The data below was collected while performing a titration of acetic acid (in vinegar) to determine its concentration using sodium hydroxide. 150 ml of vinegar with unknown molarity 45.5 ml of .5 M NaOH was added to reach equivalence point Find the molarity of the vinegar.

Answer 1

Let's call acetic acid HAc.
HAc | NaOH ==> NaAc + H2O

mols NaOH = M x L = ?
mols HAc = mols NaOH
M HAc = mols HAc/L HAc = ?

Answer 2

For a 1:1 rxn ratio of monoprotic acid reacting with a GpIA strong base use ...
(Molarity x Volume)acid = (Molarity x Volume)base

Molarity Acid
= [(M x V)base]/(Vol Acid)
= [(0.5M)(45.5 ml)]/(150 ml)
= 0.16M HOAc