Question

cetic acid (HC2H3O2) is an important component of vinegar. A 10.00-mL sample of vinegar is titrated with 0.5052 M NaOH, and 16.88 mL are required to neutralize the acetic acid that is present. Write a balanced equation for this neutralization reaction. What is the molarity of the acetic acid in this vinegar

Answer 1

The molarity of the acetic acid in this vinegar is 0.853 M

Step-by-step explanation:

Step 1: Data given

Volume of vinegar sample = 10.00 mL

Concentration of NaOH = 0.5052 M

16.88 mL are required to neutralize the acetic acid

Step 2: The reaction

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

Step 3: Calculate molarity of cetic acid (HC2H3O2)

C1*V1 = C2*V2

⇒with C1 = the molarity ofHC2H3O2= TO BE DETERMINED

⇒with V1 = the volume of HC2H3O2 = 10.00 mL = 0.01 L

⇒with C2 = the molarity of NaOH = 0.5052 M

⇒with V2 = the volume of NaOH = 16.88 mL = 0.01688L

C1 = (C2*V2)V1

C1 = (0.5052 * 0.01688 ) / 0.01

C1 = 0.853 M

The molarity of the acetic acid in this vinegar is 0.853 M