Answer to Question #75676 in General Chemistry for Fatima Dar

Question

A 30.84 mL sample of 0.128 M NaOH is needed to reach an endpoint for a titration of a 5.441 gram sample of Vinegar. Calculate the % acetic acid in the vinegar sample. (Show all work)

Answer 1

First calculate the moles of acetic acid, then the weight of acetic acid, then divide by the weight of vinegar.
1. moles of acetic acid is equivalent to moles of NaOH required to neutralize it:
moles of NaOH = volume x molarity (note molarity is expressed in moles per liter)
moles of NaOH = 0.03084 liters * 0.128 Moles/liter = 0.0039475 moles.
2. wt of acetic acid corresponding to 0.0039475 moles:
mw of acetic acid (CH3COOH) = 2*12.01 + 4*1 + 2*16.00=60.02 grams/mole
wt = 60.02 grams/mole*0.0039475 moles=0.2369 grams
3. % acetic acid = wt of acetic acid/total wt of sample *100
% acetic acid = 0.2369 grams/5.441 grams * 100= 4.35%