A vinegar contains acetic acid, CH3COOH. Titration of 5.000g of vinegar with 0.100, NaOH requires 33.0 ml to reach equivalent point. What is the weight percentage of CH3COOH in vinegar?

Answer 1

CH3COOH + NaOH ==> CH3COONa + H2O

mols NaOH =- M x L = 0.1 x 0.033 = 0.033
mols CH3COOH = mols NaOH from above.
g CH3COOH = mols CH3COOH x molar mass CH3COOH.
% CH3COOH = (g CH3COOH/mass sample)*100 = ?

Answer 2

0.033/0.198*100=16.67%

Answer 3

It's been said in Vinegar. So it suppose to be 0.198/5 * 100

Answer 4

16.67