posted by Rema .
You were given 25.00 ml of an acetic acid solution of unknown concentration. You find that it requires 29.60 ml of a 0.1050 M NaOH solution to exaclty neutralize this sample.
A) What is the molarity of the acetic acid solution?
B) what is the percentage of acetic acid in the solution? Assume the density of the solution is 1 g/ml.
HAc + NaOH ==> NaAc + H2O so it is a 1:1 titration mixture. Therefore,
mLacid x M acid = mL base x M base.
Substitute and solve for the one unknown.
B) Determine moles acetic acid, then grams.
moles = M x L
grams = moles x molar mass
%acetic acid = (grams acetic acid/mass sample)*100 = ??