A 30.84cm³volume of 0.128M NaOH is required to reach the phenolpthalein endpoint in the titration of a 5.441g sample of vinegar. Calculate the percent acetic acid in the vinegar. (Molar mass of acetic acid is 60.0g/mol)

Answer 1

First, calculate the moles of NaOH used in the titration:

Volume of NaOH = 30.84 cm³ = 0.03084 L
Molarity of NaOH = 0.128 M

moles of NaOH = volume x molarity
moles of NaOH = 0.03084 L x 0.128 mol/L
moles of NaOH = 0.00394 mol

Since acetic acid and NaOH react in a 1:1 ratio, the moles of acetic acid in the vinegar is also 0.00394 mol.

Next, calculate the mass of acetic acid in the vinegar:

Mass of acetic acid = moles x molar mass
Mass of acetic acid = 0.00394 mol x 60.0 g/mol
Mass of acetic acid = 0.2364 g

Finally, calculate the percent acetic acid in the vinegar:

Mass of vinegar sample = 5.441 g

Percent acetic acid = (mass of acetic acid / mass of vinegar sample) x 100%
Percent acetic acid = (0.2364 g / 5.441 g) x 100%
Percent acetic acid = 4.35%

Therefore, the percent acetic acid in the vinegar is 4.35%.