posted by Jeffrey
A 31.43 mL volume of 0.108 M NaOH is required to reach the phnolphthalein endpoint in the titration of a 4.441 g sample of vinegar. Calculate % acetic acid in the vinegar.
The reaction is:
CH3COOH + NaOH = CH3COONa + H2O
Thus one mole of NaOH reacts with one mole of acetic acid.
31.43 ml of 0.108M NaOH contains 0.03143*0.108=0.003394M.
1M of acetic acid weighs (12+3+12+2*16+1)=60 g
0.003394M contains .003394*60=0.2037g
This amount is contained in about 4,441 g of water. Therefore the percentage is 0.2037/4.441=4.59%