid = 2.088 × 0.9463 × 60 = 118.6 grams