Vinegar contains acetic acid CH3COOH. Titration of 5.000g vinegar with 0.100M NaOH requires 33.0cm cube to reach the equivalence point(a)what is the weight percentage of CH3COOH in vinegar?(b)if the vinegar has a density of 1.005g cm cube what is the molarity of CH3COOH in vinegar?
CHCOOH + NaOH ==> CH3COONa + H2O
millimoles NaOH = mL x M = 33.0 mL x 0.100 M = 3.30
1 mol NaOH titrates 1 mol CH3COOH so millimoles CH3COOH = 3.30
grams CH3COOH = mols x molar mass = 0.00330 x 60 = 0.198
% w/w = (g CH3COOH/g sample)*100 = (0.198 g CH3COOH/5.0 g sample)*100 = 3.96% w/w CH3COOH.
molarity = 1000 mL x 1.005 g/mL x 0.0396 x 1/60 = ?M