Question

A vinegar sample was analyzed and found to be 5.88% acetic acid by volume. The density of pure acetic acid is 1.049g/mL. What volume of 0.200M NaOH would be required to titrate a 2ml aliquot of this vinegar?

Some help would be greatly appreciated!
thanks

Answer 1

My assumption, when you say 5.88% by volume that you mean 5.88% v/v or 5.88 mL acetic acid/100 mL solution. If that is so, then each mL solution contains 0.0588 mL acetic acid.
mass = volume x density so
mass = 0.0588 mL x 1.049 g/mL = ??

Change that to moles. moles = g/molar mass acetic acid.
Then L x M = moles. Plug in moles acetic acid and M of NaOH and calculate L NaOH, then convert to mL.
Check my thinking. I get something like 5 mL but check this carefully.