Asked by Anonymous

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?
Answered by DrBob222
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 = ?
Answered by Mercy
0.033/0.198*100=16.67%
Answered by David
It's been said in Vinegar. So it suppose to be 0.198/5 * 100
Answered by Joshua Rimamsomte
16.67
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