Question

If 3.60 mL of vinegar needs 43.0 mL of 0.130 M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.30 qt sample of this vinegar?

Answer 1

There is 114.53 grams of acetic acid in 1.30 qt of sample

Step-by-step explanation:

Step 1: Data given

Volume of vinegar = 3.60 mL

Volume of 0.130 M NaOH = 43.0 mL

1 qt = 0.946353 L

1.30 qt = 1.23026 L = 1230 mL

Step 2: Calculate moles of NaOH

Moles NaOH = molarity * volume

Moles NaOH = 0.130 M * 0.043 L

Moles NaOH = 0.00559 moles

Step 3: Calculate concentration of vinegar

Concentration = moles / volume

Concentration = 0.00559/ 0.00360 L

Concentration = 1.55 M

Step 4: Calculate mass of acetic acid in 1.30 qt sample

1.30 qt * 0.946353 L * 1.55 M *60.06 g/mol =114.53 grams

There is 114.53 grams of acetic acid in 1.30 qt of sample