Question

Assume the vinegar is 5.00% acetic acid by mass and the density of vinegar is 1.00 g.ml. Assume you have a 0.100M NaOH solution. Calculate the volume in mL of NaOH needed to titrate 5.00 mL of vinegar.

Answer 1

5% by mass means = 5 grams acetic acid in 100 grams solution.

density = 1 g/ml   Volume of solution = 100 ml

Molarity of vinegar M1 = (5/60)*(1000/100) = 0.833 M

Volume of vinegar used for titration V1 = 5 ml

Molarity of NaOH M2 = 0.1 M , Vol of NaOH V2 = ?

from dilution formula

M1V1 = M2V2

V2 = (0.833*5)/0.1 = 41.65 ml